What is the relationship between Charles law and the kinetic theory of gases?

Question

5332 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-06T07:01:04

From the kinetic theory, one derives the pressure equation,

$p = \frac{m n v^{2}}{2}$ $p = (mnv^2)/2$ where $m$ $m$ is mass of a molecule, $n$ $n$ is no. of molecules in unit volume, and $v$ $v$ is rms speed.

Thus, $n = \frac{N}{V}$ $n = N/V$ where, $N$ $N$ is total number of gas molecules.

Therefore, one may write,

$p V = \frac{m N v^{2}}{3}$ $pV = (mNv^2)/3$

Now, $\frac{m v^{2}}{2} = E$ $(mv^2)/2 = E$ where $E$ $E$ is the kinetic energy of a molecule.

Thus, $p V = \frac{2 N E}{3}$ $pV = (2NE)/3$

Now from the kinetic interpretation of temperature,

$E = \frac{3 k T}{2}$ $E = (3kT)/2$ where $k$ $k$ is the Boltzmann constant.

Thus, $p V = N k T$ $pV = NkT$

Now since, $N$ $N$ and $k$ $k$ are constants, then for a fixed $p$ $p$ ,

$\frac{V}{T} =$ $V/T =$ constant

This is Charles' law from kinetic theory.