Chapter 13, Problem 124AP

Interpretation Introduction

Interpretation:

The vapor pressure of pure A and pure B is to be calculated from the given moles of A and B and total vapour pressure.

Concept Introduction:

The pressure exerted by a solvent is equal to the product of that solvent’s mole fraction and the pressure exerted by the pure solvent. It is expressed as

$P_1={\chi }_1{P}_1{}^{\circ }$

Here, $P_1{}^{\circ }$ is the vapor pressure of the pure solvent, $P_1$ is the pressure exerted by a solvent over a solution, and ${\chi }_1$ is the mole fraction of the solvent.

The total pressure is calculated by the addition of partial pressure of A and partial pressure of B. It is expressed as

$P_{total}=P_A+P_B$

or

$P_{total}={\chi }_A{P}^{\circ }{}_A+{\chi }_B{P}^{\circ }{}_B$

The mole fraction of compound can be calculated as

${\chi }_A=\frac{n_A}{n_A+n_B}$

Answer to Problem 124AP

Solution: $1.9\times {10}^2\text{mmHg}$ and $\text{4}\text{.0}\times {\text{10}}^2\text{mmHg}$

Explanation of Solution

Given information: Moles of A: $1.2\text{ mole}$.

Moles of B: $2.3\text{ mole}$.

Total vapor pressure is $331\text{ mmHg}$.

Total vapor pressure upon addition of another mole of B is $347\text{ mmHg}$.

Calculate the mole fraction of A by using the expression given below:

${\chi }_A=\frac{n_A}{n_A+n_B}$

Here, $n_A$ is the number of moles of A and $n_B$ is the number of moles of B.

Substitute $1.2\text{ mol}$ for $n_A$ and $\text{2}\text{.3 mol}$ for $n_B$

$\begin{array}{c}{\chi }_A=\frac{1.2mol}{1.2mol+2.3mol}\\ =0.3429\end{array}$

Calculate the mole fraction of B by subtracting the mole fraction of A from $1$

$\begin{array}{c}{\chi }_B=1-0.3429 \\ =0.6571\end{array}$

The total pressure is calculated by the addition of partial pressure of A and partial pressure of B. It is expressed as

$P_{total}=P_A+P_B$

or

$P_{total}={\chi }_A{P}^{\circ }{}_A+{\chi }_B{P}^{\circ }{}_B$ …...…... (1)

Substituting $P_{total}$ and the mole fractions calculated gives the equation below:

$333\text{ mmHg}=0.3429P^{\circ }{}_A+0.6571P^{\circ }{}_B$

Now, solving for $P^{\circ }{}_A$, we get

$\begin{array}{c}{P}^{\circ }{}_A=\frac{331\text{ mmHg}-0.6571P^{\circ }{}_B}{0.3429}\\ =965.3\text{ mmHg}-1.916P^{\circ }{}_B\end{array}$ …...…... (2)

Now, consider the solution with the additional mole of B.

Calculate the mole fraction of A and B:

$\begin{array}{c}{\chi }_A=\frac{1.2mol}{1.2mol+3.3mol}\\ =0.2667\end{array}$

$\begin{array}{c}{\chi }_B=1-0.2667 \\ =0.7333\end{array}$

Substituting $P_{total}$ and the mole fractions in equation (1) gives the equation as follows:

$347\text{ mmHg}=0.2669P^{\circ }{}_A+0.7333P^{\circ }{}_B$ …...…... (3)

Substituting Equation (2) into Equation (3) gives the equation below:

$\begin{array}{c}347\text{ mmHg}=0.2667\left(965.3\text{mmHg}-1.916P^{\circ }{}_B\right)+0.7333P^{\circ }{}_B\\ 0.2223P^{\circ }{}_B=89.55\text{ mmHg}\\ P^{\circ }{}_B=402.8\text{ mmHg}\\ =\text{4}\text{.0}\times {\text{10}}^2\text{mmHg}\end{array}$

Substitute the value of $P^{\circ }{}_B$ into Equation (1) to solve for ${\text{P}}_A^{\circ }$

$$ $\begin{array}{c}{P}^{\circ }{}_A=965.3\text{mmHg}-1.916\left(402.8\text{mmHg}\right)\\ =193.5\text{mmHg}\\ =1.9\times {10}^2\text{mmHg}\end{array}$

Conclusion

The vapor pressure of pure A and B is $1.9\times {10}^2\text{mmHg}$ and $\text{4}\text{.0}\times {\text{10}}^2\text{mmHg}$, respectively.