Question

The average molecular speed in a sample of He gas at a certain temperature is 1.37x10 m/s. m/s at the same temperature. The average molecular speed in a sample of Kr gas is The rate of effusion of CH4 gas through a porous barrier is observed to be 3.73 x 10 mol/h Under the same conditions, the rate of effusion of N2 gas would be mol/h

Answer 1

1)

vavg = sqrt (k*T/M)

here : T is temperature in Kelvin

M is molar mass

k is constant

If we take ratio, above expression becomes

vavg(Kr)/vavg(He) = sqrt (M(He)/M(Kr))

we have:

M(Kr) = 83.8 g/mol

M(He) = 4.003 g/mol

vavg(Kr)/vavg(He) = sqrt (4.003/83.8)

vavg(Kr)/vavg(He) = sqrt (0.0478)

vavg(Kr)/vavg(He) = 0.2186

vavg(Kr)/1370.0 = 0.2186

vavg(Kr) = 299 m/s

Answer: 2.99*10^2 m/s

2)

Rate of effusion is inversely proportional to square root of molar mass

rate = k*sqrt(1/M)

M is molar mass

If we take ratio, above expression becomes

rate(N2)/rate(CH4) = sqrt (M(CH4)/M(N2))

we have:

M(N2) = 28.02 g/mol

M(CH4) = 16.042 g/mol

rate(N2)/rate(CH4) = sqrt (16.042/28.02)

rate(N2)/rate(CH4) = sqrt (0.5725)

rate(N2)/rate(CH4) = 0.7567

rate(N2)/3.73*10^-4 mol/h = 0.7567

rate(N2) = 2.82*10^-4 mol/h

Answer: 2.82*10^-4 mol/h