Question

show that R.M.S. velocity of gas molecules is directly proportional to square root of its absolute temperature

Answer 1

9Let gas is one mole.
So,
P = (1/3) × ρ × v(rms)^2
where v(rms) = root mean square velocity

P = (1/3) × (M/V) × v(rms)^2
P = (1/3) × [(m)(Na)/ V] × v(rms)^2
we know that M = (m)(Na)
where Na = Avogadro's number
V = Volume

Multiply and divide by 2

P = (2/3) × Na × [(1/2) × m × v(rms)^2
Now, for 1 mole PV = RT
So, RT = (2/3) × Na × [(1/2) × m × v(rms)^2
(1/2) × m × v(rms)^2 = (3/2) × (R/Na) × T

(1/2) × m × v(rms)^2 = (3/2) × K × T
Because we know that R/Na = K
where K = Boltzman constant
(1/2) × m × v(rms)^2 ∝ T
v(rms)^2 ∝ T (because (1/2)m = constant)
So, v(rms) ∝ √T
Hence, proved.