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) Evalite the temperature of the gas.
(b) Evaluate the average kinetic energy per atom.
( c) Evaluate the total mass of helium gas in the box.

Answer

A, B, C, D

Explanation

Volume of the box = `1 m ^3`, pressure of the gas = ` 100 N //m^2`. Let (T) be the temperature of the gas.
(a) Time between two consecutive collisions with one wall = `1/ (500) s`.
This time should be equal to `(2 l)/v_(r m s`, where (l) is the side of the cube.
`(2 l)/v_(r m s) = 1/(500)`
or `v_(r m s) = 1000 m//s`
`:. sqrt((2 R T)/M) = 1000`
or `T = ((1000)^2 M )/(3 R)= ((10^6)(4 xx 10^-3))/(3(25/3)`
= `160 K`
(b) Average kinetic energy per atom = `3/2 kT`
= `3/2 (1 .38 xx 10^-23)(160) J`
= `3.312 xx 10^-21 J`
(c) From `pV = nRT = m/M RT`
we get mass of helium gas in the box,
`m (pVM)/(RT)`
Substituing the values, we get
`m = ((100)(1) (4 xx 10^-3))/((25/3)(160)`
=` 3.0 xx 10^-4 kg`.
(##DCP_V03_C20_S01_026_S01##).