Chapter 13.1, Problem 7RQ

Interpretation Introduction

Interpretation:

The new volume of the gas needs to be calculated.

Concept Introduction:

Avogadro’s law: The volume of a given amount of gas is directly related to the number of moles at constant temperature and pressure.

The relationship between the volume of a gas and the number of moles present in the gas is expressed as-

  $\begin{array}{l}\mathrm{V}\propto \mathrm{n}\\ \mathrm{o}\mathrm{r},\frac{\mathrm{V}}{\mathrm{n}}\mathrm{}=\mathrm{}\mathrm{C}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}\end{array}$

Where, T and P are constants.

Avogadro’s law can also be expressed in mathematical form as-

  $\frac{{\mathrm{V}}_1}{{\mathrm{n}}_1}\mathrm{}=\mathrm{}\frac{{\mathrm{V}}_2}{{\mathrm{n}}_2}$

In the above equation, V₁ is initial volume and n₁ is initial number of moles, while V₂ is final or new volume and n₂ is final or new number of moles.

Answer to Problem 7RQ

The new volume of Ar gas is 2.353 L.

Explanation of Solution

Given Information:

  $\begin{array}{l}{\mathrm{V}}_1=10.0\text{ L}\\ \begin{array}{l}{\mathrm{n}}_1=4.25\text{ mol}\hfill \\ {\mathrm{n}}_2=1\text{ mol}\hfill \end{array}\end{array}$

Calculation:

  $\begin{array}{c}\frac{{\mathrm{V}}_1}{{\mathrm{n}}_1}=\frac{{\mathrm{V}}_2}{{\mathrm{n}}_2}\\ {\mathrm{V}}_2=\frac{{\mathrm{V}}_1\times {\mathrm{n}}_2}{{\mathrm{n}}_1}\mathrm{}\\ =\frac{10.0\mathrm{L}\times 1.0\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}}{4.25\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}}\\ =2.353\mathrm{L}\end{array}$