Answer
$T_{2}=531\text{ K}$
$ϕ_{2}=0.37%$
Work Step by Step
Analysis At the inlet,$$
\begin{aligned}
& P_{v, 1}=\phi_1 P_{\mathrm{g}, 1}=\phi_1 P_{\mathrm{sat} @ 20^{\circ} \mathrm{C}}=(0.90)(2.3392 \mathrm{kPa})=2.105\ \mathrm{kPa} \\
& \omega_2=\omega_1=\frac{0.622 P_{v, 1}}{P-P_{v, 1}}=\frac{(0.622)(2.105 \mathrm{kPa})}{(100-2.105) \mathrm{kPa}}=0.0134 \mathrm{~kg} \mathrm{H}_2 \mathrm{O} / \mathrm{kg} \text { dry air }
\end{aligned}
$$ Since the mole fraction of the water vapor in this mixture is very small, $$
T_2=T_1\left(\frac{P_2}{P_1}\right)^{(k-1) / k}=(293 \mathrm{~K})\left(\frac{800 \mathrm{kPa}}{100 \mathrm{kPa}}\right)^{0.4 / 1.4}=531 \mathrm{~K}
$$ The saturation pressure at this temperature is $$
P_{g, 2}=P_{\text {sat }} @ 258^{\circ} \mathrm{C}=4542 \mathrm{kPa} \text { (from EES) }
$$ The vapor pressure at the exit is $$
P_{v, 2}=\frac{\omega_2 P_2}{\omega_2+0.622}=\frac{(0.0134)(800)}{0.0134+0.622}=16.87\ \mathrm{kPa}
$$ The relative humidity at the exit is then $$
\phi_2=\frac{P_{v, 2}}{P_{g, 2}}=\frac{16.87}{4542}=\mathbf{0 . 0 0 3 7}=\mathbf{0 . 3 7 \%}
$$
