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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.Ans) we know by grahams law of effusion thatrate of effusion(r) is propotional to 1/(density)^1/2hence rate(r)=k/(d)^1/2r=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) ] ^1/2r1/r2= (M2/M1)r1/r2=(1.16/4.4)^1/2r1/r2=1compostion of the mixture leaking is equal in propotion.
For a reversible process involving ideal gases in a closed system, show that ΔS = C v ln(T2/T1) for a constant volume process ΔS = Cp ln(T2/T1) for a constant pressure proc
The set of quantum numbers n= 3, l=0, m=0, s= -1/2 belongs to the element: (1) Mg (2) Na (3) Ne (4) F Ans: Mg
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