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1The change is an isothermal change and so ΔU = 0.

2The point on the adiabatic has the same volume as that on the isothermal but greater pressure. Hence from P 1 T 1=P 2 T 2, it has higher temperature.

3The work done is zero since the volume does not change.

4ΔU = Q − W = −250 − (−250) = 0. So the temperature stays the same.

5It stays the same since no heat is exchanged.

6The largest entropy belongs to the state with 3H and 3T. For this state Ω = 20 and so S = k ln Ω = 1.38 × 10−23 × ln 20 = 4.1 × 10−23 J K−1.

7The useful work is 0.40 × 500 =200 J so the heat rejected is 300 J.

8The efficiency of this engine is 70%. A Carnot engine operating between the same temperatures would have efficiency 1 – 400/600 = 0.33 or 33% (it is lower, and so is impossible).