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 \[C\to A\] is:

A)  \[-5\,J\]            

B)  \[-20\,J\]

C)   \[-15\,J\]          

D)  \[-25\,J\]

Correct Answer: A

Solution :

Key Idea: Process is cyclic, so change in internal energy is zero. We know that work done by gas is given by \[W=P\Delta V\] Where P is pressure and \[\Delta V\] the change in volume. Total work done by gas is: \[W={{W}_{AB}}+{{W}_{BC}}+{{W}_{CA}}\] ?..(i) According to first law of thermodynamics \[\Delta Q=\Delta U+W\]            .....(ii) Where \[\Delta Q\]is heat supplied, \[\Delta V\] the change in internal energy which is zero and W the work done. \[\therefore \] \[{{W}_{AB}}=P\Delta V=10(2-1)=10J.\] \[{{W}_{BC}}=P\Delta V\] \[=2\times 0=0.\] Thus, equations (i) and (ii) will be written as \[5=0+(10+0+{{W}_{CA}}).\] \[\Rightarrow \] \[{{W}_{CA}}=-5J.\]