# A container with a volume of 4 L contains a gas with a temperature of 770^o K. If the temperature of the gas changes to 420^o K without any change in pressure, what must the container's new volume be?

Apr 3, 2016

${V}_{2} = 2 , 18 L$

#### Explanation:

$\text{given data:}$

${V}_{1} = 4 L$
${T}_{1} = {770}^{o} K$
${T}_{2} = {420}^{o} K$
V_2=?

$\text{solution:}$
${V}_{1} / {T}_{1} = {V}_{2} / {T}_{2}$

$\frac{4}{770} = {V}_{2} / 420$

${V}_{2} = \frac{4 \cdot 420}{770}$

${V}_{2} = 2 , 18 L$