Chapter 10, Problem 66QP

Interpretation Introduction

Interpretation:

The partial pressures of the gasespresent in the given mixture of gases are to be calculated.

Concept introduction:

The mole fraction of an individual gas for the combination of gases is the ratio of the moles of the individual gas to the total number of moles of the mixture.

${\chi }_i=\frac{n_i}{n_{\text{total}}}$

Here, ${\chi }_i$ is the mole fraction, $n_i$ is the mole fraction of the individual gas, and $n_{\text{total}}$ is the total number of moles.

Also, the mole fraction of an individual gas for the combination of gases can be calculated from the ratio of the partial pressure of the individual gas with the total pressure of the combination.

${\chi }_i=\frac{P_i}{P_{\text{total}}}$

Here, ${\chi }_i$ is the mole fraction, $P_i$ is the partial pressure of the individual gas, and $P_{\text{total}}$ is the total pressure.

Answer to Problem 66QP

Solution: The partial pressure for ${\text{CH}}_4$ is $0.54\text{ atm}$, for ${\text{C}}_2{\text{H}}_6$ is $0.44\text{ atm}$, and for ${\text{C}}_3{\text{H}}_8$ is $0.51\text{ atm}$.

Explanation of Solution

Given information:

Number of moles of ${\text{CH}}_4$ $n_{{\text{CH}}_4}=0.31\text{ mol}$

Number of moles of ${\text{C}}_2{\text{H}}_6$ $n_{{\text{C}}_2{\text{H}}_6}=0.25\text{ mol}$

Number of moles of ${\text{C}}_3{\text{H}}_8$ $n_{{\text{C}}_3{\text{H}}_8}=0.29\text{ mol}$

Total pressure $P_T=1.50\text{ atm}$

From Dalton’s law for the combination of gases, the total number of moles can be calculated as

$\begin{array}{c}{n}_{\text{total}}={\displaystyle \sum n_i}\\ =n_{{\text{CH}}_4}+n_{{\text{C}}_2{\text{H}}_6}+n_{{\text{C}}_3{\text{H}}_8}\end{array}$

Substitute $0.31\text{ mol}$ for $n_{{\text{CH}}_4}$, $0.25\text{ mol}$ for $n_{{\text{C}}_2{\text{H}}_6}$, and $0.29\text{ mol}$ for $n_{{\text{C}}_3{\text{H}}_8}$ in the above equation as follows:

$\begin{array}{c}{n}_{\text{total}}=0.31\text{ mol}+0.25\text{ mol}+0.29\text{ mol}\\ =0.85\text{ mol}\end{array}$

Calculate the mole fraction of ${\text{CH}}_4$ as follows:

${\chi }_{{\text{CH}}_4}=\frac{n_{{\text{CH}}_4}}{n_{\text{total}}}$

Substitute $\text{0}\text{.31 mol}$ for $n_{{\text{CH}}_4}$ and $\text{0}\text{.85 mol}$ for $n_{\text{total}}$ in the above equation as follows:

$\begin{array}{c}{\chi }_{{\text{CH}}_4}=\frac{0.31\text{ mol}}{0.85\text{ mol}}\\ =0.36\end{array}$

Thus, the mole fraction of ${\text{CH}}_4$ is $0.36$.

Calculate the partial pressure for ${\text{CH}}_4$ as follows:

$P_{{\text{CH}}_4}={\chi }_{{\text{CH}}_4}\times P_{\text{total}}$

Substitute 0.36 for ${\chi }_{{\text{CH}}_4}$ and $1.50\text{ atm}$ for $P_{\text{total}}$ in the above equation as follows:

$\begin{array}{c}{P}_{{\text{CH}}_4}=0.36\times 1.50\text{ atm}\\ =0.54\text{ atm}\end{array}$

Thus, for ${\text{CH}}_4$, the partial pressure is $0.54\text{ atm}$.

Calculate the mole fraction of ${\text{C}}_2{\text{H}}_6$ as follows:

${\chi }_{{\text{C}}_2{\text{H}}_6}=\frac{n_{{\text{C}}_2{\text{H}}_6}}{n_{\text{total}}}$

Substitute $\text{0}\text{.25 mol}$ for $n_{{\text{C}}_2{\text{H}}_6}$ and $\text{0}\text{.85 mol}$ for $n_{\text{total}}$ in the above equation as follows:

$\begin{array}{c}{\chi }_{{\text{C}}_2{\text{H}}_6}=\frac{0.25\text{ mol}}{0.85\text{ mol}}\\ =0.29\end{array}$

Thus, the mole fraction of ${\text{C}}_2{\text{H}}_6$ is $0.29$.

Calculate the partial pressure for ${\text{C}}_2{\text{H}}_6$ as follows:

$P_{{\text{C}}_2{\text{H}}_6}={\chi }_{{\text{C}}_2{\text{H}}_6}\times P_{\text{total}}$

Substitute 0.29 for ${\chi }_{{\text{C}}_2{\text{H}}_6}$ and $1.50\text{ atm}$ for $P_{\text{total}}$ in the above equation as follows:

$\begin{array}{c}{P}_{{}_{{\text{C}}_2{\text{H}}_6}}=0.29\times 1.50\text{ atm}\\ =0.44\text{ atm}\end{array}$

Thus, for ${\text{C}}_2{\text{H}}_6$, the partial pressure is $0.44\text{ atm}$.

Calculate the mole fraction of ${\text{C}}_3{\text{H}}_8$ as follows:

${\chi }_{{\text{C}}_3{\text{H}}_8}=\frac{n_{{\text{C}}_3{\text{H}}_8}}{n_{\text{total}}}$

Substitute $\text{0}\text{.29 mol}$ for $n_{{\text{C}}_3{\text{H}}_8}$ and $\text{0}\text{.85 mol}$ for $n_{\text{total}}$ in the above equation as follows:

$\begin{array}{c}{\chi }_{{\text{C}}_3{\text{H}}_8}=\frac{0.29\text{ mol}}{0.85\text{ mol}}\\ =0.34\end{array}$

Thus, the mole fraction of ${\text{C}}_3{\text{H}}_8$ is $0.34$.

Calculate the partial pressure for ${\text{C}}_3{\text{H}}_8$ as follows:

$P_{{\text{C}}_3{\text{H}}_8}={\chi }_{{\text{C}}_3{\text{H}}_8}\times P_{\text{total}}$

Substitute 0.34 for ${\chi }_{{\text{C}}_3{\text{H}}_8}$ and $1.50\text{ atm}$ for $P_{\text{total}}$ in the above equation as follows:

$\begin{array}{c}{P}_{{}_{{\text{C}}_3{\text{H}}_8}}=0.34\times 1.50\text{ atm}\\ =0.51\text{ atm}\end{array}$

Thus, for ${\text{C}}_3{\text{H}}_8$, the partial pressure is $0.51\text{ atm}$.

Conclusion

Thepartial pressure for ${\text{CH}}_4$ is $0.54\text{ atm}$, for ${\text{C}}_2{\text{H}}_6$ is $0.44\text{ atm}$, and for ${\text{C}}_3{\text{H}}_8$ is $0.51\text{ atm}$.