# A container with a volume of 25 L contains a gas with a temperature of 150^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?

Apr 2, 2016

$\frac{160}{3} L$

#### Explanation:

From Ideal Gas Equation; $P V = n R T$

Pressure, number of moles and Gas constant are all fixed.

$V = n R \frac{T}{P}$ or $\frac{V}{T} = n \frac{R}{P}$

As long as pressure and the amount of gas remains unchanged, the ratio of $V$ and $T$ is always the same.

${V}_{1} / {T}_{1} = {V}_{2} / {T}_{2}$
$\frac{25 L}{150 K} = \frac{{V}_{2}}{320 K}$
${V}_{2} = \frac{160}{3} L$