![An idea] gas described by T_i = 300.0 K, P_i = 1.1](http://img.homeworklib.com/questions/416ce5c0-174f-11ec-95e2-d7b200311f98.png?x-oss-process=image/resize,w_560)
Homework Answers
when heated at constant volume, work done= 0 ( for constant volume process work done =0)
Temperature during constant volume process can be calculated from
T2= P2*T1/P1= 9.8*300/1.1 =2673K
during isothermal expansion, the pressure changes to 1.1 bar from 9.8 bar(P3) and P2V2= P3V3
V3= P2V2/P3= 9.8*10/1.1=89.09L
work done during isothermal expansion = -RT ln(V2/V3)= -RTln (9.8/1.1)= -48622.19 joules
it is then restored to original state by constant pressure and work done = 1.1*(10-89.09) bar.=-8699.9 joules
total work = =48622.19-8699.9=-57322.09 joules
When the cycle were run in opposite direction ,work done remain at 57322.09 joules
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