A container with a volume of 5 L contains a gas with a temperature of 360^o K. If the temperature of the gas changes to 640^o K without any change in pressure, what must the container's new volume be?

$= 8.89 L$
${V}_{2} / {V}_{1} = {T}_{2} / {T}_{1}$
${V}_{2} = {V}_{1} \times {T}_{2} / {T}_{1} = 5 \times \frac{640}{360} = 8.89 L$