`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]`.

Explanation

From gas law, we have
`PV = (m)/(M) RT`
or `PV = (m' N)/(m'N_(A)) RT` (m', mass of one mole, N, total number of molecules)
or `PV = NkT` `(as(R )/(N_(AV)) = k)`
or `P = n_(0) kT`
Here `n_(0) = N//V` the number of molecules per unit volume of gas. Students should note that Eq. (i) relates the gas pressure, temperature and molecular number density. This expression can be used as a standard equation in serveal numerical cases as an alternative form of gas law.
Thus, here molecular number density is given as
`n_(o) = (P)/(kT)`
`= ((1 atm))/((1.38 xx 10^(-23)) xx 300)`
`(10^(5))/(1.38 xx 10^(-23) xx 300) = 2.44 xx 10^(25) m^(-3)`
The rms speed of gas molecules is given as
`v_(rms) = sqrt((3 RT)/(m))`
`= sqrt((3 xx 8.314 xx 300)/(32 xx 10^(-3)) = 483.4 m//s`
Now the number of collision per square metre of vessel wall is given as
`N = (n_(0) v_(rms))/(6) = (2.44 xx 10^(25) xx 483.4)/(6) = 1.97 xx 10^(27) m^(-2)`