Question

To what temperature must a gas at $300\mathrm{K}$ be cooled in order to reduce its volume to 1/3rd of its original volume, pressure remaining constant?

Answer 1

The initial volume of a gas $\left({\mathrm{V}}_1\right)=\mathrm{V}\left(\mathrm{let}\right)$ and the initial temperature of a gas $\left({\mathrm{T}}_1\right)=300\mathrm{K}$ at constant pressure.

The initial volume of a gas $\left({\mathrm{V}}_2\right)=\frac{\mathrm{V}}{3}$ and the initial temperature of a gas $\left({\mathrm{T}}_2\right)=?$ at constant pressure.

Since the pressure is kept constant.

According to Charles law,

$\frac{{\mathrm{V}}_1}{{\mathrm{T}}_1}=\frac{{\mathrm{V}}_2}{{\mathrm{T}}_2}$

$\Rightarrow \frac{\mathrm{V}}{300}=\frac{{\displaystyle \raisebox{1ex}{$\mathrm{V}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{{\mathrm{T}}_2}$

$\Rightarrow \frac{{\displaystyle \raisebox{1ex}{$\mathrm{V}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\times 300}{\mathrm{V}}={\mathrm{T}}_2$

Hence, the final answer is ${\mathrm{T}}_2=100\mathrm{K}$