91Ó°ÊÓ

We need to find the final temperature of the gas.

Considering the Gay- Lussac's Law that if a constant volume and amount of gas are maintained, the pressure will be increased. In the temperature-pressure relationship which is known as Gay-Lussac's law, the pressure of a gas decreases, and a decrease in temperature decreases the pressure of the gas till the volume of the gas doesn't change. i.e.

$\frac{P_1}{T_1}=\frac{P_2}{T_2}...\left(1\right)$

Here we have the quantities available, that is

the initial temperature $T_1=22°=\left(22+273\right)K=295K$

initial pressure $P_1=744mmHg$

the final pressure $P_2=766mmHg$

Now, calculating the final temperature $T_2$from equation$\left(1\right)$

$$\begin{gathered} \frac{P_1}{T_1}=\frac{P_2}{T_2} \\ {T}_2=\frac{P_2×T_1}{P_1} \end{gathered}$$

Putting the values:

localid="1652608669621" $$\begin{gathered} T_2=\frac{766mmHg×295K}{744mmHg} \\ {T}_2=\frac{225970K}{744} \\ =304K \end{gathered}$$

The final temperature is$31°C$.$T_2=304K=\left(304-273\right)°C=31°C$