Question

A long rectangular box is filled with $C{l}_2$ (atomic mass=35.45) which is known to contain only $C{l}_{35}$ and $C{l}_{37}$ . If the box can be divided a partition and the two types of chlorine molecule are put into two compartments respectively. Where should are partition be made if the pressure on both sides is to be same?The ratio of pressure to original Pressure is x.
So, the value of x is:

Answer 1

let $n_1$ and $$n_2$bethemoleof$Cl_{35}$and$Cl_{37}$ in mixture then

Average molar mass$=\frac{35\ast n_1+37\ast n_2}{n_1+n_2}$

$35.45=\frac{35\ast n_1+37\ast n_2}{n_1+n_2}$

$\frac{n_1}{n_2}$=3.44

$P{V}_1=n_{1}RT;P{V}_2=n_{2}RT$

$\frac{v_1}{V_2}=\frac{n_1}{n_2}$

At constant P and T,

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

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

So, value of x is 1.