Answer to Question #160595 in Molecular Physics | Thermodynamics for Bawe

Question

Assuming the density of nitrogen at s.t.p to be 1.251 kgm^-3, find the root mean square velocity of nitrogen molecules at 127 oC.

Answer 1

$T = 127 + 273 = 400 \; K \\ M(N_2) = 28\; g/mol = 28 \times 10^{-3} \;kg/mol \\ V_{rms} = \sqrt{\frac{3RT}{M}} \\ = \sqrt{\frac{3 \times 8.314 \times 400}{28 \times 10^{-3}}} \\ = \sqrt{356314.28} \\ = 596.9 \;m/s$

Answer: 596.9 m/s