A container with a volume of 6 L contains a gas with a temperature of 250^o K. If the temperature of the gas changes to 320 ^o K without any change in pressure, what must the container's new volume be?

May 10, 2018

The new volume is $= 7.68 L$

Explanation:

Apply Charles' Law

$\text{ Volume (L) "/" temperature (K) "= " constant}$, "(at constant pressure")

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

The initial volume is ${V}_{1} = 6 L$

The initial temperature is ${T}_{1} = 250 K$

The final temperature is ${T}_{2} = 320 K$

The final volume is

${V}_{2} = {T}_{2} / {T}_{1} \cdot {V}_{1} = \frac{320}{250} \cdot 6 = 7.68 L$