A thermodynamic system is taken from an original state A to an intermediate state B by a linear process as…

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

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

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

Hence, the total work done is given by

$\begin{array}{rcl}W& =& W_{\mathrm{AB}}+W_{\mathrm{BC}}\\ & =& 2400 \mathrm{J}-1600 \mathrm{J}\\ & =& 800 \mathrm{J}\end{array}$