Calculate the R M S speed of nitrogen molecules at 27϶C

1.	Calculate the R M S speed of nitrogen molecules at 27϶C
Answer» We have R.M.S μ = $(3RT/M)^{1/2}$ in this M = MOLAR mass And in the case of Nitrogen GAS, T = 27 + 273 = 300 K.R = 8.314 (kg $m^{2}$ / $s^{2}$ )/ K mol Since molar mass of Nitrogen gas = 28 gm/mol = 2.8 x $10^{-2}$ kg/mol Hence μ = $(3RT/M)^{1/2}$ = [ 3 x 8.314 x 300 / 2.8 x $10^{-2}$ ] $]^{1/2}$ = [267235.7 $]^{1/2}$ = 516.9m/s So the r.m SPEED for oxygen gas at 0°C (273K) = 516.9 m/s

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