`1 g` mole of oxygen at `27^@ C` and `1 ` atmosphere pressure is enclosed in a vessel.
(a) Assuming the molecules to be moving with `v_(r m s)`, find the number of collisions per second which the molecules make with one square metre area of the vessel wall.
(b) The vessel is next thermally insulated and moves with a constant speed `v_(0)`. It is then suddenly stoppes. The process results in a rise of temperature of the gas by `1^@ C`. Calculate the speed `v_0.[k = 1.38 xx 10^-23 J//K` and `N_(A) = 6.02 xx 10^23 //mol]`.

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)`