Chapter 13.1, Problem 6PP

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 6PP

The new volume of the gas is $2.57L$

Explanation of Solution

Given information:

The initial volume is ${\mathrm{V}}_1=3L$

The initial temperature is ${\mathrm{T}}_1=80°C$

The final temperature is ${\mathrm{T}}_2=30°C$

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

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

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

On converting Celsius into kelvin.

$\begin{array}{l}T(K)=T(°C)+273\\ T_1(K)=80+273\\ =353K\\ T_2(K)=30+273\\ =303K\end{array}$

On substituting the values in the formula:

$\begin{array}{c}{\mathrm{V}}_2=\frac{303\times 3}{\begin{array}{l}353\\ \end{array}}\\ =2.57L\end{array}$

Therefore, the new volume is $2.57L$

Conclusion

The volume of the gas is $2.57L$.