question_answer

An ideal gas is taken through the cycle   \[A\to B\to C\to A,\] as shown in figure. If the net heat supplied to the gas in the cycle is 5J, the work done by the gas in the process.\[C\to A\] A is :

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 \[\Delta Q=\Delta U+W\] \[5=\Delta ({{W}_{AB}}+{{W}_{BC}}+{{W}_{CA}})\] \[5=10+0+{{W}_{CA}}\] \[\therefore \]  \[{{W}_{CA}}=-5J\]