Answer
$w_{}=0.01133\text{ kgH}_2O_2\text{/kg dry air}$
$\phi=76.5\%$
$V=12.8\text{ m}^{3}$
Work Step by Step
(a) The specific humidity can be determined form its definition,$$
\omega=\frac{m_v}{m_a}=\frac{0.17 \mathrm{~kg}}{15 \mathrm{~kg}}=\mathbf{0 . 0 1 1 3 3 \mathrm { kg } \mathrm { H_2 }} \mathbf{O} / \mathrm{kg} \mathrm{dry} \mathrm{air}
$$ (b) The saturation pressure of water at $20^{\circ} \mathrm{C}$ is $$
P_g=P_{\text {sat } @ 200^{\circ C}}=2.339\ \mathrm{kPa}
$$ Then the relative humidity can be determined from $$
\phi=\frac{\omega P}{(0.622+\omega) P_g}=\frac{(0.01133)(100 \mathrm{kPa})}{(0.622+0.01133)(2.339 \mathrm{kPa})}=0.7650=76.5 \%
$$ (c) The volume of the tank can be determined from the ideal gas relation for the dry air $$
\begin{aligned}
P_v & =\phi P_g=(0.7650)(2.339 \mathrm{kPa})=1.789\ \mathrm{kPa} \\
P_a & =P-P_v=100-1.789=98.211\ \mathrm{kPa} \\
V & =\frac{m_a R_a T}{P_a}=\frac{(15 \mathrm{~kg})(0.287 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(293 \mathrm{~K})}{98.211 \mathrm{kPa}}=12.8\ \mathbf{m}^3
\end{aligned}
$$
