Question #160595
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.
1
Expert's answer
2021-02-09T13:33:30-0500

T=127+273=400  KM(N2)=28  g/mol=28×103  kg/molVrms=3RTM=3×8.314×40028×103=356314.28=596.9  m/sT = 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


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