Chapter 13, Problem 102A

Apply Calculate the pressure of $(4.67\times {10}^{22})$ moleculesof CO gas mixed with $2.87\times {10}^{24}$ molecules of
   $N_2$ gas ina 6.00-L container at 34.8°C.

Interpretation Introduction

Interpretation:

The pressure of the mixture of gases is to be calculated for given molecules of the gases.

Concept introduction:

Ideal gas equation is used to calculate pressure of the gas at given temperature and volume.

According to this law,

$\mathrm{PV}=nRT$

Where

$\mathrm{P}$ is pressure of the gas

$\mathrm{V}$ is volume of the gas

$\mathrm{n}$ is the number of moles of the gas

$\mathrm{R}$ is the ideal gas constant

$\mathrm{T}$ is the temperature of the gas

Answer to Problem 102A

The pressure of the mixture is $\text{20}\text{.4}\text{atm}$

Explanation of Solution

Ideal gas equation helps to calculate pressure of a gas at given temperature and volume. According to this law,

$\mathrm{PV}=nRT$

Or

$\text{P}\text{=}\frac{\text{nRT}}{\text{V}}$

Given data is:

$\begin{array}{l}\text{Number of molecules of CO=4}{\text{.67×10}}^{\text{22}}\text{molecules}\\ {\text{n}}_{\text{CO}}\text{=}\frac{\text{4}{\text{.67×10}}^{\text{22}}}{\text{6}{\text{.023×10}}^{\text{23}}}\\ \text{=0}\text{.0776}\\ {\text{Number of molecules of N}}_{\text{2}}\text{=2}{\text{.87×10}}^{\text{24}}\text{molecules}\\ {\text{n}}_{{\text{N}}_{\text{2}}}\text{=}\frac{\text{4}{\text{.67×10}}^{\text{24}}}{\text{6}{\text{.023×10}}^{\text{23}}}\\ \text{=4}\text{.77}\\ {\text{n}}_{\text{total}}\text{=0}\text{.0776+4}\text{.77}\\ \text{=4}\text{.846}\text{mol}\\ \text{T=34}\text{.8}\text{°C}\\ \text{=34}\text{.8+273}\\ \text{=307}\text{.8}\text{K}\\ \text{V=6}\text{L}\end{array}$

$\text{R=}0.0821Latm{K}^{-1}mo{l}^{-1}$

On substituting the values in the formula:

$\begin{array}{c}\text{P=}\frac{\text{4}\text{.85×0}\text{.0821×307}\text{.8}}{6}\\ \text{=20}\text{.4}\text{atm}\end{array}$

Therefore pressure of the mixture is $\text{20}\text{.4}\text{atm}$

Conclusion

The pressure of the mixture is $\text{20}\text{.4}\text{atm}$.