# A gas at 362 K occupies a volume of 0.67 L. At what temperature will the volume increase to 1.12 L?

Apr 11, 2017

The temperature at which the volume increases to $\text{1.12 L}$ is $6.1 \times {10}^{2}$$\textcolor{w h i t e}{.} \text{K}$.

#### Explanation:

This is an example of Charles' law , which states that the volume of a gas, held at constant pressure and amount, varies directly with the temperature in Kelvins. This means that when the volume increases, the temperature increases and vice-versa.

The equation to use is:

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

where ${V}_{1}$ is the initial volume, ${T}_{1}$ is the initial temperature, ${V}_{2}$ is the final volume, and $\text{T_2}$ is the final temperature.

Given/Known
${T}_{1} = \text{362 K}$
${V}_{1} = \text{0.67 L}$
${V}_{2} = \text{1.12 L}$

Unknown: ${T}_{2}$

Solution
Rearrange the equation to isolate ${T}_{2}$, then substitute the known values into the equation and solve.

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

${T}_{2} = \left(1.12 \text{L"xx362"K")/(0.67"L}\right) = 6.1 \times {10}^{2}$$\textcolor{w h i t e}{.} \text{K}$
($\text{605 K}$ rounded to two significant figures)