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.

Answer

(407 ms)

Explanation

Number of moles `= (2 xx 10^(21))/(6 xx 10^(23)) = 0.33 xx 10^(-2), p = 7.57 xx 10^(3) Nm^(-2)`
Now, p V = n R T `rArr T = (pV)/(nR) = (7.57 xx 10^(3) xx 10^(-3))/(0.33 xx 10^(-2) xx 8.314) = 276 K`
`rArr u_(rm s) = sqrt((3RT)/(M)) = sqrt((3 xx 8.314 xx 276)/(28 xx 10^(-3))) ms^(-1) = 496 ms^(-1)`
Also, `(u_(m pa))/(r_(rm s)) = 0.82 rArr u_(m pa) = 0.82 xx u_(rm s) = 0.82 xx 496 ms^(-1) = 407 ms^(-1)`
`C_(rm s) = 496 ms^(-1), C_(m p) = 407 ms^(-1)`