A container with a volume of 8 L contains a gas with a temperature of 120^o K. If the temperature of the gas changes to 360 ^o K without any change in pressure, what must the container's new volume be?

Jun 28, 2016

The new volume is 24 L.

Explanation:

This is an isobar transformation (with constant pressure).
In this transformation when the temperature increases, the volume increases. This means that the ratio between volume and temperature is constant.

This law is called Charle's law and mathematically it is expressed as

${V}_{i} / {T}_{i} = {V}_{f} / {T}_{f}$.

The ratio between the initial volume and initial temperature is equal to the ratio between the final volume and final temperature.

The initial volume is $8$ L, the initial temperature is $120$ K. The final volume is unknown, the final temperature is $360$ K.

$\left(8 \text{ L")/(120" K")=V_f/(360" K}\right)$

${V}_{f} = \left(8 \text{ L"*360" K")/(120" K}\right)$

${V}_{f} = 24 \text{ L}$.