# A container with a volume of 8 L contains a gas with a temperature of 270^o C. If the temperature of the gas changes to 360 ^o K without any change in pressure, what must the container's new volume be?

May 11, 2016

${V}_{2} = 5.30 L$

#### Explanation:

${V}_{1} = 8 L$
${T}_{1} = {270}^{o} C = 270 + 273 = {543}^{o} K$

${T}_{2} = {360}^{o} K$
V_2=?

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

$\frac{8}{543} = {V}_{2} / 360$

${V}_{2} = \frac{8 \cdot 360}{543}$

${V}_{2} = 5.30 L$