A gas bulb of `1 L` capacity contains `2.0xx10^(11)` molecules of nitrogen exerting a pressure of `7.57xx10^(3)Nm^(-2)`. Calculate the root mean square (rms) speed and the temperature of the gas molecules. If the ratio of the most probable speed to the root mean square is `0.82`, calculate the most probable speed for these molecules at this temperature.

Explanation

Given `P =7.57 xx 10^(3) Nm^(-2), V = 1 litre = 10^(-3) m^(3)`
`R = 8.314 J, n = (2 xx 10^(21))/(6.023 xx 10^(23)` mole
Using `PV =nRT`
`7.57 xx 10^(3) xx 10^(-3) = (2 xx 10^(21))/(6.023 xx 10^(23)) xx 8.314 xx T`
`T =274.2 K`
`:. u_(rms) = sqrt(((3RT)/(M))) =sqrt(((3 xx 8.314 xx 274.2)/(28 xx 10^(-3)))`
(M in kg)
`= 494.22 m sec^(-1)`
Now `u_(MP)/(u_(rms))=0.82`
`u_(MP) = 405.26 m sec^(-1)` .