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Molar Heat Capacity- Cv
• For ideal monatomic gas– U = 3/2 nkT– Per atom this is just 1/2kT per degree of
freedom i.e. 3/2kT = K.E.x + K.E.y + K.E.z
Molar Heat Capacity- Cv
• For ideal monatomic gas– U = 3/2 nkT– Per atom this is just 1/2kT per degree of
freedom i.e. 3/2kT = K.E.x + K.E.y + K.E.z
– Per mole Cv = (đQ/T)v = (dU/T)v = 3/2R
– Consider 1st Law adiabat đQ = dU + PdV = 0
– dU + PdV = 0
– CvdT + PdV = 00PV
PdV
PV
dTCv
Molar Heat Capacity- Cv
• For ideal gas
0PV
PdV
PV
dTCv
0V
dV
RT
dTCv
0 V
dV
T
dT
R
Cv
0 V
dV
T
dT
CC
C
vp
v
constVTCC
C
vp
v
)ln()ln(
constVT vp
v
CC
C
constTV 1
v
p
C
CconstPV
Molar Heat Capacity- Cv
• For ideal diatomic gas –still non interacting but now dumbell shape
– U translational per atom = 3/2kT as before
– Molecule now can rotate– 1/2kT per rotational degree of freedom– Molecule can vibrate– has potential energy– has vibrational kinetic energy
• Utotal = 3/2kT+ kT + kT = 7/2kT
Equipartition Theorem
• Classical mechanics predicts that:
• When a substance is in equilibrium, there is an average energy of ½kT per molecule (or ½RT per mole) associated with each degree of freedom.
Molar Heat Capacity- Cv
• For ideal diatomic gas – classical physics predicts
• Utotal = 7/2RT per mole
• What do we observe?Cv
T
5R/2
3R/2
7R/2No classical explanationOnly explained by Q.M.Only discrete values of rotationand vibration allowed.Below some temperatures notEnough energy so some degreesof freedom “Frozen Out”
Properties of materials
• It is well known that most materials expand on heating. We define the volume thermal expansivity as:-
• Moduli of elasticity are the stress/strain or force/unit area divided by the fractional deformation:-
PT
V
V
1
TV
PVK
Cyclic Processes
• We have defined processes that can take us full circle in an indicator diagram.
• Consider two paths a and b between points 1 and 2.
• The A’s are the areas under paths a and b.
P
V
a
b
1
2
aa
a APdVW 2
121
bbb
b APdVPdVW 2
1
1
212
Cyclic Processes
• We end in the same place as we start so the state is the same U=0 around loop.
U12 = Q1a2 + W1a2
U21 = Q2b1 + W2b1
U12 + U21= 0• (Q1a2 + Q2b1) = - (W1a2+W2b1)
• Nett heat input = - (nett work done on the system)• Nett heat input = Aa- Ab = work done by system• We have transformed heat into work – an engine
V
Pa
b
1
2
Cyclic Processes
• We end in the same place as we start so the state is the same U=0 around loop.
U12 = Q1a2 + W1a2
U21 = Q2b1 + W2b1
U12 + U21= 0• (Q1a2 + Q2b1) = - (W1a2+W2b1)
• Nett heat input = - (nett work done on the system)• Nett heat input = Aa- Ab = work done by system• We have transformed heat into work – an engine
V
Pa
b
1
2
Carnot Cycle – Ideal Gas
• Need a cycle that we can solve• Define path as two adiabats and two
isotherms• Isotherms a and c• Adiabats b and d• Where does heat flow?
• Work done by system = area enclosed
V
P a
b
1
2
3
4c
d
Carnot Cycle – Ideal Gas
• dU=đQ + đW• On isotherm a dU = 0• đQ = -đW = PdV
• On isotherm c
V
P a
b
1
2
3
4c
d
1
22
1
2
121 ln
V
VNRT
V
dVNRTPdVWQ H
aH
aaa
3
44
3
4
343 ln
V
VNRT
V
dVNRTPdVWQ C
cC
ccc
Carnot Cycle – Ideal Gas• dU=đQ + đW• We know U for cycle = 0
• On adiabat b and d đQ = 0V
P a
b
1
2
3
4c
d
dWQQdQ ca
3
4
1
2 lnlnV
VNRT
V
VNRTdQ CH
3
4
1
2 lnlnV
VT
V
VTNRdQ CH
0 dWdQdU
Carnot Cycle – Ideal Gas• On adiabat b and d• We know
dW
V
VTTNRdQ CH
1
2ln
constTV 1
1
4
1
1
VTVT CH
1
3
1
2
VTVT CH
4
3
1
2
V
V
V
V
3
4
1
2 lnlnV
VT
V
VTNRdQ CH
V
P a
b
1
2
3
4c
d
Carnot Cycle – Ideal Gas
• We have calculated the work done by the system:
• Because we can find
• This is the definition of the Kelvin temperature scale. It is important as it is only defined by the temperatures of the reservoirs.
4
3
1
2
V
V
V
V
1
2lnV
VTTNR CH
C
H
C
H
T
T
Q
Q
||
||
Carnot Cycle – Ideal Gas• We can calculate the efficiency of the
cycle. We turn heat from the hot reservoir into useful work and discard the remainder into the cold reservoir.
• Efficiency = Work out / Heat energy put in
1
2lnV
VTTNRW CH
H
C
H
CH
a T
T
VV
NRT
VV
TTNR
Q
W
1
ln
ln
1
2
1
2
Carnot Cycle – Ideal Gas• The Carnot Cycle works between two
reservoirs at differing temperatures. Its efficiency is uniquely defined by the two temperatures.
• It is reversible – we can go round the system backwards. It then uses work to extract heat from a cold reservoir to a hot reservoir. This may be considered a refrigerator!
H
C
T
T1
Carnot Cycle – Ideal Gas• In a refrigerator the coefficient of
performance is given by how much heat can be removed from the cold reservoir per unit of work put in.
• Note this can be > 1
CH
CC
TT
T
W
Q
||
Engines and Refrigerators
Cold TC
Hot TH
RWork
QH
QC
Cold TC
Hot TH
E Work
QH
QC
Carnot Cycle• We considered a cycle
with 2 adiabats and 2 isotherms
• Found heat flows in and out on adiabats.
• Work output is = nett heat input.
• Efficiency is fundamental and fixed by temperature of two reservoirs.
V
P a
b
1
2
3
4c
d
3
4lnV
VNRTQ Cc
1
2lnV
VNRTQ Ha
H
C
T
T1
Why is efficiency fixed?
• 0th law gave us– “If two bodies A and B are in thermal
equilibrium with a third body C then A and B are in thermal equilibrium with each other.”
• 1st law gave us– Conservation of energy dU = đW + đQ
• None of these deals with the inherent direction of processes. Empirical evidence suggests that direction is important. Hot bodies cool over time, cold bodies heat up.
The Second Law of Thermodynamics
• Equivalent statements on handouts.
The Kelvin-Planck Statement “It is impossible to construct a device that, operating in a cycle, will produce no effect other than the extraction of heat from a single body at uniform temperature and the performance of an equivalent amount of work.”
The Second Law of Thermodynamics
The Kelvin-Planck Statement “It is impossible to construct a device that, operating in a cycle, will produce no effect other than the extraction of heat from a single body at uniform temperature and the performance of an equivalent amount of work.”
The Second Law of Thermodynamics
The Kelvin-Planck Statement • Cycle requires that the state of the
working substances is the same at the start and the end of the process.
• An isothermal compression U = 0 but
V
x dx
FA
Vacuum
1
2lnV
VNRTWQ Ha
The Second Law of Thermodynamics
The Kelvin-Planck Statement “It is impossible to construct a device that, operating in a cycle, will produce no effect other than the extraction of heat from a single body at uniform temperature and the performance of an equivalent amount of work.”
The Second Law of Thermodynamics
The Kelvin-Planck Statement • No effect other tells us that in addition to
the rejection of heat to a body of lower temp. the only other effect on the surroundings is via the work delivered by the engine.
• This means that the bodies delivering and accepting the heat must do so without delivering any work. In other words they are sources of heat – reservoirs.
The Second Law of Thermodynamics
The Kelvin-Planck Statement “It is impossible to construct a device that, operating in a cycle, will produce no effect other than the extraction of heat from a single body at uniform temperature and the performance of an equivalent amount of work.”
The Second Law of Thermodynamics
The Kelvin-Planck Statement
• Single Suppose Q1 + Q2 heat was supplied to an engine from bodies at T1 and T2 (T1 > T2 for instance). The cyclical engine could deliver work W = Q1 + Q2 which appears to violate the statement.
• However, Q2 could be negative.
The Second Law of Thermodynamics
The Kelvin-Planck Statement
• Single Suppose Q1 + Q2 heat was supplied to an engine from bodies at T1 and T2 (T1 > T2 for instance). The cyclical engine could deliver work W = Q1 + Q2 which appears to violate the statement.Hot T1
E Work
Q1
Hot T2
Q2
The Second Law of Thermodynamics
The Kelvin-Planck Statement
• Single Suppose Q1 + Q2 heat was supplied to an engine from bodies at T1 and T2 (T1 > T2 for instance). The cyclical engine could deliver work W = Q1 + Q2 which appears to violate the statement.
• However, Q2 could be negative and there is no violation. This type of engine is excluded by specifying single.
The Second Law of Thermodynamics
The Kelvin-Planck Statement “It is impossible for an engine, working in a cycle, to exchange heat with a single reservoir, produce an equal amount of work, and have no other effect.”
Cold TC
Hot TH
EWork
QH
The Second Law of ThermodynamicsThe Kelvin-Planck Statement
“It is impossible for an engine, working in a cycle, to exchange heat with a single reservoir, produce an equal amount of work, and have no other effect.”
• We know that we can transfer energy by mechanical work (think stirring a liquid) to raise the internal energy or heat a system. This statement asserts that the reverse is not true. Turning work into heat may therefore be irreversible.
• We cannot transform heat directly into work. There is something different about these forms of energy!
The Second Law of Thermodynamics
The Kelvin-Planck Statement “A process whose only effect is the complete conversion of heat into work is impossible.”
• Only here covers all the previously discussed eventualities.
The Second Law of Thermodynamics
The Clausius Statement “It is impossible to construct a device that, operating in a cycle, produces no effect other than the transfer of heat from a colder to a hotter body.”
• No effect is key here as the system must remain unchanged Cold TC
Hot TH
R
QH
QC
Equivalence of Statements
RWork
Q2 + Q1
Q2
Cold TC
Hot TH
E
Q1
Assuming Kelvin statement is false
Equivalence of Statements
RWork
Q2 + Q1
Q2
Cold TC
Hot TH
E
Q1
Composite Refrigerator
Q2
Q2
Cold TC
Hot TH
Assuming Kelvin statement is false implies Clausius also false
Equivalence of Statements
R
Q1
Q2
Cold TC
Hot TH
E
Q2
Assuming Clausius statement is false
Q2
Equivalence of Statements
R
Work = Q1 –Q2
Q1
Q2
Cold TC
Hot TH
E
Q2
Composite Engine
Q1-Q2
Cold TC
Hot TH
Q2
Assuming Clausius statement is false implies Kelvin also false
Maximum Efficiency – Carnot’s Theorem
E
Q3
Q2
Cold TC
Hot TH
C
Q1
Q4
WE Wc
•Assume an engine more efficient (E) than Carnot engine (c).
•1st Law still applies.
•Q2=Q1-WE Q4=Q3-Wc
•With matched W and a more efficient engine:
•WE/Q1 > Wc/Q3
•So Q3 > Q1
•But Carnot engine is reversible….
Maximum Efficiency – Carnot’s Theorem
E
Cold TC
Hot TH
C
Q1
Q4=Q3-W
WE = Wc
•With the Carnot refrigerator the system acts as a composite
•Heat flow out of the cold reservoir = Q4-Q2
•Q4-Q2 = Q3-Q1
•But Q1 < Q3 !
•So violates Clausius statements as it pumps heat from cold to hot with no external work being done.
c
Q3
Q2=Q1-W
Carnot Cycle• We considered a cycle
with 2 adiabats and 2 isotherms
• Found heat flows in and out on isotherms. V
P a
b
1
2
3
4c
d
3
4lnV
VNRTQ Cc
1
2lnV
VNRTQ HH
C
H
C
H
T
T
Q
Q
C
H
C
H
T
T
Q
Q
||
||
0C
C
H
H
T
Q
T
Q
General Cycle
V
P a
b
1
2
3
4c
d
0C
C
H
H
T
Q
T
Q
• We can build any closed reversible cycle out of a whole sequence of tiny reversible Carnot cycles.
• As the isotherms become very close the jagged contour approximates the actual path.
General Cycle0
2
2
1
1 T
Q
T
Q
• Consider each tiny cycle individually - on the isotherms at T1 and T2 heat đQ1 and đQ2 flows.
• The total effect of each tiny cycle is therefore the sum of each cycle.
• In the limit that each individual heat flow becomes tiny the sum becomes an integral. The R remind us that this is for reversible paths.
02
2
1
1 T
Qd
T
Qd
0i i
i
T
Qd
0 T
Qd RR
General Cycle - Entropy• The closed path integral
around a reversible cycle is zero.
• It has all the hallmarks of a state function lets check….
0 T
Qd RR
Pa
b
1
2
01
2
2
1
T
Qd
T
Qd
T
Qd Rb
Ra
RR
02
1
2
1
T
Qd
T
Qd
T
Qd Rb
Ra
RR
2
1
2
1 T
Qd
T
Qd Rb
Ra
2
1
)1,2(T
QdS R
Entropy• Entropy is a state function
because it does not matter which path the system takes between 1 and 2.
• Only differences in Entropy are defined – the reference point is usually taken at the lowest temperature that can be achieved. Ideally absolute zero.
• Therefore at finite temp Tf the entropy is the integral of đQR/T from the reference temp up to Tf.
• All heat must be added reversibly.
• For tiny reversible changes.
2
1
)1,2(T
QdS R
T
QddS R
Entropy and the 1st Law• The differential form of the first Law for
reversible processes can be written as:
• Heat capacities can now be redefined:
• Cv = (đQ/T)v = (TS/T)v = T (S/T)v
• Cp = (đQ/T)p = T (S/T)P
PdVTdSWdQddU
Example
• Imagine heating a beaker of water slowly by putting it in contact with a reservoir whose temperature increases very slowly so that the system passes through a succession of equilibrium states at constant pressure.
• The heat flow in going from T to T+dT is đQR = CpdT
• Hence entropy change:
• dS = đQR/T = CpdT/T
• For 1kg of water with Cp=4.2kJK-1 going from 20°C to 100°C S = 1.0110-3 JK-1
i
fp
T
T
pif T
TC
T
dTCTTS
f
i
ln),(
Real Engines and Irreversibility
E
Q3
Q2
Cold TC
Hot TH
C
Q1
Q4
WE Wc
•Assume an engine less efficient (E) than Carnot engine (c).
•1st Law still applies.
•Q2=Q1-WE Q4=Q3-Wc
•With matched W and a less efficient engine:
•WE/Q1 < Wc/Q3
•So Q3 < Q1
•But Carnot engine is reversible….
Real Engines and Irreversibility
E
Cold TC
Hot TH
C
Q1
Q4=Q3-W
WE = Wc
•With the Carnot refrigerator the system acts as a composite
•Heat flow out of the cold reservoir = Q4-Q2
•Q4-Q2 = Q3-Q1
•But Q1 > Q3
•So doesn’t violate Clausius statements as it heat flows from hot to cold with no external work being done.
E c (= for reversible)
Q3
Q2=Q1-W
Irreversible Cycle• For Carnot Cycle
• For Irreversible engine E<c
H
C
H
Cc T
T
Q
Q 1
||
||1
H
C
EH
CE T
T
Q
Q
1
||
||1
H
C
EH
C
T
T
Q
Q
||
||
H
C
H
C
T
T
Q
Q
In General
0C
C
H
H
T
Q
T
Q
H
C
H
C
T
T
Q
Q
0C
C
H
H
T
Qd
T
Qd
• Around entire closed path this must be true so:
• Clausius inequality• < for irreversible• = for reversible
0TQd
General Cycle• Remember this inequality is for
irreversible processes involved in the cycle. Our expression previously defined for entropy is for reversible cycles.
• A reversible process is one that can be reversed by an infitessimal adjustment of the system coordinates (P,V,T etc) and takes place so slowly that the system can be considered to be passing through a succession of equilibrium states.
• I.E. Quasistatic and reversible by infitessimal adjustment.
0TQd
General Cycle - Entropy• Take a cycle in which path a is
irreversible between points 1 and 2 which are both equilibrium states.
• Path b from 2 to 1 is reversible.
Pa
b
1
2
01
2
2
1
T
Qd
T
QdR
ba
02
121 SS
T
Qda
2
121 T
QdSS a
2
112 T
QdSS a T
QddS
The Entropy Statement
• The equality applies only for reversible processes.
• Entropy is created by irreversible processes!
• For a thermally isolated system đQ = 0 so any irreversible process occurring makes dS > 0.
• Entropy only ever increases!
T
QddS
The Entropy
• All changes in a thermally isolated system must lead to an increase in entropy (or stays the same if reversible).
• As such a system approaches equilibrium the entropy must increase therefore the final equilibrium configuration is the one with the maximum entropy.
• At the maximum of entropy there are no changes as entropy cannot decrease.
• The always increasing nature defines a natural direction to time.
• This all follows from the simple observation that heat flows from hot to cold!
T
QddS
Example- expansion of an ideal gas
• Imagine an ideal gas in volume V. A second empty vessel of the same volume is placed adjacent. A hole is formed between them and the gas expands to fill both.
PdVTdSWdQddU
Gas in V Gas in 2V
Example- expansion of an ideal gas
• Process is clearly irreversible, the internal energy doesn’t change and the process is not quasistatic.
0 PdVTdSWdQddU
TdSPdV NRTPV
dSdVV
NR
2ln12
2
NRSSV
dVNR
V
V
