1. Lecture 5 – The Second Law (Ch. 2)
Wednesday January 16th
Brief review of last class (adiabatic processes)
The ideal gas and entropy
The second law
The Carnot cycle
A new function of state - entropy
Reading: All of chapter 2 (pages )
Homework 1 due this Friday (18th)
Homework 2 due next Friday (25th)
Homework assignments available on web page
Assigned problems, Ch. 1: 2, 6, 8, 10, 12
Assigned problems, Ch. 2: 6, 8, 16, 18, 20

2. Heat capacity dH = dU + PdV + VdP = đQ + VdP
Using the first law, it is easily shown that:
Always true
For an ideal gas, U = f n(q) only. Therefore,
Enthalpy, H = U + PV, therefore:
dH = dU + PdV + VdP = đQ + VdP
Always true
Ideal gas:

3. Calculation of work for a reversible process
(1)
Isobaric (P = const)
Isothermal (PV = const)
Adiabatic (PVg = const)
Isochoric (V = const)
(2)
(3)
(4) V
For a given reversible path, there is some associated physics.
đQ + đW

4. Configuration Work on an ideal gas
Note: for an ideal gas, U = U(q ), so W = -Q for isothermal processes.
It is also always true that, for an ideal gas,
Adiabatic processes: đQ = 0, so W = DU, also PVg = constant.

5. Chapter 2

6. 100% Conversion of Heat to Workq2 Q M
W = Q
Heat in equals heat out; energy is conserved! However, common sense tells us this will not work (or it will in a while).

7. Something is clearly missing from the first law!
100% transfer of heat to from cold to hot body
q2 > q1
Something is clearly missing from the first law! Q1 M Q1
q1 < q2
Heat in equals heat out; energy is conserved! But we know this never happens in the real world.

8. The Second Law of Thermodynamics
Clausius’ statement: It is impossible to construct a device that operates in a cycle and whose sole effect is to transfer heat from a cooler body to a hotter body.
Kelvin-Planck statement: It is impossible to construct a device that operates in a cycle and produces no other effect than the performance of work and the exchange of heat from a single reservoir.
Carnot’s theorem: No engine operating between two reservoirs can be more efficient than a Carnot engine operating between the same two reservoirs.

9. Conversion of Heat to Work (a heat engine)
Heat reservoir at
temperature q2 > q1
Q  heat
W  work
both in Joules
*Be careful with the signs for heat!
Efficiency (h*): * Q2
Heat
Engine Q1
Cold reservoir at
temperature q1 < q2

10. V The Carnot Cycle ab isothermal expansion bc adiabatic expansion
cd isothermal compression
da adiabatic compression
W2 > 0, Q2 > 0 (in)
W' > 0, Q = 0
W1 < 0, Q1 < 0 (out)
W'' < 0, Q = 0 V
Stirling’s engine is a good approximation to Carnot’s cycle.

11. V The Carnot Cycle ab isothermal expansion bc adiabatic expansion
cd isothermal compression
da adiabatic compression
W2 > 0, Q2 > 0 (in)
W' > 0, Q = 0
W1 < 0, Q1 < 0 (out)
W'' < 0, Q = 0 V
Stirling’s engine is a good approximation to Carnot’s cycle.

12. The ‘absolute’ temperature (Kelvin) scale
T(K) = T(oC)
Triple point
of water: K
Based on the ideal gas law

13. An experiment that I did in PHY3513
P T
13.8 0

14. The Second Law of Thermodynamics
Clausius’ statement: It is impossible to construct a device that operates in a cycle and whose sole effect is to transfer heat from a cooler body to a hotter body.
Kelvin-Planck statement: It is impossible to construct a device that operates in a cycle and produces no other effect than the performance of work and the exchange of heat from a single reservoir.
Carnot’s theorem: No engine operating between two reservoirs can be more efficient than a Carnot engine operating between the same two reservoirs.

15. Consider two heat engines
Carnot
T2 > T1
T1 < T2 M'
Q2'
Q1'
W' = Q2' + Q1'
Hypothetical
Efficiency = h' > h M Q2 Q1
T2 > T1
T1 < T2
W = Q2 + Q1
Efficiency = h

16. Connect M' to M and run M in reverse
T2 > T1
W = W'
Therefore, using:
Q2' Q2 M' M W'
Q1' Q1
T1 < T2
M is a Carnot engine, so we are entitled to run it in reverse

17. This would be equivalent to:M Q1
T1 < T2
This violates Clausius’ statement of the 2nd law!

18. Conversion of Heat to Work (a heat engine)Q1 Q2
Q2 + Q1 M' M W M W Q1 Q1
T1 < T2
This violates Kelvin-Planck statement of the 2nd law!

19. The Clausius Inequality and the 2nd Law
Heat reservoir at
temperature T2 > T1
Efficiency (h*): Q2
Heat
Engine Q1
Heat reservoir at
temperature T1 < T2

20. The Clausius Inequality and the 2nd Law
Divide any reversible cycle into a series of thin Carnot cycles, where the isotherms are infinitesimally short: P v
We have proven that the entropy, S, is a state variable, since the integral of the differential entropy (dS = dQ/T), around a closed loop is equal to zero, i.e. the integration of differential entropy is path independent!

21. The Clausius Inequality and the 2nd Law
Divide any reversible cycle into a series of thin Carnot cycles, where the isotherms are infinitesimally short: P v
There is one major caveat: the cycle must be reversible. In other words, the above assumes only configuration work (PdV) is performed.
If the cycle additionally includes dissipative work, it is not clear how to include this in the above diagram.