# The volume of a sample of a gas at 273 Kelvin and is 100.0 L. If the volume is decreased to 50.0 L at constant pressure, what will be the new temperature of the gas?

Mar 4, 2016

The new temperature will be $\textcolor{b l u e}{\text{137 K}}$.

#### Explanation:

This is an example of Charles' law, which states that the volume of a given amount of a gas is directly proportional to its Kelvin temperature. The formula to use for this law is ${V}_{1} / {T}_{1} = {V}_{2} / {T}_{2}$.

Given
${V}_{1} = \text{100.0 L}$
${T}_{1} = \text{273 K}$
${V}_{2} = \text{50.0 L}$

Unknown
${T}_{2}$

Solution
Rearrange the formula to isolate ${T}_{2}$. Substitute the given values into the formula and solve.

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

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

${T}_{2} = \left(273 \text{K" xx 50.0cancel"L")/(100.0cancel"L}\right)$

${T}_{2} = \text{136.5 K"="137 K}$ rounded to three significant figures

The proportionality between volume and temperature can be easily seen in this problem. As the volume decreased by half, the Kelvin temperature decreased by half.

A very good tutorial on the gas laws can be found at http://chemistry.bd.psu.edu/jircitano/gases.html.