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CHECK YOURSELF ANSWERS
1The change is an isothermal change and so ÎU = 0.
2The point on the adiabatic has the same volume as that on the isothermal but greater pressure. Hence from P1T1=P2T2, it has higher temperature.
3The work done is zero since the volume does not change.
4ÎU = Q â W = â250 â (â250) = 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 Î© = 20 and so S = k ln Î© = 1.38 Ã 10â23 Ã ln 20 = 4.1 Ã 10â23 J Kâ1.
7The useful work is 0.40 Ã 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 â 400/600 = 0.33 or 33% (it is lower, and so is impossible).
