Calculate the rms speed of a nitrogen molecule at 25âc
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We have, r.m.s, μ = (3RT/M) where, M = molar mass
In case of Nitrogen gas , T = 27 + 273 = 300 K. R = 8.314 (kg m / s )/ K mol
Since molar mass of Nitrogen gas = 28 gm/mol = 2.8 x10 Kg/mol
Hence, μ = (3RT/M) = [3 x8.314 x300 / 2.8
x10 ] = [267235.7] =516.9 m/s
Hence, the r.m. speed for oxygen gas at O C (273K) = 516.9 m/s
We have, r.m.s, μ = (3RT/M) where, M = molar mass
In case of Nitrogen gas , T = 27 + 273 = 300 K. R = 8.314 (kg m / s )/ K mol
Since molar mass of Nitrogen gas = 28 gm/mol = 2.8 x10 Kg/mol
Hence, μ = (3RT/M) = [3 x8.314 x300 / 2.8
x10 ] = [267235.7] =516.9 m/s
Hence, the r.m. speed for oxygen gas at O C (273K) = 516.9 m/s
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