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Semester 2
Academic session 2011/2012
Characterized by the 4 laws of thermodynamics
Key Concepts
Pressure-Volume Work (P-V Work)
First Law of Thermodynamics
Perfect Gas
Molecular Nature of Internal Energy
Work (w) is defined as the force (F) that produces the movement of an
object through a distance (d):
Work is done when and object, e.g. a system's wall, moves against an
opposing force. This is equivalent to an ordered motion done by the system
on the surroundings or vice versa.
Work = force ×distance
Work also has units of J, kJ, cal, kcal, Cal, etc.
Pressure-volume work
The example of PV work – in the cylinder of an automobile engine
• The combustion of the gasoline causes gases within the cylinder to expand, pushing the
piston outward and ultimately moving the wheel of the car. The relationship between a
volume change (ΔV) and work (w):
W= -P ΔV
Where P is external pressure
The units of PV work are L·atm; 1 L·atm = 101.3 J.
• If the gas expands, ΔV is positive, and the work term will have a negative sign (work
energy is leaving the system).
• If the gas contracts, ΔV is negative, and the work term will have a positive sign (work
energy is entering the system).
• If there is no change in volume, ΔV = 0, and there is no work done. (This occurs in
reactions in which there is no change in the number of moles of gas.)
Pressure-volume (P-V) work
The work done in a volume
change is called P-V work
By removing weights
and decreasing the
pressure and allowing
the volume to adjust
according to Boyle's
law with no heat
By holding the
pressure constant
and increasing the
volume by heating
the gas
using Charle’s law
Initial state
Final state
Final state
Work done by gas
The units of PV work are
L·atm; 1 L·atm = 101.3 J.
if ΔV < 0, then W > 0.;
increases in volume means work
done BY the system on the
if ΔV > 0, then W < 0.;
decreases in volume means
work done by the environment
ON the system.
Work and heat are not state functions
On a graph of pressure versus volume, the work is the area
under the curve that describes how the state is changed from
State 1 to State 2.
A curved black line from State 1 to State 2 represents a change
brought about by removing weights and decreasing the
pressure and allowing the volume to adjust according to either
Boyle’s law (the line is curved and the amount of work done
on the gas is shown by the red shaded area below this curve)
Charle’s law (the resulting change in state proceeds from State
1 to an intermediate State "a" on the graph by heating. State
"a" is at the same pressure as State 1, but at a different
volume. If we then remove the weights, holding a constant
volume, we proceed on to State 2. The work done in this
process is shown by the yellow shaded area).
Using either process we change the state of the gas from State 1 to State 2. But the
work for the constant pressure process is greater than the work for the curved line
The work done by a gas not only depends on the initial
and final states of the gas but also on the process used to
change the state. Different processes can produce the
same state, but produce different amounts of work.
Notice that not only does the work done by the gas depend on the process, but also
the heat transferred to the gas. In the first process, the curved line from State 1 to
State2, no heat was transferred to the gas; the process was adiabatic. But in the
second process, the straight line from State 1 to State "a" and then to State 2, heat
was transferred to the gas during the constant pressure process.
The heat transferred to a gas not only depends on the
initial and final states of the gas but also on the process
used to change the state.
Why energy is a state function but
heat and work are not?
State Function is a thermodynamic parameter whose value does not depend on the thermodynamic process...It depends only
on the initial and final states.
State functions are characterized by the idea that no matter what path you take to get from point A to point B, the difference
between B and A remains the same. An example of how this works can be illustrated using gravitational potential energy. Say
you lift a box 5 meters vertically straight up into the air from the surface of the ground (which we shall set to a height of 0 m).
You increase its energy by an amount mg h where h = 5m. Now say that you lift a box from the ground level up 1 meter
above the ground, walk to the right 5 meters, lift the box 1 more meter, walk to the left 2 meters, raise the box 7 meters, walk
to the left 1 more meter, and lower the box 4 meters. How much did the gravitational potential energy change? The same
amount as before, mg h where h = 5m, because the box is still only 5 meters above where it started in either case and the
potential energy depends only on the final height minus the initial height (it depends only on the final and initial position).
In contrast, you expended a lot more energy (more work) moving the box around than you did just standing in place and lifting
it straight up. So the work you did is NOT a state function. This function depends on the path. When you lifted the box straight
up you did less work than your body did lugging the box around left and right while lifting it up and down. So the potential
energy is a state function, but the work you did is a path function.
Now in thermodynamics you usually apply these ideas to the 1st law or conservation of energy:
In this example, Q and W are path functions...the exact amount of work you do or the heat transferred to the system depend
on how you add the heat or do the work (these individual values will vary depending on whether or not you did these things in
either a direct or roundabout way). However, if you add these two path functions together the overall result, U, is a state
function. It only depends on
U (final) – U (initial) , regardless of whether or not you did direct or roundabout Q's and W's.
What are Reversible and Irreversible
There are two main types of thermodynamic processes: the reversible and irreversible. The reversible
process is the ideal process which never occurs, while the irreversible process is the natural process that
is commonly found in the nature. Therefore:
• The reversible process is an idealization.
• All real processes on Earth are irreversible.
P-V Work
A reversible process
is one that can be
halted at any stage
and reversed. In a
reversible process,
the system is at
equilibrium at every
stage of the process.
An irreversible
process is one where
it cannot be halted at
any stage and
reversed and the
system is not always
at equilibrium at
every stage of the
What is a Reversible Process?
The process in which the system and surroundings can be
restored to the initial state from the final state without
producing any changes in the thermodynamics properties
of the universe is called a reversible process. In the figure
below, let us suppose that the system has undergone a
change from state A to state B. If the system can be
restored from state B to state A, and there is no change in
the universe, then the process is said to be a reversible
process. The reversible process can be reversed completely
and there is no trace left to show that the system had
undergone thermodynamic change.
For the system to undergo reversible change, it should occur infinitely slowly due to infinitesimal
gradient. During reversible process, all the changes in state that occur in the system are in
thermodynamic equilibrium with each other. Thus there are two important conditions for the
reversible process to occur. Firstly, the process should occur in infinitesimally small time and secondly
all of the initial and final state of the system should be in equilibrium with each other. If during the
reversible process the heat content of the system remains constant, i.e. it is adiabatic process, then
the process is also isentropic process, i.e. the entropy of the system remains constant. The
phenomenon of undergoing reversible change is also called reversibility. In actual practice the
reversible process never occurs, thus it is an ideal or hypothetical process.
What is an Irreversible Process?
The irreversible process is also called the natural process because all the processes
occurring in nature are irreversible processes. The natural process occurs due to the
finite gradient between the two states of the system. For instance, heat flow between
two bodies occurs due to the temperature gradient between the two bodies; this is in
fact the natural flow of heat. Similarly, water flows from high level to low level, current
moves from high potential to low potential, etc.
Here are some important points about the irreversible process:
1. In the irreversible process the initial state of the system and surroundings cannot be
restored from the final state.
2. During the irreversible process the various states of the system on the path of
change from initial state to final state are not in equilibrium with each other.
3. During the irreversible process the entropy of the system increases decisively and it
cannot be reduced back to its initial value.
4. The phenomenon of a system undergoing irreversible process is called as
Gas confined in a
cylinder by means of a
moveable piston
The cylinder of an
automobile engine
Calculation of PV work for reversible and irreversible processes
A reversible process is one that
can be halted at any stage and
reversed. In a reversible
process the system is at
equilibrium at every stage of
the process. An irreversible
process is one where these
conditions are not fulfilled.
If pressure has changed
continuously then the work is
If the work done irreversibly
by changing pressure to Pfinal
when for the work we have:
The work done by a reversible process represents the maximal work that the system can
perform changing between the same original and final states.
Example for P-V WORK
Inflating balloon requires the inflator to do pressure-work on the surroundings. If
balloons is inflated from a volume of 0.100L to 1.85L against an external pressure of
1.00 atm, how much work is done (in joules)?
ΔV = V1 -V2
= 1.85L -0.100L
= 1.75L
W= -P ΔV
= -1.00 atm x 1.75L
= -1.75L.atm
Convert to Joule:
-1.75L.atm x 101.3 J= -177 J
The work is negative because it is being done by the
system as its volume increases due to the expansion
of the gas into the much bigger volume.
Study Example 2.2 page 44
Find the work  for processes (a) and (b) of Fig. 2.3 if P1 3.00 atm, V1
500 cm3, P2 1.00 atm, and V2 2000 cm3.
Figure 2.3
The work w done on the system in a reversible process (the heavy lines) equals minus the shaded area under the P-versus-V
curve. The work depends on the process used to go from state 1 to state 2.
Use this equation:
Simple as the previous example. Answer:  = - 152 J and  = -456 J
Processes (a) and (b) are expansions. Hence the system does positive work on its
surroundings, and the work w done on the system is negative in these processes.
Heat is a exchange of thermal energy between a system and its
surroundings caused by temperature difference.
Notice the distinction between heat and temperature. Temperature is a
measure of the thermal energy of a sample matter. Heat is transfer of the
thermal energy.
Heat may be defined as energy in transit from a high temperature object to
a lower temperature object.
Anytime two substances with different temperatures come in contact with
each other, there is an energy transfer. One substance loses heat energy and
the other substance gains heat energy.
Heat energy flows from a hotter substance to a colder substance.
For example, if the ice cube in the diagram is placed in the
container of water, there is an energy transfer. The hotter
substance loses heat energy and the colder substance
gains heat energy.
The water and its container lose heat energy and become
The ice cube gains heat energy and becomes warmer (this
causes the ice cube to melt).
According to the law of energy conservation the
total heat energy lost by the water and its container
is equal to the total heat energy gained by the ice.
Heat can be transferred reversibly or irreversibly. A reversible transfer of heat requires
that the temperature difference between the two bodies be infinitesimal. When there
is a finite temperature difference between the bodies, the heat flow is irreversible.
Heat capacity
when system absorbs heat (q) its temperature changes by ΔT:
Experimental measurements demonstrate that the heat absorbed by a system and its
corresponding temperature change are directly proportional: qαΔT. The constant of
proportionality between q and ΔT is called the heat capacity.
Heat capacity (C) is the amount of heat (q) a substance must absorb to raise its
temperature(ΔT) by 1 °C.-Heat capacity has units of J/°C (or J/K), and is an extensive
property, depending on the sample size.
Specific Heat Capacity
The specific heat (c, or specific heat capacity, Cs) of an object, is the quantity of heat
required to change the temperature of 1 gram of a substance by 1°C (or K):
Specific heat has units of J/g°C, and is an intensive property, which is independent of the
sample size.
Calculate the amount of heat needed to increase the temperature of 250g of water
from 20oC to 56oC.
q = m x Cg x (Tf - Ti)
m = 250g
Cg = 4.18 J oC-1 g-1 (from table)
Tf = 56oC
Ti = 20oC
q = 250 x 4.18 x (56 - 20)
q = 250 x 4.18 x 36
q = 37 620 J = 38 kJ
What is the difference between heat
capacity and specific heat?
Heat capacity is the amount of heat required to raise the temperature of any quantity
of substance by 1 degree centigrade.
Specific heat is the amount of heat required to raise the temperature of 1kg (MASS)
of substance by 1 degree substance.
The unit for specific heat is J/(g*K). The unit for heat capacity is J/K.
Difference between specific heat capacity and heat capacity is identical to difference
between concentration and amount of substance - one is intensive, other extensive
Heat capacity at constant pressure
(or isobaric heat – constant P
An isobaric process is a constant-pressure process. In general, none of the three
quantities ΔU, Q and dW is zero in an isobaric process
For closed syst. in equilib., P-V work only
ΔH = 
Heat capacity at constant volume
(or isochoric heat capacity)-constant V
An isochoric process is a constant-volume process. When the volume of a
thermodynamic system is constant, it does zero work on the surroundings. Then W = 0,
For closed syst. in equilib., P-V work only
w is zero
In an isochoric process, all the energy added as heat remains in the system as an
increase in internal energy. Heating a gas in a closed constant-volume container is an
example of an isochoric process. (Note that there are types of work that do not involve
a volume change. For example, we can do work on a fluid by stirring it. In some cases,
“isochoric” is used to mean that no work is done).
V f = V i : vertical line on p V diagram
Is the law of conservation of Energy.
1st Law = Conservation of Energy
The first law of thermodynamics is simply an
expression of the conservation of energy principle.
The principle of the conservation of energy states
that energy can neither be created nor
destroyed. But it can change from one type of
energy to another (for example kinetic to potential)
but the total amount remains fixed (The total energy
of a closed system remains constant.)
In thermodynamics, energy is classified into three different types:
1. work (W)
2. heat (Q)
3. internal energy (ΔU)
This allows us to write a simple form for conservation of energy (or the first
law of thermodynamics) as
(the change in the internal energy of a closed system is equal to the amount
of heat supplied to the system, minus the amount of work performed by the
system on its surroundings)
The standard unit for all these quantities would be the joule.
Internal Energy
Potential &
Kinetic Energy
• Energy is defined as the ability or capacity to do work on some form of
matter (the amount of work one system is doing on another). There are
several forms of energy:
• Potential energy is the energy that a body possesses as a consequence of
its position in a gravitational field (e.g., water behind a dam).
Kinetic energy is the energy that a body possesses as a consequence of its
motion (e.g., wind blowing across a wind generator). It is dependent upon
an object's mass and velocity (e.g., moving water versus moving air).
• Internal energy is the total energy (potential and kinetic) stored in
molecules. It is the energy associated with the random, disordered motion
of molecules; it refers to the invisible microscopic energy on the atomic
and molecular scale.
The First Law of Thermodynamics states that energy lost during one process must equal
the energy gained during another, i.e., all energy is conserved.
Unit of Energy
• Energy is measured in Joules (J) or Calories (cal).1 J = 1 kg m2s-2
• A calorie (cal) is the amount of energy needed to raise the temperature
of 1 g of water by 1°C.
1 cal = 4.184 J
The internal energy, U of a system is the sum of the kinetic and potential energies of
all the particles that compose the system or the total energy of a system (the energy
associated with the random, disordered motion of molecules).
It involves energy on the microscopic scale.
The total (internal) energy in a system includes potential and kinetic energy.
Binding energies – atomic bonds
(potential energy from
intermolecular forces)
• translational kinetic energy
• vibrational and rotational kinetic energy
For example, a room temperature glass of water sitting on a table has no apparent
energy, either potential or kinetic . But on the microscopic scale it is a seething mass
of high speed molecules traveling at hundreds of meters per second. If the water were
tossed across the room, this microscopic energy would not necessarily be changed
when we superimpose an ordered large scale motion on the water as a whole.
• There are two ways to change the internal energy: with work and heat.
• Internal energy of an object can be changed by the following methods:
It increases if energy is added to the system.
• i.e. by heating or by doing work on the system.
It decreases if energy is removed from the system or work is done
by the system.
• i.e. Thus heat and work changes the internal energy of an object.
Reaction between carbon and oxygen to form carbon dioxide
C(s) + O2(g) CO2(g)
if the reactants have a higher internal
energy than a products, ΔUsys is
negative and energy flows out of the
system into the surroundings.
if the reactants have a lower internal
energy than a products, ΔUsys is positive
and energy flows into the system from
the surroundings.
Internal energy is the state function, which means
that its value depends only the state of the system,
not the how the system arrive at the state.
Altitude is a state function. The change in altitude during climbing depends only on
the difference between the final and initial altitudes.
Problem 1: A gas expands against a constant pressure of 1 atm from a volume
of 10 L to 20 L. During this process, the system absorbs 600 J of heat from the
surroundings. Calculate the internal energy of the system.
ΔU = q + w
= 600 J + (-PΔV) = -413 J
= 600 J + {- 101.3 Pa (20-10) L}
= 600 J + (-1013 J) = - 413 J
Problem 2 :q amount of heat is transferred to the system from the
surroundings and w amount of work is done by the system. Write the
expression for the internal energy.
ΔU = q – w
Study Example 2.3 pg. 50
Calculate U when 1.00 mol of H2O goes from 25.0°C and 1.00 atm to 30.0°C and
1.00 atm. Densities of water are 0.9970 g/cm3 at 0°C and 0.9956 g/cm3 at
Refer to sildes ‘Specific Heat Capacity’, ‘P-V Work’, and the First Law equation to solve !
Answer: ΔU = q + w = 90 cal.
Now, you’ve learn about heat & work, so
what is the difference between HEAT &
Heat is an energy transfer between system and
surroundings due to a temperature difference.
Work is an energy transfer between system and
surroundings due to a macroscopic force acting
through a distance.
Heat and work, both, are energy. The difference is just the amount of
ordered motion during the energy transfer.
Work and Heat
Energy transfer between system and surrounding occurs either in the form of
work or heat.
Work (W) → refers to mechanical work
W = Fd
Heat (Q) → refers to energy
transferred from a hot to a cold
A system can exchange energy with its surroundings through heat and work:
According to the first law thermodynamic, the change in the internal energy of the
system(ΔU) must be the sum of the heat transferred (q) and the work done (w):
Sign of conventions for q, w, and ΔU
Enthalpy vs. Energy
Enthalpy (H) is the heat flow in or out of a system at constant pressure (i.e., carried out in
open containers at or near atmospheric pressure):
where E = energy, P = pressure, and V = volume.
Enthalpy depends on the amount of substance present.
For constant-pressure process:
ΔH =  const. P, closed syst., P-V work only
For constant-volume process:
ΔU = qV
const. V , closed syst., P-V work only
What is the relationship between the
change in energy and enthalpy?
Energy transfer occurs as heat when little or no work gets done. This is the case in
three scenarios:
1. Reactions that do not involve gases (done at constant pressure with little or no
volume change).
2. Reactions in which the number of moles of gas does not change (when Δn = 0
then ΔV = 0).
1. Reaction in which the volume (i.e., moles) changes but the work is negligible
compared to the heat.
What is Joule Thompson's
It is an experiment in which the Joule-Thomson coefficient is measured. Basically,
you are expanding a gas under adiabatic conditions to ensure constant enthalpy
and you will notice that there will be a temperature change (most likely cooling).
We look at the Joule Expansion to learn how to relate derivatives, such as du/dV
under constant temperature and du/dT under constant volume. The Joule
Expansion can be used to find these quantities.
Hydrogen chloride gas readily dissolves in water, releasing 75.3 kJ/mol of heat in the
process. If one mole of HCl at 298 K and 1 atm pressure occupies 24.5 liters, find
ΔU for the system when one mole of HCl dissolves in water under these conditions.
In this process the volume of liquid remains practically unchanged, so ΔV = –24.5 L.
The work done is
w = –PΔV = –(1 atm)(–24.5 L) = 24.6 L-atm
(The work is positive because it is being done on the system as its volume decreases
due to the dissolution of the gas into the much smaller volume of the solution.)
Using the conversion factor 1 L-atm = 101.33 J mol–1 and substituting in the 1st Law Eqn
we obtain:
ΔU= q +PΔV = –(75300 J) + [101.33 J/L-atm) × (24.5 L-atm)] = –72.82 kJ
In other words, if the gaseous HCl could dissolve without volume change, the heat
released by the process (75.3 kJ) would cause the system’s internal energy to diminish
by 75.3 kJ. But the disappearance of the gaseous phase reduces the volume of the
system. This is equivalent to compression of the system by the pressure of the
atmosphere performing work on it and consuming part of the energy that would
otherwise be liberated, reducing the net value of ΔU to –72.82 kJ.
Perfect gas and the first law
Perfect gas obeys both the following equations:
For a closed system in equilibrium, the internal energy (and any other state function)
can be expressed as a function of temperature and volume. However eqn. above
states that for a perfect gas U is independent of volume. Therefore U of a perfect gas
depends only on temperature. So:
We now apply the first law to a perfect gas:
Isothermal process in a perfect gas
An isothermal process is one that occurs at
constant temperature. The gas-piston container in
our example could expand isothermally if it were
kept immersed in a large hot-water bath while the
gas expanded. Since the temperature doesn't
change during an isothermal process, there is no
change in internal energy. The first law then tells
you that the work done by the gas is just equal to
the heat that flows into the system from the bath.
Adiabatic process in a perfect gas
An adiabatic process is one that occurs
without the exchange of heat with the
surroundings. If the gas-piston system were
insulated so that heat could not get in or out,
any expansion or compression would occur
adiabatically. Since Q = 0 for an adiabatic
process, the first law tells you that the
change in internal energy is just equal to the
work done on the system.  dq = 0 and q = 0
When a gas expands adiabatically, the work done in the expansion comes at the
expense of the internal energy of the gas, causing the temperature of the gas to drop.
The figure below shows P-V diagrams for these two processes.
The figure compares two processes that begin with the same state and involve
expansion to the same final volume. For the isothermal process, the product of P·V
remains constant since T remains constant. Since the temperature must decrease for
the adiabatic process, it follows that the final pressure must be less for this process.
Thus the adiabat lies below the isotherm. Let's look at one more example that
incorporates many of the AP points of emphasis.
Study Example 2.5 pg. 60
A cylinder fitted with a frictionless piston contains 3.00 mol of He gas at P 1.00 atm
and is in a large constant-temperature bath at 400 K. The pressure is reversibly
increased to 5.00 atm. Find w, q, and U for this process.
Refer to the slide of Isothermal process in a perfect gas to solve !
W = 1.61x104 J
q = - 1.61x104 J
The molecular description of internal energy is outside the scope of
thermodynamics, but a qualitative understanding of molecular energies is helpful.
Consider first a gas. The molecules are moving through space. A molecule has a
translational kinetic energy mv2, where m and v are the mass and speed of the
molecule. A translation is a motion in which every point of the body moves the
same distance in the same direction.
If each gas molecule has more than one atom, then the molecules undergo
rotational and vibrational motions in addition to translation. A rotation is a motion
in which the spatial orientation of the body changes, but the distances between all
points in the body remain fixed and the center of mass of the body does not move
(so that there is no translational motion).
Besides translational and rotational energies, the atoms in a molecule have
vibrational energy. In a molecular vibration, the atoms oscillate about their
equilibrium positions in the molecule. A molecule has various characteristic ways
of vibrating, each way being called a vibrational normal mode.
Figure 2.14 shows translational, rotational, and vibrational motions in CO2.
The limitations of the first law of
No restriction on the direction of the flow of heat: the first law establishes
definite relationship between the heat absorbed and the work performed by a
system. The first law does not indicate whether heat can flow from a cold end to
a hot end or not. For example: we cannot extract heat from the ice by cooling it
to a low temperature. Some external work has to be done.
Does not specify the feasibility of the reaction: first law does not specify that
process is feasible or not for example: when a rod is heated at one end then
equilibrium has to be obtained which is possible only by some expenditure of
Practically it is not possible to convert the heat energy into an equivalent
amount of work.
To overcome this limitations, another law is needed which is known as second law of
thermodynamics. The second law of thermodynamics helps us to predict whether the
reaction is feasible or not and also tell the direction of the flow of heat. It also tells that
energy cannot be completely converted into equivalent work.