Chapter 13.1, Problem 10PP

Interpretation Introduction

Interpretation:

The increase intemperature of the gas is to be calculated in $°C$ for a given change in pressure when the volume and amount of gas are kept constant.

Concept introduction:

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

According to this law,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

Where

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

${\mathrm{P}}_1$ is initial pressure of the gas

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

${\mathrm{P}}_2$ is the final pressure of the gas

Answer to Problem 10PP

The increase in temperature of the gas is $273°C$

Explanation of Solution

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

If ${\mathrm{P}}_1$ is the pressure of the gas at temperature ${\mathrm{T}}_1$ and pressure is changed to ${\mathrm{P}}_2$ Then temperature is changed to ${\mathrm{T}}_2$. The formula used to calculate ${\mathrm{T}}_1$ is

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

Or

${\mathrm{T}}_2=\frac{T_1{P}_2}{P_1}$

Where

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

${\mathrm{P}}_1$ is initial pressure of the gas

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

${\mathrm{P}}_2$ is the final pressure of the gas

Now initial pressure is $30.7KPa$

initial temperature is $0°C$

Final pressure

$\begin{array}{l}=2\times 30.7KPa\\ =61.4KPa\end{array}$

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

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

Therefore

$\begin{array}{c}{T}_1\left(K\right)=0+273\\ =273K\end{array}$

On substituting the values in the formula:

$\begin{array}{c}{T}_2=\frac{T_1{P}_2}{P_1}\\ =\frac{273\times 61.4}{30.7}\\ =546K\end{array}$

If result is to be reported in $°C$, converting Kelvin to $°C$ :

$\mathrm{T}(°C)=T(K)-273$

Therefore

$\begin{array}{c}{T}_2\left(°C\right)=546-273\\ =273°C\end{array}$

Therefore increase in temperature is

$\begin{array}{c}\Delta \mathrm{T}=546-273\\ =273°C\end{array}$

Conclusion

The increase in temperature of the gas is $273°C$