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

Question

2117 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-09-02T04:53:29

The volume should double......why? Well, old $Charles Law$ $"Charles Law"$ says......

........ $V \propto T$ $VpropT$ , at constant pressure...and $T$ $T$ is quoted in $absolute temperature.....$ $"absolute temperature....."$

And thus $V = k T$ $V=kT$ , $k$ $k$ is a so-called $constant of proportionality$ $"constant of proportionality"$ that we must ourselves define......

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

And thus $\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$ $V_1/T_1=V_2/T_2$ , pressure, and quantity of gas constant.

And so we solve for $V_{2} = \frac{V_{1} \times T_{2}}{T_{1}}$ $V_2=(V_1xxT_2)/T_1$ , and clearly this has units of volume as required.........

And finally we solve the problem, given that $T_{2} = 2 T_{1}$ $T_2=2T_1$

$V_2=(V_1xx2cancelT_1)/cancelT_1=2V_1$ . Capisce?