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

Volume of the box = ` 1m^(3)`, Pressure of the gas =` 100N//m^(2)`.
Let `T` be the temperature of the gas
(a) Time between two consecutive collections with one wall = ` (1)/(500)"sec"`
This time should be equal to `(2l)/(V_(rms))` , where l is the side of the cube.
`(2l)/(V_("rms")) = (1)/(500) implies V_("rms") = 1000m//s therefore sqrt((3RT)/(M)) = 1000 implies T ((1000)^(2)M)/(3R) = ((10)^(6)(4xx10^(3)))/(3(25/3)) = 160K`
(b) Average kinetic energy per atom `(3)/(2)kT = (3)/(2) [(1.38xx10^(-23) xx 160)] J = 3.312 xx ^(-21)J`
(c) From `PV` = `nRT = (m)/(M)RT` , Mass of helium gas in the box m=`(PVM)/(RT)`
Substituting the values , `m= ((100)(1) (4xx10_(-3)))/((25)/(3) (160)) = 3.0 xx 10^(-4)`kg