# At constant pressure, how should the volume of a gas enclosed in a piston evolve when the absolute temperature is DOUBLED?

Sep 2, 2017

The volume should double......why? Well, old $\text{Charles Law}$ says......

#### Explanation:

........$V \propto T$, at constant pressure...and $T$ is quoted in $\text{absolute temperature.....}$

And thus $V = k T$, $k$ is a so-called $\text{constant of proportionality}$ that we must ourselves define......

And so $\frac{V}{T} = k$, but this should hold under ALL scenarios where pressure and amount of gas are constant.....

And thus ${V}_{1} / {T}_{1} = {V}_{2} / {T}_{2}$, pressure, and quantity of gas constant.

And so we solve for ${V}_{2} = \frac{{V}_{1} \times {T}_{2}}{T} _ 1$, and clearly this has units of volume as required.........

And finally we solve the problem, given that ${T}_{2} = 2 {T}_{1}$

${V}_{2} = \frac{{V}_{1} \times 2 {\cancel{T}}_{1}}{\cancel{T}} _ 1 = 2 {V}_{1}$. Capisce?