Chapter 13, Problem 32A

Interpretation Introduction

Interpretation: -Volume of the gas at different temperature is to be calculated using the variable values given.

Concept Introduction: - Volume of the gas at different temperature can be calculated by using combined gas law.

  $\frac{{\mathrm{P}}_1{\mathrm{V}}_1}{{\mathrm{T}}_1}=\frac{{\mathrm{P}}_2{\mathrm{V}}_2}{{\mathrm{T}}_2}$

Answer to Problem 32A

Volume of the sample is $464.2\mathrm{m}\mathrm{L}$

Explanation of Solution

Given information:

  $\begin{array}{l}{\mathrm{P}}_1=1.05\mathrm{a}\mathrm{t}\mathrm{m}\\ {\mathrm{V}}_1=459\mathrm{m}\mathrm{l}\\ {\mathrm{T}}_1=27^0\mathrm{C}+273=300\mathrm{K}\end{array}$

  $\begin{array}{l}{\mathrm{P}}_2=0.997\mathrm{a}\mathrm{t}\mathrm{m}\\ {\mathrm{T}}_2=15^0\mathrm{C}+273=288\mathrm{K}\\ {\mathrm{V}}_2=?\end{array}$

When state of a gas represented by ${\mathrm{P}}_1,{\mathrm{V}}_1\text{ and }{\mathrm{T}}_1$ changes to ${\mathrm{P}}_2,{\mathrm{V}}_2\text{ and }{\mathrm{T}}_2$ , then, according to Ideal gas law,

  $\frac{{\mathrm{P}}_1{\mathrm{V}}_1}{{\mathrm{T}}_1}=\frac{{\mathrm{P}}_2{\mathrm{V}}_2}{{\mathrm{T}}_2}$

Any one of the variables of this gas can be found out if other fine variable is known.

Put the values in the formula,

For calculating ${\mathrm{V}}_2$

  $\frac{{\mathrm{P}}_1{\mathrm{V}}_1}{{\mathrm{T}}_1}=\frac{{\mathrm{P}}_2{\mathrm{V}}_2}{{\mathrm{T}}_2}$

On rearranging,

  $\frac{{\mathrm{P}}_1{\mathrm{V}}_1{\mathrm{T}}_2}{{\mathrm{T}}_1{\mathrm{P}}_2}={\mathrm{V}}_2$

Putting the values,

  $\begin{array}{l}{\mathrm{V}}_2=\frac{{\mathrm{P}}_1{\mathrm{V}}_1{\mathrm{T}}_2}{{\mathrm{T}}_1{\mathrm{P}}_2}\\ {\mathrm{V}}_2=\frac{1.05\mathrm{a}\mathrm{t}\mathrm{m}\mathrm{}\times \mathrm{}459\mathrm{m}\mathrm{l}\mathrm{}\times \mathrm{}288\mathrm{K}\mathrm{}}{300\mathrm{K}\mathrm{}\times \mathrm{}0.997\mathrm{a}\mathrm{t}\mathrm{m}}\\ {\mathrm{V}}_2=\frac{1.05\mathrm{}\times \mathrm{}459\mathrm{}\times \mathrm{}288}{300\mathrm{}\times \mathrm{}0.997}\\ {\mathrm{V}}_2=\frac{1388016}{2991}=464.2\mathrm{m}\mathrm{l}\end{array}$

Volume of the sample is 464.2mL.

Conclusion

Volume of the sample is 464.2mL.