Six moles of an ideal gas performs a cycle as shown in figure. If the temperatures are TA=600K,TB=800K,TC=2200K and TD=1200K
, then the work done for complete cycle is (in kJ)
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Solution
Processes A to B and C to D are parts of straight line graphs of the form y=mx
Also P=nRVT(n=6)
⇒ P∝T. So volume remains constant for the graphs AB and CD
So no work is done during processes for A to B and C to D i.e. WAB=WCD=0andWBC=P2(Vc−VB)=nR(Tc−TB)=6R(2200−800)=6R×1400J AlsoWDA=P1(VA−VD)=nR(TA−TB)=6R(600−1200)=−6R×600J
Hence, work done in complete cycle W=WAB+WBC+WCD+WDA=0+(6R×1400)+0−(6R×600)=6R×800=6×8.3×800=40kJ