Chapter 5, Problem 5.43PAE

43 In an experiment, a mixture of gases occupies a volume of 30.00 L at a temperature of ${122.5}^{\circ }$C. The mixture contains 14.0 g of water, 11.5 g of oxygen, and 37.3 g of nitrogen. Calculate the total pressure and the partial pressure of each gas.

Interpretation Introduction

Interpretation:

To find out the partial pressure and total pressure of the gas mixture

Concept introduction:

The pressure is exerted by an individual gas in a mixture is known as its partial pressure. Dalton’s law of partial pressure can be expressed in terms of the mole fraction of a gas in the mixture. The mole fraction of a gas is the number of moles of that gas divided by the total moles of gas in the mixture.

Given:

Volume $\text{30}\text{.00}\text{L}$, Temperature $\text{122}\text{.5}{}^{\text{o}}\text{C}$

The mixture contains $\text{14}\text{.0}\text{g}$ of water, $\text{11}\text{.5}\text{g}$ of oxygen, and $\text{37}\text{.3}\text{g}$ of nitrogen

Answer to Problem 5.43PAE

Solution:

The total pressure of gas mixture is $2.667\text{atm}$. The partial pressure of nitrogen, water and oxygen are $1.439atm$, $0.839atm$ and $0.388\text{atm}$ respectively.

Explanation of Solution

Formula used:

$\begin{array}{l}{n}_{tot}=n_{N_2}+n_{H_{2}O}+n_{O_2}\\ P_{total}=\frac{n_{tot}RT}{V}\\ P_x=x_x{P}_{tot}\end{array}$

$P_x$ is the partial pressure of respective gas.

$P_{tot}$ is the total pressure

$x_x$ is the mole fraction of respective gas

R- Gas constant

T -temperature

V- Volume

$n_{tot}$ is the total number of moles of gas

Calculation:

$\begin{array}{l}{1}^{st}step:\hfill \\ \text{First we must find how many moles of each gas}.\hfill \end{array}$

$\begin{array}{l}{\text{n}}_{{\text{N}}_{\text{2}}}\text{=}\frac{\text{1 mol}}{\text{28}\text{.014 g/mol}}\text{×37}\text{.3 g}\text{of}\text{nitrogen}\\ \boxed{{\text{n}}_{{\text{N}}_{\text{2}}}\text{=1}\text{.3314 mol}{\text{N}}_{\text{2}}}\\ {\text{n}}_{{\text{H}}_{\text{2}}\text{O}}\text{=}\frac{\text{1 mol}}{\text{18}\text{.03 g/mol}}\text{×14}\text{.0 g}\text{of}\text{water}\\ \boxed{{\text{n}}_{{\text{H}}_{\text{2}}\text{O}}\text{=0}\text{.7764 mol}{\text{H}}_{\text{2}}\text{O}}\\ {\text{n}}_{{\text{O}}_{\text{2}}}\text{=}\frac{\text{1 mol}}{\text{31}\text{.98 g/mol}}\text{×11}\text{.5 g}\text{of}\text{oxygen}\\ \boxed{{\text{n}}_{{\text{O}}_{\text{2}}}\text{=0}\text{.3595 mol}{\text{O}}_{\text{2}}}\\ {\text{n}}_{\text{tot}}{\text{=n}}_{{\text{N}}_{\text{2}}}{\text{+n}}_{{\text{H}}_{\text{2}}\text{O}}{\text{+n}}_{{\text{O}}_{\text{2}}}\\ {\text{n}}_{\text{tot}}\text{=1}\text{.3314+0}\text{.7764+0}\text{.3595}\\ \boxed{{\text{n}}_{\text{tot}}\text{=2}\text{.4673 mol}\text{gas}}\end{array}$

$\begin{array}{l}\text{Total}\text{pressure,}{\text{P}}_{\text{tot}}\text{=}\frac{{\text{n}}_{\text{tot}}\text{RT}}{\text{V}}\\ {\text{P}}_{\text{tot}}\text{=}\frac{\text{2}\text{.4673 mol×0}\text{.0820 (L}\text{atm}\text{mol}\text{K)×(273+122}\text{.5) K}}{\text{30}\text{.00 L}}\\ {\text{P}}_{\text{tot}}\text{=}\frac{\text{2}\text{.4673×0}\text{.0820×395}\text{.5}}{\text{30}\text{.00}}\\ {\text{P}}_{\text{tot}}\text{=}\frac{\text{80}\text{.0170}}{\text{30}\text{.00}}\\ \boxed{{\text{P}}_{\text{tot}}\text{=2}\text{.667 atm}}\\ \text{Partial}\text{pressure,}{\text{P}}_{\text{x}}{\text{=x}}_{\text{x}}{\text{P}}_{\text{tot}}\\ {\text{P}}_{{\text{N}}_{\text{2}}}{\text{=x}}_{{\text{N}}_{\text{2}}}{\text{P}}_{\text{tot}}\\ {\text{P}}_{{\text{N}}_{\text{2}}}\text{=}\left(\frac{{\text{n}}_{{\text{N}}_{\text{2}}}}{{\text{n}}_{\text{tot}}}\right){\text{P}}_{\text{tot}}\\ {\text{P}}_{{\text{N}}_{\text{2}}}\text{=}\left(\frac{\text{1}\text{.3314}}{\text{2}\text{.4673}}\right)\text{(2}\text{.667)}\\ \boxed{{\text{P}}_{{\text{N}}_{\text{2}}}\text{=1}\text{.439 }\text{atm}}\\ {\text{P}}_{{\text{H}}_{\text{2}}\text{O}}\text{=}\left(\frac{{\text{n}}_{{\text{H}}_{\text{2}}\text{O}}}{{\text{n}}_{\text{tot}}}\right){\text{P}}_{\text{tot}}\\ {\text{P}}_{{\text{H}}_{\text{2}}\text{O}}\text{=}\left(\frac{\text{0}\text{.7764}}{\text{2}\text{.4673}}\right)\text{(2}\text{.667)}\\ \boxed{{\text{P}}_{{\text{H}}_{\text{2}}\text{O}}\text{=0}\text{.839 }\text{atm}}\\ {\text{P}}_{{\text{O}}_{\text{2}}}\text{=}\left(\frac{{\text{n}}_{{\text{O}}_{\text{2}}}}{{\text{n}}_{\text{tot}}}\right){\text{P}}_{\text{tot}}\\ {\text{P}}_{{\text{O}}_{\text{2}}}\text{=}\left(\frac{\text{0}\text{.3595}}{\text{2}\text{.4673}}\right)\text{(2}\text{.667)}\\ \boxed{{\text{P}}_{{\text{O}}_{\text{2}}}\text{=0}\text{.388 }\text{atm}}\end{array}$

Conclusion

The total pressure of gas mixture is $\text{2}\text{.667 atm}$.

The partial pressure of

1. Nitrogen $\text{1}\text{.439 atm}$,
2. Water $\text{0}\text{.839 atm}$ and
3. Oxygen $\text{0}\text{.388 atm}$.