Chapter 13.1, Problem 5PP

Interpretation Introduction

Interpretation:

The temperature of the gas is to be calculated in $°C$ 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 5PP

The new temperature of the gas is $332.13°C$

Explanation of Solution

Given information:

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

The initial temperature is $T_1=89°C$

The final volume is ${\mathrm{V}}_2=1.12L$

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 temperature is given in degree Celsius, so firstly convert it into Kelvin.

$\mathrm{T}\left(K\right)=T\left(°C\right)+273$

Therefore

$\begin{array}{c}{T}_1\left(K\right)=89+273\\ =362K\end{array}$

On substituting the values in the formula:

$\begin{array}{c}{T}_2=\frac{1.12\times 362}{\begin{array}{l}0.67\\ \end{array}}\\ =605.13K\end{array}$

But result is to be reported in $°C$. So now converting Kelvin to $°C$ :

$\mathrm{T}(°C)=T(K)-273$

Therefore

$\begin{array}{c}{T}_2\left(°C\right)=605.13-273\\ =332.13°C\end{array}$

Therefore, new temperature is $332.13°C$

Conclusion

The temperature of the gas is $332.13°C$