If `10^(22)` gas molecules each of mass `10^(-26)kg` collide with a surface (perpendicular to it) elastically per second over an area `1m^(2)` with a speed `10^(2)m//s`, the pressure exerted by the gas molecules will be of the order of :

Answer

`2.347 xx 10^(4) Nm^(-2)`

Explanation

`P = F//A = 2 Nmv cos theta//A = 2 xx 10^(24) xx 3.32 xx 10^(-27) xx 10^(3) cos 45^(@) // 2 xx 10^(-4) = 2347 xx 10^(4) Nm^(-2)`