Question 14.6: Equilibrium in Binary Mixtures of Two Phases In absorption r...

Apply Raoult’s law to obtain the partial pressure of each component in the vapor mixture.

Assumptions
The solution is an ideal solution so that Raoult’s model can be applied. The vapor phase can be modeled by the IG mixture model.

Analysis
From the PC system-state TESTcalc, obtain p_{ NH _{3}, sat @ 20^{\circ} C }=857 kPa \text { and } p_{ H _{2} O , sat @ 20^{\circ} C }=2.34 kPa .
Using Eq. (14.34), the partial pressures of the components in the vapor phase are given as:

p_{k}=y_{k}^{\prime} p_{k, \text { sat@T }} ; \quad \text { where } \quad p=\sum p_{k}=\sum y_{k}^{\prime} p_{k, \text { sat } @ T} ; \quad[ kPa ]                           (14.34)

\begin{aligned}&p_{ NH _{3}}=y_{ NH _{3}}^{\prime} p_{ NH _{3}, sat @ 20^{\circ} C }=(0.5)(857)=428.5 kPa \\&p_{ H _{2} O }=y_{ H _{2} O }^{\prime} p_{ H _{2} O , sat @ 20^{\circ} C }=(0.5)(2.34)=1.17 kPa\end{aligned}

Therefore, the mixture pressure is:

p=p_{ NH _{3}}+p_{ H _{2} O }=429.67 kPa

The molar composition can be obtained from Eq. (14.33):

p_{k}=y_{k} p ; \quad \text { and }, \quad p=\sum p_{k} ; \quad[ kPa ]                             (14.33)

\begin{aligned}&y_{ NH _{3}}=p_{ NH _{3}} / p=428.5 / 429.67=0.997 \\&y_{ H _{2} O }=1-y_{ NH _{3}}=0.003\end{aligned}

Discussion
In the phase diagram of Figure 14.20, State-1 and State-2 represent the liquid and vapor states of the initial equilibrium mixture. If the temperature of the mixture is increased (at a constant total pressure), the phase diagram requires both the liquid and vapor states to migrate the left (towards States 3 and 4, respectively), indicating a lowered mole fraction of ammonia in both the liquid and vapor mixtures.