Answer
$$3.16 \mathrm{\ kJ} $$
Work Step by Step
The work done by the gas is
$$
\begin{aligned} W &=\int_{i}^{f} p d V=\int_{V_{i}}^{V_{i}}(5.00 \mathrm{kPa}) e^{\left(V_{i}-V\right) / a} d V=(5.00 \mathrm{kPa}) e^{V / a} \cdot\left[-a e^{-V_{a}}\right]_{V_{i}}^{V_{f}} \\ &=(5.00 \mathrm{kPa}) e^{1.00}\left(1.00 \mathrm{m}^{3}\right)\left(e^{-1.00}-e^{-2.00}\right) \\ &=3.16 \mathrm{\ kJ} \end{aligned}
$$
