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parag28289

a long rectangular box is filled with Cl₂(atomic wieght 35.45) which is known to contain only 35_cland 37_cl. if the box could be divided by a partition and the two types of Cl₂molecules put in the two compartments respectively. calculate where should be the partition mae if pressure on both side are to be equal. is this pressure same as original pressure?

malay

we first find the relative no. of mole of cl-35 and cl-37
since pressure and temperature are , the the no. of moles will decide the volume for each gas.

answer is
volume for cl-35 is 77.5% of the rectangular box

malay

This is the explanation:
Let no. of moles of cl-35=x
no. of moles of cl-37=y
total mass present in the system=35x+37y
total no. of moles=x+y
effective molar weight of the mixture(i.e. chlorine)=(35x+37y)/(x+y)
but this is given to be 35.45
hence
35.45=(35x+37y)/(x+y)
1.55y=.45x
y/x=.45/1.55
y/x=9/31

now the two gases are separated.
they are at same temperature, and have same pressure.
pV=nRT
V/n=RT/p
hence V/n is same for both gases
let for cl35, V be V1 and for cl37 it be V2
V1/x=V2/y
V1/V2=x/y
V1/V2=31/9
total volume= V1+V2=V=initial volume of the box
hence
V1/V=31/40
V2/V=9/40

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