Download physical chemistry lecture 3

Basic concept:
1. Work, heat and energy
2. The internal energy
3. Expansion work
4. Heat transaction
5. Adiabatic changes
Thermochemistry:
1. Standard enthalpy changes
2. Standard enthalpies of
formation
3. The temperature dependence
of reaction enthalpies.
First Law
Thermodynamic
State of function and exact differentials:
1. Exact and inexact differentials
2. Changes in internal energy
3. The Joule – Thomson effect
Work
• Work (w) is defined as the force (F) that produces the movement of an object
through a distance (d):
Work = force × distance
w=Fxd
• Work also has units of J, kJ, cal, kcal, Cal, etc.
• The two most important types of chemical work are:
– the electrical work done by moving charged particles.
– the expansion work done as a result of a volume change in a system,
particularly from an expanding or contracting gas. This is also known
as pressure-volume work, or PV work.
• PV work occurs when the force is the result of a volume change against an
external pressure.
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 (∆V) and work (w):
W= -P ∆V
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.)
Expansion Work
• 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.)
b) Free expansion is expansion against zero opposing force.
W=0
No work is done when a system expand freely. Expansion of this kind occurs when a gas
expands into a vacuum.
c) Expansion against constant pressure – now suppose that the external pressure is
constant throughout expansion.
Therefore, if we write the change in
volume as ∆V=Vf -Vi
d) Reversible Expansion
A reverse change in thermodynamic is a change that can be reversed by an
infinitesimal modification of a variable.
The key “infinitesimal” sharpen the everyday meaning of the word “reversible” as
something that can change direction.
d) Isothermal reversible expansion
• When the volume is greater than the initial volume, as in expansion, the logarithm
in above equation is positive and hence w<0. In this case the system has done work
on the surroundings and there is a corresponding reduction in its internal energy.
• From the equation also show that more work is done for a given change of volume
when temperature is increased:
at higher temperature the greater pressure of the confined gas needs a higher
opposing pressure to ensure reversibility and the work done is
correspondingly greater.
Table 1
Type of work
Type of work
Variables
Conventional unit
Volume expansion
Pressure (P), volume (V)
Pa m3 = J
Stretching
Tension (ɤ), length (l)
N m= J
Surface expansion
Surface tension (ɤ), area (σ)
(Nm-1)(m2) = J
Electrical
Electrical potential(φ),
Electrical change (q)
VC=J
Example 1:
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)?
Sort: you are given the initial and final
volumes of the volumes of the balloon
and the pressure against which its
expands. The balloon and its contents are
the system
Given: V1 = 0.100L, V2 = 1.85L, P=1.00 atm
Find: W
Strategy: the equation W= -P ∆V specifies
the amount of work done during a volume
Change against an external pressure.
Conceptual Plan:
P,∆V → W
W= -P ∆V
Solve: To solve the problem, compute the Solution:
value of ∆V and substitute it, together with ∆V = V1 - V2
P, into the equation.
= 1.85L - 0.100L
= 1.75L
W= -P ∆V
= - 1.00 atm x 1.75L
= -1.75L.atm
The units of the answer (L.atm) can be
-1.75 L.atm x 101.3 J = -177 J
converted to joules using 101.3J = 1L.atm
1L.atm
Check : The units (J) are correct for work. The sign it should be for an expansion: work is done
on the surroundings by the expanding balloon.
Example 2:
A sample of argon of mass 6.56 g occupies 18.5 dm3 at 305K.
(a)Calculate the work done when the gas expands
isothermally against a constant external pressure of 7.7 kPa
until its volume has increased by 2.5 dm3.
(b) Calculate the work that would be done if the same
expansion occurred reversibly.
(a)
(b)
HEAT
 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.
Temperature Change and Heat Capacity – when system absorbs heat (q) its temperature
changes by ∆T:
Heat (q)
System
∆T
Experimental measurements demonstrate that the heat absorbed by a system and its
corresponding temperature change are directly proportional : q α ∆T. The constant of
proportionality between q and ∆T is called the heat capacity.
Heat Capacity
• Heat capacity (C) is the amount of heat (q) a substance must absorb to raise
its temperature(∆T) by 1 °C.
- Heat capacity has units of J/°C (or J/K), and is an extensive property,
depending on the sample size.
• Two type of heat capacity:
1) Heat capacity at constant volume:
2) Heat capacity at pressure constant
Specific Heat
• 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.
• The molar heat capacity (Cm), is the quantity of heat required to change
the temperature of 1 mole of a substance by 1 °C (or K):
Heat Transactions
In general the change in internal energy of a system:
dwe is work in addition(e for “extra”) to the expansion work
dwexp = 0, when a system kept a constant volume cannot do any expansion work.
If the system also incapable of doing any other kind of work (if it is not, for instance,
an electrochemical cell connected to an electric motor), then dwe = 0 too. Under these
circumstances :
Heat transfer at volume
We express this relationship by writing, dU=dqv where the subscript implies a change
at constant volume. For measurement change,
A bomb Calorimeter
The amount of heat absorbed by the water in
the calorimeter is equal to the energy that is
released by the reaction (but opposite in
sign).
qcal : heat absorbed by the entire calorimeter assembly
Ccal: heat capacity of entire calorimeter assembly
If no heat escape from the container:
During the summer, you have experienced the
effect of water’s high specific heat capacity.
Example:
Sacramento (an inland city) and San Francisco
(a coastal city) can be 18oC (30oF).
San Francisco enjoy cool 20oC (68oF) while
Sacramento bakes at nearly 38oC (100oF).
Why the large temperature difference?
San Francisco sit on peninsula, surrounded by the
Pacific Ocean. Water – absorbed much the sun’s
heat without undergoing a large increase in
temperature.
While the Sacramento, with its low heat capacity
undergoes a large increase in temperature as it
absorbs a similar amount of heat.
heat = mass x specific heat x temperature
The high heat capacity of water surrounding
San Francisco results in relatively cool summer
temperatures.
q= m x Cs x ∆T
The specific heat can be used to quantify the relationship between the amount of heat
added to a given amount of the substance and the corresponding temperature increase.
The equation is:
heat = mass x specific heat x temperature
q= m x Cs x ∆T
Example 2:
Suppose you find a copper penny (minted pre-1982
when pennies were almost entirely were almost
entirely copper) in the snow. How much heat is
absorbed, by the penny as it warms, from the sun
temperature of the snow, which is -8.0oC, to the
temperature of your body, 37oC? Assume the penny is
pure copper and has a mass of 3.10g.
Sort: You are given the mass of copper
as well as its initial and final
temperature. You are asked to find the
heat required for the given temperature
change
Given:
m = 3.10 g copper
Ti = -8.0 oC
Tf = 37.0 oC
Find : q
Strategy: The equation q = m x Cs x ∆T
Conceptual Plan:
give the relationship between the
Cs , m, ∆T → q
amount of heat (q) and the temperature q = m x Cs x ∆T
change (∆T )
Relationship Used:
q = m x Cs x ∆T
Cs = 0.385 J/g.oC
Solve: Gather the necessary quantities
for the equation in the correct units and
Substitute these into the equation to
compute q.
Solution:
∆T = Tf – Ti
= 37.0oC – (-8.0oC) = 45.0oC
q = m x Cs x ∆T
= 3.10 g x 0.385 J x 45.0oC = 53.7J
g. oC
Check : The unit (J) are correct for heat. The sign of q is positive, as it should be since the
penny absorbed heat from surrounding.
Energy
Energy is define as the ability to do work.
Work is done when a force is exerted through a distance.
Force through distance; work is done.
Energy is measured in Joules (J) or Calories (cal).
1 J = 1 kg m2 s-2
Energy may be converted from one to another, but it is neither created nor
destroyed (conversion of energy).
In generally, system tend to move from situations of high potential energy (less
stable) to situations having lower energy (more stable).
Unit of Energy
• 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 nutritional unit Calorie (Cal) is actually a kilocalorie (kcal):
1 Cal = 1000 cal = 1 kCal = 4184 J
Potential and Kinetic Energy
• Kinetic energy (EK) is the energy due to the motion of an object with mass m
and velocity v:
EK = ½ mv2
– Thermal energy, the energy associated with the temperature of an object, is
a form of kinetic energy, because it arises from the vibrations of the atoms
and molecules within the object.
• Potential energy (EP) is energy due to position, or any other form of “stored”
energy. There are several forms of potential energy:
– Gravitational potential energy
– Mechanical potential energy
– Chemical potential energy (stored in chemical bonds)
Internal Energy
•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.
•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.
•Some examples include energy (and many other thermodynamic terms),
pressure, volume, altitude, distance, etc.
• An energy change in a system can occur by many different combinations of
heat (q) and work (w), but no matter what the combination, ΔU is always the
same — the amount of the energy change does not depend on how the change
takes place.
∆U = q + w
Altitude is a state function. The change in altitude during climbing depends only on
the difference between the final and initial altitudes.
Example: reaction between carbon and oxygen to form carbon dioxide
C(s) + O2 (g)
CO2(g)
In this reaction, energy is released from the system to the surroundings.
– The reactants have a higher E than the products, so ΔU for the system is
negative.
– This energy is gained by the surroundings, where ΔU is positive.
(ΔUsystem = -ΔUsurroundings)
ΔU < 0 (negative)
System
∆Usys < 0 (negative)
Surroundings
Energy flow
∆Usurr > 0 (positive)
If the reaction is reversed, energy is absorbed by the system from the
surroundings.
– The reactants have a lower E than the products, so ΔU for the system is
positive.
– This energy is lost by the surroundings, where ΔU is negative.
ΔU > 0 (positive)
The difference , ∆U is positive and energy glows into the system and out of
surroundings
System
∆Usys > 0 (positive)
Surroundings
Energy flow
∆Usurr < 0 (negative)
Summarizing:
 if the reactants have a higher internal energy than a products, ∆Usys is negative
and energy flows out of the system into the surroundings.
 if the reactants have a lower internal energy than a products, ∆Usys is positive and
energy flows into the system from the surroundings.
A system can exchange energy with its surroundings through heat and work:
Heat (q)
System
Surroundings
Work (w)
According to the first law thermodynamic, the change in the internal energy of the system
(∆U) must be the sum of the heat transferred (q) and the work done (w):
∆U = q +w
Sign of conventions for q, w, and ∆U
q (heat)
+ system gain thermal energy
- System loses thermal energy
w (work)
+ work done on the system
- Work done by the system
∆U
+ energy flows into the system
(change in
internal energy)
- Energy flows out of the system
• For an isolated system, with no energy flowing in or out of the system, the
internal energy is a constant.
– First Law of Thermodynamics (restated): The total internal energy of an
isolated system is constant.
• It is impossible to completely isolate a reaction from its surroundings, but it is
possible to measure the change in the internal energy of the system, ΔU, as
energy flows into the system from the surroundings or flows from the system
into the surroundings.
∆U = Uf - Ui
Enthalpy
•Most reactions are not done in sealed containers: they are carried out in open vessels
at constant pressure, with the volume capable of changing freely, especially if the
reactants or products of the reaction involve gases.
• In these cases, ΔV ≠ 0, and the energy change may be due to both heat transfer and
PV work.
• In order to eliminate the contribution from PV work, a quantity called enthalpy, H, is
defined as internal energy (U) plus the product of pressure and volume:
H = U + pV
p: pressure, V:volume
Because p,V,U are all state function, the enthalpy is a state function too.
• The change in enthalpy (ΔH) is:
∆H = qP
The Measurement of Enthalpy Change
1) Calorimeter
Monitoring the temperature change that accompanies a physical or chemical change
occurring at constant pressure.
Example: thermally insulated vessel open to the atmosphere: the heat released in the
reaction is monitor by measuring the change in temperature of the content.
2) Bomb Calorimeter
Measuring the internal energy change.
Hm = Um +pVm ≈ Um
3) Differential Scanning Calorimeter
The most sophisticated way to measure enthalpy changes.
Example 3:
When 2.0 mol CO2 is heated at the constant pressure of
1.25 atm, it’s temperature increase from 250 K to 277K.
Given the molar heat capacity of CO2 at constant
pressure is 37.11JK-1, calculate q, ∆H and ∆U.
Adiabatic Changes
The change in internal change of a perfect gas
when the temperature is change from Vi to Vf.
Step 1:
The volume changes and the temperature is
constant at its initial value. The system expands
at the constant temperature; there is no change
in internal energy if the system consist of a
perfect gas.
Step 2:
The temperature of the system is reduced at
constant volume.
Because the expansion is adiabatic, we know that q = 0; because ∆U=q + w , it then
follows that ∆U = wad. Therefore, by equating the two expansion we have obtained for
∆U we obtained
Wad = Cv ∆T
Renungan:
“Termodinamik merupakan satu subjek yang aneh. Kali pertama anda mempelajarinya,
anda tidak faham langsung. Kali kedua anda mempelajarinya, anda fikir anda
memahaminya
kecuali satu dua perkara. Kali ketiga anda mempelajarinya, anda tahu bahawa anda tidak
memahaminya, akan tetapi anda sudah terbiasa dengan subjek itu maka subjek itu tidak
akan menjadi masalah lagi kepada anda”
Arnorld Sommerfeld (1868-1951)