Chapter 13.1, Problem 9PP

Interpretation Introduction

Interpretation:

The temperature of the gas is to be calculated in $°C$ for a given 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 9PP

The initial temperature of the gas is $-137.6°C$

Explanation of Solution

Given information:

The initial pressure ${\mathrm{P}}_2=1.12atm$

The final temperature is ${\mathrm{T}}_2=36.5°C$

The final pressure is $P_2=2.56atm$

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

${\mathrm{T}}_1=\frac{T_2{P}_1}{P_2}$

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

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

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

Therefore

$\begin{array}{c}{T}_2\left(K\right)=36.5+273\\ =309.5K\end{array}$

On substituting the values in the formula:

$\begin{array}{c}{T}_1=\frac{T_2{P}_1}{P_2}\\ =\frac{309.5\times 1.12}{2.56}\\ =135.4K\end{array}$

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

$\mathrm{T}(°C)=T(K)-273$

Therefore

$\begin{array}{c}{T}_1\left(°C\right)=135.4-273\\ =-137.6°C\end{array}$

Therefore, initial temperature is $-137.6°C$

Conclusion

The temperature of the gas is $-137.6°C$