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Thermodynamics
B.Sc (Prog.)IInd Year
Paper No : CHPT – 103
Chapter No and Name: 1, Thermodynamics
Author :Dr Adarsh Gulati
College/ Department: Miranda House University of Delhi
Author: Dr Neeti Misra
College/ Department :Acharya Narendra Dev College
University of Delhi
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Thermodynamics
Introduction
The word thermodynamics is a combination of two Greek words thermos (heat) and
dynamics (power or motion).
Thermodynamics is a physical science that involves the study of the changes in the
energy of the system when the temperature, pressure, or volume of a system, are
changed or when a chemical transformation takes place. It deals with the relationship of
energy with heat and work and the interconversion of various forms of energy.
Therefore, thermodynamics is sometimes also known as Energetics.
It has simple statements based on human experiences. Several laws of physical
chemistry can be directly deduced from the laws of thermodynamics. None of its
deductions has ever been proved to be wrong in experimental testing. Whenever a
conflict has arisen between the laws of thermodynamics and any other deduction, the
mistake is found in the other deduction.
Chemical Thermodynamics is that branch of thermodynamics that deals with
energy changes accompanying chemical transformations. It investigates the feasibility of
the change and tries to find the answer to the question-Why does a chemical reaction
take place?
3.1.1 Characteristics of Thermodynamics
•
•
•
•
•
Thermodynamics is an exact science as it has a mathematical base.
Usefulness of thermodynamics lies in correlating the observable properties of
substances with some of the nonobservable ones.
It points out the feasibility of changes, extent and direction in which they
may proceed. These changes may be physical processes or chemical reactions.
It is not necessary to know the nature of the fundamental particles or the
mechanism of the phenomenon. The deductions are made using everyday
experiences, some basic laws and mathematical tools.
Classical thermodynamics can be studied by dealing with the energy changes
taking place at the macroscopic level without emphasizing the changes that occur
at molecular level.
Did you know?
3.1.1 Characteristics of Thermodynamics
Each molecule of the system contributes to the total property of the system. These
properties can be pressure, energy, volume, or any other property. Classical
thermodynamics does not study the individual contribution of each molecule, but the
average behaviour of a large number of molecules.
The branch of science, which deals with the relation between properties of single
molecule and bulk properties of the system, is known as Statistical Thermodynamics.
3.1.2 Limitations of Thermodynamics
•
•
•
•
•
•
It deals with materials in bulk without any reference to the internal structures.
It gives the feasibility, direction, and extent of a change but not the time taken
for such a change to take place. The time taken for the completion of a process is
studied under chemical kinetics.
It deals with stationary or equilibrium states only.
It deals with materials in bulk without any reference to the internal structures.
It gives the feasibility, direction, and extent of a change but not the time taken
for such a change to take place. The time taken for the completion of a process is
studied under chemical kinetics.
It deals with stationary or equilibrium states only.
3.1.3 Important Thermodynamic Terms
System: It is that part of the universe on which theoretical or experimental
investigations are carried out. A system is said to be a part of the universe to which we
direct our attention for theoretical or experimental investigations and which is separated
from the rest of the universe by a real or imaginary well-defined boundary. For example,
if the effect of pressure on the volume of the gas has to be studied, then the system is
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the gas taken in a container. Similarly, any solution taken in a beaker is known as the
system when its properties are being studied.
Surroundings: The rest of the universe in the neighbourhood of the system is known as
its surroundings. It lies outside the system and is also called the environment of the
system. The system can exchange matter or energy or both with its surroundings.
A system and its surrounding are together known as the Universe.
Boundary: Anything that separates the system from the surroundings is known as a
boundary. It is also known as the wall of the system. The system interacts with its
surroundings across the boundary. There are various types of boundaries depending
upon their nature:
•
•
•
•
•
•
Real or imaginary
Rigid or non rigid
Conducting (diathermic) or nonconducting (adiabatic)
Permeable, impermeable or semi permeable
Electrically conducting or nonconducting
Transparent, translucent, or opaque.
A given boundary can belong to various categories. For example, the walls of a beaker
constitute a real, rigid, impermeable and diathermic boundary. The various categories of
boundaries have been summarized in Fig. 3.1.
Figure 3.1: Important categories of boundaries
3.1.4 Types of Systems Based on the Type of Boundary
The system exchanges matter and energy with the surroundings across the boundary.
Therefore, the boundary decides the relationship of the system with its surroundings and
hence it also decides the category to which the system belongs.
i.
ii.
iii.
Open system: When the system has a permeable boundary then both the
matter and the energy can move across the boundary. So, the exchange of
matter (mass, m) and energy (U) can take place between the system and the
surroundings. This type of system is known as an open system. For the open
system Δm ≠ 0 and ΔU ≠ 0.
Closed system: When a system has an impermeable but diathermic boundary
then the matter cannot flow across the boundary but energy exchange can take
place between the system and the surroundings. Such a system is known as a
closed system. For this system Δm = 0 and ΔU ≠ 0.
Isolated system: When a system has an adiabatic boundary, then matter and
energy exchange cannot take place between the system and the surroundings.
Such a system is known as an isolated system. For this system Δm = 0 and ΔU
= 0.
The physical state and other important properties of the system are decided by the
pressure (p or P), temperature (T), volume (V), and the amount (number of moles, n) of
the system. These four variables define the state of the system. Therefore, these
variables are known as state variables.
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3.1.5 State of the System
Thermodynamic Process: It is any operation or process that brings about a change in
the state of the system. Expansion, compression, heating and cooling are some of the
examples of a thermodynamic process. The initial state of the system is defined by the
pressure, temperature, volume, and amount of the system before the change takes
place. The final state of the system is defined by the values of pressure, temperature,
volume, and amount of the system after the change has taken place. A process is said to
have taken place whenever there is a change in any one or more of the state variables.
For any state variable Y, if the change in the state variable is large it is represented by
ΔY (delta Y). If the change is infinitesimal (very small) it is represented by dY.
3.1.6 Thermodynamic Processes
Classification of Processes Based upon the Change in Thermodynamic Variables
Thermodynamic processes are classified into various categories depending upon the
thermodynamic properties that are undergoing a change during the process. These
categories are as follows:
i.
ii.
iii.
iv.
v.
Isothermal process: It is a process during which the temperature of the system
remains constant, that is, ΔT = 0.
Adiabatic process: It is a process during which the system does not exchange
heat with the surroundings, that is, q = 0.
Isobaric process: It is a process during which the pressure of the system
remains constant, that is, there is no change in the pressure, ΔP = 0.
Isochoric process: It is the process during which the volume of the system
remains constant, that is, ΔV = 0.
Cyclic process: It is the process during which the system after undergoing a
series of changes comes back to the initial state. The series of changes taking
place is known as a cycle.
Path: The sequence of steps taken by a system during a thermodynamic process
starting from the initial state, through an intermediate state to the final state is known
as a path. The path can consist of a single or more steps.
Classification of Processes Based upon Path
Depending upon the path followed by the system, there are two types of processes:
reversible and irreversible processes.
Reversible process: A process is said to be reversible if the system undergoes a
change in the state through a specified sequence of intermediate states, each one of
which is an equilibrium state. In addition, the system can be restored to the initial state
by following the same sequence of steps in the reverse order. That is, a process is said
to be reversible if it can proceed in forward and backward directions by a small change
in state variables. When the system comes back to the initial state the surroundings are
also restored to their initial state. That is, the process is said to be reversible only if the
system can retrace its steps without any net change in the system and the
surroundings. The process should be reversible in both forward and backward directions.
Each reversible process is slow but each slow process may or may not be reversible.
The chemical reaction in a galvanic cell at the null point is an example of a reversible
process. At null point the external potential applied is equal but in the opposite direction
to the cell emf. When the external potential is decreased by an infinitesimal amount the
reaction goes in the forward direction and when the external potential is infinitesimally
greater than the cell emf the reaction proceeds in the backward direction.
Irreversible process: A process is said to be irreversible if the change takes place
without the system being in equilibrium with its surroundings at any stage during the
change. The system cannot retrace its path without the help of an external source.
Hence, the system can attain its initial state by some change in the surroundings.
Most of the natural processes and the spontaneous processes are irreversible processes.
3.1.7 Equilibrium and Steady State
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Equilibrium: A system exists in a state of equilibrium if its macroscopic properties such
as temperature, pressure, volume, and mass remain constant without the help of an
external source.
Steady state: When the macroscopic properties of a system are kept constant with the
help of an external source the system is said to be in a steady state.
Ethanol kept in a closed container having adiabatic walls is an example of a system
at equilibrium. Initially some ethanol in the liquid state evaporates and goes to the
vapour state. After some time equilibrium is attained between the liquid and vapour
phases of ethanol. The volume of ethanol, its pressure and the temperature inside the
container do not change with time.
An example of steady state is the water taken in a tank that has an inlet and an
outlet of water. The rate at which the water flows in is maintained equal to the rate at
which the water flows out. As a result the level of water in the tank does not change
with time.
Various types of equilibrium are as follows:
Thermal equilibrium: Temperature of each part of the system is the same.
Mechanical equilibrium: The external forces on the system are equal to internal forces
in the system. That is, there are no unbalanced forces in the system.
The pressure acting on the system is equal to the pressure exerted by the system.
Chemical equilibrium: The composition of each part of the system is same and does
not change with time.
A system is said to be in thermodynamic equilibrium if it is in thermal,
mechanical, and chemical equilibrium. Even if one of the equilibrium is not achieved the
system is not in thermodynamic equilibrium.
Extensive properties are those properties
that depend upon the size of the system. They
are additive in nature. This means that the
system is divided into small parts then the total
value of a property is obtained by adding up
the value of that property in all the parts of the
system. Some of the important extensive
properties are: mass, volume, amount (number
of moles), energy, enthalpy, entropy, free
energy, and heat capacity. You are familiar
with some of these
Figure 3.2: Properties of an
ideal
gas in the
four
compartments
properties; the rest will be introduced in the subsequent chapters of thermodynamics.
Intensive properties are those properties that do not depend upon the size of the
system. These properties are not added up to get the total value of a given property of
the system. They have the same value in all the parts of the system. Some of the
important intensive properties are: pressure, temperature, density, concentration, mole
fraction, refractive index, surface tension, viscosity, and specific heat capacity.
Let a container containing an ideal gas be divided into four compartments namely A,
B, C, D as shown in Fig. 3.2.
Let the pressure in each compartment be pA, pB, pC, pD, respectively. Let the
temperature be TA, TB, TC, TD; volume be VA, VB, VC, VD; and the amount be nA, nB, nC,
nD. When the partition is removed without changing the state of the system then, the
total volume of the container is:
V = V A+ V B+ V C + V D
The total amount of the gas is
n = n A+ n B+ n C + n D
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However, the total pressure of the container is not equal to the sum of the pressure of
each compartment. Similarly, the temperature of the container is also not equal to the
sum of the temperatures of each compartment. Thus, it is seen that the volume and
amount depend upon the size of the system while temperature and pressure do not
depend upon the size of the system.
Another illustration of extensive and intensive properties is of the 1 mol dm–3sodium
chloride solution taken in a one litre flask. This solution is divided unequally among four
students and each student measures the mass, volume, density, temperature, surface
tension and viscosity of the solution. What is the expected result?
All the four students will report almost the same values of density, temperature, surface
tension, and viscosity of the solution within experimental error. The mass and the
volume of the solutions will be different but the volumes of solutions of each student will
add up to one litre. Again it is observed that the intensive properties, that is, density,
temperature, surface tension, and viscosity of the solution do not depend upon the size
of the system while the extensive properties depend upon the size.
Extensive Properties
Intensive Properties
mass, volume, amount, heat capacity, internal energy,
enthalpy, entropy, free energy
pressure, temperature, density, surface tension, viscosity,
refractive index, molarity, molar volume, specific heat
capacity, molar heat capacity, molar energy
3.1.8 Intensive and Extensive Properties
The following rules help in determining whether a given property is extensive or
intensive. These are based on the facts that many of the physical properties are related
to other physical properties.
i.
The sum of two extensive properties is an extensive property.
ii.
Dividing an extensive property by another extensive property results in an
intensive property. For example, molarity = amount of subtance/volume of
solution, in litres.
Here amount and volume are extensive properties while molarity is intensive
property. Other examples are
Pressure = Force/ Area
Density = Mass / Volume
All the molar properties are intensive properties.
iii.
The product of an intensive property and an extensive property is an extensive
property and the product of two extensive properties is also an extensive
property.
3.1.9 State Variables
The state of the system depends upon the pressure (p or P), temperature (T), volume
(V) and the amount of substance (number of moles, n) in the system. The state of the
system changes when one or more of these variables change. Therefore, these variables
are known as state variables. It is generally possible to set up an algebraic equation
showing the relationship between the defining variables of state of the system. Such
equation or equations are called equations of state for the system. For example, for an
ideal gas the equation of state is:
PV=nRT
3.1.10 State Property or State Function
The property of a system that depends on the state variables is known as a state
property. The change in the state property does not depend upon how the process is
carried out. That is, the value of a state property depends only on the state of the
system and not on the process by which this state is reached. A function that depends
only on the state variables is known as a state function.
Thermodynamic properties like of a system internal energy (U), enthalpy (H),
entropy (S), Helmholtz energy(A) and free energy (G) change with the change in
variables like temperature, pressure, volume and amount of substance in the system.
The changes in these properties depend upon the initial and final states of the system.
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They do not depend upon the path followed during the change. Therefore, these
thermodynamic properties are also state properties or state functions.
Characteristics of state functions: Let z be a state function that depends upon two
variables x and y. This is written as
z = f (x,y)
The various characteristics of state functions are as follows:
i.
ii.
State functions can be differentiated.
They have exact differentials, that is, they can be integrated within the limits.
iii.
iii.
∆z =
The order of differentiation of z with respect to x and y is immaterial. This
may be explained as follows. The total change in z is represented by dz.
The change in z takes place due to change in x (dx) and change in y (dy).
It is written as
Here,
= rate of change of z with change in x when y is kept constant =
partial
derivative of z with respect to x at constant y
= rate of change of z with change in y when x is kept constant =
partial
derivative of z with respect to y at constant x
dx = infinitesimal change in x
dy = infinitesimal change in y
Let
= M and
=N
If z is state function and dz is exact differential then the following differentials are
equal.
This means that,
.
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It can also be written as
The above equality is known as Euler’s theorem of exactness. If this relation
is valid for a function then that function is a state function.
Some other characteristics of state functions are:
iv.
v.
If the subscripts are the same the partial differentials can be multiplied.
However,
vi.
iv.
For any three state functions x, y, z, the cyclic rule is,
In a cyclic process the total differential of a state function is equal to zero.
Example
Show that the volume of an ideal gas is a state function.
Solution:
Volume depends upon temperature and pressure. This can be written as
V=f
(T,P)
....(3.1)
The change in volume, dV, due to change in temperature, dT and change in pressure, dP
is given by the equation,
....(3.2)
Here,
= total change in volume due to change in temperature dT and
= total change in volume due to change in pressure dP.
If V is a state function dV must be exact differential, that is, the Euler’s theorem of
exactness is valid.
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Therefore,
For an ideal gas,
PV =
nRT
....(3.3)
Differentiation of Eq.(3.3) with respect to P at constant T gives
However,
= 0,
therefore,
=0
or
Substituting the expression of V in terms of T and P in the above equation we get
...(3.4)
Differentiating Eq. (3.4) with respect to T at constant P we get the second derivative
....(3.5)
Differentiating Eq. (3.3) with respect to T at constant P we get
....(3.6)
Differentiating Eq. (3.6) with respect to P at constant T we get the second derivative
....(3.7)
Comparing Eqs. (3.5) and (3.7) it is seen that the right hand sides are equal, therefore,
the left hand sides must also be equal. Hence, it is proved that
Therefore, dV is an exact differential and V is a state function.
We know from our experience that change in volume of an ideal gas during a process is
independent of the path followed. If the pressure and temperature of an ideal gas are
changed then the final volume can be calculated. The change in pressure and
temperature can be carried out in three ways such that in all the three cases the final
state is the same. It can be shown by the following calculations that the final volume is
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the
same
in
all
the
three
cases:
3.1.11 Inexact (Path) Function
It is a thermodynamic property that depends upon how the process is carried out, that
is, it depends upon the path taken by the system during the thermodynamic process.
Work and heat are path functions. Distance travelled is a path function while
displacement is a state function.
Path functions are inexact differentials, that is, they cannot be integrated.
There is no such thing as initial work and final work, therefore, w1 and w2 do not mean
anything. w is the work done when the system goes from an initial state to the final
state. Similarly, since q is a path function we only talk about the total heat exchange
between the system and the surroundings during a process and not the initial heat and
the final heat. dq and dw where ever they are mentioned do not mean change in heat or
change in work. dq stands for small amount of heat exchanged by the system. dw
denotes small amount of work done.
The correct symbol for small amount of heat exchanged is đq and for small amount of
work done is đw, where, đ represents an inexact differential. However, in this book we
will use dw and dq as symbols.
Example
Check the following equations for exactness.
i.
ii.
iii.
iv.
dK = -(RT/P)dP + RdT
dΦ = (x2 –y2)dx + 2xydy
dz = xy2dx + yx2dy
dV = (R/P)dT –(RT/P2)dP
Solution:
The equation of the type dz = Mdx + Ndy
i.
is exact if
dK = -(RT/P)dP + RdT
Here M =
dK = MdP + NdT
and N =R
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For this equation to be exact
=
=
=0
Since
ii.
≠
, dK is not an exact differential
dΦ = (x2–y2)dx + 2xydy
M = (x2–y2)
and N = 2xy
Differentiating M with respect to y at constant x and N with respect to x at
constant y we get,
= -2y
Since
iii.
and
≠
= 2x
dΦ is not exact differential
dz = xy2dx + yx2dy
M = xy2 and N = yx2
Differentiating M with respect to y at constant x and N with respect to x at
constant y we get,
= 2xy
Since
iv.
and
=
= 2xy
, dz is exact differential
dV = (R/P)dT –(RT/P2)dP
and N = -(RT/P2)
M = R/P
=
Since
=
, dV is an exact differential
3.1.12 Laws of Thermodynamics
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These laws of thermodynamics are based upon observations. The laws were numbered
as they were stated. The last law to be given was the zeroth law. Since it is the most
fundamental, it should come before first law and as zero comes before one it was named
zeroth law.
Zeroth law of thermodynamics: If two systems A
with a third system C, then A and B will be in thermal
placed. That is, if the temperature of system A and
temperature of system B, and of system C, are equal
will be same as that of system B i.e.,
If TA = TC
and B are in thermal equilibrium
equilibrium when with each other
of system C, are equal and the
then, the temperature of system
and TB = TC, then TA = TB
First law of thermodynamics:
ΔU = q + w
The change in internal energy (ΔU) of the system is equal to the heat exchanged (q) by
the system and the work done (w) on/by the system. The internal energy of an isolated
system is constant.
Second law of thermodynamics: It is impossible to transfer heat from a colder
system to a warmer system without other changes occurring simultaneously in the two
systems or in their environment. The state of maximum entropy is the most stable state
for an isolated system.
Third law of Thermodynamics: The entropy of a pure and perfectly crystalline solid is
zero at zero Kelvin of temperature.
These laws and their applications will be discussed in details in the subsequent
sections of this unit.
3.2 First Law Of Thermodynamics
The change in the energy of the system is related to the work done and the heat
exchanged between the system and its surroundings. The first law of thermodynamics
deals with the relation between the change in the energy, work and heat. Therefore, an
understanding of these terms is important before studying the first law of
thermodynamics.
3.2.1 Internal Energy (U)
The total energy of the system is due to energy possessed by the constituents of the
system, such as atoms, ions, and molecules as well as their motion and position. The
internal energy of the system does not include the kinetic energy due to motion of the
system and potential energy due to position of the system. The potential energy and the
kinetic energy constitute the external energy of the system.
The energy possessed by all the constituents of the system is the internal energy. It is
represented by the symbol U.
Internal energy is the total energy of all the molecules. It includes translational energy
(Ut), vibrational energy (Uv), rotational energy (Ur), electronic energy (Ue), and
intermolecular interaction energy (Uint)
U = Ut + Uv + Ur + Ue + Uint
Internal energy is also known as intrinsic energy.
Total energy = External energy + Internal energy of system
The SI unit of energy is Joule (J).
1 J = 1 kg m2 s–2 = 107 erg
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Other units : calories (cal) 1 cal = 4.184 J
1 eV = 1.602 x 10–19 J
1 L atm = 1.013 x 102 J
1 Watt hour = 3.6 x 103 J
In thermodynamic studies, the emphasis is not on the individual energy of the
molecules, but on the overall contribution of all the molecules to the energy of the
system. Therefore, thermodynamics generally deals with the energy at the macroscopic
level and not at the microscopic level.
Internal energy is a state function. It depends upon temperature (T), pressure (P),
volume (V) and the amount (number of moles present in the system, n). For a closed
system the amount of the system is constant, so U will depend upon T, P and V. Since
temperature, volume and pressure are related to each other, internal energy is not
written as a function of all the three variables but only as a function of two variables.
Therefore, for a closed system, the internal energy is a function of T and V
only.
U=
f(T,V)
(3.8)
The total change in the internal energy (dU) of the system takes place due to the
change in temperature, dT and the change in volume, dV. It is written as
dU=
(3.9)
where, (∂U/∂T)V is the rate of change of internal energy with temperature at constant
volume and (∂U/∂T)V dT is the change in internal energy due to small change in the
temperature (dT) at constant volume.
(∂U/∂V)T is the rate of change of internal energy with volume at constant temperature
and (∂U/∂V)T dV is the change in internal energy due to small change in the volume (dV)
at constant temperature. Since U is a state function, dU is exact differential and Euler’s
reciprocity
relation
is
valid
(3.10)
Characteristics of Internal Energy
•
•
•
•
•
•
•
Internal energy is represented by U
The SI unit of U is Joule.
U is an extensive property.
U is a state function
The change in U is equal to the difference in internal energies of the final state
and initial state of the system. that is,
ΔU = U2–U1.
Here, Δ stands for change.
In a cyclic process, there is no net change in internal energy.
Internal energy of the system can be changed when system exchanges heat with
its surroundings or when work is done on the system or by the system. The
internal energy can also change when the system undergoes physical or chemical
process.
The experiments show that the internal energy of the system increases when the work is
done on the system or when the heat is absorbed by the system. Therefore, we should
know more about work and heat.
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In thermodynamics, work is defined as any quantity that flows across the boundary of a
system during a change in its state and is completely convertible into the lifting of a
weight in the surroundings. The symbol of work is w. SI unit of work is Joule (J).
It has the following characteristics.
i.
ii.
Work appears only at the boundary of a system.
Work appears only during a change in state, that is, only during a
thermodynamic process.
Work is manifested by an effect in the surroundings.
During an adiabatic process, work results in the change in the internal energy of
the system. When the work is done by the system, the system uses its internal
energy. Therefore, internal energy of the system decreases. When work is done
on the system, the internal energy of the system increases.
Work is a path function, that is, it is an inexact function. If the system is changed
from the initial state to the final state by two different processes, then the work
done will be different for the two processes.
iii.
iv.
v.
3.2.2 Work
Mathematical expression for work
Mechanical work is the result of displacement of a body when the force is applied on it.
It is related to the force , F and displacement by the following equation.
w = F * Δx
(3.11)
When the system does mechanical work, it can be related to the change in the volume
of the system. This can be illustrated by considering a system consisting of a gas taken
in a cylinder with a piston at the top. This piston is assumed to be massless and
frictionless; therefore, no work is done to overcome friction when the piston moves up
and down. An external pressure (P) is applied on the system. When the gas expands,
the piston moves up by a distance Δx. The force is related to the external pressure by
the equation:
Force = Pressure x Area =
PxA
Here, A is the area of cross section of the base of the piston. Substituting the expression
of force in Eq.(3.11)
w = Px A x Δx
Since,
A x Δx = ΔV
w =PΔV
(area x distance = volume)
(3.12a)
Figure 3.3: Change in volume of a gas on applying an external pressure
When the displacement is infinitesimal, i.e., dx the change in volume is represented by
dV and the work done is dw. Therefore,
dw = PdV
Sign Convention in Thermodynamics
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(3.12b)
Thermodynamics
In thermodynamics studies, the emphasis is on the system. Any change that results in
an increase in the energy of the system is taken as positive. During compression, the
internal energy of the system increases, therefore, the work done on the system is
taken as positive. Similarly, the work done by the system is done at the cost of the
internal energy of the system, therefore, the work done by the system is taken as
negative. Hence, the Eqs.(3.12a) and (3.12b) are modified as
w = −PΔV
and
(for finite change)
dw = − PdV
(for infinitesimal change)
It can be shown that the above formula will give a positive value of w during
compression, when the work is done on the system and a negative value of w during
expansion when the work is done by the system.
Figure 3.4: Compression and expansion of a gas taken in a cylinder
Initially some weights are kept on the piston so that the external pressure is equal to
the internal pressure and the piston does not move. Then the weight on the piston is
increased to compress the gas and move the piston down. Here, work is done on the gas
and it can be calculated using Eq. (3.12a).
w = –pexΔV
Animation
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pex is the external pressure acting on the gas. During compression, the volume of the
gas decreases, so, ΔV is negative. Therefore, the value of w is positive when the work is
done on the gas.
To expand the gas some weights are removed from the piston and the gas
expands lifting the piston with the remaining weights. During expansion, work is done by
the gas to lift the weights. Volume of the gas increases and so ΔDV is positive.
Therefore, the value of w is negative when work is done by the gas.
3.2.3 Heat
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In thermodynamics heat is defined as a quantity that flows across the boundary of a
system, during a change in its state by virtue of the difference in temperature between
the system and its surrounding. Heat flows from a region of higher temperature to a
region of lower temperature. The symbol of heat is q. SI Unit of heat is Joule (J).
•
•
•
•
Heat appears only at the boundary of the system
Heat appears only during the change in state
Heat is manifested by an effect in the surroundings
It is a path function
Sign Convention
The heat absorbed by the system is represented by q. When the system absorbs heat,
that is, heat flows from the surroundings to the system, the internal energy of the
system increases provided no work is done. Therefore, q is positive.
When the system loses heat, that is, heat flows from the system to the surroundings,
the internal energy of the system decreases provided no work is done. Therefore, q is
negative.
q is positive when heat is absorbed by the system.
q is negative when heat is lost by the system.
Proof that q is a Path Function
According to the first law of thermodynamics,
dU = dq + dw
When only the work of expansion or compression is done then
dw = – PdV
where P is the external pressure
dU = dq –PdV
or
dq = dU + PdV
For an ideal gas, P = nRT/ V and dU = CvdT (Eq.3.21). Substituting these values in the
expression for dq
dq = CvdT +
dV
Let M = Cv and N =
Therefore,
dq = MdT + NdV
For dq to be an exact differential, the order of differentiation is immaterial. That is,
Differentiating M with respect to V at constant T
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From the definition,
CV =
.
Therefore, the above equation becomes
Since, U is a state function and dU is an exact differential, therefore,
Substituting the value in the expression of (∂M/∂V)T we get
For an ideal gas
=0
, therefore,
0
(3.13a)
Also, N =
Differentiating with respect to T at constant V we get,
(3.13b)
On comparing Eqs(3.13a) and (3.13b) it is found that
≠
Therefore, dq is not an exact differential and q is not a state function.
According to the law of conservation of energy, energy can neither be created nor
destroyed. If it disappears from one form it must appear in another form. This statement
is true for those processes also that involve work done and heat transfer.
According to the definitions of heat and work done discussed earlier, both heat and
work done affect the internal energy of the system. However, they appear only during
the change. They are not forms of energy. They are the manner in which energy
manifests or shows itself during a change.
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When the energy is supplied from outside by heating the system, a part of it is used
to do work and the rest remains with the system. Since the total energy should be
conserved during any change, the change in internal energy of a system must be related
to heat exchanged by the system and work done on/by the system.
That is, for a finite change
w
where,
ΔU = q +
(3.14a)
ΔU = U2 –U1.
U1 is the internal energy of initial state and U2 is the internal energy of the final state. q
is the heat absorbed by the system and w is the work done on the system
When the process involves infinitesimal change
dU = dq
+dw
(3.14b)
Equation (3.14a) and (3.14b) are the respective mathematical forms of the first law of
thermodynamics for finite change and infinitesimal change.
In Eq.(3.14a), dq and dw do not represent exact differentials. That is,
dq ≠ q2 – q1.
This is due to the fact that the heat of the initial state and heat of the final state has no
meaning. dq represents a small amount of heat absorbed by the system. Similarly, dw is
the small amount of work done on the system during the change.
When only mechanical work, i.e., work of expansion or compression is done then
Therefore, Eq. (3.14b) becomes,
dU= dq – PdV
(3.15)
Example
An ideal gas absorbs 500 J of heat and its internal energy increases by 260 J. On the basis
of first law of thermodynamics explain what happened to rest of the heat.
Solution:
Heat absorbed by the system, q = 500 J
The change in internal energy, ∆U = 260 J
According to the first law of thermodynamics
ΔU = q + w
or
w = ΔU – q
w = 260 J – 500 J = – 240 J
Since sign of the value of w is negative, 240 J of work is done by the system. So out of
the 500 J of heat absorbed, 240 J was used to do work by the gas and 260 J was used to
increase the internal energy of the gas.
3.2.5 Relation Between Internal Energy and Heat
When no work is done and the system does not undergo any phase transformation then
the heat transfer between the system and its surroundings results in the change in the
temperature of the system. As a result, the volume and pressure will also change. The
conditions can be maintained in such a way that either the volume or the pressure is
kept constant. Heat exchange is also accompanied by the change in energy of the
system. This energy change will be different for the two conditions, that is when volume
is kept constant and when the pressure is kept constant.
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Heat Exchange at Constant Volume
ΔU = q + w = q – PΔV
When the volume change is constant, i.e. ΔV = 0,
Hence, PΔV is also zero. The heat exchanged, q can be written as qv.
Therefore, ΔU = qv
This implies that, the change in internal energy of a system is equal to
heat exchange between the system and its surroundings at constant volume.
Since internal energy is a state function heat exchange at constant volume is also
a state function.
3.2.6 Enthalpy and Heat Exchange at Constant Pressure
In the previous section heat exchange at constant volume has been related to the
change in internal energy. However, in the actual practice, the experiments are carried
out at constant pressure in the laboratory. The energy changes are measured in terms
of heat exchange taking place between the system and its surroundings. Since the heat
exchange takes place at constant pressure it cannot be equal to the change in internal
energy. Therefore, a new energy term is needed that will account for the heat exchange
at constant pressure. This new energy term is called enthalpy and the symbol for it is
H. It was earlier also known as heat content. The symbol H is taken from heat content.
H is related to the internal energy, pressure and volume by the equation
H=U+
PV
(3.16)
Since U, P and V are state functions U + PV must also be a state function. Since U and
PV have dimensions of energy, it must also have dimensions of energy. The change in
enthalpy is given by dH.
The differential of Eq. (3.16) gives
dH = d( U+ PV)
From the first law of thermodynamics,
dq = dU + PdV
Therefore,
dH = dq +VdP
(3.17)
When the pressure is constant, dP = 0 , VdP = 0 and dq = dqp.
Therefore,
dH = dqp.
(3.18)
For finite change,
∆H = qp
where,
∆H = H2 – H1
Therefore, the heat exchanged at constant pressure is equal to the change in enthalpy of
the system. Whenever the system exchanges heat at constant pressure (qp), there will
be a change in temperature of the system. Some part of heat, qp will be used to increase
the internal energy and rest is used up for the change in volume. Hence, change in
enthalpy results in change in internal energy as well as change in volume of the system.
For solids and liquids the volume change is negligible, therefore, enthalpy and internal
energy changes are practically same.
Relationship between ∆H and ∆U for an ideal gas
H = U + PV
∆H = ∆U + ∆PV
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For an ideal gas,
PV = nRT
Therefore,
∆H = ∆U + ∆(nRT)
At constant temperature,
∆H = ∆U + (∆n)RT
Here, ∆n is the change in the amount of the gas. If the temperature changes but the
amount of gas is constant then,
∆H = ∆U +nR ∆T
Characteristics of enthalpy
•
•
•
•
•
•
•
The enthalpy change is equal to the heat exchanged by the system at constant
pressure
It is a state function.
H = U + PV
For solids and liquids, the enthalpy change(ΔH) is taken as equal to the change in
the internal energy (ΔU)
Its units are same as the units of energy
When the system absorbs heat at a constant pressure, ΔH is positive and the
process is known as an endothermic process
When the system loses heat to the surroundings at a constant pressure, ΔH is
negative and the process is known as an exothermic process.
Since enthalpy is a state function, it depends upon T, V, P and the amount of the
system (n). Since, temperature, pressure and volume are related to each other by the
equation of state, for a closed system, enthalpy can be written as a function of
temperature and pressure.
H = f (T, P)
The change in H due to the change in the temperature and the pressure is given by the
equation
(3.19)
where, (∂H/∂T)P is the rate of change of enthalpy with temperature at constant
pressure and (∂H/∂P)T is the rate of change of enthalpy with pressure at constant
temperature. Since H is a state function dH is exact differential and Euler’s reciprocity
relation is valid.
3.2.7 Heat Capacity
Heat capacity is defined as the heat needed to cause unit rise in the temperature of a
substance.
C = dq / dT
Unit of heat capacity
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SI unit is J K-1.
Heat capacity is an extensive property as it depends upon the size of the system.
Specific heat capacity: It is defined as the heat needed to cause a unit rise in the
temperature of a unit mass of a substance. It is denoted by s. Its SI unit is
J K-1 kg-1.
Molar heat capacity: It is defined as the heat needed to cause a unit rise in the
temperature of one mole of a substance. Molar heat capacity is denoted by Cm. Thus,
Molar heat capacity =
Cm =
The SI unit of molar heat capacity is J K-1mol-1.
Specific heat capacity and molar heat capacity are intensive properties.
When the heat is absorbed at a constant pressure, the heat capacity is known as
heat capacity at constant pressure,denoted by Cp. It is also known as an isobaric
heat capacity. It can be mathematically expressed as
Cp = (∂q /∂T)p
Heat exchanged at a constant pressure, dqp = dH. Therefore,
Cp =
(3.20)
When heat is absorbed at constant volume the heat capacity is known as heat capacity
at constant volume, CV. It is also known as isochoric heat capacity.
CV = (∂q /∂T)v
Since,
dqV = dU
CV =
(3.21)
Using Eqs (3.19) and (3.20), we get
(3.22)
Using Eqs (3.9) and (3.21), we get
(3.23)
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Relationship between Cp and CV for an ideal gas:
H = U + PV
Differentiating with respect to temperature at constant pressure we get
=
+
=
+P
From Eqs. (3.9)
dU =
Dividing by dT and imposing the constant pressure condition we get
=
+
For an ideal gas (∂U/∂V)T = 0 (proved after the derivation of thermodynamic equation
of state in chapter 4).Therefore,
=
and
=
+P
For an ideal gas
PV = nRT
Differentiating with respect to T at constant P we get
P
= nR
Therefore,
or
Cp =
=
−
+ nR
= nR
and Cv =
,
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Therefore,
−
= Cp − CV= nR
Dividing the equation by n we get,
Cp,m – CV,m = R
Here, Cp,m = Cp/n and CV,m = CV/n.
This equation can be derived from the thermodynamic equation of state. The
derivation is given in section 3.5.6.
The ratio between Cp and Cv is represented by γ (gamma), i.e.,
Cp, m /CV, m = γ.
The value of γ is useful in calculation of relationship between P,V and T during adiabatic
process. The values of γ for various gases are given here.
For a monotomic gas such as He:
Cp, m /CV, m = γ = 1.67
For a diatomic gas such as H2:
Cp, m /CV, m = γ = 1.40
For a triatomic gas such as CO2:
Cp, m /CV, m = γ = 1.33
For an ideal gas,
Therefore, γ is always greater than 1.
3.3 Work Done During Expansion or Compression of an Ideal Gas
Let the system consist of an ideal gas. The mechanical work on the gas can be done
either reversibly or irreversibly. During mechanical work, the pressure and the
volume of the gas change. These changes can take place at constant temperature, that
is, under isothermal conditions or under adiabatic conditions. Therefore, the various
processes can be
i.
ii.
iii.
iv.
Isothermal reversible volume change
Isothermal irreversible volume change
Adiabatic reversible volume change
Adiabatic irreversible volume change
Since work is a path function, it should depend upon the nature of process. In the
following sections expressions for work done will be derived for all the four processes
and it will be proved that the amount of work done in all the four cases will be different.
Work Involved in Reversible Process
The reversible work of expansion or compression can be illustrated by the following set
up.
An ideal gas is taken in a container fitted with a massless and frictionless piston. On
the top of the piston a container with a large number of weights are kept.
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The weights exert pressure on the gas. This pressure is known as external pressure, Pex.
Initially, the external pressure is balanced by the pressure exerted by the gas on the
piston, that is, Pin.
Pex = Pin
This state is an equilibrium position. During expansion, the weight on the piston is
decreased slightly,as a result the external pressure becomes infinitesimally smaller than
Pin by the amount dP. The new external pressure will be P'ex
P'ex = Pin – dP
That is, the pressure exerted by the gas is infinitesimally different from the new external
pressure on the gas. The system is said to be infinitesimally removed from the
equilibrium state. To maintain equilibrium, the gas expands infinitesimally and the piston
moves up. If the process of decreasing the external pressure infinitesimally is continued,
then after a large number of steps, there is a significant change in the volume of the
gas. At any stage during the expansion, the internal pressure is practically equal to
external pressure, P'ex. So each step is an equilibrium step. The work done by the gas at
each step is (– dw). According to Eq. (13.12b) dw is given by the expression
–dw = P'exdV = (Pin – dP)dV ≈ Pin dV
as dPdV is very small it is neglected. Pin keeps on changing after each stage of
expansion.
During compression the weight on the piston is increased progressively.Due to this
the external pressure increases slightly after each addition and becomes greater than
the internal pressure. To maintain the equilibrium, the gas compresses and the piston
moves down. When all the weights have been replaced, the gas comes back to the
original position. The work done during each step of compression will be equal to dw.
Work involved in irreversible process
When the gas expands or compresses rapidly the process is irreversible in nature.
During the expansion, the pressure on the gas is decreased by a finite amount. That is,
the new external pressure is less than the pressure exerted by the gas on the piston,
P'ex < Pin
The equilibrium is disturbed and the gas expands without maintaining the equilibrium
with the surroundings. The work done by the gas is given by –dw.
–dw = P'exdV
The gas expands against constant external pressure at any given stage of expansion.
The total work done by the system (−w) is equal to
−w = P'ex∆V
or
w = −P'ex∆V
Similarly, during compression, the external pressure on the gas is increased such that
the new external pressure is much greater than the pressure exerted by the gas on the
piston. So, the gas compresses without maintaining the equilibrium with the
surroundings.
The reversible as well as irreversible work done can take place under isothermal as
well as adiabatic conditions depending upon the boundary of the container. During
isothermal process the container is made up of diathermal boundary while during
adiabatic process the container is made up of adiabatic boundary.
3.3.1 Work Done During Isothermal Volume Change for an Ideal Gas
During the isothermal process the temperature of the system remains constant. The
system can exchange heat with the surroundings to maintain the temperature. The
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amount of the gas remains constant.
Reversible process: The process is carried out in an infinite number of steps. The work
done in a step is given by the equation
dw = –PexdV
Here, Pex is the external pressure on the gas.
Let n be the amount of an ideal gas at temperature T. Let the initial and the final state
of the gas be as shown in the following.
Initial State
Pressure
P1
Volume
V1
Final State
P2
V2
In the reversible volume change the external pressure differs from the internal
pressure by the expression
Pex = (P ± dP).
where, + and − signs are meant for the compression and expansion, respectively and P is
the internal pressure, i.e., the pressure of the gas.
Substituting this in the expression of dw, we get,
dw = −(P ± dP)dV = –PdV −(± dPdV)
Since dPdV is a product of two differentials it may be neglected. Hence,
dw = –PdV
Total work done during the process is obtained by integrating the above equation
….(3.24)
Since the pressure is also changing during the process, P cannot be taken out of the
integral. However, for an ideal gas,
P = nRT / V
….(3.25)
(3.26)
Note: During the expansion V2 > V1 . Therefore, log (V2/ V1) is greater than zero. Since
n, R and T are always positive; the work done by the ideal gas during isothermal
reversible expansion is always negative. For compression V2 < V1,therefore, log (V2/ V1) is
less than zero and the work done is positive.
At a constant temperature for an ideal gas,
P 1V 1 = P 2V 2
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Therefore, Eq.(3.26), in terms of pressure becomes
(3.27)
Graphical Representation of Isothermal Reversible Expansion of a Gas
Figure 3.5(a) illustrates the magnitude of the work involved in one step in which the
volume change given by the difference of volume of gas at point d and volume of gas at
point c is infinitesimally small against an external pressure of P1. It is obvious that this
work is equal to the area of the trapezoid abcd. Similarly, the magnitude of the work done
by the gas during next step will be the area of the adjacent trapezoid. Thus, the total
magnitude of the work done by the gas during various steps of a reversible process will
be equal to the sum of the areas of all the trapezoids from the initial state A to the final
state B Fig. 3.5(b). This is equal to the area of the region ABCD.
Figure 3.5: Work done by an ideal gas under reversible isothermal conditions
Graphical Representation of Isothermal Reversible Compression
During isothermal compression, the pressure is increased slowly from P2 to P1 in an
infinitesimal number of steps and at each stage the volume decreases slightly. According
to the Figure 3.6, the change starts from point A (P2, V2) and goes till the point B(P1,V1).
The total work done on the gas is equal to the area of the region ABCD.
Figure 3.6: Pressure–volume work done during reversible isothermal compression
Comparison of Isothermal Reversible Work During Expansion and Compression
On comparing Fig. 3.5(b) and 3.6 it is seen that the magnitude of the work done is same
if P1, P2 and V1, V2 in the two figures are same. However, the sign of work done is
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different.
Work done during isothermal reversible expansion =
Work done during isothermal reversible compression =
Note: The work done on an ideal gas to bring it back to its initial state is equal to the
work done by the gas during expansion for an isothermal reversible process. Therefore, in
a cyclic process from A to B and back from B to A, the total work done is zero.
Irreversible Volume Change
In this case the pressure may be changed in one step or more than one step. If the
volume is changed from V1 to V2 in one step, then the total work done is given by the
expression
If the gas is allowed to expand unrestrictedly, then the external pressure is equal to the
final pressure of the gas after expansion. However, the expansion can be done against
the external pressure smaller than the final pressure of the gas and the desired expansion
may be achieved by a set of stops. If Pex = P2 then the expression of work is given by
(3.28)
Since P2 is constant we get
(3.28)
Note: For expansion V2 > V1 and w is negative indicating that the work is done by the
gas. For compression V2 < V1 and w is positive indicating that the work is done on the
gas.
Since for an ideal gas,
V2 = nRT / P2 and
V1 = nRT / P1,
The expression of work in terms of pressure is
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(3.30)
Graphical representation of isothermal irreversible expansion
During an irreversible process, the external
pressure is decreased from P1 to P2 and the
gas expands from V1 to V2. The total work
done is equal to –P2(V2 – V1).
From Figure 3.7,
(V2 – V1) = CD and P2 = DE.
Therefore, the magnitude of the work done
will be equal to
|w| = ED x CD = Area of the region BCDE
Figure 3.7: Pressure–volume work done
by
ideal
gas
under irreversible
isothermal conditions
Irreversible Expansion When the Expansion of the Gas is Restricted
In this case, the pressure on the gas is decreased by a finite amount and then the gas is
allowed to expand isothermally under an irreversible condition. However, the expansion is
stopped by blocking the movement of the piston with the stoppers as shown in Fig. 3.8.
As a result, the final pressure of the gas is not equal to the external pressure exerted on
the gas.
Here, final pressure, P2 is greater than external pressure. The final volume (V'2) will be
less than the situation when the expansion of gas is not stopped (V2).
Figure 3.8: Irreversible expansion when P2> Pex
The gas expands against the external pressure. Therefore, the work done by the gas will
be equal to
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It is important to note that the expression of work done during the expansion involves the
external pressure and not the final pressure of the gas. However, during the unrestricted
expansion the final pressure is equal to the external pressure. The work done by the gas
during restricted expansion will be less as compared to the case when the gas is allowed
to undergo unrestricted expansion.
Graphical Representation of Isothermal Irreversible
Compression
The work done on the gas during isothermal
irreversible compression is given by the
expression –P1(V1– V2).
From the Figure 3.9,
P1 = C'D' and
(V1– V2) = A'D'.
Therefore, the work done on the gas is equal to
|w| = C'D' x A'D' = area of the region A'B'C'D'.
Figure 3.9: Pressure–volume work done
during
irreversible
isothermal
compression
Comparison of Isothermal Irreversible Work During Expansion and Compression
The gas expands from State I to State II irreversibly and then it is compressed
irreversibly to State III.
State I (P1,V1,T1)
State II (P2,V2,T1)
Sta
te III (P1,V1, T1)
As can be seen, State I and State III are identical. Therefore, when the process is
complete the system comes back to the initial state. On comparing the magnitude of the
work done during irreversible expansion and irreversible compression, it can be seen
that the work done on the gas to bring it back to the initial state is different from the
work done by the gas during expansion.
From Fig. 3.7 and 3.9, the area of the region A'B'C'D' > area of the region BCDE.
The work done on the gas is greater than the work done by the gas during
irreversible process.
Comparison of Work Done During Reversible and Irreversible Expansion
The work done by the gas is the useful work that can be obtained from the gas. The
magnitude of the work done by the gas has been determined graphically in Figs 3.5(b)
and 3.7. On comparing the two figures it is observed that
area of the region ABCD > area of the region BCDE.
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Therefore, the work done by the gas during reversible expansion is more than work
done by the gas during an irreversible expansion.
Work Done at a Constant Volume (Isochoric Process)
When the volume is constant ΔV= 0. The process is isochoric; the gas neither expands
nor contracts. The gas does not do any mechanical work.
Work Done at Zero External Pressure
The external pressure is zero if the gas is surrounded by vacuum. Under such conditions
the gas undergoes expansion into vacuum. This expansion is known as free expansion.
Since dw = – Pext dV and Pext = 0, therefore, dw = 0. No work is done by the gas during
free expansion.
The change in the internal energy depends upon the changes in the temperature and
volume
During an isothermal process, the temperature is constant. Therefore, dT = 0 and the
above equation becomes
(∂U/∂V)T is known as the internal pressure. It depends upon the intermolecular
interactions. Since in the ideal gases, the intermolecular interactions are not present,
(∂U/∂V)T is equal to zero.
Therefore, the change in internal energy during an isothermal expansion or
compression of an ideal gas is equal to zero.
Measurement of
- Joule’s experiment
The apparatus consists of two bulbs A and B joined by a stopcock S that can be opened
and closed. Bulb A is filled with air and bulb B is evacuated. The bulbs are kept in a
water bath that has a stirrer and thermometer in it. The water bath has adiabatic walls
so no heat is lost to the surroundings. The initial temperature of the water bath is noted.
Then the stop cock is opened so that, air from bulb A moves into bulb B that has
vacuum. The setup is shown in Fig. 3.10.
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Figure 3.10: Set up for Joule's experiment
The air is expanding against vacuum so no work is done.
dw = 0
The final temperature of the water bath is noted.
It was observed by Joules that the temperature of the surroundings did not change
showing that no heat loss or gain took place during the process. Therefore, for this
process
dq= 0.
dU = dq + dw = 0 in Joule’s experiment
=0
Since temperature was constant during the experiment dT = 0. This implies that
dV = 0
However, since the gas expanded, dV ≠ 0. Therefore,
=0
Limitations of Joule’s Experiment
The temperature change in Joule’s experiment is not zero. It is so small that it was not
recorded by the thermometers available during Joule’s time. However, for gases such as
hydrogen and helium the temperature change is almost negligible. As these two gases
have very weak forces of attraction, their internal pressure, (∂U/∂V)T , will also be
negligible. The result of Joule’s experiment is valid only for ideal gas and not for real
gases.
Enthalpy Change During Isothermal Volume Change of an Ideal Gas
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H = U + PV
For an ideal gas PV = nRT. Therefore,
H = U + nRT
For a closed system, n being constant, the differential of the above equation gives
dH = dU + nRdT
(3.31)
When the temperature is constant, dU = 0 and dT = 0. Therefore, for ideal gas at
constant temperature dH = 0
The change in enthalpy during isothermal expansion or compression of an ideal gas
is equal to zero.
Since H = f(T,P), the change in enthalpy is given by
(3.19)
For isothermal process, dH = 0 , therefore,
Since the temperature is constant, dT = 0. Therefore,
During isothermal volume change, dP ≠ 0.Hence,
= 0 for an ideal gas
(3.32)
Heat Exchange During Isothermal Process
According to the first law of thermodynamics
dU = dq + dw
For an isothermal process involving an ideal gas, dU = 0.
Therefore,
dq = –dw
and hence,
q=–w
For an isothermal reversible volume change for an ideal gas
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q=
(3.33)
For an isothermal irreversible volume change for an ideal gas,
q=
(3.34)
Here it is assumed that the external pressure on the gas is equal to the final pressure of
the gas.
The process during which there is no exchange of heat between the system and its
surrounding is known as an adiabatic process. For a finite change, q = 0.
Therefore, from the First law of thermodynamics
ΔU = w
(3.35)
For an infinitesimal change, dq = 0, therefore,
dU = dw = –PdV
Since dw = dU, the work in adiabatic process is accompanied with the change in energy
(hence its temperature) of the system. Let n be the amount of an ideal gas. Let the
initial and the final state of the gas be shown as the following
Initial State
Pressure
P1
Volume
V1
Temperature
Final State
P2
V2
T1
T2
Reversible process :The gas expands in an infinite number of steps. The work done
during any step of the process is given by
dw = –PexdV
(3.36)
Hence,
For an ideal gas, P is replaced by
.
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Therefore,
Since, both T and V change, w cannot be calculated directly by integration. However,
this can be determined by using the expression, dw= dU.
Since, U = f ( T, V), and (∂U/∂T)V= CV, the expression of dU is
For an ideal gas, since(∂U/∂V)T= 0, therefore,
dU = CVdT
(3.37)
On integrating Eq. (3.37) we get
Hence, the expression for the work is
Note: For the expansion, w is negative, and thus, T2 < T1, i.e., the expansion is
accompanied with the decrease in temperature. For compression the reverse is
observed.
For calculating the enthalpy change, the equation H = U + PV is considered. At a
constant pressure, for a finite process the enthalpy change is given by the equation
ΔH = ΔU + PΔV = CV ΔT + nRΔT
ΔH = (CV + nR) ΔT
For an ideal gas (CV + nR) = Cp, therefore,
ΔH = CpΔT
(3.38)
Relationship Between the Temperature, Pressure and Volume for an Adiabatic Reversible
Process
During an adiabatic expansion and compression of an ideal gas, the temperature,
pressure and volume change simultaneously. Therefore, the relationship between any
two of these variables cannot be calculated directly from the ideal gas equation.
Derivation of T–V Relation for an Adiabatic Reversible Process
Cv dT = – PdV
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(3.39)
Thermodynamics
For an ideal gas, P =
Cv dT = –
, therefore,
dV
=–
(3.40)
nR = Cp– Cv and Cp/ Cv = γ
Therefore,
nR / Cv = Cp– Cv/ Cv = Cp/ Cv – 1 = γ – 1
d ln T = –( γ – 1) d ln V
d ln T = – d ln
On integrating between the limits T = T
ln
1
, T = T 2and V = V1, V = V2 , we get
= ln
(3.42)
On taking antilog we get
=
or,
T2 = T1
(3.43)
Derivation of T– P relation for an Adiabatic Process
Cv dT = – PdV
For an ideal gas, PV = n RT
Differentiating the Equation we get,
P dV + V dP = nRdT
or,
– PdV = V dP – nRdT =
dP – nRdT
(3.43)
From Eqs (3.41) and (3.43) we get
(Cv+ nR)
= nR
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or,
d ln T =
d ln P = dln (P ) (γ –1) / γ
where Cp/ Cv = γ and γ – 1 = nR / Cv = (Cp – Cv)/ Cv
On integrating between the limits T = T
ln
1
,T=T
2
and P = P1, P = P2 ,we get
= ln
On taking antilog we get
=
T2 = T1
(3.44)
Derivation of V – P Relation for an Adiabatic Process
CvdT = – PdV
(3.41)
For ideal gas PV = n RT and
dV + V dP = n R dT
dV +
or,
dP = n R dT
dT = T
+T
= T (d ln V + d ln P)
On substituting dT in to Eq. (3.41) we get
CV x T (d ln V + d ln P) = – PdV = –
or,
dV = – nRT d ln V
(CV + nRT) d ln V = – CV d ln P
d ln (V)γ = – d ln P
or,
On integrating between the limits V = V1 , V = V2 and P = P1, P = P2 then taking
antilog we shall get
=
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and
P1
= P2
= constant
PVγ= constant
or,
(3.45)
Another method of Derivation of T-P and P-V Relations for an Adiabatic Process
T-P relation can be derived directly from T-V relation and ideal gas equation.
According to Eq. (3.42)
T2 = T1
(3.42)
For an ideal gas
(3.46)
(3.47)
Substituting the value of Eq. (3.47) in Eq. (3.42) we get
Rearranging the equation,
or
or
or
Similarly, P-V relation can be derived from Eq. (3.42) and Eq. (3.46) by substituting the
value of T2/T1 in terms of P and V.
From Eq. (3.42)
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Therefore,
or
P1
= P2
= constant
Expression for an Adiabatic Reversible Work Done in Terms of P and V
For an ideal gas,
If the initial temperature is not known then, this expression can be written in terms of P,
V and γ as follows:
Since
(3.48)
3.3.4 Comparison of Isothermal Reversible Work Done and Adiabatic Reversible
Work Done for an Ideal Gas
Adiabatic and isothermal reversible work done by an ideal gas cannot be compared
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directly as the volume change for the same external pressure will be different in the two
cases. That is, if an ideal gas is expanded from same initial state under the two
conditions, then the final state will be different even if the final pressure is same.
Initial State
Final State
Isothermal reversible expansion
P1,V1, T1
Piso , V2 ,T1
Adiabatic reversible expansion
P1,V1, T1
Pad ,V2 ,T2
For isothermal expansion
P1V1 = Piso V2
or
(3.49)
For adiabatic expansion P1
= Pex
(3.50)
Comparing Eq.s (3.49) and (3.50)
Taking logarithm,
For an ideal gas γ > 1,
Therefore,
or
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Figure 3.11: Pressure – volume work done during reversible isothermal
and adiabatic reversible expansion
Also, during adiabatic expansion, the final temperature is less than the initial
temperature. Therefore, Vad will be less than Viso. This can be graphically
represented as shown in Figure 3.11.
Similarly, adiabatic reversible and isothermal reversible processes can be
compared for the case when the final volume after expansion is same in both the
cases. In such case the final pressure for adiabatic and isothermal expansion will
be different.
Initial State
Final State
Isothermal reversible expansion
P1,V1, T1
Piso , V2 ,T1
Adiabatic reversible expansion
P1,V1, T1
Pad ,V2 ,T2
For isothermal expansion P1V1 = Piso V2
(3.51)
For adiabatic expansion P1
= Pad
(3.52)
During expansion V2 > V1 , also for an ideal gas γ > 1.
Therefore,
>
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or
>
Therefore, Piso > Pad
The
final
pressure
during
isothermal reversible expansion is
greater than the final pressure
during
adiabatic
reversible
expansion. This can be graphically
represented as shown in Fig. 3.12.
The magnitude of the work done
by the gas is equal to the area
under the curve. As seen from the
Fig. 3.11 and 3.12, the area under
the adiabatic curve is less than the
area under the isothermal curve.
Hence, the magnitude of work done
is larger in isothermal reversible
expansion as compared to the
corresponding work in adiabatic
reversible expansion.
Figure 3.12: Pressure – volume work during
reversible isothermal and adiabatic
reversible expansion
Expression for Final Temperature in Adiabatic Irreversible Process (T2irr)
For irreversible adiabatic process
Cv(T2irr – T1) = – Pex(V2– V1)
If Pex = P2, then
Cv(T2irr – T1) = – P2(V2– V1)
For ideal gas,
PV = nRT
Therefore,
Cv(T2irr – T1) =
Rearranging the terms we get
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(3.53)
3.3.5 Process Involving Change in the State of the System
The change in the state of the system can be vaporization, fusion, sublimation and solid-solid transition
constant temperature; therefore, they are isothermal processes. However, the change in internal energ
(∆H) are not zero as during the change in state of the system internal energy is involved.
The work done during the phase change = w = −P∆V
When the state change involves the vapour state then the change in volume is appreciable. Hence,
However, when only condensed states are involved in the process then the change in volume is very sm
negligible.
The heat exchanged during the state change is equal to heat of transition. For solid → liquid, q is eq
vapour, q is equal to heat of vapourization. Since the change in state takes place at constant pressure,
enthalpy change for the process. That is,
q = qp = ∆H
∆U = q + w
∆U = q −P∆V
The change in thermodynamic properties when an ideal gas undergoes volume changes
are summarized in Table 3.1.
Table 3.1 Summary of Changes in the Thermodynamic
Properties
During Volume Changes Process Work
Process
Work Done(w)
Isothermal
reversible
volume
change
Isothermal
irreversible
volume
change
Adiabatic
reversible
volume
change
Adiabatic
irreversible
volume
change
Heat(q)
Change in
Change In
Internal Enthalpy(ΔH)
Energy(ΔU)
q=–w
0
0
0
0
q=0
Cv(T2– T1)
Cp(T2– T1)
q=0
Cv(T2– T1)
Cp(T2– T1)
q=
– P2(V2– V1)
Cv(T2– T1) =
– P2(V2– V1)
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aa
Example
One mole of an ideal monoatomic gas at 300 K and 1atm is allowed to expand to the
final pressure of 0.1 atm under three different conditions (i) isothermal reversible (ii)
isothermal irreversible against external pressure of 0.1 atm (iii) adiabatic reversible.
Calculate work done for each process. R = 8.314 J K-1 mol-1.
Solution:
The given data are :
n = 1 mol; T1 = 300 K;
P1 = 1.0 atm P2 = 0.1 atm
(i) Isothermal reversible process
T1 = T2 = 300 K
=–
5744.14 J
(ii) Isothermal irreversible process
=–
2244.78 J
(iii) Adiabatic reversible process
and
T2 = T1
Here γ = Cp / Cv and for monoatomic ideal gas Cp,m = 5R/ 2 and Cv,m = 3R/2 = 1.5R
γ = 5/3
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(γ – 1)/ γ = 1– 3/5 = 2/5
T2 = 300 K
=119.4 K
= (1 mol)( 1.5 x 8.314 J K-1 mol-1)( 119.4 K – 300 K ) = – 2252.26 J
Example
One mole of an ideal gas is heated at constant pressure from 0°C to 100°C. Calculate w,
q, ΔU and ΔH. Given Cp,m = 3.5 R.
Solution:
On heating the gas will expand against constant pressure. Therefore, the process is
irreversible.
The given data are T1 = 0°C = 273 K ; T2 = 100°C = 373 K ; n = 1mol
The expression of work is
w = −Pex(V2 − V1)
For an ideal gas V = nRT / P. Hence
w = −Pex{(nRT2/ P2) − (nRT1/ P1) }
Since the expansion occurs at constant pressure, Pex = P1 = P2.
Hence,
w = – nR(T2 – T1) = –(1 mol )( 8.314 J K-1mol-1)(373 K – 273 K ) = –831.4 J
∆U = nCV,m(T2 –T1)
Cp,m – Cv,m = R
Cv,m = Cp,m – R = 3.5 R –R = 2.5 R
∆U = 1 x 2.5 x 8.314 JK-1mol-1( 373 K – 273 K ) = 2078.5 J
q = ∆U – w = 2078.5 J –(–831.4 J) = 2909.9 J
∆H = nCp,m(T2 –T1) = (1 mol)( 3.5 x 8.314 J K-1 mol-1)( 373 K – 273 K ) = 2909.9 J
Example
One mole of an ideal gas is held by a piston at 273 K and under a pressure of 10 kPa.
The pressure is suddenly released to 0.4 kPa and the gas is allowed to expand
isothermally. Calculate q, w, ∆U and ∆H for the process.
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Hint: The change is irreversible.
Solution:
Since the gas is expanding against constant pressure the process is irreversible in
nature. The work done by an ideal gas under isothermal irreversible conditions is given
by the expression
V1 = nRT/ P1 and V2 = nRT / P2
The data given are: T = 273 K ; P1 = 10 kPa = 104 N m-2
P2 = 0.4 kPa = 0.4 x 103 N m-2 ; n = 1mol
=–
2178.93 J
For an ideal gas under isothermal conditions ∆U = 0 and ∆H = 0
According to first law of thermodynamics
∆U = q + w = 0
or q = –w = 2178.93 J
Example
Two moles of benzene are converted into its vapour at its boiling point 80°C, by
supplying heat. The vapours expand against a constant pressure of 1 atm. The enthalpy
of vapourization of benzene is 395 J g−1. Calculate q, w, ∆U and ∆H for the process.
Solution:
Given : Benzene(l) → Benzene(v)
n = 2 mol, T = 80°C = (273 + 80) K = 353 K
Molar mass of benzene = 78 g mol−1
Enthalpy of vapourization = 395 J g−1
Enthalpy of vapourization per mol = 395 J g−1 x 78 g mol−1
Total heat involved = q = n (395 Jg−1 x 78 g mol−1 )
= (2 mol )( 395 Jg−1 )( 78 g mol−1) = 61620 J
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w = −P∆V = P( Vv−V l)
Since, V v≫ V l , P( V v−V l)≈ P Vv = nRT.
Hence,
w = − nRT =(2 mol )( 8.314 J K mol−1 )( 353 K) = −5869.68 J
∆U = q + w = 61620 J + (−5869.68 J) = 55750.32 J
∆H =q = qp = 61620 J
3.4 Thermochemistry
Thermochemistry is a part of thermodynamics that deals with the changes in energy
during a chemical reaction or a physical transformation. Generally, these changes are
mentioned in terms of the heat exchanged during the process.
Heat is the energy in transit and it flows from a system at a higher temperature to
one at lower temperature. It is also defined as the energy transferred across the
boundary between a system and its surroundings by the virtue of the temperature
difference. By convention, the heat absorbed or evolved during a process is called the
heat exchanged during the process. Heat exchanges can also take place during the
physical or the chemical processes such as the change of state, a chemical reaction, or
even a simple process such as dissolution. Heat and energy are used interchangeably
because heat exchange at a constant volume is equal to the change in the internal
energy and the heat exchange at constant pressure is known as the change in enthalpy.
Thus,
qv = ΔU and qp = ΔH
Thermochemistry enables us to predict the amount of heat that would be evolved or
absorbed in a process without actually performing a tedious set of experiments in the
laboratory. We can also calculate the energy changes for the processes that are not
feasible experimentally. This is done by taking advantage of the fact that although heat
is a path function, the heat exchanges become state functions when the measurements
are made under the conditions of constant volume or constant pressure.
Under these conditions they are dependent only on the initial and final states of the
system and hence can be added, subtracted or multiplied by an integer according to the
algebraic laws.
The measurements of the heat exchanges at a constant volume are made in a bomb
calorimeter and the heat exchange measurements at a constant pressure can be made
under ordinary laboratory conditions in a normal calorimeter. Since the constant
pressure is easy to maintain and the heat measurements are also simpler in a normal
calorimeter, therefore, the heat exchanges measurements in the terms of enthalpy
changes are preferred. Heat involved in a process is calculated by measuring the change
in the temperature and is given by the expression
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q=m x s
x ΔT
where, m = mass of subsantace, s = specific heat capacity of subsantace and T = T2 –
T1
3.4.1 Measurement of Change in Internal energy of the system (ΔU)
The change in the internal energy during a chemical reaction is measured using a bomb
calorimeter. The reaction is carried out in a container whose volume remains constant.
This container is made up of a metal, which is generally steel lined with gold. It is known
as the bomb (Fig. 3.13). In this experiment, a known mass of the reaction materials are
taken in the bomb. The reaction material is the system. The bomb is immersed in a
vessel containing a known mass of water. The calorimeter and the water act as its
surroundings. The water is stirred by the rotating blades and its initial temperature (T1)
is recorded.
As the reaction takes place, there is a change in the chemical energy. If the
energy is released during a process, it is transferred to the surroundings as heat. This
heat is used to warm water and the calorimeter. When the heat is absorbed during the
reaction, the calorimeter gives the heat to the system, so the temperature of the
calorimeter decreases. When the temperature of water reaches a constant value, the
final reading of thermometer (T2 ) is noted.
The change in temperature of water and the calorimeter (surroundings) = ∆T
= T2 – T1
The heat exchanged between the system (reaction) and its surroundings is
calculated from the equation
q (water + calorimeter)= (m x s + C) x ΔT = q (surroundings)
where, m = mass of water, s = specific heat capacity of water and C = heat capacity of
the calorimeter.
Figure 3.13: Bomb Calorimeter
The heat liberated in the reaction is calculated as the heat appearing in the
surroundings. During the process the volume remains constant so no work is done.
Thus,
∆U = q (system) = – q (surroundings) at constant volume
Difficulties in the experimental determination of ΔU
If a large quantity of heat is liberated in a reaction then there is a danger of explosion
since the experiment is performed at constant volume in a bomb calorimeter. Why?
Because the heat liberated during the reaction will make the gaseous products to
expand. Since the vessel is closed, the gas cannot expand. This will cause a build up of
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high pressure inside the bomb calorimeter. If the walls of the vessel are not strong
enough to withstand the high pressure developed, an explosion may take place.
3.4.2 Measurement of Change in Enthalpy of the System (ΔH)
In the laboratories since it is easier to maintain a constant pressure as compared to the
constant volume, most of the thermochemistry experiments are carried out under
constant pressure conditions. The heat exchange is measured using a normal
calorimeter that has adiabatic walls. The simple calorimeter is a glass vessel, usually a
glass beaker which is well insulated. It is fitted with a glass stirrer and a thermometer.
The set up is kept inside a Dewar flask or thermos flask for insulation. The glass vessel
(beaker) is kept inside it on a rubber cork. The reactants are mixed in it and the change
in the temperature is measured, using a 0.1oC thermometer. For accurate work, more
sensitive devices such as a Beckmann thermometer, platinum resistance thermometer
or thermocouple may be used. The experimental set up is shown in Fig. 3.14.
Figure 3.14: Calorimeter setup for the measurement of enthalpy change
Some times the Dewar flask itself is used as the vessel in which the reaction is carried
out. This set up is commonly used in the laboratory for determination of enthalpy
reactions involving solutions.
The enthalpy change is determined by measuring the temperature change before and
after the process. The heat gained or lost by the system is manifested by the change in
temperature of the solution and the container which are assumed to be the
surroundings. It is also assumed that no heat is lost during the exchange of heat
between the system and the surroundings. Therefore,
qsystem = − qsurroundings
During the process if the system gains heat then the surroundings will lose heat and the
final temperature recorded by the thermometer will be less than the initial temperature.
Similarly, if the system loses heat then the heat lost is gained by the surroundings and
the temperature rises.
qsurroundings
=
heat
the
exchanged
by
the
solution
+
heat
exchanged
by
calorimeter.
Since the process takes place at a constant pressure, the heat exchanged is equal to the
enthalpy change. Therefore,
∆Hsystem = − ∆Hsurrounding
∆Hsurrounding = ∆Hsolution + ∆Hcalorimeter
At a constant pressure
∆H = Cp ∆T , here ∆T = T2 −T1
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∆Hsolution= Cp (solution) x ∆T = m x s x ∆T
where, m is the mass of the solution and s is the specific heat capacity of the solution. If
the solution is aqueous solution, s is taken as equal to the specific heat capacity of
water. For the calorimeter,
∆Hcalorimeter = Cp (calorimeter) x ∆T
Therefore, the ∆Hsurrounding becomes equal to
∆Hsurrounding= Cp (solution) x ∆T + Cp (calorimeter) x ∆T
Since the temperature change is same for the solution and calorimeter,
∆Hsurrounding= {Cp (solution) + Cp (calorimeter)} x ∆T
or
∆Hsystem = −{Cp (solution) + Cp (calorimeter)} x ∆T
(3.54)
Sign Convention
If the heat is absorbed by the system the process is known as an endothermic process.
Since the energy of the system increases when it absorbs heat, ΔU and ΔH are given
positive sign.
If the system loses heat the process is known as an exothermic process. ΔU and ΔH
have negative sign since the energy of the system decreases.
3.4.3 Standard State
The standard state for a substance is defined for the substance in its pure state and the
most stable form of that substance at a standard pressure of one bar and a specified
temperature. Although the standard state definition is valid for all temperatures,
conventionally the data is mentioned for 298.15 K. However, in most cases the
temperature for which the standard state is defined is mentioned in the brackets.
For gases the standard state is the pure gas behaving ideally at one bar and
specified temperature. For liquids the standard state is the pure liquid at one bar and
specified temperature.
For solid it is the pure and most stable crystalline form of aggregation of solid at
one bar and specified temperature. For example, graphite is taken as the stable form of
carbon and rhombic form is the most stable form for sulphur.
For solution it is one mol dm−3 solution at one bar and the specified temperature.
The standard states can be summarized in Table 3.2.
Table 3.2 Description of Standard States
S.No.
Phase of the
substance
1.
Gas
2.
Liquid
Pure liquid at 1 bar and a specified temperature
3.
Solid
Pure and in the most stable state of
aggregation or stable allotropic form of solid at
1 bar and a specified temperature
4.
Solution
One mol dm−3 solution at 1 bar and a specified
temperature
Condition
Pure gas at 1 bar and a specified temperature
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3.4.4 Heat of Reaction
Heat of a reaction is defined as the heat exchanged between the system and the
surroundings during the complete transformation of reactants to products at a constant
temperature and pressure for the given stoichiometrically balanced equation. During a
chemical process some bonds break and others are formed so there is a net change in
the energy of the system. This results in the heat exchange during a process. This heat
exchange can be measured if the energy of initial state and final state is known. The
heat exchanges are measured either at a constant pressure or at a constant volume.
When the reaction is carried out at a constant volume, the heat exchange is
measured in terms of the internal energy change. It is known as the heat of reaction at
constant volume, qv or energy of reaction. The symbol is Δr U. The energy change is
given by the equation
Δr U = Ufinal − Uinitial
When the reaction is carried out at a constant pressure, the heat exchange is measured
in terms of the enthalpy change. It is known as heat of reaction at constant pressure, qp
or enthalpy of reaction. The symbol is Δr H.
The enthalpy of reaction is given by the equation
Δr H = Hfinal −Hinitial
Since the measurements are generally made under constant pressure conditions, the
heat of reaction is normally referred as enthalpy of reaction.
The enthalpy of reaction is the enthalpy change during a reaction. Since enthalpy is a
state function, the change in enthalpy is equal to the difference in the enthalpies of the
final state and the initial state. For a general reaction
ν A A + ν BB → ν C C + ν DD
The reactants are the initial state and the products form the final state assuming that
the reaction has been completed. Therefore,
ΔrH = Total enthalpy of products – total enthalpy of reactants
=
ΔrH = (νCHm,C + νD Hm,D) –( νA Hm,A + νBHm,B)
Here, Hm,A = molar enthalpy of A
Hm,B = molar enthalpy of B
Hm,C = molar enthalpy of C
Hm,D = molar enthalpy of D
νA, νB, νC and νD are stoichiometric coefficients of A, B, C and D, respectively in the
balanced chemical equation
To get the complete information about the reaction regarding its enthalpy changes,
the reaction is written along with the following information:
i.
The physical state of each species is mentioned in the brackets after its molecular
formula in the balanced chemical equation.
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ii.
The total enthalpy change for the process is written along with the sign of
enthalpy change.
The unit of enthalpy change of the reaction is kJ mol−1.
The temperature and pressure at which the reaction is carried out is written.
iii.
iv.
For example, during the formation of CO2, one mole of carbon in its solid state reacts
with one mole of oxygen gas to form one mole of carbon dioxide. The heat produced at
constant pressure and 298 K is 393.5 kJ. This information is represented as
C(s) + O2(g) → CO2(g)
Δr H(298 K,1 atm) = −393.5 kJ mol−1
When no pressure is mentioned, it is assumed that the reaction is taking place at
atmospheric pressure, that is, 1 atm.
When the reactants and the products are in their respective standard states, the
enthalpy of reaction is known as standard enthalpy of reaction Δr Hθ. If the reaction
takes place at 298 K the standard enthalpy of reaction is written as Δr Hθ(298 K)
Δr Hθ(298 K) = Δr H( 298 K, 1 bar)
While defining the heat of a reaction, the nature of the reaction has not been mentioned.
Therefore, the term can be used for all the types of reactions. However, these reactions
can be classified into various categories and for each category the heat exchange is
given a specific name. Since, these heat exchanges are generally studied at a constant
pressure, these will be referred to in terms of enthalpy changes in this chapter. The
reactions involving the transformations are reported for 1 mol of the substance to
standardize the values of the enthalpy changes.
3.4.5 Enthalpy of Formation
There are many ways in which a substance can be formed and for each such reaction the
heat of reaction is different. In many of the reactions more than one product is formed.
The values of the heat of reactions for all the reactions in which a given substance is
formed cannot be taken as the heat of formation of that substance. In order to define a
single value of heat of formation, the chosen reaction should be such that only one mole
of the substance is formed from its constituent elements and no other product is formed.
So the enthalpy of formation can be defined as the amount of heat exchanged at
constant temperature and pressure during the formation of one mole of the
substance from its constituent elements. It is represented by Δf H. Its unit is kJ
mol−1.
For example, the enthalpy of formation of CO2 is equal to the enthalpy change for
the following reaction
C(s) + O2(g) → CO2(g)
Δf H = −393.5 kJ mol−1
Similarly, the enthalpy of formation of water is equal to the enthalpy change for the
reaction
H2(g) + ½O2(g) → H2O(l)
Δf H = −285.8 kJ mol−1
Note that although water is formed during the neutralization process also, it is not taken
as the formation reaction of water as it is not formed from its elements in this reaction.
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There are many ways in which a substance can be formed and for each such reaction the
heat of reaction is different. In many of the reactions more than one product is formed.
The values of the heat of reactions for all the reactions in which a given substance is
formed cannot be taken as the heat of formation of that substance. In order to define a
single value of heat of formation, the chosen reaction should be such that only one mole
of the substance is formed from its constituent elements and no other product is formed.
So the enthalpy of formation can be defined as the amount of heat exchanged at
constant temperature and pressure during the formation of one mole of the
substance from its constituent elements. It is represented by Δf H. Its unit is kJ
mol−1.
For example, the enthalpy of formation of CO2 is equal to the enthalpy change for
the following reaction
C(s) + O2(g) → CO2(g)
Δf H = −393.5 kJ mol−1
Similarly, the enthalpy of formation of water is equal to the enthalpy change for the
reaction
H2(g) + ½O2(g) → H2O(l)
Δf H = −285.8 kJ mol−1
Note that although water is formed during the neutralization process also, it is not taken
as the formation reaction of water as it is not formed from its elements in this reaction.
Standard enthalpy of formation is defined as the amount of heat exchanged at a given
temperature and a pressure of one bar during the formation of one mole of the
substance in its standard state from its constituent elements in their respective standard
states. It is represented by Δf Hθ.
Enthalpy of formation of an element
By convention the enthalpy of formation of an element in its standard state is taken as
zero at all temperatures. Hence, whether the element is a solid, a liquid or a gas its
standard enthalpy of formation is taken as zero. For example, hydrogen gas, solid zinc,
rhombic sulphur, liquid bromine and graphite all
have Δf Hθ(298 K) = 0.
While dealing with the ionic species in the solution, the standard enthalpy of
formation of H+ ions is also taken as zero.
3.4.6 Laws of Thermochemistry
These laws are based on the fact that heat exchanges are measured either at constant
volume or under constant pressure. Under these conditions heat is a state function, that
is, ΔU or ΔH respectively. These values are independent of the path a process takes and
depend only on the initial and final states. These laws are hence the direct consequence
of law of conservation of energy.
Laplace Lavoisier Law-First Law of Thermochemistry
The enthalpy of formation of any compound is equal in magnitude and of opposite sign
to the enthalpy of dissociation of that compound at the given temperature and pressure.
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Dissociation is the reverse of formation. So when the reaction is reversed, the sign
of enthalpy change is also reversed but the magnitude of the enthalpy change remains
same. This law has been generalized now for all the types of reactions. When a reaction
is reversed, the sign of the enthalpy of reaction is reversed.
Hess’s Law of Constant Heat Summation-Second Law of Thermochemistry
The total enthalpy change of a reaction is the same regardless of whether the reaction is
completed in one step or in several steps. For example, for a change from reactant to
product that can take place in four steps or a single step, the total enthalpy change will
be same.
Single step process
Reactant → Product ΔH
Multiple step process
Reactant → A
ΔH1
A → B
ΔH2
B → C
ΔH3
C →
ΔH4
Product
According to Hess’s law
ΔH = ΔH1 + ΔH2 + ΔH3 + ΔH4
This can also be shown as in Fig. 3.15
Figure 3.15: Illustratation of Hess’s law
The following rules are derived from the two laws of thermochemistry
•
•
•
•
We can reverse any equation and the sign of ΔH is also reversed.
The chemical equations can be added or subtracted. The ΔH of the resulting
equation is sum or difference of the ΔH of the chemical equations.
The chemical equation can be multiplied or divided by an integer and the ΔH is
also multiplied or divided by that integer.
The identical terms on both the sides of the equation are cancelled.
These laws help us to theoretically calculate the enthalpy changes of processes which
cannot be easily carried out in a laboratory, for example, to determine the enthalpy
change for the process
C(graphite)+ ½ O2 (g)→ CO(g) ΔH
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This process when carried out in the laboratory would also generate CO2. However, the
following two processes can be carried out quantitatively.
(i) C(graphite)+ O2 (g)→ CO2(g)
(ii) CO(g) + ½ O2 (g)→ CO2(g)
ΔH1 = −393.3 kJ mol−1
ΔH2 = −282.8 kJ mol−1
If the Eqs (ii) is reversed, we get,
(iii) CO2(g) → CO (g)+ ½ O2 (g)
ΔH3 = − ΔH2
Add Eqs. (i) and (iii)
C( graphite)+ O2 (g) + CO2(g) → CO2(g) + CO (g)+ ½ O2 (g)
or
C(graphite)+ ½ O2 (g)→ CO(g)
ΔH = ΔH1 + ΔH3 = ΔH1 – ΔH2
ΔH = −393.3 kJ mol−1+ (−282.8 kJ mol−1) = −110.5 kJ mol−1
Hence, the enthalpy change for a reaction that cannot be carried out quantitatively in
the laboratory can be theoretically calculated.
Calculation of Enthalpy of a Reaction from Enthalpies of Formation of Reactants
and Products
We have seen that the enthalpy of a reaction is calculated from the molar enthalpies of
the products and the reactants. However, ∆rH can also be calculated from the enthalpies
of formation values of all the reactants and the products involved in the reaction. Let us
consider the following reaction:
(A) CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
By definition,
Δr Hθ;= Total enthalpy of products – total enthalpy of reactants
= ∑ Hθ(products) - ∑ Hθ (reactants)
Δr Hθ = {Hθm(CO2) + 2Hθm(H2O)} – {Hθm(CH4) – 2Hθm(O2)}
The reaction A can be carried out in the following steps:
(i) C(graphite)+ 2H2 (g) → CH4(g)
(ii) O2(g) → O2(g)
(iii) C(graphite)+ O2 (g)→ CO2(g)
(iv) H2 (g) +½ O2(g)→ H2O(l)
∆f H (CH4)
∆f H (O2)
∆f H (CO2)
∆f H (H2O)
According to Hess’s law,
Eq. (A) = {Eq. (iii) + 2Eq.(iv)} –{Eq. (i) + 2Eq. (ii)}
Therefore, ∆H(A) = {∆fH (CO2) + 2∆fH (H2O)} −{∆fH (CH4) + 2∆fH (O2)}
Therefore, for a reaction
ν A A + ν BB → ν C C + ν DD
The enthalpy of reaction can be written as
ΔrH = {νC Δf Hθ(C) + νD Δf Hθ(D)}−{ νAΔf Hθ(A) + νB Δf Hθ(B)}
(3.55)
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= Total enthalpy of formation of products – Total enthalpy of formation of reactants
Hence, from the tabulated Δf H values of the substances at a given temperature, the
enthalpy of any reaction can be calculated using the above formula. The standard
enthalpy of reaction can be calculated from the tabulated values of Δf Hθ.
Relationship between Δr U and ΔrH
H = U + PV
..(3.16)
For calculating the enthalpy change the equation becomes,
ΔH =ΔU + PΔV
For a chemical reaction the equation becomes
ΔrH = ΔrU + Δr (PV)
(3.56)
For the reactions involving solids and liquids the volume change with the change in the
pressure is negligible and therefore, the change in PV is negligible in comparison to ΔrU.
The Eq. (3.56) gives
Δ r H = Δ rU
For the following reaction involving gases,
νA A(g) + νBB(g) → νCC(g) + νDD(g)
∑ V(reactants)= νA Vm(A) + νB Vm(B)
∑ V(products)= νC Vm(C) + νD Vm(D)
Δr (PV) = Δr (PνgVm )
If all the gaseous species are assumed to behave as an ideal gas, then,
Vm = RT / P
Therefore,
Δr (PV)= Δ(νgRT)
At a constant temperature,
Δr (νgRT ) = RT Δνg
where,
Δν = νC + νD − νA − νB
Therefore, on substituting the value of Δr (PV) in Eq. (3.56),
ΔrH = ΔrU + RTΔνgas
For heterogeneous reactions, the same rule applies; however, here Δνgas is equal to the
change in the stoichiometric number of gaseous species involved in the balanced
chemical equation.
Since
ΔrH = qp and ΔrU = qv
qp = qv + RTΔνgas
(3.57)
The Eq. (3.57) gives the relationship between heat exchanges at a constant pressure
and at a constant volume for reactions involving gases.
Example
Calculate ΔrU for the following reactions at 1 atm and 298K.
(i) N2(g) + 3H2(g) → 2NH3(g)
Δr H = −92.38 kJmol−1
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(ii) C(graphite) + ½O2 → CO(g)
(iii) C(graphite) + O2 → CO2 (g)
Δr H = −110.5 kJmol−1
Δr H = −393.5 kJmol−1
Solution:
From Eq. (3.57), ΔrU = ΔrH − RTΔνgas
R = 8.314 J K−1mol−1
T = 298 K
RT = 8.314 J K−1mol−1x 298 K = 2477 J mol−1 = 2.477 kJ mol−1
(i) N2(g) + 3H2(g) → 2NH3(g)
Δr H = −92.38 kJ mol−1
Δνgas = 2 – (1+3) = −2
Δr U = −92.38 kJ mol−1 – (− 2 x 2.477 kJ mol−1 ) = −87.426 kJ mol−1
(ii) C(graphite) + ½O2 → CO(g) ΔrH = −110.5 kJ mol−1
Δνgas =1 – ½ = ½
Δr U = −110.5kJmol−1 –(½ x 2.477kJ mol−1) = −111.738kJ mol−1
(iii) C(graphite) + O2 → CO2 (g) ΔrH = −393.5 kJ mol−1
Δνgas =1 –1=0
ΔrU = Δr H = −393.5 kJ mol−1
Enthalpy of Combustion
It is the heat exchange that takes place when one mole of a substance is burnt
completely in the presence of oxygen at a given temperature and pressure. For most of
the organic compounds the products of combustion are H2O and CO2. Other elements
undergo combustion to form their respective oxides. It is denoted by ΔcH and the unit is
kJmol−1. The combustion is always an exothermic process.
For example,
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
ΔcHθ(298 K)= −890 kJ mol−1
C2H6 (g) +
ΔcHθ(298 K)= −1560 kJ mol−1
O2(g) → 2 CO2 (g) + 3H2O(l)
The enthalpy of combustion is always measured for the combustion of one mole of a
compound. Therefore, while writing the balanced chemical equation, the stoichiometric
number of the compound undergoing combustion is taken as one and the rest of the
reactants and the products are balanced accordingly.
Example
Apply Hess’s law to calculate the Δf Hθ(C4H10) using the following thermodynamic data:
ΔcHθ(C4H10,g)= −2878.7 kJ mol−1
Δf Hθ(H2O,l) = −285.8 kJ mol−1
Δf Hθ(CO2,g) = −393.5 kJ mol−
Solution:
The required equation is
4C(graphite) + 5H2 (g) → C4H10(g)
Δf Hθ
Given
(i) H2(g) + ½ O2 (g) → H2O(l)
Δf Hθ(H2O,l) = −285.8 kJ mol−1
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Δf Hθ(CO2,g) = −393.5 kJ mol−1
(ii) C(graphite)+ O2 (g) → CO2(g)
(iii) C4H10(g) + 13/2 O2(g) → 4CO2(g) + H2O(l)
ΔcHθ(C4H10,g)= −2878.7 kJ mol−1
Multiply Eq. (ii) by 4 and equation (i) by 5 and add the resulting equations
(iv) 4C(graphite)+ 4O2 (g) + 5H2(g) + 5/2 O2 (g)→ 4CO2(g) + 5H2O(l)
Subtract Eq(iii) from Eq(iv)
4C(graphite) + 5H2 (g) → C4H10(g)
This is the required equation. Therefore,
ΔfHθ(C4H10) = 4 x ΔfHθ(CO2,g) + 5 x ΔfHθ(H2O,l) − ΔcHθ(C4H10,g)
ΔfHθ(C4H10) = 4 x (−393.5 kJ mol−1 ) + 5 x (−285.8 kJ mol−1 ) – (−2878.7 kJ mol−1)
ΔfHθ(C4H10) = (−1574 −1429 + 2878.7) kJ mol−1
ΔfHθ(C4H10) = −124.3 kJ mol−1
Enthalpy of Transition
It is the amount of heat exchanged when one mole of a substance is transformed from
one allotropic form to the other allotropic form at a given temperature and pressure. The
symbol is ∆trsH.
C(graphite) → C(diamond)
Enthalpy of Vapourization
It is the amount of heat absorbed to convert one mole of a liquid to its vapour state at a
given temperature and pressure.
H2O (l) → H2O(g)
ΔvapHθ(298 K) = 44.0 kJ mol−1
Enthalpy of Fusion
It is the amount of heat absorbed to convert one mole of a solid to its liquid state at a
given temperature and pressure.
H2O (s) → H2O(l)
ΔfusHθ(298 K)= 6.0 kJ mol−1
Enthalpy of Sublimation
It is the amount of heat absorbed to convert one mole of a solid to vapour state at a
given temperature and pressure.
H2O (l) → H2O(g)
ΔsubHθ(298 K)= 50.0 kJ mol−1
Enthalpy of Atomization
It is the amount of heat required to convert one mole of a substance into its constituent
atoms in the gaseous state.
C(graphite)
H2(g)
CH4 (g)
C(g)
2H(g)
C(g) + 4H(g)
∆aH(C) =716.68 kJ mol−1
∆aH(H)= 436 kJ mol−1
∆aH = 1664 kJ mol−1
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For an element in its solid state, the enthalpy of atomization is equal to its enthalpy of
sublimation.
Mg(s)
ΔaHθ = 150.2 kJ mol–1
Mg(g)
Integral Enthalpy of Solution
It is the amount of heat exchanged when one mole of a solute is dissolved in a given
amount of solvent at a specified temperature and pressure.
When the solvent is water the dissolution of solute X is represented as
X + n H2O → X.n H2O
This process results in the solution with the solute solvent ratio of 1: n
For example,
(i) HCl (g) + 10 H2O(l) → HCl.10H2O(aq)
ΔH = −69.01 kJ mol−1
(ii) HCl (g) + 40 H2O(l) → HCl.40H2O(aq)
ΔH = −72.79k J mol−1
(iii) HCl (g) + 200 H2O(l) → HCl.200H2O(aq) ΔH = −73.961 kJ mol−1
These values of ΔH show the general dependence of the heat of solution on the amount
of the solvent. As more and more solvent is used the value of heat of solution changes.
As the amount of the solvent increases the resulting solution becomes more dilute and
ultimately it becomes so dilute that further addition of solvent produces no enthalpy
change. This solution is known as infinitely dilute solution
Integral Enthalpy of Dilution
The difference in the enthalpies of solution for Eqs. (i) and (ii) is 3.78 kJ mol−1. That is,
for the reaction
HCl.10H2O + 30 H2O(l) → HCl.40H2O(aq) ΔH = −3.78 kJ mol−1
This process represents dilution of the solution by adding 30 mol of water. The
accompanying enthalpy change is known as the enthalpy of dilution. It is defined as the
amount of heat exchanged when a given amount of solvent is added to the solution of
known concentration at a constant temperature and pressure.
Integral enthalpy of solution is defined as the change in enthalpy when a solution
containing one mole of solute is diluted from one concentration to another concentration
at a constant temperature and pressure. It depends upon the original concentration of
the solution and the moles of solvent added.
Differential Enthalpy of Solution
It is defined as the enthalpy change taking place when one mole of a solute is dissolved
in a very large amount of solution of known concentration so that there is no appreciable
change in the concentration of the solution.
Mathematically, it is written as,
H2,m =
Here n1, n2 are the amounts of solvent and solute, respectively. Differential enthalpy of
solution is determined by plotting the graph of ∆H versus n2 keeping the amount of
solvent constant as shown in Fig. 3.16.
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Figure 3.16: Differential enthalpy of solution
Δsol H = Enthalpy of solution =
=
To determine the enthalpy of solution at a given concentration, a tangent is drawn at the
given concentration. The slope of the tangent is equal to
.This is
known as differential enthalpy of solution. From the graph it is seen that for lower values
of amount of solute, that is, for dilute solution, the graph is almost linear and hence the
slope is constant. So the differential enthalpy of solution is almost constant for dilute
solutions. For higher concentrations the change is no longer a straight line, it becomes
curved and ultimately it is parallel to the x-axis. At this point the slope of the tangent to
the curve is zero. The solution has become saturated and no more amount of solute can
be dissolved in it. Thus, the differential enthalpy of solution is zero for a saturated
solution.
Enthalpy of neutralization
It is defined as the enthalpy change taking place when one mole of H+ ions are
neutralized completely by OH− in dilute solutions at a constant temperature and
pressure. The reaction for neutralization is written as
H+(aq) + OH−(aq) → H2O
The enthalpy of neutralization is denoted by ΔneutH and its value for this reaction is
−57.3 kJ mol−1. This reaction is exothermic in nature. The question arises that there are
a large number of acids and bases what will be the value of neutralization for each acid
base pair? The neutralization process can be classified on the basis of strengths of acid
and base. It is observed that the enthalpy of neutralization of strong acid – strong base
pair is independent of the nature of acid and base.
For example,
HCl (aq) + NaOH (aq) → H2O + NaCl(aq)
ΔneutH = −57.3 kJ mol−1
HNO3 (aq) + KOH (aq) → H2O + KNO3(aq)
ΔneutH = −57.3 kJ mol−1
Since the strong acids, strong bases and their salts are completely ionized in water,
these reactions can be written as
H+ (aq) + Cl−(aq) + Na+(aq) + OH− (aq) → H2O + Na+(aq) + Cl−(aq)
Since Na+ and Cl− are present on both sides of the equation they cancel out and the net
reaction is
H+ (aq) + OH− (aq) → H2O
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Similarly, all the strong acid strong base pair have the same net neutralization reaction.
Therefore, their enthalpy of neutralization is also same provided the solutions are dilute.
The enthalpy of neutralization also depends upon the concentrations of the acid
and base solutions. If the solutions are concentrated then the enthalpy change during
the process will involve ΔneutH for strong acid-strong base pair as well as enthalpy
change due to dilution. Therefore this value is different from
−57.3 kJ mol−1.
When either the acid or the base or both are weak then the enthalpy of
neutralization depends upon the nature of acid base pair. The reason is that the weak
acid and the weak base are not completely ionized in water. Their ionization requires
some heat which comes from heat evolved during neutralization reaction. For example,
for the neutralization of acetic acid with NaOH
CH3COOH(aq) + NaOH(aq) → CH3COONa (aq) + H2O
∆neutH = ?
This reaction takes place in two steps
CH3COOH(aq) → CH3COO−(aq) + H+(aq)
H+ (aq) ) + OH− (aq) → H2O
ΔionH
ΔH = −57.3 kJ mol−1
Overall reaction is
CH3COOH(aq) + OH− (aq) → H2O + CH3COO−(aq)
The overall enthalpy change is calculated using Hess’s law
ΔneutH = ΔionH + ΔH = ΔionH+ ( −57.3 kJ mol−1)
Heat is absorbed during ionization, so the enthalpy of neutralization of a weak acid is
less than −57.3 kJ mol−1. Since ΔionH depends upon the strength of acid the enthalpy
change will be different for different weak acids
Every reaction involves breaking and making of bonds between the atoms. If the
energies involved in these processes can be determined the net amount of enthalpy
change for the reaction Δr H can be estimated.
3.4.7 Bond Energies
Energy is given to the system to break a bond and the energy is released whenever a
bond is formed. Therefore, bond formation is an exothermic process while bond
dissociation is an endothermic process. For example, when one mole of hydrogen gas is
formed from its atoms in gaseous state, 436 kJ of energy is released; however, 436 kJ
of energy is required to break one mole of H−H bonds.
H(g) + H(g) → H2(g)
ΔrH = -436 kJ mol-1
H2(g) → H(g) + H(g)
ΔrH = 436 kJ mol-1
The energy required to break a bond is known as the bond dissociation energy. Since
the measurements are made at a constant pressure, the heat required to break the
bond is known as the bond dissociation enthalpy and the heat released during bond
formation is known as the bond formation enthalpy.
In our studies, the use of bond dissociation enthalpy is preferred over bond
formation enthalpy since it is easier to break the bonds to form gaseous atoms.
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Measurement of the bond formation enthalpy requires the difficult task of isolating
gaseous atoms and then combining them to form the molecules.
In a simple diatomic molecule only one type of bond is broken. However, in a polyatomic
molecule the energy required to break a bond will depend upon the presence of other
atoms or group of atoms. For example, the bond dissociation enthalpies of the two O−H
linkages in the water molecule are different.
H−O−H(g)→ H(g) + OH(g)
ΔH- (298 K) = 498 kJ mol-1
O−H(g) → O(g) + H(g)
ΔH- (298 K) = 430 kJ mol-1
Similarly, breaking up of an O−H bond in ethyl alcohol molecule or acetic acid molecule
would involve different quantitiest of heat. However, while tabulating the data of
enthalpy required for breaking the bond, generally the net average of various values is
taken. Therefore, there are two types of enthalpies of dissociation, one is the exact
value for a specific bond and the other is the average value for a bond. These are known
by different names; bond dissociation enthalpy and bond enthalpy.
Bond dissociation enthalpy: It is the enthalpy required to break a specific bond in a
specific molecule at a given pressure. It depends upon the presence of other groups in
the molecule.
Bond enthalpy: It is the average of dissociation enthalpies of a given bond in a series
of different types of dissociating species.
In the case of diatomic molecules, the bond enthalpy and bond dissociation enthalpy
are the same.
The convention is to represent bond enthalpy by є (A−B).
Relationship between Bond Enthalpy and Bond Energy
The enthalpy of a reaction is related to the internal energy change of that reaction
by the equation
Δr H = ΔrU + RTΔvgas
Similarly, for the dissociation of A-B bond
A-B(g) → A(g) + B(g)
The bond enthalpy is related to the bond energy by the equation
є (A-B) = ΔU (A-B) + RTΔvgas
For this reaction, Δvgas = 2-1 = 1 and at 298 K
RTΔvgas = 2.48 kJ mol-1.
This value is almost negligible compared to bond enthalpy values. So the bond
enthalpy is approximately equal to the bond energy and the two are used
interchangeably.
Calculation of Bond Enthalpy
In a simple molecule such as methane all the bonds are identical. It is assumed that the
enthalpy required to break each bond is equal and this is equal to bond enthalpy of C−H
bond. So the total enthalpy required to break all the bonds and form gaseous atoms is
equal to four times the bond enthalpy of C−H bond. That is 4 × ∆H (C−H) = ∆H for the
reaction
CH4(g) → C(g) + 4H(g)
∆H
∆H for this reaction can be calculated using the following data:
(i) C (graphite)
C(g)
∆H1 = 716.68 kJ mol-1
(ii) ½ H2 (g)
H(g)
∆H2 = 217.97 kJ mol-1
(iii) C (graphite) + 2H2 (g) → CH4(g)
∆H3 = -74.75 kJ mol-1
Using the laws of thermochemistry these equations are modified to get the required
value.
Multiplying Eq. (ii) by 4 gives
(iv) 2H2(g)
4H(g)
∆H4 = 4 × ∆H2
Reversing Eq. (iii) we get
(v) CH4 (g) → C (graphite) + 2H2 (g)
∆H5 = -∆H3
Adding Eqs. (i), (iv), and (v) gives the required equation
CH4 (g) → C (g) + 4H (g)
Therefore, ∆H = ∆H1 + ∆H4 + ∆H5
∆ H = ∆H1 + 4 × ∆H2 − ∆H3
∆ H = 716.68 kJ mol−1 + 4 × 217.97 kJ mol−1 - (−74.75 kJ mol−1)
= (716.68 + 871.88 + 74.75) kJ mol−1
= 1663.31 kJ mol−1
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Since C−H bond is present in other molecules also, the average value is calculated
by taking into account C−H bonds in different molecules. The bond enthalpy of C−H
bond is taken as 413 kJ mol−1.
Similarly, the C−C, C=C and C
C bond enthalpies can be calculated by using
the enthalpies of formation of ethane, ethene and ethyne and bond enthalpy of C−H
bond. Thus, the bond enthalpy data of polyatomic molecules can also be tabulated.
Applications of Bond Enthalpies
(i) Determination of enthalpy changes accompanying a chemical reaction.
Bond enthalpies can be used for determining the enthalpies of reaction.
∆ r H = Total bond enthalpies of reactants − Total bond enthalpies
of
products
.. (3.58)
During a reaction some bonds of the reactants dissociate and bonds of the products are
formed. The enthalpy of a reaction is the total enthalpy change during bond formation
and bond dissociation. The enthalpy change during bond dissociation is equal to the
bond enthalpy. Since the bond formation is the reverse of bond dissociation, the
enthalpy change during bond formation is negative of bond enthalpy. Therefore, the
total bond enthalpies of the products are subtracted from the total bond enthalpies of
the reactants to get the enthalpy of reaction. This can be illustrated by the following
example:
C2H4 (g) + H2 (g) → C2H6 (g)
or
The bonds being broken on the reactant side are C
C and H−H bonds in ethene and
hydrogen gas, respectively and bond formed are C−H bonds and C−C. Given,
Є (C
C) = 610 kJ mol−1, Є (H−H) = 436 kJ mol−1,
Є (C−C) = 348 kJ mol−1, Є (C−H) = 413 kJ mol−1,
Total enthalpy needed to break the bonds = Total bond enthalpies of reactants
=Є (C
C) + Є (H−H) = (610 + 436) kJ mol−1 = 1046 kJ mol−1
Enthalpy released during the formation of bonds = Total bond enthalpies of products
=Є (C−C) + 2 Є (C−H) = (348 + 2 × 413) kJ mol−1 = 1174 kJ mol−1
ΔrH = {Є (C
C) + Є (H−H)} − {Є (C−C) + 2 Є (C−H)}
= 1046 kJ mol−1 − 1174 kJ mol−1
ΔrH = −128 kJ mol−1
(ii) Determination of enthalpy of formation
If the reaction involves the formation of a compound from its elements then
the bond enthalpy data can be used to calculate the enthalpy of formation of
the compound by the formula
Δ f H θ = Total enthalpy of atomization of the reactants Bond enthalpies of products
(3.59)
For example, in the following reaction
2C (graphite) + 3H2 (g) + ½ O2 (g) → C2H5OH (l)
The steps involved are
2C (graphite)
3H2 (g)
½ O2 (g) → O (g)
2C (g)
Δ r H= 2ΔaH (C) = 716.68 kJ mol−1
6H (g)
Δ r H = 3Δ a H (H) = 436 kJ mol−1
Δr H = ½ ΔaH (O) = 498.4 kJ mol−1
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Enthalpy change during bond breakage = Total enthalpy of atomization
= 3Δ a H (H) + 2Δ a H (C) + ½ Δ a H (O)
= 3 × 436 kJ mol−1 + 2 × 716.68 kJ mol−1
+ 249.2 kJ mol−1
= 2990.88 kJ mol−1
The bond enthalpy values are
Є (C−C) = 348 kJ mol−1, Є (C−H) = 413 kJ mol−1,
Є (C−O) = 351 kJ mol−1, Є (O−H) = 363 kJ mol−1
The enthalpy released during the formation of bonds in C2H5OH = Bond enthalpy of
C2H5OH
=Є (C−C) + 5 Є (C−H) + Є (C−O) + Є (O−H)
= (348 + 5× 413 + 351 + 363) kJ mol−1 = 3227 kJ mol−1
Δ f H (C2H5OH) = Total enthalpy of atomization − Bond enthalpy
= (2990.38 − 3227) kJ mol−1 = − 236.52 kJ mol−1
The drawback of this method is that it will give same value of Δ f H θ for all the isomers
of a compound, which is not correct.
(iii) Determination of resonance energy
Generally, the values of bond enthalpies of formation calculated from bond
enthalpy data coincide with those determined experimentally. However, in
certain cases some discrepancies are observed. For example, in the case of
enthalpy of formation of benzene the calculated and experimental values are
significantly different.
6C (g) + 3H2 (g) → C6H6 Δ f H (benzene)
Δ f H (benzene) = Total enthalpy of atomization - Total bond enthalpies
Δ f H (benzene) = {3ΔaH (H2) + 6ΔaH (C)}−{6e (C−H) + 3Є (C−C) + 3Є (C
Δa H (H2) = 436 kJ mol−1 ΔaH (C) = 716.68 kJ mol−1
Є (C−H) = 413 kJ mol−1
Є (C−C) = 348 kJ mol−1
Є (C
C)}
C) = 610 kJ mol−1
Δf H (benzene) = (3 × 436 + 6 × 716.68) − (6 × 413 + 3 × 348 + 3 × 610) kJ mol−1
Δf H (benzene) = (5607.9 − 5352) kJ mol−1 = 256 kJ mol−1
However, the experimentally determined value is only 82.9 kJ mol−1. The difference in
the actual and calculated Δ f H is 173 kJ mol−1. This shows that the benzene molecule is
more stable than the Kekule structure. The greater stability of the benzene molecule can
be explained on the basis of the two resonating structures of benzene.
The difference in actual enthalpy of formation and calculated enthalpy of formation
is known as resonance energy.
Example
Calculate the enthalpy change for the following reaction using the
given thermodynamic data.
CH4 (g) + 2O2 → CO2 (g) + 2H2O (l)
Bond
Є/kJ mol−1
C−H
413
O
O
498
C
O O−H
732
464
Solution :
ΔrH = Bond enthalpy of reactants - Bond enthalpy of products
Bond enthalpy of reactants = ΔH (reactant) = 4 Є (C−H) + 2 Є (O=O)
= (4 × 413 + 2 × 498) kJ mol−1 = 2648 kJ mol−1
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Bond enthalpy of products = ΔH (product) = 4Є (O−H) + 2Є (C
O)
= (4 × 464 + 2 × 732) kJ mol−1 = 3320 kJ mol−1
Δr H = 2648 kJ mol−1 − 3320 kJ mol−1 = − 672 kJ mol−1
The actual value is 802 kJ mol-1. The difference arises because CO2 has a resonating
structure with the resonance energy of 130 kJ mol-1.
Note: During the calculation of Δ r H, bond dissociation enthalpies are used. However,
in the products, bonds are being formed. Since bond formation process is the reverse
of bond dissociation, therefore, a negative sign is used before the bond enthalpies
of products.
3.4.8 Variation of Enthalpy of Reaction with Temperature (Kirchhoff’s Equation)
The enthalpy and internal energy values vary with variation in temperature. Hence,
the literature values that are given for a particular temperature cannot be directly
used for study at a different temperature. It is therefore essential to understand the
variation of these values with temperature.
In 1885 the scientist Kirchhoff gave a relationship between the heat of the reaction
and temperature. It is derived as follows.
Let us take a general reaction
vAA (g) + vBB (g) → vCC (g) + vDD (g)
Δ r H = (vCHm,C + vDHm,D) - (vAHm,A + vBHm,B)
To study the effect of variation of temperature on Δ r H, we should know the rate of
change of Δ r H with temperature. This is obtained by differentiating the equation
with respect to T at constant P
= vCCP,m (C) + vDCP,m (D) - vACP,m (A) - vBCP,m(B)
(3.60)
d(∆H) = ΔCP dT
Here,
Kirchhoff’s law
(3.61)
= molar heat capacity at constant pressure
and
∆CP = change of heat capacity in a chemical reaction.
From the Kirchhoff’s Law we can calculate the enthalpy change at any temperature
if its value at a specific temperature is known.
To calculate the heat of the reaction at constant volume, ∆U at varying
temperatures we can derive the Kirchhoff’s equation as
or
dΔU = ∆CV dT
On integrating the Kirchhoff’s law equations the ∆H and ∆U can be calculated at
different temperatures
Case I: Heat capacity is independent of temperature change
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Similarly,
∆H(T2) − ∆H(T1) = ∆CP(T2−T1)
(3.63)
∆U(T2) − ∆U(T1) = ∆CV(T2−T1)
(3.64)
Case II: Heat capacity is dependent on temperature change
Dependence of molar heat capacity on temperature is given by the equation
CP,m= α +β T + γT2
(3.65)
Here α, β and γ are constants for a given substance. Therefore,
∆ CP = (vCαC + vDαD - vAαA - vBαB) + (vCβC + vDβD - vAβA - vBβB) T
+ (vCγC + νDγD − vAγA − νBγB) T
2
∆CP = ∆α + ∆βT + ∆γT2
(3.66)
Therefore, Kirchhoff’s law can be written as
d(∆H) = (∆α + ∆βT + ∆γT2 ) dT
On integrating
Equation 3.67 is used to calculate precisely the enthalpy of a reaction at any
temperature T2, if the value at T1 is known.
3.5 The Second And Third Laws Of Thermodynamics
Some processes take place on their own without any help from the external agencies.
Such processes are known as spontaneous processes. For example, expansion of gas
kept in an open container and cooling of a hot liquid kept in a beaker take place
spontaneously, while air can never compress itself and fill a balloon on its own.
Similarly, a body at room temperature can never take heat from the surroundings and
become hot. These processes take place with the help of an external agency. Such
processes that do not take place on their own but require the help of an external agency
are known as nonspontaneous processes.
Limitations of the First Law of Thermodynamics
The first law of thermodynamics is about conservation of energy. It is very useful in
calculating energy changes taking place during various processes. However, it gives no
idea about the direction of energy flow. The first law of thermodynamics is unable to
predict whether a process will be spontaneous or nonspontaneous.
To make this prediction we have to look into the most important property of the system,
which is energy. It is a known fact that energy flows from a high energy state to a low
energy state. The system having lower energy is more stable than the system having
higher energy. Then the question that can be asked is whether all the processes in which
energy of the system is decreasing are always spontaneous? Are all exothermic
processes always spontaneous?
It is observed that it is not necessary that all spontaneous processes are exothermic.
Various cases are known where a spontaneous process is endothermic. For example, a
gas expands in vacuum spontaneously although there is no energy change. Similarly,
water evaporates spontaneously into water vapour but the process is endothermic in
nature.
There is another reason why an exothermic process cannot be taken as the sole
criterion of spontaneity and that is the law of conservation of energy. If the energy of
the system decreases during a process then the energy of the surrounding must
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increase since the total energy must be conserved. So if the system undergoes an
exothermic process spontaneously then the change in the surroundings should also be
spontaneous although the energy of the surrounding will increase. The second law of
thermodynamics decides the direction of a spontaneous process. It introduces a new
concept of entropy that governs the criterion of spontaneity.
3.5.1 Various Statements of the Second Law of Thermodynamics
The second law of thermodynamics has been stated in various ways by different
scientists.Although these statements are different but each one of them predicts the
direction in which a change will take place spontaneously.
i.
ii.
iii.
iv.
v.
vi.
vii.
viii.
ix.
x.
Heat cannot pass spontaneously from a colder to a warmer body. (Clausius)
No change is possible in which the sole result is the absorption of heat from a
reservoir and its complete conversion into work. (Kelvin)
It is impossible to transfer heat from a colder system to a warmer system without
other simultaneous changes occurring in the two systems or in their
environment. (P. S. Epstein)
Every system that is left to itself will on the average change towards a condition
of maximum probability.
The state of maximum entropy is the most stable state of an isolated system.
(Enrico Fermi)
In any irreversible process the total entropy of all bodies concerned is increased.
In an adiabatic process the entropy either increases or remains unchanged.
Entropy is time’s arrow. (A. Edington)
There is general tendency in nature for energy to pass from more available form
to less available form.
The entropy of an isolated system increases during a spontaneous process.
3.5.2 Entropy
Whenever a change occurs, the internal energy is distributed into various forms. It has
been found that those processes are spontaneous in which the energy has been
distributed into a more dispersed form, that is there is more chaotic or random
distribution of energy. This random distribution of energy makes it less useful for doing
work. For example, when a ball bounces it loses some energy at each bounce. The
energy lost is either in the form of frictional energy or it is given to the floor molecules
as thermal energy. The total energy is conserved but the energy is in the distributed
state and hence is of less use.
A new concept of entropy was introduced by Clausius as a measurement of
randomness or chaos. The word entropy has been made from energy (en) and trope
(tropy) which means chaos. The symbol for entropy is S. It was given in honour of
famous scientist Sadi Carnot.
According to one of the statements of second law of thermodynamics, heat cannot be
completely converted into work. This has been attributed to entropy and it can be
explained by taking the example of a gas kept in a cylinder with a piston that can move
without friction.
When the gas kept in the cylinder is heated its kinetic energy increases. The gas
molecules move faster. They collide with each other and with the walls of thecylinder.
When they collide with the piston, it is pushed forward and the gas expands. This results
in useful work done by the gas. Consider a hypothetical condition when all the gas
molecules move in ordered manner towards the piston. Then each molecule will strike
the piston and contribute to the total work done by the gas as shown in Fig. 3.17. Under
this situation the entire heat absorbed can be converted into work. In reality the
molecules are moving randomly in all directions. Therefore, all the molecules do not
strike the piston.
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Figure 3.17: Conversion of heat into
collisions are ordered and when the collisions are random
work
when
the
Animation
http://www.illldu.edu.in/mod/resource/view.php?id=5641
Hence, all the collisions of the gas molecules do not lead to expansion. Since only a
fraction of collisions result in useful work, only a fraction of heat absorbed is converted
into useful work and the rest is the unavailable heat, that is, the heat is not used for
doing work. The fraction of heat used to do useful work decreases and the amount of
unavailable energy increases as the randomness of the gas molecules increases. Hence,
the extent of randomness, which has been termed entropy, can also be related to heat.
Mathematical form of Entropy
The change in entropy is defined as
(3.68) Note, that
while calculating the entropy change, heat transfer during the reversible
processes are only considered. However, the entropy change for an irreversible
process is not related to the heat transfer during an irreversible process. q is a path
function. It can be shown that the entropy is a state function.
Let us see whether dqrev /T is an exact differential or not. If dqrev/T is an exact
differential then dS will also be an exact differential and hence the entropy will be a
state function. This can be determined by considering an ideal gas that undergoes a
reversible change.
Proof that dqrev/T is an Exact Differential
Initial state: P1, V1, T1
Final state: P2, V2, T2.
According to the first law of thermodynamics for an infinitesimal change
dU = dq + dw
(3.69)
For mechanical work,
dw = −PdV
Since the process is reversible, dq = dqrev
Hence, Eq. (3.69) is written as dU = dqrev − PdV
(3.70)
For an ideal gas, P = nRT/V and dU = CVdT. Substituting these values in Eq. (3.70) and
rearranging the equation we get
CVdT = dqrev − (nRT/V )dV
dqrev = CV d T + (nRT/V ) dV
(3.71)
To determine the total heat exchanged during the process Eq. (3.71) has to be
integrated between the initial and the final state. Since both T and V are changing
(nRT/V) cannot be integrated. To solve this problem, Eq. (3.71) is divided by T.
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(3.72)
Now the right-hand side of the Eq. (3.72) can be integrated.
CV is assumed to be independent of temperature for small temperature change, so it is
taken as constant. n and R are also constant. Therefore,
(3.73)
Since right-hand side of Eq. (3.73) can be integrated, it is an exact differential.
Therefore, left-hand side of Eq. (3.73), that is, dqrev/ T must also be an exact
differential. From Eq. (3.69),
dqrev / T = dS.
Therefore, dS is an exact differential and hence S is a state function. Equation (3.72)
can be written as
dS =
(3.74)
Multiplying dqrev by 1/T converts an inexact differential into exact differential. Therefore,
1/T is known as integrating factor.
(3.75)
From above equation it is seen that the entropy depends upon temperature and volume.
Hence,
S = f (T, V)
Although here the proof has been given for ideal gas, dqrev /T is exact differential for all
types of systems. Dividing the Eq. (3.75) by n,
(3.76)
Here ∆mS = molar entropy change and CV,m = molar heat capacity.
Proof that Entropy is a State Function
In the previous section we have seen that entropy depends upon temperature and
volume
S = f (T, V)
The total change in the entropy is given by
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dS=
also
(3.77)
dS=
(3.74)
Comparing the coefficients of Eqs. (3.77) and (3.74),
=
M
(3.78)
and
N
(3.79)
=
Differentiating Eq. (3.78) with respect to V at constant T
=0
for an ideal gas, therefore,
0
Differentiating Eq. (3.79) with respect to T at constant V
(3.80)
(3.81)
Comparing Eqs. (3.80) and (3.81),
=0
or
Euler’s reciprocity relation is valid. Hence, dS is exact differential so S is a state function.
Here we see that
= Rate of change of entropy with temperaature at constant volume.
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Since heat capacity and T are always positive, therefore, the rate of change of entropy
with temperature is positive. That is, entropy increases with increase in
temperature.
= Rate of change of entropy with volumee at constant volume.
Since nR/V is always positive, the entropy increases as the volume of the system
increases.
Relationship between Internal Energy and Entropy
dU = dq + dw
(First law of thermodynamics)
dU = dq − PdV
For reversible process,
dq = dqrev = TdS
dU = TdS − PdV
(3.82)
This is also known as the combined form of first law and second law for reversible
processes. This expression indicates that the internal energy depends upon the change
in entropy and volume.
Relationship between Enthalpy and Entropy
H = U + PV
Differentiating the equation we get
dH = dU + PdV + VdP
(3.83)
From the combined form of the first law and the second law of thermodynamics
dU = TdS − PdV
Substituting the above expression in Eq. (3.83) we get
dH = TdS − PdV + PdV + VdP
dH = TdS + VdP
(3.84)
3.5.3 Entropy Change for Various Processes
Entropy Change for an Isolated System
According to the first law of thermodynamics,
dUrev = dqrev + dwrev and dUirr = dqirr + dwirr
If the initial states and final states of the system are same for both the processes, then
dUrev = dUirr
dqrev + dwrev = dqirr + +dwirr
(3.85)
dwrev < dwirr
Therefore,
dqrev > dqirr
or
(3.86)
= dS
Therefore,
dS >
This equation is known as the Clausius inequality.
Since,
dqirr = dU − dwirr
(3.87)
dS >
For an isolated system dq = 0, dU = 0.
Therefore, dw = 0
dqrev = 0
(3.88)
(dS)rev = 0
dqirr = 0
(dS)irr > 0
or
dS (isolated) ≥ 0
(3.89)
Although entropy is a state function it appears that dS is different for reversible and
irreversible processes for an isolated system. This can be explained if the nature of the
isolated system is considered. An isolated system includes the system as well as its
immediate surroundings. Therefore, the entropy change for an isolated system includes
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the entropy change for the system as well as its surroundings.
dS (isolated) = dS (system) + dS (surroundings)
Since,
Universe = System + Surroundings
dS (isolated) = dS (universe)
For the state functions, the initial state and the final state of the system are considered
and not that of the surroundings. Since the initial states and final states of the system
are same for both the processes, the entropy change for the system is same for
reversible and irreversible process. The changes in the surroundings are different for
reversible and irreversible processes so the entropy changes for the surroundings will be
different. As a result, the total entropy change for the universe will be different for
reversible and irreversible process.
dS rev (system) = dS irr (system)
(3.90)
For a reversible process, ddrev (system)=
(3.91)
dqrev(surroundings) = −dqrev (system)
and
Therefore,dSrev(surrounding) =
3.92a)
dSrev (universe) = dSrev (system) + dSrev (surrounding)
Substituting the values of Eqs. (3.91) and (3.92a) in Eq. (3.92b)
(3.92b)
dSrev(universe)=
In view of Eq. (3.92) the above equation becomes
dSrev(universe)=
For an irreversible process,
Heat absorbed by the system = dqirr (system)
From Eq. (3.87),
(3.93)
Since the surroundings are large, the addition or removal of a small amount of heat in
the surrounding is taken as a reversible process. Therefore,
ΔSirr(surrounding)=
(3.94)
The entropy change for the universe is given by the equation,
dSirr (universe) = dSirr (system) + dS irr (surrounding)
On substituting the value of ∆Sirr (surrounding) from Eq. (3.94) in the above equation,
dSirr(universe) = dSirr(system)
From Eqs. (3.93) and (3.95) it is seen that dSirr (universe) > 0.
(3.95)
Entropy Change for Simultaneous Reversible Heating and Expansion of an Ideal
Gas
Both the temperature and the volume of the gas changes from T1, V1 to T2, V2. dU = TdS
− PdV (Combined form of first law and second law for reversible process) Rearranging
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the equation we get
TdS = dU + PdV
For an ideal gas the change in U is given by
Also, for an ideal gas
(3.96)
nRT/V
Substituting the values of dU and P in combined form of first and second law,
Dividing the equation by T
(3.97)
Integrating using the limits T = T1 to T = T2 and V = V1 to V = V2 and assuming that CV
is independent of temperature
(3.98)
Changing the natural log to log10
(3.99)
Entropy Change for an Isothermal Reversible
Volume Change for an Ideal Gas
(3.98)
For an isothermal process, T1 = T2.
Therefore,
Equation 3.98 becomes
or
(3.100)
For an ideal gas at constant temperature
therefore, Eq. (3.100) can be
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written in terms of pressure as
(3.101)
Entropy Change for an Isochoric Reversible Heating of an Ideal Gas
For an isochoric process the volume remains constant, V1 = V2.
Therefore,
The Eq. (3.98) becomes
or
When the Temperature and the Pressure are Changed Simultaneously
TdS = dH − VdP (3.103)
For an ideal gas the change in H is given by the equation
dH=CpdT
Since for an ideal gas V = nRT/P
(3.102)
Substituting the values of dH and V in Eq. (3.103), we get
TdS = cpdT
or
dS
(3.104)
Integrating using the limits T = T1 to T = T2 and P = P1 to P = P2 and assuming that Cv is
independent of temperature
(3.105)
Converting the natural logarithm to log10
(3.106)
From Eq. (3.104) it is seen that the entropy depends upon temperature and pressure.
Entropy Change for Isobaric Reversible Heating of an Ideal Gas
For isobaric process pressure is constant P1 = P2 and ln (P2/P1) = 0. Therefore, Eq.
(3.105) will become
ΔS=Cp
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or
Entropy Change for a Reversible Adiabatic Process
For an adiabatic process, dqrev = 0
For a reversible process, dSrev = dqrev /T.
Therefore,
dSrev (adiabatic) = 0
(3.107)
Another proof
From Eq. (3.98)
For an adiabatic process as studied in the Section on the First Law of
thermodynamics
Taking natural logarithm
(3.108)
Rearranging,
Substituting in Eq. (3.98),
Example
Calculate the w, q, ∆U, ∆H, ∆Ssystem, ∆Ssurr, ∆Suniverse for isothermal expansion of
two moles of an ideal gas at 298 K from 0.5 dm3 to 1.0 dm3 under (i) reversible
conditions (ii) irreversible condition when final pressure is equal to the external
pressure.
Solution :
Given: Ideal gas
n = 2 mol
T = 298 K
V1 = 0.5 dm3
V2 = 1.0 dm3
R = 8.314 J K−1mol−1
The process is isothermal
(i) Reversible process
For an ideal gas ∆U = CV∆T and for an isothermal process, ∆T = 0, therefore,
∆U = 0
Since from the first law of thermodynamics ∆U = q + w, therefore,
q = −w = 3434.9 J
∆H = CP∆T = 0
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(ii) Irreversible process
= -2 mol × 8.314 J K−1mol−1× 298 K (1 - 0.5) = -2477.57 J
For an isothermal process ∆T = 0, therefore, ∆U = 0
or
q = − w = 2477.57 J
∆ H = CpΔT = 0
Entropy Change During Phase Transformation
The change of entropy accompanying the transformation of one mole of a substance
from one physical state into another physical state at its transition temperature is called
entropy of a phase transformation or entropy of transition. It is denoted by ∆trsS.
The phase transformations can be of four types:
(i) Fusion/Melting Solid
(ii) Vapourization Liquid
(iii) Sublimation Solid
Liquid at melting point (Tf)
Vapour at boiling point (Tb)
Vapour at sublimation point (Tsub)
(iv) Solid − solid α (s)
β (s) at transition temperature
For example, when heat is supplied to water, its temperature starts increasing. The
water starts boiling at 100°C. As long as the liquid water and water vapour are at
equilibrium the temperature does not rise even though water is being heated. This is
because the heat supplied is being used to convert liquid into vapour. The phase
transformation at transition temperature is a reversible process that takes place at
constant pressure.
When the pressure is constant, qtrs = ∆trsH
(3.109)
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(i)
(ii)
(iii)
The transition temperatures are taken in Kelvin. Since the order of entropy for the three
states of matter is
Svapour > Sliquid > Ssolid
The entropy of fusion (∆fusS ), entropy of vaporization (∆vapS ) and entropy of
sublimation (∆subS ) have positive values.
Trouton’s Rule
For non associating and nondissociating compounds the molar entropy of vaporization at
the boiling point is constant and is approximately equal to
85 J K−1mol−1. This implies that the difference in entropies of liquid phase and the
vapour phase are same for these liquids.
ΔvapS = Svap − Sliq ≈ 85 J K-1mol-1
Liquid
∆vapS/J K−1mol−1
Methane
83
Benzene
87
Cyclohexane
85
Diethyl ether
84
For the liquids that show either association or dissociation the value of ΔvapS are
different.
For ethanol ∆vapS is 110 J K-1mol-1 and for water it is 109 J K-1mol-1. These values
imply that either the vapour phase of these liquids is more random than the liquids that
obey Trouton’s rule or the liquid phase is more ordered. On examining the two liquids it
is found that for both ethanol and water the liquid phase is more ordered due to
hydrogen bonding. Therefore, the difference in the randomness of liquid phase and
vapour phase is more than that in case of liquids that follow Trouton’s rule.
For acetic acid the value of ∆vapS is 62 J K-1mol-1, indicating that the difference
in randomness of liquid phase and vapour phase of acetic acid is less. This may be due
to more randomness of liquid phase or less randomness of vapour phase. Since in the
liquid phase, intermolecular hydrogen bonding exists between acetic acid molecules, the
liquid phase is ordered phase. This implies that the second assumption is true, that is
the vapour phase is less random. There is some hydrogen bonding existing between
acetic acid molecules in the vapour phase also. Thereby making the vapour phase of
acetic acid more ordered and hence with lower entropy.
Entropy Change for an Irreversible Process
The entropy change for an irreversible process cannot be calculated directly. It is
calculated by considering the heat transfers that would have taken place if the change
took place in a reversible manner. This is based on the fact that since entropy is a state
function, it is independent of the path. Therefore, for two different processes in which
the initial states and the final states of the system are same, the entropy changes will be
equal.
Example
Calculate the entropy change for the following process
H2O (l, − 10°C, 1 atm) → H2O (s, − 10°C, 1 atm)
Given: ∆fusH (273 K) = 6.01 kJ mol-1
CP,m (H2O, l) = 75.42 J K-1mol-1 CP,m (H2O, s) = 37.20 J K−1mol−1
Solution 3.12:
Since phase transformation is not taking place at transition temperature, entropy
change cannot be directly calculated. The process is divided into three steps
H2O (l, −10°C, 1 atm)
H2O (l, 0°C, 1 atm)
H2O (l, 0°C, 1 atm) ∆Si
H2O (s, 0°C, 1 atm) ∆Sii
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H2O (s, 0°C, 1 atm)
∆S = ∆Si + ∆Sii + ∆Siii
Since pressure is constant
H2O (s, − 10°C, 1 atm) ∆Siii
ΔSi = 2.303Cp
ΔSi = 2.303Cp,m(H2O, l)
= 2.303 Χ 75.42 J K-1mol-1Χ 0.0162
= 2.814 J K-1mol-1
ΔSiii = 2.303Cp
= 2.303 × 37.20 J K−1 mol−1× (− 0.0162)
ΔSiii = −1.388 J K−1mol−1
ΔS = 2.814 J K−1mol−1 + (−22.015 J K−1mol−1) + (−1.388 J K−1mol−1)
ΔS = −20.589 J K−1mol−1
Example
One mole of an ideal gas is allowed to expand isothermally at 300 K until its volume is
doubled. Calculate the entropy change of the system and the universe if (a) the
expansion is carried out reversibly (b) the expansion is a free expansion.
Solution:
The data given are n =1 mol, T = 300 K and V2 = 2 V1
(a) For an isothermal reversible expansion of an ideal gas, the entropy change is
given by the expression
∆S = qrev/T
qrev= −w =2.303 nRT log (V2/V1) = 2.303 (1 mol) (8.314 J K−1mol−1) (300 K) log
(2V1/V1) = 1728.99 J
∆Ssys = (1728.99 J)/(300 K) = 5.76 J K−1
∆Ssys = −∆Ssurr
∆Suniv = ∆Ssys + ∆Ssurr = 0
(b) The free expansion is an irreversible process so the expression ∆S = qrev/T
cannot be used. However, since entropy is a state function, the ∆Ssys is same
whether the process is reversible or irreversible. Therefore,
∆Ssys = 5.76 J K−1
For an isothermal free expansion,
qsys = −qsurr = 0
Therefore,
∆Ssurr = qsurr /T = 0
Therefore,
∆Suniv = ∆Ssys = 5.76 J K−1
3.5.4 Free Energy Functions
One of the purposes of studying thermodynamics is to predict the feasibility of a
process, that is, to predict the direction in which a process will be spontaneous. We have
seen that the sign of internal energy alone cannot be the criterion for determining the
spontaneity of a process. The second law of thermodynamics introduced the concept of
entropy. Whenever a process occurs spontaneously it is an irreversible process. It is
observed that for such a process the entropy change of the universe is positive. This
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implies that to predict the spontaneity of a process it is important to determine the
entropy change of the universe. This involves measurement in entropy change of the
system as well as that of surroundings.
The problem arises in the measurement of entropy change of the surroundings since the
surroundings are not clearly defined.For most of the spontaneous processes it is
observed that the system either tries to attain the state of minimum energy or
maximum entropy. So it is necessary to define a new function which takes into account
both the properties. Gibbs free energy (G) and Helmholtz free energy (A) are two
functions that take into account the composite effect of energy and entropy change of
the system.
Helmholtz Free Energy or Helmholtz Energy or Work Function
Helmholtz energy is also known as work function. Its symbol is A which comes from
German word for work, arbeit. It is related to internal energy and entropy by the
equation
A = U − TS
(3.110)
Since U, T and S are state functions, A is also a state function. Its SI unit is J.
dA = dU − TdS − SdT
At constant temperature, dT = 0 therefore,
dA = dU −TdS
For a finite change
ΔA = ∆U − T ∆S
dU = TdS + dwrev (Combined form of first law and second law for reversible process)
dA = TdS + dwrev − TdS
or
dA = dwrev
∆A = wrev
Therefore, the change in Helmholtz energy is equal to the work done during isothermal
reversible process. When the Helmholtz energy decrease, the value of wrev will be
negative indicating that the work is done by the system. Therefore, the decrease in
Helmholtz energy is equal to work done by the system during isothermal reversible
process.
Gibbs Free Energy or Gibbs Energy
The symbol of Gibbs energy is G. It is related to the enthalpy and entropy by the
relation
G = H − TS
(3.111)
Since H, T and S are state functions; therefore, G must also be a state function.
Differential of Eq. (3.111) is
dG = dH − TdS − SdT
If the temperature is constant, dT = 0 therefore,
dG = dH − TdS
(3.112)
At constant pressure,
dH = dq
Therefore,
(dG)T, P = dq − TdS
(3.113)
Let us see how Gibbs energy can be used as criterion of spontaneity
Whether the process is reversible or irreversible
TdS = dqrev
Therefore,
(dG)T, P = dq − dqrev
For a reversible process each step is an equilibrium step and dq = dqrev.
Therefore,
(dG)T, P = dqrev − dqrev = 0
So for the equilibrium state,
dG = 0
For an irreversible process
dq = dqirr .
Therefore,
(dG)T, P (irr) = dqirr − dqrev
We know that dqrev > dqirr ,
Therefore,
dqirr − dqrev < 0
and
(dG)T, P < 0
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(3.114)
Thermodynamics
Since all the spontaneous processes are irreversible processes, for all spontaneous
process taking place at constant temperature and pressure the above relation is valid.
• G is a state function.
• For a spontaneous process the change in free energy is always negative.
• For equilibrium state dG = 0.
• For nonspontaneous process dG > 0.
• The SI unit of G is J.
For finite change at constant temperature and pressure
ΔG = ΔH − TΔS
(3.115)
Here ΔG is the free energy change in the system
ΔH is the enthalpy change in the system
ΔS is the entropy change in the system
The process will be spontaneous, that is the change in free energy of the system, Δ∆G
will be negative if either of the following conditions is true
• ΔH < 0 and ΔS > 0
• ΔH > 0 but ΔH < TΔS
• ΔH < 0 and ΔS < 0 but |ΔH| > |TΔS|
When ΔH > 0 and ΔS < 0, the process will always be nonspontaneous.
This can be elaborated as follows:
On the basis of the enthalpy change there are two types of processes
(i) Exothermic process for which the enthalpy change is negative, that is, ΔH < 0
(ii) Endothermic process which the enthalpy change is positive, that is, ΔH > 0
Each type of the process is also accompanied by the change in entropy ΔS which
is positive when the entropy increases during a process and negative when the
entropy decreases during a process. Therefore, there are the following four cases:
(i) Exothermic process, ΔH < 0
(a) ∆S > 0, that is, the entropy increases during the process.
Therefore, TΔS > 0
ΔH − TΔS < 0
Therefore ΔG is always negative and the process is always spontaneous.
(b) ΔS < 0, that is, the entropy decreases during the process.
Therefore, TΔS < 0
ΔH − TΔS will be less than zero and ΔG will be negative when
|ΔH| >|TΔS|
That is, the process will be spontaneous when ΔH is less than T∆S
(ii) Endothermic process, ΔH > 0
(a) ΔS > 0, that is, the entropy increases during the process.
Therefore, TΔS > 0
The process will be spontaneous when
ΔH − TΔS < 0
or |ΔH| <|TΔS|
(b) ΔS < 0
that is the entropy decreases during the process.
Therefore, TΔS < 0
∆H − TΔS will always be positive, that is, ∆G > 0
Therefore, the process is always nonspontaneous.
Gibbs Energy and Work
Gibbs energy is known as free energy because ∆G is energy available with the
system for doing non expansion work. It can proved as follows.
G = H − TS
(3.111)
and
H = U + PV
Substituting the expression of H in Eq. (3.111),
G = U + PV − TS
The differential of the equation is,
dG = dU + PdV +VdP − TdS − SdT
According to the combined form of first law and second law of thermodynamics, for
a reversible process
dU = TdS + dwrev
Therefore, on substituting the value of dU in the expression for dG
dG = TdS + dwrev + PdV + VdP − TdS − SdT
For isothermal and isobaric process dT = 0 and dP = 0
Therefore,
(dG)T,P = dwrev + PdV
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dwrev is the total work done. This includes all the types of work like P - V work, electrical
work and chemical work. That is,
dwrev = dwmechanical + dwnet
Mechanical work is the work done during the volume change, that is, P − V work.
dwmechanical = − PdV
Therefore,
(dG)T,P = dwrev − dwmechanical = dwnet
Since the chemists are mainly interested in electrical or chemical work that can be
obtained from the system dwrev − dwmechanical is equal to useful work or the net
work involved in the system.
(dG)T,P = dwnet
For a finite change
(3.116)
(ΔG)T,P = wnet
When the free energy decreases (ΔG)T,P is negative and hence the value of w will be
negative. Negative value of w indicates that the work is done by the system. It can be
concluded that decrease in free energy is equal to net work that can be obtained from
the system at constant temperature and pressure.
Relationship between Gibbs Energy Change and Entropy Change of the
Universe
For a spontaneous process entropy of the universe increases but Gibbs energy
decreases. Since both the functions decide the spontaneity of a process they are related
to each other. Let us see how they are related to each other.
From Eq. (3.115)
ΔG = ΔH − TΔS
At constant pressure, ΔH = q. Therefore,
ΔG = q − TΔS
For a system,
ΔG (system) = q (system) − TΔS (system)
Since,
Also,
q (system) = − q (surr)
ΔG (system) = − q (surr) − TΔS (system)
q (surr) = TΔS (surr)
ΔG (system) = − TΔS (surr) − TΔS (system)
Taking T common in the right - hand side of the equation,
ΔGsystem = − T [ΔS (surr) + ΔS (system)] = − TΔS (universe)
or
(3.117)
Relation between A and G
From Eq. (3.111)
G = H − TS
and
H = U + PV
Substituting the value of H we get
G = U + PV − TS
From Eq. (3.110),
A = U −TS
Therefore,
G = A+ PV
(3.118)
For a change in state from
P1, V1, T1, G1, A1 to P2, V2, T2, G2, A2
For initial state
G 1 = A 1+ P 1V 1
For final state
G 2 = A 2 + P 2V 2
ΔG = G2 − G1 = (A2 − A1) + (P2V2 − P1V1)
For an ideal gas
PV = nRT
Therefore,
ΔG = ΔA + nR (T2 − T1)
ΔG = ΔA + nRΔT
(3.120)
For an isothermal process, ΔT = 0. Therefore, ΔG = ΔA for an ideal gas in a closed
system under isothermal conditions.
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Variation of Helmholtz Energy with Temperature
and Volume
A = U − TS
Differential of the Eq. (3.110) gives
dA = dU − TdS − SdT
Since,
dU = TdS − PdV
Substituting the value of Eq. (3.84) in Eq. (3.120),
dA = TdS − PdV − TdS − SdT
or
dA = − SdT − PdV
This relation shows that A depends upon temperature and volume
A = f (T,V )
The total change in A is given by the equation
(3.110)
(3.120)
(3.82)
(3.121)
(3.122)
Comparing Eqs. (3.121) and (3.122) for the coefficients of dT and dV
(3.123)
and
(3.124)
This indicates that since entropy is always positive, (∂A/∂T)V is negative. That is, A
decreases on heating if the volume is kept constant. (∂A/∂V)T is also negative, so A
decreases as the volume is increased at constant temperature.
Variation of A with T at Constant Volume: Gibbs–Helmholtz Equation
A = U − TS
From Eq. (3.123),
(3.110)
Substituting the value of Eq. (3.123) in Eq. (3.110),
or
Dividing by T2
(3.125)
Differentiating A/T with respect to T at constant V
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(3.126)
Comparing right - hand sides of Eqs. (3.125) and (3.126)
Gibbs−Helmholtz equation
For finite changes in U and A
and
Helmholtz Energy Change During Isothermal Volume Change for Ideal Gas
From Eq. (3.121), at constant temperature
dA = − PdV
For an ideal gas
P = nRT/V
Substituting the value of P in the expression for dA
Integrating
Variation of G with Temperature and Pressure
G = H − TS
(3.111)
Differential of this equation gives
dG = dH − TdS − SdT
(3.127)
According to combined form of first law and second law of thermodynamics, for
reversible process
dH = TdS + VdP
Substituting the value in equation we get
dG = TdS + VdP − TdS − SdT
or
dG= − SdT + VdP
(3.128)
This relation shows that G depends upon temperature and pressure, that is,
G = f(T,P)
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(3.129)
Comparing the coefficients of dT and dP in Eqs. (3.128) and (3.129)
(3.130)
and
(3.131)
These are very important relations and are used extensively in thermodynamic
studies.
= rate of change of Gibbs energy with temperature when the pressure is
constant
S = absolute entropy
Since absolute entropy is always positive, this relation implies that (∂G/∂T)P is
negative. That is, when temperature is increased the Gibbs energy decreases.
= rate of change of Gibbs energy with pressure when temperature is
kept constant. V = volume of the system. Since volume is a positive quantity (∂G/∂P)V
is positive. That is Gibbs energy increases when the pressure is increased.
Derivation of Gibbs–Helmholtz Equation: Variation of G With Temperature
G = H − TS
(3.111)
From Eq. (3.130),
Substituting the value of Eq. (3.130) in Eq. (3.111)
(3.132)
This is one form of Gibbs−Helmholtz equation. Rearranging the Eq. (3.132)
Dividing Eq. (3.132) by T
2
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(3.133)
Differentiating G/T with respect to T at constant P
(3.134)
Comparing the Eqs. (3.133) and (3.134),
(3.135)
This is another form of Gibbs–Helmholtz equation
For finite change in G and H the Gibbs–Helmholtz relation is written as
and
Integrated form of Gibbs–Helmholtz Equation
or
On integrating,
(3.136)
Gibbs Energy Change Accompanying Isothermal Volume Change in an Ideal Gas
(3.129)
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For isothermal process dT = 0. Therefore,
dG = VdP
For ideal gas, V = nRT/P
Integrating between limits of initial state and final state
(3.130)
For an ideal gas at constant temperature
Substituting the value in Eq. (3.130), and converting natural logarithm to log10
(3.131)
Various Thermodynamic Relations Between U, H, S, T, V and P
dU = TdS − PdV
dH = TdS + VdP
dG = − SdT + VdP
dA = − SdT − PdV
These equations are starting point for the derivation of four important equations known
as Maxwell’s relations.
3.5.5 Maxwell’s Relations
These relations give the variation of one thermodynamic property with another
under various conditions.
dU = TdS − PdV
(3.84)
Since U is a state function and dU is exact differential, Euler’s reciprocity relation is
valid. Therefore,
(First Maxwell relation)
dH = TdS + VdP
(3.132)
(3.85)
Since H is a state function and dH is exact differential Euler’s reciprocity relation is
valid. Therefore,
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(Second Maxwell relation)
dG = − SdT + VdP
G = f (T,P) and
(3.133)
(3.112)
(3.134)
Comparing the coefficients of dT and dP in Eqs. (3.112) and (3.134 )
Since,
Differentiating with respect to P at constant T
Similarly,
and differentiating this equation with respect to T at constant P
Since G is a state function and dG is exact differential Euler’s reciprocity relation is
valid, that is the order of differentiation is immaterial. Therefore,
Hence,
(Third Maxwell relation)
dA = − SdT −PdV
A = f (T,V)
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(3.135)
(3.121)
Thermodynamics
(3.136)
Comparing Eqs. (3.121) and (3.136)
Differentiating
= - S with respect to V at constant T
Differentiating
= -P with respect to T at constant V
Since A is a state function and dA is exact differential Euler’s reciprocity relation is
valid, the order of differentiation is immaterial. Therefore,
or
(Fourth Maxwell relation)
(3.137)
The four Maxwell relations are
(i)
(ii)
(iii)
(iv)
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Significance of Maxwell's Relations
Maxwell relations can be used to calculate those properties which are difficult to be
measured experimentally. For example, the variation of entropy with volume cannot be
determined experimentally. However, variation of pressure with temperature at constant
volume can be experimentally studied, so (∂S/∂V)T can be calculated using the fourth
Maxwell relation. Similarly, the variation of entropy with pressure can be determined
experimentally by studying the variation of volume with temperature at a constant
pressure.
TdS = dU + PdV
Since
(3.82)
U = f (T, V)
(3.138)
3.5.6 Derivation of Thermodynamic Equation of State
Substituting the value of Eq. (3.138) in Eq. (3.84)
Dividing the equation by T
;
Since
(3.139)
S = f (T,V)
(3.140a)
Comparing the coefficients of dT and dV in Eqs. (3.139) and (3.140a)
Differentiating
Differentiating the equation
constant T
with respect to V at constant T
with respect to V at
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Since S is a state function and dS is exact differential, the order of differentiation is
immaterial. Therefore,
or,
or
(3.140b)
Since U is a state function,
Therefore, Eq. (3.140b) becomes
Rearranging the equation,
(Thermodynamic equation of state)
(3.141)
Example
Using the thermodynamic equation of state, prove that internal pressure is
zero for ideal gas.
Solution :
Internal pressure =
According to thermodynamic equation of state
For ideal gas
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PV = nRT
Differentiate with respect to T at constant V
Dividing by V
Substitute the value in thermodynamic equation of state
3.5.7 Third Law of Thermodynamics
In the previous section we have studied a relation that can be used to calculate absolute
entropy of a substance. However, this method involves the entropy of a substance at
absolute zero, that is, 0 K. The entropy at absolute zero is determined with the help of
third law of thermodynamics which states that Entropy of pure and perfectly crystalline
substance is zero at absolute zero.
This statement has a logical explanation. The best organized physical state of a
system is that of a single crystal since in this state each particle has definite
arrangement in space with respect to other particles. At very low temperatures only
vibrational motion of the particle (atom or molecule) remains and at absolute zero all
the particles vibrate with lowest possible energy. Therefore, the arrangement is most
ordered. For a perfectly crystalline substance there is only one possible arrangement of
particles, so there is no randomness. Hence, the entropy is zero at absolute zero.
The value of absolute entropy can be calculated using the formula
S = k ln W = 2.303 k log W
Here k = Boltzmann constant = R/NA
(3.42)
And W= Thermodynamic probability = number of ways of distributing the particles
in various energy levels. This formula is based upon statistical thermodynamics which is
beyond the scope of present syllabus.
W depends upon the number of molecules and the energy which in turn depends
upon the temperature. When the number of possible arrangement of particles is large,
the value of W is large. As a result entropy has greater value, that is, there is more
randomness in the system. At absolute zero for perfectly crystalline substance, there is
only one possible arrangement of particles, therefore, W = 1
S = 2.303 k log 1 = 0
W is never equal to zero since for any given substance there is at least one possible
arrangement of particles.
3.5.8 Absolute Values of Entropy of a Substance
So far we have talked about the change in the entropy that takes place during a
process. However, we have not mentioned the absolute values of entropies of various
substances. The question that can be asked is: the standard enthalpy of formation of an
element is taken as zero, so can the standard entropy of an element be assumed to be
zero? Logically speaking, the standard entropy of an element cannot be taken as zero.
The reason is that the entropy is the measure of randomness of a system and at normal
temperatures there is some randomness in the system due to motion of atoms and
molecules. So as long as some randomness is there in the system, the absolute entropy
cannot be zero.
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We have seen that the entropy decreases with the decrease in temperature, then
what will be the value of entropy of a substance at absolute zero?
Since the entropy decreases with the lowering of temperature, the absolute entropy of a
substance should be a minimum value at absolute zero, that is, at 0 K. Therefore, the
absolute value of entropy at any temperature will be equal to the sum of the entropy of
the substance at 0 K and the entropy changes taking place as the substance is heated to
the given temperature. The change in entropy due to heating is given by
dS =
During the entire change the pressure is assumed to be constant, therefore
dP = 0.
dS =
When the temperature is changed from 0 K to T the entropy change is calculated by
integrating the equation
(3.143)
If the temperature change is small CP can be taken as independent of temperature
otherwise it depends upon temperature. If the temperature change is large then the
process will also involve change in the phase of the substance. For phase transformation
entropy change is given by
For example, for the following transition,
X (T = 0 K, α)
X (T = Ttrans, α)
X (Ttrans, β )
X (Tm, b)
X (Tm, l)
X (T, l)
Absolute value of entropy of X in liquid phase at temperature T is calculated using the
following expression:
Difficulty in Measurement of Absolute Entropy
For calculation of absolute entropy we need to know the heat capacity of the substance
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at absolute zero, which is difficult to measure. So the method of extrapolation is used.
The value of Cp/T vs T is plotted at various temperatures. The curve is extended up to 0
K. The area under the curve from 0 K to temperature T gives the absolute entropy at
temperature T.
Figure 3.18: Variation of CP / T with temperature
Characteristics of Absolute Entropy
• Entropy of perfectly crystalline substance is zero at absolute zero.
• Substances that are not perfectly crystalline have positive entropy at absolute zero.
• At higher temperatures the absolute entropy of all the substances is positive.
• Entropy depends upon mass of the molecules. Heavier molecules have
greater entropy.
• Entropy depends upon the number of atoms in a molecule. More are the number
of atoms in a molecule greater is the entropy.
• Entropy depends upon the physical state of the substance.
Sgas> Sliquid>Ssolid.
Example
Arrange F2, Cl2, Br2 and I2 in the increasing order of entropy at room temperature.
Solution 3.15:
The increasing order of entropy is I2 < Br2 < F2 < Cl2.
Since iodine is solid at room temperature it has least entropy followed by bromine
which is in liquid state and so has greater entropy than iodine. Chlorine and fluorine
both are gases but chlorine has greater molecular mass than fluorine so chlorine has
maximum entropy in case of halogens
Residual Entropy
The entropy possessed by a substance at absolute zero is known as residual entropy.
The substances that are not perfectly crystalline at absolute zero possess residual
entropy. For example, ice, CO, N2O, H2. The value of residual entropy generally lies
between 3 to 6 J K −1 mol −1. In these molecules there is more than one possible
arrangement of atoms in the crystal. For example, in case of CO some of the possible
arrangements of CO molecules in the crystal are
CO CO CO
CO CO CO
CO CO CO
OC OC OC
OC OC OC
OC OC OC
CO CO CO
CO OC CO
CO CO OC
Residual entropy is also found in substances that exist as super cooled liquids at
absolute zero. When these substances crystallize at absolute zero some entropy change
takes place indicating that super cooled liquids have positive entropy at absolute zero.
Summary
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Thermodynamics
Thermodynamics is a physical science that involves the study of the system and its
surroundings. A system is that part of the universe on which theoretical or experimental
investigations are carried out. The rest of the universe in the neighbourhood of the
system is known as its surroundings. A system and its surrounding are together known
as the Universe. Anything that separates the system from the surroundings is known as
a boundary. For the open system Δm ≠ 0 and ΔU ≠ 0. For closed system Δm = 0 and
ΔU ≠ 0. For an isolated system Δm = 0 and ΔU = 0. Thermodynamic process is any
operation or process that brings about a change in the state of the system. Extensive
properties are those properties that depend upon the size of the system. They are
additive in nature. Intensive properties are those properties that do not depend upon
the size of the system. The property of a system that depends on the state variables viz,
T, P, V and n is known as a state property. State functions can be differentiated and
they have exact differentials. There are four important laws of thermodynamics;
Zeroth law of Thermodynamics:
If TA = TC and TB = TC , then TA = TB
First law of Thermodynamics:
ΔU = q + w
Second law of Thermodynamics:
ΔS = qrev/T
Third law of Thermodynamics:
The entropy at absolute zero (0 K) is zero for
a pure crystalline substance.
Internal energy (U) is an extensive property and also a state function. U =
f(T,V). For an infinitesimal change work done = dw = − PdV. The value of w is
positive for work done on the gas and the value of w is negative for work done by the
gas.
The value of q is positive when heat is absorbed by the system and the value of q is
negative when heat is lost by the system. w and qare path functions.
The change in internal energy of a system is equal to heat exchange between the
system and its surroundings at constant volume. The heat exchanged at constant
pressure is equal to the change in enthalpy of the system.
Heat capacity(C) is defined as the heat needed to cause unit rise in the temperature of a
substance. Specific heat capacity(s) is defined as the heat needed to cause a unit rise
in the temperature of a unit mass of a substance.
Molar heat capacity(Cm) is defined as the heat needed to cause a unit rise in the
temperature of one mole of a substance. When the heat is absorbed at a constant
pressure, the heat capacity is known as heat capacity at constant pressure, Cp.
When the heat is absorbed at constant volume the heat capacity is known as heat
capacity at constant volume, CV.
An ideal gas can undergo volume change under various conditions (isothermal,
adiabatic, reversible and irreversible).The work done on the gas is greater than the work
done by the gas during irreversible process. The work done by the gas during reversible
expansion is more than work done by the gas during an irreversible expansion. At
constant volume, the gas does not do any mechanical work. Free expansion is the
expansion when external pressure is zero. The final pressure during isothermal
reversible expansion is greater than the final pressure during adiabatic reversible
expansion.
Thermochemistry is the study of heat transfer during chemical processes
andphysical transformations. ΔU is measured in a bomb calorimeter and ΔH is measured
in a normal calorimeter. The enthalpy change for a process is calculated using the
equation:
ΔHsystem = −{Cp (solution) + Cp (calorimeter)} ΔT
If the heat is absorbed by the system the process is known as an endothermic
process (ΔH >0). If the system loses heat the process is known as an exothermic
process(ΔH < 0)
The standard state is defined for the substance in its pure state and it is the most
stable form of that substance at a standard pressure of one bar and a specified
temperature.
Standard enthalpy of formation(Δ f Hθ) is defined as the amount of heat
exchanged at a given temperature and a pressure of one bar during the formation of
one mole of the substance in its standard state from its constituent elements in their
respective standard states. By convention, Δf H θ of an element is taken as zero at all
temperatures.
First law of thermochemistry states that the enthalpy of formation of any
compound is equal in magnitude and of opposite sign to the enthalpy of dissociation of
that compound at the given temperature and pressure. Second law of
thermochemistry states that the total enthalpy change of a reaction is the same
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Thermodynamics
regardless of whether the reaction is completed in one step or in several steps.
Δ r H= ∑ H θ(products) - ∑ H θ(reactants)
For the reactions involving solid or liquid Δ r H = Δ r U
For gaseous reactants and products Δ r H = Δ r U + RTΔνgas
Bond dissociation enthalpy is the enthalpy required to break a specific bond in a
specific molecule at a given pressure. Bond enthalpy is the average of dissociation
enthalpies of a given bond in a series of different types of dissociating species.
Kirchoff’s equation gives the relationship between enthalpy of reaction and
temperature.
Limitations of the first law of thermodynamics is that it is unable to predict
whether a process will be spontaneous or nonspontaneous. The second law of
thermodynamics decides the direction of a spontaneous process. It introduces a new
concept of entropy(S) that governs the criterion of spontaneity. Entropy is a state
function and is a measure of randomness.
Svapour > Sliquid > Ssolid
The change in entropy is defined as
For an isolated system, dS (isolated) ≥ 0.
According to Trouton’s rule for non-associating and non-dissociating compounds
the molar entropy of vaporization at the boiling point is constant.
Gibbs free energy (G) and Helmholtz free energy (A) are two functions that take
into account the composite effect of energy and entropy change of the system. The
process will be spontaneous if ΔG < 0.
ΔS(universe) =
The variation of free energy with temperature is given by Gibbs-Helmholtz
equations. Maxwell’s relations give the variation of one thermodynamic property
with another under various conditions.
The absolute entropy at any temperature will be equal to the sum of the entropy of
the substance at 0 K and the entropy changes taking place as the substance is heated to
the given temperature.
The entropy possessed by a substance at absolute zero is known as residual entropy.
The value of absolute entropy can be calculated using the formula based on statistical
thermodynamics, S = k ln W = 2.303 k log W where W = thermodynamic probability.
Exercises
3.1 Explain the following terms:
(i) System (ii) Surroundings (iii) Boundary (iv) Reversible process
(v) Irreversible process
3.2 Classify the following as extensive and intensive variables:
(i) Internal energy (ii) Heat capacity (iii) Refractive index (iv) Specific
heat capacity (v) Mole fraction (vi) Molarity (vii) Mass
(viii) Density (ix) Enthalpy (x) Pressure (xi) Temperature (xii) Dipole
moment (xiii) Entropy (xiv) Free energy
3.3 Which of the following processes are reversible and which are irreversible?
(i) Vaporization of benzene into vacuum at 60ºC.
(ii) Vaporization of water at 100ºC and 1atm pressure.
(iii) An ideal gas expands against constant pressure.
(iv) Diffusion of a gas into another gas at constant temperature and pressure
3.4 Which of the following statements are true and which are false?
(i) An isochoric process the volume remains constant.
(ii) q and w are state functions.
(iii) All the intensive properties are state functions.
(iv) In an isolated system heat and matter are both exchanged with the
surroundings.
(v) First law of thermodynamics is essentially the statement of the law of
conservation of energy.
(vi) State functions are inexact differentials.
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Thermodynamics
3.5 What are state functions? State and explain the Euler’s theorem of exactness
with the help of an example.
3.6 Given that
z = a x2y + by + cxy2
where a, b, c are constants. Show that
(i) dz is an exact differential and z is a state function.
(ii)
3.7 Derive the cyclic rule
3.8 Derive an expression for q, w, ∆U and ∆H for an ideal gas
undergoing isothermal irreversible expansion against constant pressure.
3.9 Three moles of an ideal gas at 27°C expand isothermally and reversibly from
20 dm3 to 60 dm3. Calculate ∆U, w, q and ∆H.
3.10 One mole of argon at 25°C and 1 atm is allowed to expand reversibly to a
volume of 50 dm3. Assuming the gas to be ideal, calculate final pressure if
(i) the expansion is isothermal
(ii) the expansion is adiabatic, (γ = 1.67)
3.11 Calculate ∆U and ∆H for one mole of an ideal gas (CV,m = 1.5 R) being heated
reversibly at a constant pressure of 101.325 kPa from 298 K to 373 K.
3.12 Derive the following relation for an ideal gas undergoing adiabatic
reversible process:
TV γ-1 = constant.
TP -R/CP = constant.
3.13 Justify the following statements:
(i) For an ideal gas the reversible work of compression from volume V1 to V2
is less than the corresponding work involved in the irreversible process.
(ii) For an ideal gas the reversible work done in going from volume V1 to V2
is greater in isothermal process than in adiabatic process.
(iii) Final pressure in an adiabatic expansion of an ideal gas is less than that
of the isothermal expansion of the same volume.
3.14 Describe the Joule’s experiment. What are its limitations?
3.15 An ideal gas with an initial volume of 22.4 litres is compressed adiabatically
until its volume is 11.2 litres. During this process 1350 J of work is performed
on the gas and the temperature rises from 0°C to 160° C. What is
the change in the internal energy during the process?
3.16 0.35 mole of an ideal monatomic gas is expanded adiabatically from a volume
of 1 litre at 400°C to a volume of 5 litres against a constant external
pressure of 0.5 atm. What is the final temperature of the gas and what is the
enthalpy change during the process?
3.17 Two litres of an ideal diatomic gas at 300 K and 5 atm expands until the
pressure of the gas is 2 atm. Calculate w, q, ∆U and ∆H for the process if
the expansion is
(i) isothermal and reversible
(ii) adiabatic and reversible
Given that CV,m = 5R/2.
3.18 Explain Hess’s law of constant heat summation.
3.19 Explain with examples the following:
(i) Standard enthalpy of formation
(ii) Standard enthalpy of neutralization
(iii) Standard enthalpy of combustion
(iv) Standard enthalpy of formation of an element
(v) Integral enthalpy of solution
3.20 For a reaction involving only condensed phase show that ΔH = ΔU.
3.21 Explain why the enthalpy of neutralization of all strong acid–strong base pair
is same.
3.22 Explain why does the enthalpy of neutralization for weak acid–strong base
pair depends upon the nature of weak acid.
3.23 Distinguish between bond enthalpy and bond dissociation enthalpy giving
examples.
3.24 Derive the Kirchhoff’s equation showing variation of enthalpy of a reaction
with temperature. Assume that the CP values are independent
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Thermodynamics
of
temperature.
3.25 Compute the standard enthalpy of formation of methane using the
following data:
(i) CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l) ΔH (298) = - 890.35 kJ mol−1
(ii) H2 (g) + ½ O2 → 2H2O (l) ΔH (298) = - 285.84 kJ mol−1
(iii) C (graphite) + O2 (g) → CO2 (g) ΔH (298) = - 393.51 kJ mol−1
3.26 From the following data:
(i) Δf H (CH3CN) = 87.8 kJ mol-1
(ii) Δf H (C2H6 ) = - 83.6 kJ mol-1
(iii) Δsub H ( graphite) = 719 kJ mol-1
(iv) Δdiss H (N2) = 944.7 kJ mol-1
(v) Δdiss H (H2) = 434.7 kJ mol-1
(vi) C
H bond enthalpy = 413.8 kJ mol-1
Calculate (i) ε(C
C) and ε(C
C)
3.27 Calculate the enthalpy of formation of cane sugar from the following data:
(i) Δ f H (H2O) = − 286 kJ mol-1
(ii) Δ c H (C12H22O11) = − 5638.8 kJ mol-1
(iii)Δ f H (CO2) = − 393 kJ mol-1
3.28
3.29
3.30
3.31
3.32
3.33
3.34
3.35
3.36
3.37
3.38
3.39
3.40
Write a limitation of the first law of thermodynamics.
What is the condition when the thermodynamic relation dS = dq/T is applicable.
Write a relation between the free energy change and net work done.
Derive a relation between Gibbs energy (G) and Helmholtz energy (A).
Write the conditions in terms of ∆H and ∆S when a reaction is always
spontaneous.
When ∆H > 0 and ∆S < 0 a reaction is never spontaneous. Give reason.
Using equation dA = - SdT - PdV derive Maxwell equation.
State the conditions when change in Gibbs energy is equal to work function.
State Trouton’s rule. Explain why it is valid for methane but not for acetic
acid.
Derive a relation between Gibbs energy change and entropy change of the
universe.
Show that:
ΔS = nCP ln ( T2 / T1) − nR ln (P2/P1 )
Calculate entropy change when 2 moles of an ideal gas expand isothermally
and reversibly from an initial volume of 1 dm3 to a final volume of 10 dm3 at
300 K.
10 g of ice are heated to become vapours at 100°C and 1 atm. Calculate
ΔS for the process given that: Δ f H (ice, 0°C) = 334.4 J g-1:
Δ vap H (H2O, 100°C) = 2.26 kJ
Average specific heat capacity of water is 4.18 J K-1g-1.
3.41 Which one of the following conditions will always lead to a spontaneous
change and which one will be for nonspontaneous change?
(i) ΔH < 0 and ΔS > 0
(ii) ΔH > 0 and ΔS < 0
(iii) ΔH < 0 and ΔS < 0
(iv) ΔH > 0 and ΔS > 0
3.42 Derive the following relations:
(i)
(v)
(ii)
(vi)
(iii)
(vii)
(iv)
(viii)
References
http://webphysics.davidson.edu/mjb/SESAPS2000/internal_Energy.html (internal
energy simulation)
Institute of Lifelong Learning, University of Delhi
Thermodynamics
P.W.Atkins : Physical Chemistry, Oxford University Press.
G.W.Castellan: Physical Chemistry, Narosa Publishing House.
G.M.Barrow, Physical Chemistry, Tata McGraw Hill.
Glossary
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Adiabatic process- System does not exchange heat with the surroundings, that
is, q = 0.
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Boundary- Anything that separates the system from the surroundings is known
as a boundary.
Bomb calorimeter- The measurements of the heat exchanges at a constant
volume are made in a bomb calorimeter.
Bond Energies- Energy is given to the system to break a bond and the energy is
released whenever a bond is formed. Therefore, bond formation is an exothermic
process while bond dissociation is an endothermic process.
Bond formation enthalpy- The heat released during bond formation at a
constant pressure is known as the bond formation enthalpy.
Bond dissociation enthalpy- It is the enthalpy required to break a specific
bond in a specific molecule at a given pressure. It depends upon the presence of
other groups in the molecule.
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Bond enthalpy- It is the average of dissociation enthalpies of a given bond in a
series of different types of dissociating species.
Chemical Thermodynamics- Deals with energy changes accompanying
chemical transformations.
Closed system- When the matter cannot flow across the boundary but energy
exchange can take place between the system and the surroundings.
Cyclic process- The system after undergoing a series of changes comes back to
the initial state. The series of changes taking place is known as a cycle.
Chemical equilibrium- The composition of each part of the system is same and
does not change with time.
Differential enthalpy of solution-It is defined as the enthalpy change taking
place when one mole of a solute is dissolved in a very large amount of solution of
known concentration so that there is no appreciable change in the concentration
of the solution.
Extensive properties-Extensive properties are those properties that depend
upon the size of the system.
Enthalpy- The heat exchange at constant pressure is called enthalpy and the
symbol is H.
Entropy- Entropy is a measure of randomness. More is the randomness greater
is the entropy of the system. The symbol is S.
Endothermic process -If the heat is absorbed by the system the process is
known as an endothermic process. ΔU and ΔH are given positive sign.
Exothermic process- If the system loses heat the process is known as an
exothermic process. ΔU and ΔH have negative sign since the energy of the
system decreases.
Enthalpy of Formation- The amount of heat exchanged at constant
temperature and pressure during the formation of one mole of the substance
from its constituent elements.
Enthalpy of transition-It is the amount of heat exchanged when one mole of a
substance is transformed from one allotropic form to the other allotropic form at
a given temperature and pressure.
Enthalpy of vapourization-It is the amount of heat absorbed to convert one
mole of a liquid to its vapour state at a given temperature and pressure.
Enthalpy of Fusion-It is the amount of heat absorbed to convert one mole of a
solid to its liquid state at a given temperature and pressure.
Enthalpy of Sublimation-It is the amount of heat absorbed to convert one mole
of a solid to vapour state at a given temperature and pressure.
Enthalpy of atomization-It is the amount of heat required to convert one mole
of a substance into its constituent atoms in the gaseous state.
Enthalpy of neutralization-It is defined as the enthalpy change taking place
when one mole of H+ ions are neutralized completely by OH− in dilute solutions
at a constant temperature and pressure.
Equilibrium-If the properties of the system, such as temperature, pressure,
volume and mass remain constant without the help of an external source.
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Thermodynamics
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Extensive properties-Extensive properties are those properties that depend
upon the size of the system
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First law of Thermodynamics-The change in internal energy (ΔU) of the
system is equal to the heat exchanged (q) by the system and the work done (w)
on/by the system.
First law of thermo chemistry- The enthalpy of formation of any compound is
equal in magnitude and of opposite sign to the enthalpy of dissociation of that
compound at the given temperature and pressure.
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Gibbs energy- Gibbs energy is known as free energy because ΔG is energy
available with the system for doing non expansion work
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Heat- Is defined as a quantity that flows across the boundary of a system, during
a change in its state by virtue of the difference in temperature between the
system and its surrounding.
Heat capacity-Heat capacity is defined as the heat needed to cause unit rise in
the temperature of a substance.
Heat Capacity at constant pressure(Cp)-It is also known as an isobaric heat
capacity.
Heat capacity at constant volume (CV )- It is also known as isochoric heat
capacity.
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Isolated system: When the matter and energy exchange cannot take place
between the system and the surroundings.
Isothermal process- The temperature of the system remains constant, that is,
∆T = 0.
Isobaric process- The pressure of the system remains constant, that is, there is
no change in the pressure, ∆P = 0.
Isochoric process- The volume of the system remains constant, that is, ∆V = 0.
Irreversible process- The system cannot retrace its path without the help of an
external source.
Intensive properties - Intensive properties are those properties that do not
depend upon the size of the system.
Inexact (Path) Functions-It is a thermodynamics property that depends upon
how the process is carried out, that is, it depends upon the path taken by the
system during the thermodynamic process.
Internal Energy (U)-The total energy of the system is due to energy possessed
by the constituents of the system such as atoms, ions and molecules as well as
their motion and position.
Integral Enthalpy of solution -It is the amount of heat exchanged when one
mole of a solute is dissolved in a given amount of solvent at a specified
temperature and pressure.
Integral enthalpy of dilution-The amount of heat exchanged when a given
amount of solvent is added to the solution of known concentration at a constant
temperature and pressure.
Law of conservation of energy - Energy can neither be created nor destroyed.
If it disappears from one form it must appear in another form.
Laws of Thermo chemistry-These laws are based on the fact that heat
exchanges are measured either at constant volume or under constant pressure.
Under these conditions heat is a state function, that is, ΔU or ΔH respectively.
Mechanical equilibrium- The external forces on the system are equal to
internal forces in the system.
Molar heat capacity-It is defined as the heat needed to cause a unit rise in the
temperature of one mole of a substance.
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Open system- When the exchange of matter (mass, m) and energy (U) can take
place between the system and the surroundings.
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Path: The sequence of steps taken by a system during a thermodynamic process
starting from the initial state to the final state.
Polarisibile-Anions are large in size than cations and, therefore, their electron
clouds are less tightly held. When two oppositely charged ions approach each
other closely, the positively charged cation attracts the outer most electrons of
the anions and repels its positively charged nucleus. This results in the distortion
or polarisation of the anion.
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Thermodynamics
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Polarizing power-The power of an ion to distort the other ion is known as its
polarizing power.
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Reversible process-When the system comes back to the initial state the
surroundings are also restored to their initial state.
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System- This is separated from the rest of the universe by a real or imaginary
well defined boundary.
Surroundings- It lies outside the system and is also called the environment of
the system.
State of the system-The pressure (p), temperature (T), volume (V) and the
amount (number of moles, n)of the systems, are four variables which define the
state of the system.
Steady state- When the properties of a system are kept constant with the help
of an external source.
State property or state function-The property of a system that depends on
the state variables is known as a state property.
Second law of Thermodynamics -It is impossible to transfer heat from a
colder system to a warmer system without other changes occurring
simultaneously in the two systems or in their environment.
Specific heat capacity- It is defined as the heat needed to cause a unit rise in
the temperature of a unit mass of a substance.
Standard enthalpy of formation- The amount of heat exchanged at a given
temperature and a pressure of one bar during the formation of one mole of the
substance in its standard state from its constituent elements in their respective
standard states.
Second law of thermo chemistry- (Hess’s law of constant heat summation)
the total enthalpy change of a reaction is the same regardless of whether the
reaction is completed in one step or in several steps.
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Thermodynamics- Is a combination of two Greek words thermos (heat) and
dynamics (power or motion).
Thermodynamic process- It is any process that brings about a change in the
state of the system. Expansion, compression, heating and cooling are some of
the examples of a thermodynamic process.
Thermal equilibrium- Temperature of each part of the system is the same.
Third law of Thermodynamics - The entropy of a pure and perfectly crystalline
solid is zero at zero Kelvin of temperature.
Thermo chemistry- Is a part of thermodynamics that deals with the changes in
energy during a chemical reaction or a physical transformation.
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Universe-A system and its surrounding are together known as the Universe.
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Work-In thermodynamics, work is defined as any quantity that flows across the
boundary of a system during a change in its state and is completely convertible
into the lifting of a weight in the surroundings.
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Zeroth law of Thermodynamics- If two systems A and B are in thermal
equilibrium with a third system C, then A and B will be in thermal equilibrium
when placed together.
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