Answer
$h_{fg}=2159.9\text{ kJ/kg}$
Work Step by Step
From the Clapeyron equation,
$$
\begin{aligned}
h_{f g} & =T \boldsymbol{v}_{f g}\left(\frac{d P}{d T}\right)_{\mathrm{sat}} \\
& \cong T\left(\boldsymbol{v}_g-\boldsymbol{v}_f\right)_{@ 300 \mathrm{kPa}}\left(\frac{\Delta P}{\Delta T}\right)_{\mathrm{sat}, 300 \mathrm{kPa}} \\
& =T_{\text {sat@300 } \mathrm{kPa}}\left(\boldsymbol{v}_g-\boldsymbol{v}_f\right)_{@ 300 \mathrm{kPa}}\left(\frac{(325-275) \mathrm{kPa}}{T_{\text {sat } @ 325 \mathrm{kPa}}-T_{\text {sat } @ 275 \mathrm{kPa}}}\right) \\
& =(133.52+273.15 \mathrm{~K})\left(0.60582-0.001073 \mathrm{~m}^3 / \mathrm{kg}\right)\left(\frac{50 \mathrm{kPa}}{(136.27-130.58)^{\circ} \mathrm{C}}\right) \\
& =\mathbf{2 1 5 9 . 9 k J / k g}
\end{aligned}
$$ The tabulated value of $h_{f g}$ at $300 \mathrm{kPa}$ is $2163.5 \mathrm{~kJ} / \mathbf{k g}$.
