# The absolute temperature of a gas is increased four times while maintaining a constant volume. What happens to the pressure of the gas?

Jun 3, 2017

It also increases by a factor of $4$.

#### Explanation:

We can use the temperature-pressure relationship of gases, illustrated by Gay-Lussac's law:

$\frac{{P}_{1}}{{T}_{1}} = \frac{{P}_{2}}{{T}_{2}}$

We're not given any specific values for each quantity, but we can use $1$ for the original temperature (${T}_{1}$) and $4$ for the final temperature (${T}_{2}$) to illustrate that it increased by a factor of $4$. To find out how the pressure changes, we can rearrange the equation to solve for the final pressure, ${P}_{2}$, and plug in the two temperature values to find how the pressure changed (in terms of ${P}_{1}$):

${P}_{2} = \frac{{P}_{1} {T}_{2}}{{T}_{1}}$

${P}_{2} = \frac{{P}_{1} \left(4\right)}{1}$

Therefore,

${P}_{2} = 4 \left({P}_{1}\right)$

The pressure increases by a factor of $4$, same as the temperature, which is explained by the kinetic-molecular theory.