Chapter 13, Problem 4SEPP

Interpretation Introduction

Interpretation:

The partial pressures of the components A and B, above the condensed liquid at $55°\text{C}$ are to be calculated.

Concept introduction:

The moles are calculated as follows:

$\text{Moles}=\frac{\text{Mass}}{\text{Molar mass}}$

Mole fraction can be expressed as follows:

$\text{Mole fraction=}\frac{\text{No of moles of that component}}{\text{Total no of moles of all the components}}$

The partial pressure of the components can be calculated using the expression as follows: $P_{\text{A}}\text{=}{X}_{\text{A}}\times P_{\text{A}}{}^{\circ }$

Here, $P_{\text{A}}$ is the partial pressure of component A, $X_{\text{A}}$ is the mole fraction of A, and $P_{\text{A}}{}^{\circ }$ is the vapor pressure of pure component A.

The relation of vapour and the condensed state will be as follows:

$\begin{array}{c}{X}_{\text{A}}\left(\text{vapor}\right)=X_{\text{A}}\left(\text{condensed liquid}\right)\\ =\frac{P_{\text{A}}}{P_{\text{A}}+P_{\text{B}}}\end{array}$

Answer to Problem 4SEPP

Solution: Option (c).

Explanation of Solution

Reasons for the correct option:

Molar mass of A is $\text{100 g/mol}$. Molar mass of B is $\text{110 g/mol}$. The partial pressure of pure liquid A is $98\text{ mm Hg}$ and pure liquid B is $42\text{ mm Hg}$.

The partial pressure of components can be found using the expression as follows: $P_{\text{A}}\text{=}{X}_{\text{A}}\times P_{\text{A}}{}^{\circ }$

Here, $P_{\text{A}}$ is thepartial pressure of component A, $X_{\text{A}}$ is the mole fraction of A, and $P_{\text{A}}{}^{\circ }$ is the vapor pressure of pure component A.

Suppose, mass of A and mass of B as $w\text{ g}$,

The moles of A are calculated as follows:

$\begin{array}{c}\text{Moles of A}=\frac{\text{Mass of A}}{\text{Molar mass of A}}\\ =\frac{w}{100}\text{ moles}\end{array}$

The moles of B are calculated as follows:

$\begin{array}{c}\text{Moles of B}=\frac{\text{Mass of B}}{\text{Molar mass of B}}\\ =\frac{w}{110}\text{ moles}\end{array}$

Total no of moles in the solution arecalculated as follows:

$\begin{array}{c}\text{Moles of A}+\text{Moles of B}=\frac{w}{100}+\frac{w}{110}\\ =\frac{210w}{110\times 100}\end{array}$

The mole fraction of $\text{A}$ is calculated as follows:

$\begin{array}{c}\text{Mole fraction of A}=\frac{\text{No of moles of A}}{\text{Total no of moles of A and B}}\\ =\frac{\frac{\cancel{w}}{100}}{\frac{210\cancel{w}}{110\times 100}}\\ =0.52\end{array}$

Since, sum of mole fraction of all the components of a solution is equal to unity.

So, the mole fraction of $\text{B}$ is as follows:

$\begin{array}{c}\text{Mole fraction of B}=\text{1}-\text{Mole fraction of A}\\ =\text{1}-\text{0}\text{.52}\\ =\text{0}\text{.48}\end{array}$

Now, partial pressure of A can be calculated by the expression as follows:

$\begin{array}{c}{P}_{\text{A}}\text{=}{X}_{\text{A}}\times P_{\text{A}}{}^{\circ }\\ =0.52\times 98\\ =50.96\\ =51\text{ mm Hg}\end{array}$

Similarly, the partial pressure of B can be calculated by the expression as follows:

$\begin{array}{c}{P}_{\text{B}}\text{=}{X}_{{}_{\text{B}}}\times P_{{}_{\text{B}}}{}^{\circ }\\ =0.48\times 42\\ =20\text{ mm Hg}\end{array}$

Now, if some of the vapors are condensed at $55°\text{C}$ to a liquid, the mole fraction of the condensed liquid and the mole fraction of the vapor state will be almost equal to each other.

The relation of vapour and the condensed state will be as follows:

$\begin{array}{c}{X}_{\text{A}}\left(\text{vapor}\right)=X_{\text{A}}\left(\text{condensed liquid}\right)\\ =\frac{P_{\text{A}}}{P_{\text{A}}+P_{\text{B}}}\end{array}$

Substituting $P_{\text{A}}$ as $51\text{ mm Hg}$ and $P_{\text{B}}$ as $20\text{ mm Hg}$ in the above equation as follows:

$\begin{array}{c}{X}_{\text{A}}\left(\text{condensed liquid}\right)=\frac{P_{\text{A}}}{P_{\text{A}}+P_{\text{B}}}\\ =\frac{51\cancel{\text{mm Hg}}}{\left(20+51\right)\cancel{\text{mm Hg}}}\\ =0.72\end{array}$

$\begin{array}{c}{X}_{\text{B}}\left(\text{condensed liquid}\right)=\frac{P_{\text{A}}}{P_{\text{A}}+P_{\text{B}}}\\ =\frac{\text{20 }\cancel{\text{mm Hg}}}{\left(20+51\right)\cancel{\text{mm Hg}}}\\ =0.28\end{array}$

By substituting the partial pressure of liquid A $\left(P_{\text{A}}°\right)$ is $98\text{ mm Hg}$ and pure liquid B $\left(P_{\text{B}}°\right)$ is $42\text{ mm Hg}$ and above calculated values of $X_{\text{A}}\left(\text{condensed liquid}\right)$ and $X_{\text{B}}\left(\text{condensed liquid}\right)$ in the given equation as follows:

$\begin{array}{c}{P}_{\text{A}}\text{=}{X}_{\text{A}}\times P_{\text{A}}{}^{\circ }\\ =0.72\left(98\text{ mm Hg}\right)\\ =71\text{ mm Hg}\end{array}$

$\begin{array}{c}{P}_{\text{B}}\text{=}{X}_{\text{B}}\times P_{\text{B}}{}^{\circ }\\ =0.28\left(\text{42 mm Hg}\right)\\ =12\text{ mm Hg}\end{array}$

Hence, option (c) is correct.

Reason for incorrect options:

Option (a) is incorrect because after substituting the values, the values given in the option are not obtained.

Option (b) is incorrect because after substituting the values, the values given in the option are not obtained.

Option (d) is incorrect because after substituting the values, the values given in the option are not obtained.

Hence, options (a), (b), and (d) are incorrect.