1. Thermal & Kinetic Lecture 17
The Carnot cycle (contd),
heat engine efficiency,
phase diagrams and real gases
LECTURE 17 OVERVIEW
Recap
Maximum efficiency – the Carnot cycle
Functions of state and entropy for the Carnot cycle.
Maxwell’s demon
Phase diagrams and real gases

2. Last time…. Reversible and irreversible processes. Cyclical processes.
The Carnot engine.

3. Maximum efficiency: the Carnot cycle
We can represent the Carnot cycle on a PV diagram P A B QH
Adiabatic C W
Isotherm D QL
Animation V

4. Calculating heat, work, and changes in internal energyP A B
Adiabatic QH
Isotherm C W D QL V
A  B: Isothermal process, hence the change in internal energy is……? ?
Ans: 0 ?
A  B: If the volume at point B is twice that at point A, derive an expression for the work done in the isothermal expansion. ?
A  B: …hence, the value of QH is…..?
Ans: +nRT ln(2)

5. Functions of state P The internal energy is a function of state:
path independent;
net change in a closed reversible cycle is 0
(which must be the case because we return the gas to its original state) A B C D V
Heat and work are NOT functions of state (path dependent)
Is entropy a function of state?
Ans: Yes. The gas is returned to its original state, hence the net change in entropy is also 0. (We’ll not rigorously prove that entropy is a state function).

6. ? ? Entropy in a cycle of the Carnot engine
Entropy change of high temperature reservoir, dSHigh = -QH/TH
Why is the entropy change of the high temperature reservoir a negative quantity? ?
Ans: Heat is transferred from the reservoir into the engine.
What is the entropy change of the low temperature reservoir? ?
Ans: +QL/TL
This is a reversible engine – no change in total entropy of Universe. Therefore:

7. Energy in a cycle of the Carnot engine
First law: DU = QH – QL + W  W = QL - QH
From expression for QH above, QH > QL.
Therefore the net work done is a negative quantity and the engine does work on the surroundings.

8. ? Reversible engine: Maximum efficiency
What is the efficiency of a reversible steam engine using steam at 373 K and a cold ‘reservoir’ at 273 K? ?
Ans: 27%
Using these two equations and the 2nd law of thermodynamics we can show that no engine running between TH and TL is more efficient than a reversible engine.
Imagine an inventor claims to have made a cyclic engine that has a
higher efficiency than 1 – (QL/QH) when running between TL and TH.
In that case, for a given QH, the engine must reject a smaller amount of
heat, QL, into the lower temperature reservoir (so that h is closer to 1).
This in turn would mean that
….but this is a violation of the 2nd law.

9. ! ? Reversible engine: Maximum efficiency
Hence, no other engine can be more efficient than a reversible (Carnot) engine.
Running a heat engine in reverse: a refrigerator
If we run the cycle in reverse the net result will be that we will do work on the gas.
However, we will also have transferred heat from the low temperature reservoir to the high temperature reservoir!
If I leave the ‘fridge door open will my kitchen:
(i) cool down, (ii) stay at the same temperature, or (iii) heat up? ?

10. Maxwell’s demon
The final thing we’ll consider with regard to heat engines is the concept of Maxwell’s demon.
In its original ‘incarnation’ Maxwell’s demon was a being that could sort molecules in a box according to their speed. By opening and shutting a door in a partition he could make one side of the box hot and one side cold.
Can such a demon violate the 2nd law?

11. Maxwell’s demon
Let’s consider a mechanical analog of the demon (introduced by Feynman (see the Feynman Lectures in Physics, Vol I))
We’ll try to invent a device that violates the 2nd law.
The ratchet and pawl device shown above is set up so that it will apparently generate work from a heat reservoir at a single temperature.
Vane is bombarded with molecules but ratchet only allows the shaft to turn in one direction  perpetual motion machine!
Why can’t we do this? Why won’t the ratchet and pawl device work?

12. Final result is that there is no net work done
Maxwell’s demon
Must be a spring in the pawl to get it to snap back into position.
..but if the collision is perfectly elastic then the wheel could turn the other direction when the pawl is ‘up’.
Let’s introduce damping so that the pawl stops bouncing!
..but then energy is dissipated in the wheel, the wheel heats up and both the vane and the wheel are subject to random fluctuations.
Final result is that there is no net work done

13. Phase diagrams
Relative fractions of solid, liquid and gas phases depend upon pressure, volume and temperature. P
Consider line of constant temperature, T1: at low pressures we only have gas (unsaturated vapour) in the vessel. At higher pressures we have a solid. T1 T2
Along line of constant temperature, T2: at low pressures we only have gas (unsaturated vapour) in the vessel. At higher pressures we have a liquid and then a solid.
TTP
Liquid
Solid
Gas and solid coexist in equilibrium at different temperatures.
Triple point
Gas T

14. Phase diagrams
Relative fractions of solid, liquid and gas phases depend upon pressure, volume and temperature. P TC
Above the critical temperature, TC the liquid can no longer exist and the substance is a gas (or at higher pressures, a solid).
Above TC the substance cannot be liquefied by any amount of pressure. (For example, N2 can’t be liquefied at any pressure when its temperature exceeds 126 K).
Liquid
Solid
Gas and liquid coexist in equilibrium at different temperatures: curve of saturated vapour pressure vs T
Gas
TTP T2

15. Isotherms and changes of phase
Thus far in the module we’ve only discussed isotherms for ideal gases and haven’t considered changes of phase.
What does an isotherm look like for a real substance that undergoes phase changes?
Critical point P
Isotherm at temperature T3> T2
Saturated vapour – gas starts to liquefy
Liquid
All molecules now in liquid phase
Gas
Isotherm at temperature T2 V

16. Isotherms and changes of phase
Critical isotherm: no region where
gas and liquid coexist P TC V

17. From ideal to real gases
Behaviour of real gases only approximates that of ideal gases at low pressures where the average intermolecular separation is large.
Need to modify the ideal gas equation to take account of:
nonzero volume of molecules
intermolecular forces
U(r) r E0 r0
The van der Waals equation of state takes into account these two factors:
Depends on intermolecular force,
related to E0
Due to non-zero volume
of molecules and intermolecular
repulsion at short distances

18. From ideal to real gases: isotherms
The van der Waals equation of state can be written as:
Inflection of curve at critical point:
From this we can determine
PC, VC and Tc P V