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Question

# The density of vapour of a substance at 1 at pressure and 500K is 0.36Kg m−3. The vapour effuses through a small hole at a rate of 1.33 times faster than oxygen under the same condition.Determine:(i) mol.wt (ii) molar volume (iii) compression factor(z) of the vapour (iv) which forces among the gas molecules are dominating, the attractive or the repulsive.

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Solution

## By Graham's Law of effusion states rate of effusion is inversely proportional to the molecular weight of the gas. rateofeffuusionofoxygenrateofeffusionofvapor=√molecularweightofvapormolecularweightofoxygenhence molecular weight of vapor=32×[34]2=18molar volume is defined as volume occupied by 1 mole of substance.masof1molofvapordensityofvapor0.0180.36=0.05m3PVRT=Z1×500.0821×500=Z=1.218Here Z>1 hence it is a repulsive force.

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