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

Oct 12, 2016

$\text{The final volume of gas is } {V}_{f} = 5.96 L$

#### Explanation:

${V}_{i} : \text{initial volume of gas ;} {V}_{i} = 8 L$

${T}_{i} : \text{initial temperature of gas ;} {T}_{i} = {210}^{o} C = 273 + 210 = {483}^{o} K$

${T}_{f} : \text{final temperature of gas ;} {T}_{f} = {360}^{o} K$

${V}_{i} / {T}_{i} = {V}_{f} / {T}_{f}$

$\frac{8}{483} = {V}_{f} / 360$

$8 \cdot 360 = 483 \cdot {V}_{f}$

${V}_{f} = \frac{8 \cdot 360}{483}$

${V}_{f} = 5.96 L$

Oct 12, 2016

The new volume will be 6 L.

#### Explanation:

This is an example of Charles' law, which states that the volume of a gas held at constant pressure is directly proportional to its temperature in Kelvins.

The equation to use is ${V}_{1} / {T}_{1} = {V}_{2} / {T}_{2}$

Known Values
${V}_{1} = \text{8 L}$
${T}_{1} = \text{210"^@"C"+"273.15"="483 K}$
${T}_{2} = \text{360 K}$

Unknown
${V}_{2}$

Solution
Rearrange the equation to isolate ${V}_{2}$. Substitute the known values into the equation and solve.

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

V_2=(8"L"xx360cancel"K")/(483cancel"K")="6 L" rounded to one significant figure