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Question

A monatomic ideal gas of two moles is taken through a cyclic process starting for A as shown in the figure. The volume ratios are (VB/VA)=2 and (VD/VA)=4. If the temperature TA at A is 27C, calculate:
(a) The temperature of the gas at point B,
(b) Heat absorbed or released by the gas in each process.
(c) The total work done by the gas during the complete cycle.
Express your answer in terms of the gas constant R.
1014946_0e34d55caaa940b09d7fbcad7f1e8e6b.jpg

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Solution


Taking VA=V0;VB=2V0;VD=4V0
Process AB: Isobaric process as V/T=constant
V0/(273+27)=(2V0)/TBTB=600 K=327C.
Q1=nCPT=2(5/2)R(600300)=1500R(absorbed)
Process BC (Isothermal)
Q2=W2=nRTBln(V2/V1)=831.78R(absorbed)
Process CD (Isochoric)
Q3=nCVT=900R(released)
Process DA (Isothermal)
Q4=W4=nRTAln(V2/V1)=831.78 R
Total word done: W=Q1+Q2+Q3+Q4=600R.

934579_1014946_ans_833669f6a18d424b83a299a35d32716c.jpg

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