Lecture 5 – The Second Law (Ch. 2) Wednesday January 16 th

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 25 - 48) 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 Using the first law, it is easily shown that: Always true • For an ideal gas, U = fn(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 đQ + đW P (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.

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 Work • Heat in equals heat out; energy is conserved! However, common sense tells us this will not work (or it will in a while). q2 Q M W = Q

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

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

10. 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. 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) + 273.15 Triple point of water: 273.16 K Based on the ideal gas law

13. An experiment that I did in PHY3513 P T 17.7 79 13.8 0 3.63 -195.97

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 Hypothetical T2 > T1 T2 > T1 Q2' Q2 W = Q2+Q1 M' M W' = Q2' +Q1' Q1' Q1 T1 < T2 T1 < T2 Efficiency = h' > h Carnot Efficiency = h

16. Connect M' to M and run M in reverse Therefore, using: T2 > T1 W = W' 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: • This violates Clausius’ statement of the 2nd law! T2 > T1 Q1 M Q1 T1 < T2

18. Conversion of Heat to Work (a heat engine) T2 > T1 T2 > T1 Q2 +Q1 Q1 Q2 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.