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One litre glass bubl contains 2 xx 10^(...

One litre glass bubl contains ` 2 xx 10^(21)` molecules of nitrogen at a pressure of ` 7.57 xx 10^(-3)N.m^(-2)`. Find out the RMS velocity of nitrogen.

Text Solution

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`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)`
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