Answer
$P_{2}=59.5\text{ psia}$
Work Step by Step
From the Clapeyron equation,
$$
\left(\frac{d P}{d T}\right)_{\text {sat }}=\frac{h_{f g}}{T v_{f g}}=\frac{(250 \mathrm{Btu})\left(\frac{5.404 \mathrm{psia} \cdot \mathrm{ft}^3}{1 \mathrm{Btu}}\right) /(0.5 \mathrm{lbm})}{(475 \mathrm{R})\left(1.5 \mathrm{ft}^3\right)(0.5 \mathrm{lbm})}=1.896 \mathrm{psia} / \mathrm{R}
$$ Using the finite difference approximation, $$
\left(\frac{d P}{d T}\right)_{\text {sat }} \approx\left(\frac{P_2-P_1}{T_2-T_1}\right)_{\text {sat }}
$$ Solving for $P_2$, $$
P_2=P_1+\frac{d P}{d T}\left(T_2-T_1\right)=50 \mathrm{psia}+(1.896 \mathrm{psia} / \mathrm{R})(480-475) \mathrm{R}=\mathbf{5 9 . 5}\ \mathrm{psia}
$$
