Q.
Six moles of an ideal gas performs a cycle shown in figure. If the temperature are TA=600K,TB=800K,TC=2200K and TD=1200K, the work done per cycle is approximately
Processes A to B and C to D are parts of straight line graphs passing through origin P∝T.
So, volume remains constant for the graph AB and CD
So, no work is done during processes for A to B and C to D
i.e., WAB=WCD=0
and WBC=P2(VC−VB)=nR(TC−TB) =6R(2200−800)=6R×1400J
Also, WDA=P1(VA−VD)=nR(TA−TD) =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