Chapter 13, Problem 99A

Interpretation Introduction

Interpretation:

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

Concept introduction:

The change in volume of the gas on changing 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 99A

The mass of the helium gas added is $\text{0}\text{.56}\text{g}$

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:

$\begin{array}{l}V\propto n\\ or\\ \frac{V_1}{{\mathrm{n}}_1}=\frac{V_2}{{\mathrm{n}}_2}\\ {\text{n}}_2\text{=}\frac{{\text{n}}_1{\text{V}}_{\text{2}}}{{\text{V}}_{\text{1}}}\end{array}$

Given data is:

$\begin{array}{l}{\text{V}}_{\text{1}}\text{=3}\text{.8}\text{L}\\ {\text{V}}_{\text{2}}\text{=2}\text{.8}\text{L}\\ {\text{n}}_{\text{1}}\text{=0}\text{.1}\text{mol}\end{array}$

Now applying the formula

$\begin{array}{c}{\text{n}}_2\text{=}\frac{{\text{n}}_1{\text{V}}_{\text{2}}}{{\text{V}}_{\text{1}}}\\ =\frac{0.1\times 3.8}{2.8}\\ \text{=0}\text{.14}\text{mol}\end{array}$

The mass of helium gas is given by:

$\begin{array}{c}\text{m=n×M}\\ \text{=0}\text{.14×4}\\ \text{=0}\text{.56}\text{g}\end{array}$

Therefore mass of the gas is $\text{0}\text{.56}\text{g}$

Conclusion

The mass of the gas that must be added is $\text{0}\text{.56}\text{g}$.