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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.

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 p0 and volume V0 at point a in the diagram.

Grade:upto college level

2 Answers

Deepak Patra
askIITians Faculty 471 Points
9 years ago
233-2458_1.PNG
233-263_1.PNG
Kushagra Madhukar
askIITians Faculty 628 Points
4 years ago
Dear student,
Please find the solution to your problem.
 
(a) W = – PΔV
= – po(4Vo – Vo) = – 3poVo
So, Wabc = Wab = – 3poVo
Hence, work done on the gas in the path abc will be – 3poVo
 
(b) The change in internal energy ΔEbc along the path b to c would be,
ΔEbc = 3/2nRΔTbc
= 3/2(nRTc – nRTb)
= 3/2(PcVc – PbVb)                    [PV = nRT]
= 3/2[(2po)(4Vo) – (po)(4Vo)]
= 3/2 x 4poVo
= 6poVo
The change in entropy ΔSbc along the path b to c would be
ΔSbc = 3/2 nR ln(Tc/Tb)
= 3/2 nR ln(Pc/Pb)                    [since, V remains constant from b to c hence, Tc/Tb = Pc/Pb]
= 3/2 nR ln(2po/po)
= 3/2 nR 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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