Answer
$m_{a}= 9.264\text{ kg}$
$P_{v}= 4.2469\text{kPa}$
$w=0.0262\text{ kg H}_2\text{O/kg dry air}$
$h=97.1\text{ kJ/kg dry air}$
Work Step by Step
(a) The mass of dry air can be determined from the ideal gas relation for the dry air,$$
m_a=\frac{P_a V}{R_a T}=\frac{[(105-4.2469) \mathrm{kPa}]\left(8 \mathrm{~m}^3\right)}{(0.287 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(30+273.15 \mathrm{~K})}=\mathbf{9 . 2 6 4\ \mathrm { kg }}
$$ (b) The relative humidity of air is $100$ percent since the air saturated. The vapor pressure is equal to the saturation pressure of water at $30^{\circ} \mathrm{C}$ $$
P_v=P_g=P_{\text {sat } @ 30^{\circ} \mathrm{C}}=4.2469\ \mathrm{kPa}
$$ The specific humidity can be determined from $$
\omega=\frac{0.622 P_v}{P-P_v}=\frac{(0.622)(4.2469 \mathrm{kPa})}{(105-4.2469) \mathrm{kPa}}=\mathbf{0 . 0 2 6 2} \mathbf{k g ~ \mathbf { H } _ { 2 }} \mathbf{O} / \mathbf{k g ~ d r y ~ a i r ~}
$$ (c) The enthalpy of air per unit mass of dry air is determined from $$
\begin{aligned}
h & =h_a+\omega h_v \equiv c_p T+\omega h_{g @ 30{ }^{\circ} \mathrm{C}} \\
& =\left(1.005 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)\left(30^{\circ} \mathrm{C}\right)+(0.0262)(2555.6 \mathrm{~kJ} / \mathrm{kg}) \\
& =97.1 \mathrm{~kJ} / \mathrm{kg}\ \mathbf{d r y} \text { air }
\end{aligned}
$$
