Chapter 13.2, Problem 25PP

Interpretation Introduction

Interpretation:

The volume of the gas is to be calculated on changing the mass of the gas in the container.

Concept introduction:

The change in volume of the gas on changing the mass of the mass of the gas is studied by a law named as Avogadro’s Law.

According to this law,

$\begin{array}{l}V\propto n\\ or\\ \frac{V_1}{{\mathrm{n}}_1}=\frac{V_2}{{\mathrm{n}}_2}\end{array}$

Where

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

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

Answer to Problem 25PP

The volume of the gas is $12.41L$

Explanation of Solution

Avogadro’s Law is to be used to calculate new volume of the gas on changing the mass of the gas in container. The formula given by this law is:

$\begin{array}{l}V\propto n\\ or\\ \frac{V_1}{{\mathrm{n}}_1}=\frac{V_2}{{\mathrm{n}}_2}\end{array}$

Where

${\mathrm{V}}_1$ is initial volume of the gas

${\mathrm{n}}_1$ is the initial number of moles of the gas

${\mathrm{V}}_2$ is final volume of the gas

${\mathrm{n}}_2$ is the final number of moles of the gas

Here,

${\mathrm{V}}_1$ is $19.2L$

The number of moles is calculated by using the formula:

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

Here, $\mathrm{m}$ is the given mass of gas,

and $\mathrm{M}$ is the molar mass of gas

Molar mass of helium is $4gmo{l}^{-1}$

${\mathrm{m}}_1$ is $0.860g$

As $0.205g$ is removed from the container,

$\begin{array}{c}{m}_2=0.860-0.205\\ =0.555g\end{array}$

Therefore

$\begin{array}{c}{n}_1=\frac{0.860}{4}\\ =0.215gmo{l}^{-1}\end{array}$

$\begin{array}{c}{n}_2=\frac{0.555}{4}\\ =0.139gmo{l}^{-1}\end{array}$

Now applying the formula

$\begin{array}{l}V\propto n\\ or\\ \frac{V_1}{{\mathrm{n}}_1}=\frac{V_2}{{\mathrm{n}}_2}\end{array}$

$\begin{array}{c}{V}_2=\frac{n_2{V}_1}{n_1}\\ =\frac{0.139\times 19.2}{0.215}\\ =12.41L\end{array}$

Therefore, volume of the gas is $12.41L$

Conclusion

The volume of the gas is $12.41L$.