A container with a volume of 48 L contains a gas with a temperature of 210^o K. If the temperature of the gas changes to 80 ^o K without any change in pressure, what must the container's new volume be?

Sep 15, 2016

Use Charles' Law to get a new volume of $18.3 L$

Explanation:

Charles' Law for an ideal gas is that for a given mass and constant pressure, $V \propto T$.

To solve problems, this can be written as

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

In this problem ${V}_{1} = 48 L$, ${T}_{1} = 210 K$, ${T}_{2} = 80 K$ and V_2=?

$\frac{48}{210} = {V}_{2} / 80$

Cross multiply (multiply the numerator of one side by the denominator of the other side)

$210 \cdot {V}_{2} = 48 \cdot 80$

$210 {V}_{2} = 3840$

Divide both sides by 210
$\frac{210 {V}_{2}}{210} = \frac{3840}{210}$

${V}_{2} = 18.3 L$