Chapter 13, Problem 69QAP

A tank contains a mixture of $\text{52}.\text{5 g}$of oxygen gas and $\text{65}.\text{1 g}$of carbon dioxide gas at $\text{27 }°\text{C}$. The total pressure in the tank is 9.21 am. Calculate the partial pressure (in arm) of each gas in the mixture.

Interpretation Introduction

Interpretation:

The partial pressure of oxygen and carbon dioxide gas should be calculated.

Concept Introduction:

Dalton’s law of partial pressure: In a container for mixture of gases, the total pressure is equal to the sum of partial pressures of all the gases present in the container. The partial pressure is the pressure exerted by a gas if it is the only gas present in the container.

Let a mixture of three gases with partial pressures ${\text{P}}_1$, ${\text{P}}_2$ and ${\text{P}}_3$, the total pressure of the gas will be sum of these partial pressures as follows:

${\text{P}}_{\text{T}}={\text{P}}_1+{\text{P}}_2+{\text{P}}_3$

This is the Dalton’s law of partial pressure.

The behaviour of gases is assumed to be ideal thus, partial pressure of gases can be calculated from an ideal gas equation as follows:

$\text{P}\text{V}=\text{n}\text{R}\text{T}$

Here, P is pressure, V is volume, n is number of moles, R is Universal gas constant and T is temperature of the gas.

The pressure exerted by an ideal gas depends on the number of gas particles, this does not depend on the nature of particles of gas. The two important things concluded from it will be:

1. The volume of gases is important.
2. The forces in between the particles of gas is not important.

Answer to Problem 69QAP

4.84 atm and 4.37 atm.

Explanation of Solution

Calculation:

First calculate the number of moles of each gas as follows:

$\text{n}=\frac{\text{m}}{\text{M}}$

Since, molar mass of O₂ is 32 g/mol thus, number of moles will be:

$\begin{array}{c}{\text{n}}_{{\text{O}}_2}=\frac{52.5\text{ g}}{\text{32 g/mol}}\\ =1.64\text{ mol}\end{array}$

Similarly, molar mass of ${\text{CO}}_{\text{2}}$ is 44.01 g/mol thus, number of moles will be:

$\begin{array}{c}{\text{n}}_{{\text{CO}}_2}=\frac{65.1\text{ g}}{\text{44}\text{.01 g/mol}}\\ =1.48\text{ mol}\end{array}$

Thus, total number of moles will be:

${\text{n}}_{\text{T}}={\text{n}}_{{\text{O}}_{\text{2}}}+{\text{n}}_{{\text{CO}}_{\text{2}}}$

Putting the values,

$\begin{array}{c}{\text{n}}_{\text{T}}=1.64\text{ mol}+1.48\text{ mol}\\ =3.1\text{2 mol}\end{array}$

From total pressure and number of moles, volume can be calculated using the ideal gas equation as follows:

${\text{P}}_{\text{T}}=\frac{{\text{n}}_{\text{T}}\text{R}\text{T}}{\text{V}}$

First convert the temperature from ${}^{\circ }\text{C}$ to K as follows:

$0{}^{\circ }\text{C}=273.15\text{ K}$

Thus,

$\begin{array}{c}27{}^{\circ }\text{C}=\left(25+273.15\right)\text{ K}\\ \text{=300}\text{.15 K}\end{array}$

Putting the values,

$9.21\text{ atm}=\frac{\left(3.12\text{ mol}\right)\left(0.082{\text{ L atm K}}^{-1}{\text{ mol}}^{-1}\right)\left(300.15\text{ K}\right)}{\text{V}}$

Thus,

$\begin{array}{c}\text{V}=\frac{\left(3.12\text{ mol}\right)\left(0.082{\text{ L atm K}}^{-1}{\text{ mol}}^{-1}\right)\left(300.15\text{ K}\right)}{\left(9.21\text{ atm}\right)}\\ =8.34\text{ L}\end{array}$

Now from volume, partial pressure of gases can be calculated from ideal gas equation as follows:

${\text{P}}_{{\text{O}}_{\text{2}}}=\frac{{\text{n}}_{{\text{O}}_{\text{2}}}\text{R}\text{T}}{\text{V}}$

Putting the values,

$\begin{array}{l}{\text{P}}_{{\text{O}}_{\text{2}}}=\frac{\left(1.64\text{ mol}\right)\left(0.082{\text{ L atm K}}^{-1}{\text{ mol}}^{-1}\right)\left(300.15\text{ K}\right)}{\left(8.34\text{ L}\right)}\\ =4.84\text{ atm}\end{array}$

Similarly, partial pressure of ${\text{CO}}_{\text{2}}$ can be calculated as follows:

${\text{P}}_{{\text{CO}}_{\text{2}}}=\frac{{\text{n}}_{{\text{CO}}_{\text{2}}}\text{R}\text{T}}{\text{V}}$

Putting the values,

$\begin{array}{l}{\text{P}}_{{\text{O}}_{\text{2}}}=\frac{\left(1.48\text{ mol}\right)\left(0.082{\text{ L atm K}}^{-1}{\text{ mol}}^{-1}\right)\left(300.15\text{ K}\right)}{\left(8.34\text{ L}\right)}\\ =4.37\text{ atm}\end{array}$.

Conclusion

Thus, partial pressure of oxygen and carbon dioxide gas is 4.84 atm and 4.37 atm respectively.