Answer
$$-5.0 \times 10^{3} \mathrm{J}$$
Work Step by Step
$$W_{\text {straight }}=\left(\frac{p_{i}+p_{f}}{2}\right) \Delta V$$
$W=p \Delta V$ for constant-pressure processes
$W = 0$ for constant-volume processes
Eq. $19-44$ with Eq. $19-5$ gives
$$E_{\text {int }}=n\left(\frac{f}{2}\right) R T=\left(\frac{f}{2}\right) p V$$
The internal energy change is
$$ E_{\text {int } c}-E_{\text {int } a} =\frac{5}{2}\left(p_{c} V_{c}-p_{a} V_{a}\right)=\frac{5}{2}\left(\left(2.0 \times 10^{3} \mathrm{Pa}\right)\left(4.0 \mathrm{m}^{3}\right)-\left(5.0 \times 10^{3} \mathrm{Pa}\right)\left(2.0 \mathrm{m}^{3}\right)\right) $$ $$ =-5.0 \times 10^{3} \mathrm{J}$$
