Answer
$F=134.0\text{ psia}$
$F=99.4\text{ psia}$
Work Step by Step
The Clapeyron equation is $$
\left(\frac{d P}{d T}\right)_{s a t}=\frac{h_{f g}}{T v_{f g}}
$$ Using the finite difference approximation,
$$
\left(\frac{d P}{d T}\right)_{\mathrm{sat}} \approx\left(\frac{P_2-P_1}{T_2-T_1}\right)_{\mathrm{sat}}=\frac{h_{f g}}{T v_{f g}}
$$ Solving this for the second pressure gives for $T_2=110^{\circ} \mathrm{F}$
$$
\begin{aligned}
P_2 & =P_1+\frac{h_{f g}}{T v_{f g}}\left(T_2-T_1\right) \\
& =116.7 \mathrm{psia}+\frac{154.85 \mathrm{Btu} / \mathrm{lbm}}{(560 \mathrm{R})\left(0.86332 \mathrm{ft}^3 / \mathrm{lbm}\right)}\left(\frac{5.404 \mathrm{psia} \cdot \mathrm{ft}^3}{1 \mathrm{Btu}}\right)(110-100) \mathrm{R} \\
& =134.0\ \mathrm{psia}
\end{aligned}
$$ When $T_2=90^{\circ} \mathrm{F}$
$$
\begin{aligned}
P_2 & =P_1+\frac{h_{f g}}{T v_{f g}}\left(T_2-T_1\right) \\
& =116.7\ \mathrm{psia}+\frac{154.85 \mathrm{Btu} / \mathrm{lbm}}{(560 \mathrm{R})\left(0.86332 \mathrm{ft}^3 / \mathrm{lbm}\right)}\left(\frac{5.404 \mathrm{psia} \cdot \mathrm{ft}^3}{1 \mathrm{Btu}}\right)(90-100) \mathrm{R} \\
& =99.4\ \mathrm{psia}
\end{aligned}
$$
