Chapter 13, Problem 59A

Interpretation Introduction

Interpretation:

The pressure of the gas is to be calculated for a given new temperature when the volume and amount of gas remain 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 59A

The new pressure of the gas is $\text{1}\text{.16}\text{atm}$

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 temperature is changed to ${\mathrm{T}}_2$ then pressure is changed to ${\mathrm{P}}_2$. The formula used to calculate ${\mathrm{P}}_2$ is

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

Or

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

Given data is:

${\text{P}}_{\text{1}}\text{=1}\text{.11}\text{atm}$

${\text{T}}_{\text{1}}\text{=15}\text{°C}$

${\text{T}}_2\text{=30}\text{°C}$

$\begin{array}{l}{T}_{}(K)=T_{}(°C)+273\\ T_1(K)=15+273\\ =288K\\ T_2(K)=30+273\\ =303K\end{array}$

On substituting the values in the formula:

$\begin{array}{c}{P}_2=\frac{303\times 1.11}{\begin{array}{l}288\end{array}}\\ \text{=}\text{1}\text{.16}\text{atm}\end{array}$

Therefore new pressure is $\text{1}\text{.16}\text{atm}$

Conclusion

The pressure of the gas is $\text{1}\text{.16}\text{atm}$.