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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.
a small 0.600ml bubble forms at the bottom of a lake where the temperature is 4.0 degrees Celsius and the pressure is 2.40 atm. What volume will the bubble occupy near the surface
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