Find RMS speed of N2at 27c
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We have, r.m.s, μ = (3RT/M)1/2 where, M = molar mass
In case of Nitrogen gas , T = 27 + 273 = 300 K. R = 8.314 (kg m2 /s2 )/ K mol
Since molar mass of Nitrogen gas = 28 gm/mol = 2.8 x10-2 Kg/mol
Hence, μ = (3RT/M)1/2 = [3 x8.314 x300 / 2.8 x10-2 ] 1/2 = [267235.7]1/2 =516.9 m/s
Hence, the r.m. speed for nitrogen gas at O0C (273K) =516.9 m/s
In case of Nitrogen gas , T = 27 + 273 = 300 K. R = 8.314 (kg m2 /s2 )/ K mol
Since molar mass of Nitrogen gas = 28 gm/mol = 2.8 x10-2 Kg/mol
Hence, μ = (3RT/M)1/2 = [3 x8.314 x300 / 2.8 x10-2 ] 1/2 = [267235.7]1/2 =516.9 m/s
Hence, the r.m. speed for nitrogen gas at O0C (273K) =516.9 m/s
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