Chapter 13, Problem 92A

Interpretation Introduction

Interpretation:

The mass of water vapor is to be calculated for a given pressure, temperature and volume of the gas.

Concept introduction:

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

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 92A

The mass of water vapor is $3.85g$

Explanation of Solution

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

$\mathrm{PV}=nRT$

Or

$\mathrm{n}=\frac{PV}{RT}$

Further mass of gas $\mathrm{m}$ is calculated by multiplying number of moles $\mathrm{n}$ with molar mass of the gas $\mathrm{M}$

Given data is:

$\text{V=7}\text{.0}\text{L}$

$\text{P=1}\text{.0}\text{atm}$

$\begin{array}{c}\text{T=125}\text{°C}\\ \text{=125+273}\\ \text{=398}\text{K}\end{array}$

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

On substituting the values in the formula:

$\begin{array}{c}\mathrm{n}=\frac{1.0\times 7.0}{\begin{array}{l}0.0821\times 398\end{array}}\\ =0.214mol\end{array}$

So $\mathrm{m}$ is given by

$\mathrm{m}=M\times n$

$\begin{array}{c}M=2\times 1+1\times 16\\ =2+16\\ =18gmo{l}^{-1}\end{array}$

$\begin{array}{c}\mathrm{m}=18\times 0.214\\ =3.85g\end{array}$

Therefore mass of the gas is $3.85g$

Conclusion

The mass of water vapor is $3.85g$.