First we find the volume at point $B$ . We are told that the process BC is isothermal, so by the ideal gas law
$\begin{align*}P_B V_B = P_C V_C \Rightarrow V_B = \frac{P_C V_C}{P_B} = \frac{500 \mbox{ kPa} \cdot 2 \;\mathrm{m}^3}{200 \m...$
The work done by the gas as it goes from state $A$ to state $B$ is
$\begin{align*}200 \mbox{ kPa} \left(5 \;\mathrm{m}^3 - 2 \;\mathrm{m}^3 \right) = 600 \mbox{ kJ}\end{align*}$
The work done by the gas as it goes from state $B$ to state $A$ is
$\begin{align*}\int P dV = P_B V_B \int_{V_B}^{V_C} \frac{d V}{V} = P_B V_B \ln V_C/V_B = - 1000 \mbox { kJ} \cdot \ln 5/2 \en...$
The negative sign indicates that work is actually being done on the gas. During the process C to A, there is not net work done on the gas because $P dV = 0$ . Therefore, the net work done by the gas during the complete cycle beginning and ending at A is
$\begin{align*}600 \mbox{ kJ} - 1000 \mbox{ kJ} \ln 5/2 = -316.29 \mbox { kJ}\end{align*}$
Therefore, answer (D) is correct.

Solution to 2001 Problem 37