Question 30 of 190

A thermodynamic system is taken from an original state $A$ to an intermediate state $B$ by a linear process as shown in the figure. Its volume is then reduced to the original value from $B$ to $C$ by an isobaric process. The total work done by the gas from $A$ to $B$ and $B$ to $C$ would be :

A$33800J$B$2200J$C$600J$D$800J$

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Answer: (D) $800J$

Work done during the path $A\rightarrow B$ can be written as $\begin{matrix}{W}_{AB} & = & \dfrac{1}{2}(8000+4000)dyne{cm}^{-2}\times \dfrac{1N}{{10}^{5}dyne}\times \dfrac{1{cm}^{2}}{{10}^{-4}{m}^{2}}\times \left(7-3\right){m}^{3} \\ & = & 2400J\end{matrix}$
The work done during the process $B\rightarrow C$ is given by
$\begin{matrix}{W}_{BC} & = & \left(4000dyne{cm}^{-2}\times \dfrac{1N}{{10}^{5}dyne}\times \dfrac{1{cm}^{2}}{{10}^{-4}{m}^{2}}\right)\times \left(3-7\right){m}^{3} \\ & = & -1600J\end{matrix}$
Hence, the total work done is given by
$\begin{matrix}W & = & {W}_{AB}+{W}_{BC} \\ & = & 2400J-1600J \\ & = & 800J\end{matrix}$