If the temperature of a fixed amount of a gas is doubled at constant volume, what happens to the pressure?

A substance's volume is the amount of space it takes up, whereas its mass is the amount of stuff it contains.

Let us solve the given problem.

Gay-Lussac's law (or the pressure law) explains the relationship between gas pressure and temperature when there is a constant quantity of gas in a closed and rigid container.

According to Gay-Lussac's rule, absolute pressure is exactly proportional to temperature.

\(\begin{aligned}{}{\rm{P}}\alpha {\rm{T}}\\\frac{{\rm{P}}}{{\rm{T}}}{\rm{ = constant}}\\{\rm{ }}\frac{{{{\rm{P}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{1}}}}}{\rm{ = constant ,}}\\\frac{{{{\rm{P}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{2}}}}}{\rm{ = constant}}\end{aligned}\)

Therefore,\(\frac{{{P_1}}}{{{T_1}}} = \frac{{{P_2}}}{{{T_2}}}\).

It is given that temperature is doubled at constant volume and amount of gas. Hence

Initial Pressure \(\left( {{P_1}} \right) = {\rm{P}}\)\(atm.\) Final Pressure \(\left( {{P_2}} \right) = ?\;{\rm{atm}}.\)

Initial Temperature \(\left( {{T_1}} \right) = {\rm{T}}K.\) Final Temperature \(\left( {{T_2}} \right) = 2\;{\rm{T}}K\).

\(\begin{aligned}{}\frac{{{P_1}}}{{{T_1}}} = \frac{{{P_2}}}{{{T_2}}}\\\frac{{\rm{P}}}{{\rm{T}}} = \frac{{{P_2}}}{{2\;{\rm{T}}}}\\{P_2} = \frac{{({\rm{P}}) \cdot (2\;{\rm{T}})}}{{\rm{T}}}\\ = 2{\rm{P}}\end{aligned}\)

Therefore, at constant volume and amount of gas, if temperature is doubled then pressure will also be doubled.