One litre glass bulb contains 2 xx10^(21) molecules of nitrogen at a pessure of 7.57 xx 10^(21) Newton m^(-2). Find the RMS velocity and temperature of the gas. If the ratio of most probable velocity and RMS velocity is 0.82, find the most probable velocity of nitrogen gas at the same temperature.

One litre glass bulb contains 2 xx10^(21) molecules of nitrogen at a p

1.	One litre glass bulb contains 2 xx10^(21) molecules of nitrogen at a pessure of 7.57 xx 10^(21) Newton m^(-2). Find the RMS velocity and temperature of the gas. If the ratio of most probable velocity and RMS velocity is 0.82, find the most probable velocity of nitrogen gas at the same temperature.
Answer» Solution :From the ideal equation PV = nRT `T = (PV)/(nR) =(7.57 xx 10^3 xx 1 xx 6.-23 xx 10^(23))/(1.013 xx 10^5 xx 2 xx 10^(21) xx 0.0821) = 274.6K` `T = (PV)/(Nr)= (7.57 xx 10^3 xx 1 xx 6.023 xx 10^23)/(1.013 xx 10^5 xx 2 xx 10^(21) xx 0.0821) = 274.6 K` R.M.S . Velocity (C )= `sqrt((3RT)/(M)) = sqrt((3 xx 8.314 xx 274.6)/(28 xx 10^(-3))) = 495.5 ms^(-1)` Most probable velocity `(C_P) = 0.82 xx ` RMS velocity `= 0.82 xx 494.5 = 405 ms^(-1)`.