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

Amount of the gas `= (2.0 xx 10^(21))/(6.023 xx 10^(23))` mole
Given, V = 1 litre `= 10^(-3) m^(3), P = 7.57 xx 10^(3) Nm^(-2), R = 8.314 J K^(-1) mol^(-1)`
Applying `PV = nRT " or " T = (PV)/(nR)`
`T = (10^(-3) xx 7.57 xx 10^(3))/((2.0 xx 10^(21))/(6.023 xx 10^(23) ) xx8.314) = 274.2 K`
rms speed `= sqrt((3RT)/(M)) = sqrt((3 xx 8.314 xx 274.2)/(28 xx 10^(-3))) = 494.22 ms^(-1)`
Most probable speed `= 0.8 xx rms`
`= 0.8 xx 494.22 ms^(-1) = 395.376 sm^(-1)`