One gram mole of oxygen at `27^@` and one atmospheric pressure is enclosed in vessel.
(i) Assuming the molecules to be moving the `V_(rms)`, Find the number of collisions per second which the molecules make with one square metre area of the vessel wall.
(ii) The vessel is next thermally insulated and moved with a constant speed `V_0`. It is then suddenly stopped. The process results in a rise of the temperature of the gas by `1^@C`. Calculate the speed `V_0`.

Explanation

Given `P = 1 atm = 10^(5) N//m^(2), n = 1`
`T = 27^(@)C = 27 +273 = 300 K`
Mass of molecule
`m = (M_(0))/(N_(A)) = (32)/(6.023 xx 10^(23)) = 5.3 xx 10^(-23) g`
`v = v_(rms) = sqrt((3 R T)/(m)) = sqrt((3 xx 1.38 xx 10^(-23) xx 300)/(5.3 xx 10^(-26)))`
`= 484 m//s`

Change in momentum of molecule
`Delta p = 2 m v`
if `n` molecules per second strike the wall
`F = (Deltap)/(Delta t) = n. 2m v`
`P = (F)/(A), A = 1 m^(2)`
`P = F = n'.2 m v`
`n'` :number of molecules`//s//m^(2)`
`10^(5) = n' xx 2 xx 5.3 xx 10^(-26) xx 484`
`n' = 1.95 xx 10^(27)`