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 $T_1=-15°C=\left(-15+273\right)K=258K$

initial pressure $P_1=2.35atm$

the final temperature $T_2=46°C=\left(46+273\right)K=319K$

Now, calculating the final pressure $P_2$from the equation $\left(1\right)$

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

Putting the values:

$$\begin{gathered} P_2=\frac{2.35atm×319K}{258K} \\ {P}_2=\frac{749.65atm}{258} \\ =2.9atm \end{gathered}$$

The final pressure is $2.9atm$.