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`.

Answer

`(i) 1.96 xx 10^(27)` (ii) `36m//s`

Explanation

(i) Let n = number of collisions per second per unit area , change in momentum = `2`mv
`therefore`Pressure exerted on wall = `overset(*)(n)(2mv) = P_(0)`
`implies (overset(*)(n) xx 2 xx 32 xx 10^(-3))/(6.02 xx 10^(23)) xx sqrt((3 xx 8.3 xx 300)/(0.032))= 10^(5)`
`implies overset(*)(n) = 1.95 xx 10^(27)`
(ii) If vessel is suddenly stpped then `KE` will utilized in Increase in temperature. So `(1)/(2)MV_(0)^(2) = nC_(v)DeltaT`
`implies V_(0) = sqrt((2nC_(v)DeltaT)/(M)) = sqrt((2C_(v)DeltaT)/(M_(w)) = 36 ms^(-1)`