# At 95.0^@"C", a sample of gas has a volume of "1.55 mL". What will the volume be if the gas is cooled to 0.2^@"C"?

Dec 17, 2016

The final volume will be $\text{1.15 mL}$.

#### Explanation:

Your question involves Charles' law, which states that the volume of a gas is directly proportional to its Kelvin temperature, with pressure and mass remaining constant. This means that if the volume increases, the temperature will also increase, and vice versa.

Equation

${V}_{1} / {T}_{1} = {V}_{2} / {T}_{2}$, where $V$ is volume, and $T$ is temperature in Kelvins.

Known
${V}_{1} = \text{1.55 mL}$
${T}_{1} = \text{95.0"^@"C"+273.15=368.2 "K}$
${T}_{2} = \text{0.2"^@"C"+273.15=273.4 "K}$

Unknown
${V}_{2}$

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

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

V_2=(1.55"mL"xx273.4"K")/(368.2"K")="1.15 mL"