# A container with a volume of 8 L contains a gas with a temperature of 210^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?

Jul 10, 2016

$5.13 L = {V}_{2}$

#### Explanation:

To compare the volume and temperature change when pressure is held constant use Charles' Law ${V}_{1} / {T}_{1} = {V}_{2} / {T}_{2}$

${V}_{1} = 8 L$
${T}_{1} = {210}^{o} C$ Convert to Kelvin $210 + 273 = 483 K$
V_2=?
${T}_{2} = 310 K$

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

$\frac{8 L}{483 K} = \frac{{V}_{2}}{310 K}$

$\frac{8 L}{483 \cancel{K}} \cdot \left(310 \cancel{K}\right) = \frac{{V}_{2}}{\cancel{310 K}} \cdot \cancel{310 K}$

$5.13 L = {V}_{2}$