A mol 4:1 mixture of He and CH4 is contained in vessel at 20 bar pressure.Due to a hole in the vessel the gas mixture leaks out.the composition of the mixture,effusing out initially is about

we know by grahams law of effusion that

rate of effusion(r) is propotional to 1/(density)^1/2

hence rate(r)=k/(d)^1/2

r=k/(M/2)^1/2  =  k.(2/M)^1/2..........................since 2.vapour density(d)=molecular mass(M)

hence ratio of compostion during effusion=r1/r2=[ (2/M1) / (2/M2) ] ^1/2

r1/r2= (M2/M1)

r1/r2=(1.16/4.4)^1/2

r1/r2=1

compostion of the mixture leaking is equal in propotion.

(ANS)

hona toh 4:1 he chye

e know by grahams law of effusion that

rate of effusion(r) is propotional to 1/(density)^1/2

hence rate(r)=k/(d)^1/2

r=k/(M/2)^1/2  =  k.(2/M)^1/2..........................since 2.vapour density(d)=molecular mass(M)

hence ratio of compostion during effusion=r1/r2=[ (2/M1) / (2/M2) ] ^1/2

r1/r2= (M2/M1)

r1/r2=(1.16/4.4)^1/2

r1/r2=1

compostion of the mixture leaking is equal in propotion.

(ANS)

M(CH4)= 16M(He)= 4Let total pressure=P Then, P(he) = (4/5)PP(CH4) = (1/5)PNow...r(He)/r(CH4) = [P(He)/P(CH4)]× √[M(CH4)/M(He)]ANS : r(he)/r(CH4) = 8:1