Answer
$223.1\text{ kPa}$
Work Step by Step
According to Raoults's law, the partial pressures of ammonia and water are given by $$
\begin{aligned}
& P_{\mathrm{g}, \mathrm{NH} 3}=y_{f, \mathrm{NH} 3} P_{\mathrm{sat}, \mathrm{NH} 3} \\
& P_{\mathrm{g}, \mathrm{H} 2 \mathrm{O}}=y_{f, \mathrm{H} 2 \mathrm{O}} P_{\mathrm{sat}, \mathrm{H} 2 \mathrm{O}}=\left(1-y_{f, \mathrm{NH} 3}\right) P_{\mathrm{sat}, \mathrm{H} 2 \mathrm{O}}
\end{aligned}
$$ Using Dalton's partial pressure model for ideal gas mixtures, the mole fraction of the ammonia in the vapor mixture is $$
\begin{aligned}
y_{g, \mathrm{NH} 3} & =\frac{y_{f, \mathrm{NH} 3} P_{\text {sat }, \mathrm{NH} 3}}{y_{f, \mathrm{NH} 3} P_{\mathrm{sat}, \mathrm{NH} 3}+\left(1-y_{f, \mathrm{NH} 3} P_{\mathrm{sat}, \mathrm{H} 2 \mathrm{O}}\right)} \\
0.96 & =\frac{430.6 y_{f, \mathrm{NH} 3}}{430.6 y_{f, \mathrm{NH} 3}+0.6112\left(1-y_{f, \mathrm{NH} 3}\right)} \longrightarrow y_{f, \mathrm{NH} 3}=\mathbf{0 . 0 3 2 9 4}
\end{aligned}
$$ Then, $$
\begin{aligned}
P & =y_{f, \mathrm{NH} 3} P_{\text {sat }, \mathrm{NH} 3}+\left(1-y_{f, \mathrm{NH} 3}\right) P_{\text {sal }, \mathrm{H} 2 \mathrm{O}} \\
& =(0.03294)(430.6)+(1-0.03294)(0.6112)=14.78\text{ kPa}
\end{aligned}
$$ Performing the similar calculations for the regenerator, $$
\begin{aligned}
& 0.96=\frac{1830.2 y_{f, \mathrm{NH} 3}}{1830.2 y_{f, \mathrm{NH} 3}+10.10\left(1-y_{f, \mathrm{NH} 3}\right)} \longrightarrow y_{f, \mathrm{NH} 3}=\mathbf{0 . 1 1 7 0} \\
& P=(0.1170)(1830.2)+(1-0.1170)(10.10)=\mathbf{2 2 3 . 1\ k P a}
\end{aligned}
$$
