Question

Assuming the expression for the pressure exerted by an ideal gas, prove that (1) the kinetic energy per mole of the gas = \(\frac32\) RT (2) the rms speed of a gas molecule, u_rms = \(\sqrt{3RT/M_0}\).

OR 

Assuming the expression for the pressure exerted by an ideal gas, show that the rms speed of a gas molecule is directly proportional to the square root of its absolute temperature.

OR 

Show that the rms velocity of gas molecules is directly proportional to the square root of its absolute temperature.

Answer 1

According to the kinetic theory of gases, the pressure P exerted by a gas is

ρ = \(\frac13\)ρv²_rms = \(\frac13\)\(\frac MV\)v²_rms

∴ PV =  \(\frac13\)mv²_rms 

where v_rms is the rms speed (root-meansquare speed) of the gas molecules; M, V and ρ are the mass,

volume and density of the gas, respectively. If there are n moles of the gas and M₀ is the molar mass,

M = nM₀ , so that PV = \(\frac13\)nM₀v²_rms.........(1)

The equation of state of an ideal gas is 

PV = nRT … (2)

where T is the absolute temperature of the gas and R is the molar gas constant. 

From Eqs. (1) and (2), we get,

where the term on the left-hand side is the kinetic energy of one mole of the gas.

∴ Kinetic energy per mole of the gas = \(\frac32\)RT.......(5)

From Eq. (3),

In Eq. (6), R and M0 are constant so that v_rms ∝ √T. Thus, the rms speed of a gas molecule is directly proportional to the square root of the absolute temperature of the gas.