A thermodynamic system is taken from an original state A to an...

Work done during the path $A \to B$ can be written as $\begin{cases}{W}_{AB} & = & \dfrac{1}{2} ( 8000 + 4000 ) dyne{cm}^{- 2} \times \dfrac{1N}{{10}^{5}dyne} \times \dfrac{1{cm}^{2}}{{10}^{- 4}{m}^{2}} \times \left(7 - 3\right){m}^{3} \\ & = & 2400J\end{cases}$

The work done during the process $B \to C$ is given by

$\begin{cases}{W}_{BC} & = & \left(4000dyne{cm}^{- 2} \times \dfrac{1N}{{10}^{5}dyne} \times \dfrac{1{cm}^{2}}{{10}^{- 4}{m}^{2}}\right) \times \left(3 - 7\right){m}^{3} \\ & = & - 1600J\end{cases}$

Hence, the total work done is given by

$\begin{cases}W & = & {W}_{AB} + {W}_{BC} \\ & = & 2400J - 1600J \\ & = & 800J\end{cases}$