# A container with a volume of 5 L contains a gas with a temperature of 320^o K. If the temperature of the gas changes to 240^o K without any change in pressure, what must the container's new volume be?

Apr 28, 2016

${V}_{2} = 3 , 75 L$

#### Explanation:

$\text{first volume: } {V}_{1} = 5 L$
$\text{first temperature :} {T}_{1} = {320}^{o} K$

$\text{last temperature :} {T}_{2} = {240}^{o} K$
"last volume :"V_2=?

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

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

${V}_{2} = \frac{5 \cdot 240}{320}$

${V}_{2} = 3 , 75 L$