One litre glass bubl contains2 xx 10^(21) molecules of nitrogen at a pressure of7.57 xx 10^(-3)N.m^(-2). Find out the RMS velocity of nitrogen.

One litre glass bubl contains2 xx 10^(21) molecules of nitrogen at a p

1.	One litre glass bubl contains2 xx 10^(21) molecules of nitrogen at a pressure of7.57 xx 10^(-3)N.m^(-2). Find out the RMS velocity of nitrogen.
Answer» Solution :`C = sqrt((3PV)/(M))` Where V is molar VOLUME Volume OCCUPIED by one mole of `N_(2)` gas ` = (6 xx 10^(23))/(2 xx 10^(21)) = 30 L = 30 xx 10^(-3)m^(3) = 3 xx 10^(-2) m^(3)` `P = 7.57 xx 10^(3) N.m^(-2), M = 28g = 28 xx 10^(-3)` Kg RMS VELOCITY `C = sqrt((3PV)/(M)) = sqrt((3xx 7.57 xx 10^(3) xx 3 xx 10^(-2))/(28 xx 10^(-3))) = 494.5ms^(-1)`