Question

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

Answer 1

Given:
m=molar mass of Nitrogen=28 gm./mol =2.8 x 10 ⁻² kg/mol
Temperature=T=273 +27 =300K
R=gas constant=8.314 joules/.mol K
To calculate RMS speed :
Formula:
μ=√3RT/M
=√(3*8.314*300)/(2.8x10⁻²)
=√267235.7
=516.9m/s
∴RMS of Nitrigen gas at 27 °c is 516.9m/s

Answer 2

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