A) -5J
B) -10J
C) -15J
D) -20J
Correct Answer: A
Solution :
Work done in Process\[A\to B\] \[{{W}_{AB}}=P\Delta V\] \[=10(2-1)=10J\] Process\[B\to C\]is at constant volume so work done is zero. The process is cyclic, so change in internal energy \[\Delta U=0\] From 1st law of thermodynamics \[\Delta Q=\Delta U+W\] \[5=\Delta ({{W}_{AB}}+{{W}_{BC}}+{{W}_{CA}})\] \[5=10+0+{{W}_{CA}}\] \[\therefore \] \[{{W}_{CA}}=-5J\]You need to login to perform this action.
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