A gas bulb of 1 litre capacity contains `2.0 xx 10^(21)` molecules of nitrogen exerting a pressure of `7.57 xx 10^3 Nm^(-2)` . Calculate the root mean square speed and the temperature of gas molecules. If the ratio of most probable speed to the root mean square speed is 0.82, calculate the most probable speed for the 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)`