Question

Prove that room mean square velocity of gas molecule is directly proportional to the square root of its absolute temperature.

Answer 1

Let us consider a particle in the cube of side a, to bounce between the walls along distance a and it has velocity $v_x$ and mass m. It would then have momentum.

$M=m{v}_x$

It takes time dt to go between the two walls

$dt=\frac{a}{v_x}$

When the particle strikes the wall, it transfers momentum to the wall. Thus the force going into the wall is given by Newton's Second law as rate of change of momentum, $F=\frac{M}{dt}=\frac{m{v}_x^2}{a}$

Pressure, $P=\frac{Force}{Area}=\frac{m{v}_x^2}{a^3}=\frac{m{v}_x^2}{V}$
where V is the volume of cube.

Molecules are going in all directions, not just 'x', and the average velocity in all directions is
$\sum v^2=\sum v_x^2+\sum v_y^2+\sum v_z^2$

Where v is the root mean square speed.

Since there is an equal probability for going all ways,
$v_x^2=\frac{v^2}{3}$

Substituting this value in the above equation for n molecules of gas .
PV =$\frac{Force}{Area}=\frac{nm{v}_x^2}{a^3}=\frac{nm{v}^2}{3}$

From equation of state for an ideal gas.

PV=nRT, R is universal gas constant and T is the absolute temperature.

Equating the two

$\frac{nm{v}^2}{3}=nRT$

$v=\sqrt{\frac{3RT}{m}}$

Thus root mean square velocity of gas molecule is directly proportional to the square root of its absolute temperature.