Chapter 13, Problem 58A

Interpretation Introduction

Interpretation:

The volume of the gas is to be calculated for a given new temperature when the pressure and amount of gas remain constant.

Concept introduction:

When pressure and amount of gas are kept constant, increasing the temperature of gas increases its volume. This law is known as Charles’ Law.

According to this law,

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

Where

${\mathrm{T}}_1$ is initial temperature of the gas

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

${\mathrm{T}}_2$ is the final temperature of the gas

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

Answer to Problem 58A

The new volume of the gas is $2.68L$

Explanation of Solution

The change in volume on changing the temperature of the gas keeping the pressure and amount of gas fixed is studied by a law named as Charles’ Law. According to this law, The volume of a given mass of gas is directly proportional to the Kelvin temperature at constant pressure.

If ${\mathrm{V}}_1$ is the volume of the gas at temperature ${\mathrm{T}}_1$ and temperature is changed to ${\mathrm{T}}_2$ then volume is changed to ${\mathrm{V}}_2$. The formula used to calculate ${\mathrm{V}}_2$ is

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

Or

${\mathrm{V}}_2=\frac{T_2{V}_1}{T_1}$

Given data is:

${\text{V}}_1=2.50L$

${\text{T}}_1=22°C$

${\text{T}}_2=43°C$

$\begin{array}{l}{T}_{}(K)=T_{}(°C)+273\\ T_1(K)=22+273\\ =295K\\ T_2(K)=43+273\\ =316K\end{array}$

On substituting the values in the formula:

$\begin{array}{c}{\mathrm{V}}_2=\frac{316\times 2.50}{\begin{array}{l}295\end{array}}\\ =2.68L\end{array}$

Therefore new volume is $2.68L$

Conclusion

The volume of the gas is $2.68L$.