Chapter 7, Problem 7.27E

Interpretation Introduction

Interpretation:

The expression for $p_{\text{tot}}$ like equation 7.24 in terms of $y_2$ rather than $y_1$ is to be stated.

Concept introduction:

According to the Raoult’s law, the vapor pressure of the solution is equal to the product of vapor pressure of the solvent and its mole fraction in the solution.

According to the Dalton’s law of gases, the partial pressure of a component is equal to the product of total pressure of the gas mixture and mole fraction of that component.

Answer to Problem 7.27E

The expression for $p_{\text{tot}}$ like equation 7.24 in terms of $y_2$ is shown below.

$p_{\text{tot}}=\frac{p_1^{\ast }{p}_2^{\ast }}{p_2^{\ast }+\left(p_1^{\ast }-p_2^{\ast }\right)y_2}$

Explanation of Solution

The mole fraction of the two components in vapor phase can be calculated by using variables $y_1$ and $y_2$. These variables are calculated by using Dalton’s law. It states that the partial pressure of a component is equal to the product of total pressure of the gas mixture and mole fraction of that component.

The equation is represented as,

$y_2=\frac{p_2}{p_{\text{tot}}}=\frac{p_2}{p_1+p_2}=\frac{x_2{p}_2^{\ast }}{x_2{p}_2^{\ast }+x_1{p}_1^{\ast }}$

Where,

• $y_2$ represents the mole fraction of component $2$ in the gas phase.

• $x_2$ represents the mole fraction of component $2$ in the solution.

• $p_1^{\ast }$ and $p_2^{\ast }$ represents the partial pressures of component 1 and 2 in the solution.

Substitute the value of $x_1=1-x_2$ in the above formula,

$\begin{array}{rll}{y}_2& =& \frac{x_2{p}_2^{\ast }}{x_2{p}_2^{\ast }+\left(1-x_2\right)p_1^{\ast }}\\ y_2& =& \frac{x_2{p}_2^{\ast }}{p_1^{\ast }+\left(p_2^{\ast }-p_1^{\ast }\right)x_2}\end{array}$

The above equation is rearranged as shown below.

$\frac{p_1^{\ast }+\left(p_2^{\ast }-p_1^{\ast }\right)x_2}{x_2}=\frac{p_2^{\ast }}{y_2}$ (1)

On solving the left hand side of the equation (1),

$\frac{p_1^{\ast }}{x_2}+\left(p_2^{\ast }-p_1^{\ast }\right)=\frac{p_2^{\ast }}{y_2}$

$\frac{p_1^{\ast }}{x_2}=\frac{p_2^{\ast }}{y_2}-\left(p_2^{\ast }-p_1^{\ast }\right)$ (2)

On solving the right hand side of the equation (2),

$\frac{p_1^{\ast }}{x_2}=\frac{p_2^{\ast }-y_2\left(p_2^{\ast }-p_1^{\ast }\right)}{y_2}$

$\frac{p_1^{\ast }}{x_2}=\frac{p_2^{\ast }-y_2{p}_2^{\ast }+y_2{p}_1^{\ast }}{y_2}$ (3)

Take the common terms together and rearranging the equation (3),

$x_2=\frac{y_2{p}_1^{\ast }}{p_2^{\ast }+\left(p_1^{\ast }-p_2^{\ast }\right)y_2}$

The $p_{\text{tot}}$ is calculated by the formula as shown below.

$p_{\text{tot}}=\frac{x_2{p}_2^{\ast }}{y_2}$

Where,

• $p_{\text{tot}}$ is the total pressure of the system

• $y_2$ represents the mole fraction of component $2$ in the gas phase.

• $x_2$ represents the mole fraction of component $2$ in the solution.

Substitute the value of $x_2$ in the above formula.

$\begin{array}{rll}{p}_{\text{tot}}& =& \frac{\left(\frac{y_2{p}_1^{\ast }}{p_2^{\ast }+\left(p_1^{\ast }-p_2^{\ast }\right)y_2}\right)p_2^{\ast }}{y_2}\\ & =& \frac{y_2{p}_1^{\ast }{p}_2^{\ast }}{\left[p_2^{\ast }+\left(p_1^{\ast }-p_2^{\ast }\right)y_2\right]y_2}\\ p_{\text{tot}}& =& \frac{p_1^{\ast }{p}_2^{\ast }}{p_2^{\ast }+\left(p_1^{\ast }-p_2^{\ast }\right)y_2}\end{array}$

Hence, the expression for $p_{\text{tot}}$ in terms of $y_2$ is shown below.

$p_{\text{tot}}=\frac{p_1^{\ast }{p}_2^{\ast }}{p_2^{\ast }+\left(p_1^{\ast }-p_2^{\ast }\right)y_2}$

Conclusion

The expression for $p_{\text{tot}}$ like equation 7.24 in terms of $y_2$ is shown below.

$p_{\text{tot}}=\frac{p_1^{\ast }{p}_2^{\ast }}{p_2^{\ast }+\left(p_1^{\ast }-p_2^{\ast }\right)y_2}$