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Question

A long rectangular box is filled with Cl2 (atomic mass=35.45) which is known to contain only Cl35 and Cl37. If the box can be divided into a partition and the two types of chlorine molecule are put into two compartments respectively. Calculate where should partition to be made if pressure on both sides is to be same? The ratio of pressure to original pressure is 1/x.
So, the value of x is:

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Solution

let n1 and $$n_2bethemoleofCl_{35}andCl_{37}$ in mixture then

Average molar mass=35n1+37n2n1+n2

35.45=35n1+37n2n1+n2

n1n2=3.44

PV1=n1RT;PV2=n2RT

v1V2=n1n2

At constant P and T,

Volume of gas number of mole i.e.the mole ratio yields the volume ratio .

Therefore partition should be made in the volume ratio of 3.44: 1.Also, the pressure at this condition is same as the original condition since the volume of box and number of mole along with temperature are constant.

So, the value of x is 1.


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