one mole of an ideal monatomic gas is caused to go through the cycle shown in Fig. 24-22. (a) How much work is done on the gas in expanding the gas from a to c along path abc? (b) What is the change in internal energy and entropy in going from b to c? What is the change in internal energy and entropy in going through one complete cycle? Express all answer in terms of the pressure p 0 and volume V 0 at point a in the diagram.

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Deepak Patra · Answer

Kushagra Madhukar · Answer

Dear student, Please find the solution to your problem. (a) W = – P ΔV = – p o (4V o – V o ) = – 3p o V o So, W abc = W ab = – 3p o V o Hence, work done on the gas in the path abc will be – 3p o V o (b) The change in internal energy ΔE bc along the path b to c would be, ΔE bc = 3/2nRΔTbc = 3/2(nRT c – nRT b ) = 3/2(P c V c – PbV b ) [PV = nRT] = 3/2[(2p o )(4V o ) – (p o )(4V o )] = 3/2 x 4p o V o = 6p o V o The change in entropy ΔS bc along the path b to c would be ΔS bc = 3/2 nR ln(T c /T b ) = 3/2 nR ln(P c /P b ) [since, V remains constant from b to c hence, T c /T b = P c /P b ] = 3/2 nR ln(2p o /p o ) = 3/2 n R ln2 (c) The change in both internal energy and entropy would both be zero in a cyclic process. Thanks and regards, Kushagra

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