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 310 ^o K without any change in pressure, what must the container's new volume be?

Feb 18, 2016

use the relation ${V}_{1} / {T}_{1} = {V}_{2} / {T}_{2}$ of Charle's law
Here ${V}_{1} =$8L
${T}_{1} =$270+273 =543K
${T}_{2} =$310K
${V}_{2} =$?
${V}_{2} = 8 \cdot \frac{310}{543}$=4.57 L