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 5 J, the work done by the gas in the process. \[C\to A\] is

A)  - 5 J                                      

B)  - 10 J

C)  -15 J                     

D)         -20 J

Correct Answer: A

Solution :

Work done by gas is given by                 \[W=P\Delta V\] Total work done by gas is                 \[W={{W}_{AB}}+{{W}_{BC}}+{{W}_{CA}}\]                         ... (i) From first law of thermodynamics                 \[\Delta Q=\Delta U+W\]                                             ... (ii) Here      \[{{W}_{AB}}=P\Delta V+10(2-1)=10\,\,J\]                 \[{{W}_{BC}}=P\Delta V=2\times 0=0\] Hence, Eqs. (i) and (ii) will be written as                 \[5=0+(10+0+{{W}_{CA}})\] \[\Rightarrow \]               \[{{W}_{CA}}=-5\,\,J\]