# Question #68e61

Feb 2, 2018

$146.575 K$, or $- {126.4}^{\circ} C$

#### Explanation:

Use Charles' Law:

$V \propto T$, where $V$ is the volume of the gas and $T$ its temperature.

This can become:

$\frac{V}{T} = k$, where $k$ is a constant.

It therefore follows that ${V}_{1} {T}_{2} = {V}_{2} {T}_{1}$.

We first must convert the temperature to Kelvin: ${20}^{\circ} C + 273.15 = 293.15 K$

Input:

$2 {T}_{2} = 1 \cdot 293.15$

${T}_{2} = \frac{293.15}{2}$

${T}_{2} = 146.575 K$, or $- {126.4}^{\circ} C$