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Question

Six moles of an ideal gas undergoes a thermodynamic cycle as shown in figure. If the temperatures are TA=600 K,TB=800 K,TC=2200 K and TD=1200 K, the work done per cycle is approximately:
(Take universal gas constant, R=253 J/mole. K)


A
20 kJ
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B
30 kJ
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C
40 kJ
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D
60 kJ
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Solution

The correct option is C 40 kJ
The processes (AB) and (CD) are parts of straight line passing through origin.
PT
PT=constant
This indicates isochoric process.
Hence, the volume remains constant for the processes AB and CD.
So no work is done during processes for (A to B) and (C to D)
WAB=WCD=0
Processes BC and DA are isobaric processes,
WBC=PΔV=nRΔT
Or, WBC=nR(TCTB)
Or,WBC=6R(2200800)=(6R×1400) J
Also WDA=nRΔT=nR(TATD)
WDA=6R(6001200)=(6R×600) J
Hence, work done in complete cycle is:
W=WAB+WBC+WCD+WDA
W=0+(6R×1400)+0+(6R×600)
W=6R×800
W=6×253×800=40000 J=40 k J

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