A cubical box of side `1 m` contains helium gas (atomic weight 4) at a pressure of `100 N//m^2`. During an observation time of `1 second`, an atom travelling with the root - mean - square speed parallel to one of the edges of the cube, was found to make `500 hits` with a particular wall, without any collision with other atoms . Take `R = (25)/3 j //mol - K and k = 1.38 xx 10^-23 J//K`.
(a) Evaluate the temperature of the gas.
(b) Evaluate the average kinetic energy per atom.
( c) Evaluate the total mass of helium gas in the box.

Explanation

Given `P = 100 N//m^(2)`, side of box, `l = 1 m, V = 1 m^(3)`
(a) `v_(rms) = sqrt((3 R T)/(M_(0)))` (i)
Number of collisions/s `= 500`
Time between successive collisions `= (1)/(500) s`
`v_(rms) = ("distance")/("time") = (2 l)/((1//500)) = 1000 l = 1000 m//s` (ii)
From (i) and (ii)
`sqrt((3 R T)/(M_(0))) = 1000`
`sqrt((3 xx (25)/(3) xx T)/(4 xx 10^(-3))) = 1000 = 10^(3)`
`(25)/(4)T = 10^(6) xx 10^(-3) = 10^(3)`
`T = 160 K`
(b) Mean `K.E`.per atom `= (3)/(2) k T`
`= (3)/(2) xx 1.38 xx 10^(-23) xx 160 = 5.3 xx 10^(-21) J`
( c) `PV=nRT=(M)/(M_(0)) RT`, M: mass of gas
`PV = (PVM_(0))/(RT) = (100 xx 1 xx 4)/((25)/(3) xx 160) = 0.30 g`