Chapter 13.1, Problem 7PP

Interpretation Introduction

Interpretation:

The temperature of the gas is to be calculated for a given new volume when the pressure and amount of gas are kept 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 in Kelvin

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

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

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

Answer to Problem 7PP

The new temperature of the gas is $192.5K$

Explanation of Solution

Given information:

The initial volume ${\mathrm{V}}_1=0.67L$

The initial temperature is ${\mathrm{T}}_1=350K$

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 volume is changed to ${\mathrm{V}}_2$ Then temperature is changed to ${\mathrm{T}}_2$. The formula used to calculate ${\mathrm{T}}_2$ is

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

Or

${\mathrm{T}}_2=\frac{T_1{V}_2}{V_1}$

Where

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

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

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

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

As the volume is reduced by $45\%$ ,

Thus, the final volume

$\begin{array}{c}{\mathrm{V}}_2=0.67-\frac{45\times 0.67}{100}\\ =0.37L\end{array}$

On substituting the values in the above formula:

$\begin{array}{c}{T}_2=\frac{0.37\times 350}{\begin{array}{l}0.67\\ \end{array}}\\ =192.5K\end{array}$

Therefore, new temperature is $192.5K$

Conclusion

The temperature of the gas is $192.5K$