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
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:
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
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.
Chapter 2
100% Conversion of Heat to Work q2 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).
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.
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.
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
V The Carnot Cycle ab isothermal expansion bc adiabatic expansion cd isothermal compression da 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.
V The Carnot Cycle ab isothermal expansion bc adiabatic expansion cd isothermal compression da 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.
The ‘absolute’ temperature (Kelvin) scale T(K) = T(oC) + 273.15 Triple point of water: 273.16 K Based on the ideal gas law
An experiment that I did in PHY3513 P T 17.7 79 13.8 0 3.63 -195.97
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.
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
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
This would be equivalent to: M Q1 T1 < T2 This violates Clausius’ statement of the 2nd law!
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!
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
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!
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.