Chapter 13, Problem 93A

Interpretation Introduction

Interpretation:

The pressure of the gas is to be calculated for a given mass of the gas.

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 93A

The pressure of the gas is $\text{4}{\text{.94×10}}^{\text{-6}}\text{atm}$

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:

$\text{m}=2\times {10}^{-5}g$

$\text{V=3}\text{.50}\text{L}$

$\begin{array}{c}\text{T=22}\text{°C}\\ \text{=22+273}\\ \text{=295}\text{K}\end{array}$

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

Atomic mass of nitrogen is $14gmo{l}^{-1}$

Hence,

$\begin{array}{c}M=2\times 14\\ =28gmo{l}^{-1}\end{array}$

Further number of moles $\mathrm{n}$ is calculated by using formula:

$\begin{array}{c}\mathrm{n}=\frac{m}{M}\\ =\frac{2\times {10}^{-5}}{28}\\ =7.14\times {10}^{-7}mol\end{array}$

On substituting the values in the formula:

$\begin{array}{c}\text{P=}\frac{\text{7}{\text{.14×10}}^{\text{-15}}\text{×0}\text{.0821×295}}{\text{3}\text{.50}}\\ \text{=4}{\text{.94×10}}^{\text{-6}}\text{atm}\end{array}$

Therefore pressure of the gas is $\text{4}{\text{.94×10}}^{\text{-6}}\text{atm}$

Conclusion

The pressure of the gas is $\text{4}{\text{.94×10}}^{\text{-6}}\text{atm}$.