Question

(a) Compute the root-mean-square speed of a nitrogen molecule at 99.1°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) 1/3 times that value and (c) 2 times that value?

Answer 1

(a) Root-mean-square (r.m.s) speed of a gas = square root of ( 3RT/M)

where R = 8.314 J/K/mole
T = 99.1 + 273 = 372.1 K
M = molecular mas of N2 in kg = 28 X 10^-3 kg

Therefore,

r.m.s. speed = square root of [(3 X 8.314 X 372.1)/28 X 10^-3] = 575.7 m/s

(b) Now, suppose the requisite temperature = T1

condition is -

v1/v = 1/3

from the above expression -

v is proportional to sqrt(T)

So, sqrt(T1/T) = 1/3

=> T1 = T/9 = 372.1/9 = 41.34 K = 41.34 - 273 = -231.7 deg. C

(c) again,

sqrt(T2/T) = 2

=> T2 = 4T = 4*372.1 = 1488.4 K = 1488.4 - 273 = 1215.4 deg. C