# Question #f34ab

Aug 8, 2015

The average kinetic energy of the gas molecules will decrease, assuming constant pressure and number of moles.

#### Explanation:

For a gas, the average kinetic energy of its molecules is determined by the temperature at which it is kept.

In your case, the extent to which the change in volume will impact the gas' temperature depends on the other parameters of the gas, number of moles and pressure.

When these two parameters are constant, you can use the ideal gas law equation to write two equations that describe your gas at a volume ${V}_{1}$ and at a volume ${V}_{2} < {V}_{1}$.

$\left\{\begin{matrix}P \cdot {V}_{1} = n \cdot R \cdot {T}_{1} \\ P \cdot {V}_{2} = n \cdot R \cdot {T}_{2}\end{matrix}\right.$

Since the average kinetic energy of the gas molecules increases when temperature increases and decreases when temperature decreases, you are actually interested in comparing ${T}_{2}$ and ${T}_{1}$.

If you divide these equations, you'll get

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{P}}} \cdot {V}_{1}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{P}}} \cdot {V}_{2}} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{n \cdot R}}} \cdot {T}_{1}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{n \cdot R}}} \cdot {T}_{2}}$

Rearrange to get

${T}_{2} = {V}_{2} / {V}_{1} \cdot {T}_{1}$

If ${V}_{2} < {V}_{1}$, then ${V}_{2} / {V}_{1} < 0$ and ${T}_{2} < {T}_{1}$, which means that the average kinetic energy of the molecules has decreased.

So, when pressure and number of moles are kept constant, a gas' volume and its temperature have a direct relationship - this is known as Charles' Law.

When volume increases, temperature decreases, and when volume increases, temperature increases as well.