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 total mass of helium gas in the box.

Answer

(i) `600K` (ii) `3.3 xx 10^(-1)J` (iii) `0.3`g

Explanation

Volume of the box = `1m^(3)`
Pressure of the gas = `100N//m^(2)`
Let T be the temperature of the gas.
(i) Time between two consecutive collisions with one wall =`(1)/(500)`s
This time should be equal to `(2l)/(V_(rms))`

where l is the side of the cube,
`implies (2l)/(V_(rms)) = (1)/(500) implies V_(rms) = 1000m//s` (as l=1m)
`implies sqrt(3RT)/(M) = 1000`
` therefore T = ((1000)^(2)M)/(3R) = (10)^(6)(4 xx 10^(-3))/(3(25//3)) = 160K`
(ii) Average kinetic energy energy per atom =`(3)/(2)kT`
=`(3)/(2)(1.38 xx 10^(-23))(160)J = 3.312 xx 10^(-21)J`
(iii) From `PV = nRT = (m)/(M)RT`
We get mass of helium gas in the box, `m = (PVM)/(RT)`
Substituting the values we get `m=((100)(1)(4))/((25//3)(160)) = 0.3g`