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Compute the root-mean-square speed of a nitrogen molecule at 23.6°C. The molar mass of nitrogen molecules...

Compute the root-mean-square speed of a nitrogen molecule at 23.6°C. The molar mass of nitrogen molecules (N2) is 28.0×10-3 kg/mol. At what temperatures will the root-mean-square speed be (b) 1/4 times that value and (c) 4 times that value?

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Answer #2

T = 273 + 23.6 = 296.6 K

M = 28x10^-3 kg/mol

Vrms = = 514 m/s

b)

As V is proportional to square root of absolute temperature, it will be 1/4th if temperature is made 1/16th

T2 = 296.6/16 = 18.53 K

c)

It will be 4 times if temp is made 16 times

T3 = 296.6 x16 = 4745.6 K

answered by: anonymous
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