A container with a volume of 48 L contains a gas with a temperature of 140^o K. If the temperature of the gas changes to 190 ^o K without any change in pressure, what must the container's new volume be?

$65.143 \setminus L$

Explanation:

Assuming the gas to be ideal then using ideal gas equation then by Charles law

At constant pressure, for a closed system

$V \setminus \propto T$

$\setminus \frac{{V}_{1}}{{V}_{2}} = \setminus \frac{{T}_{1}}{{T}_{2}}$

${V}_{2} = {V}_{1} \left({T}_{2} / {T}_{1}\right)$

Given that the initial volume of gas is ${V}_{1} = 48 \setminus L$ at initial temperature ${T}_{1} = 140 \setminus K$. Now the final new volume ${V}_{2}$ at temperature ${T}_{2} = 190 \setminus K$ keeping pressure constant

${V}_{2} = 48 \left(\frac{190}{140}\right)$

$= \frac{456}{7}$

$= 65.143 \setminus L$