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# 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

Akash Kumar Dutta
98 Points
8 years ago

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)

siddharth chandil
36 Points
8 years ago

hona toh 4:1 he chye

Abhishekh kumar sharma
34 Points
7 years ago

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)

Prashant
11 Points
4 years ago
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