2g of gas A introduced in a evacuated flask at flask at 25 degree celcius. The pressure of the gas is 1atm. now 3g of another gas B is introduced in the same flasks total pressure becomes 1.5atm. The ratio of molecular mass of A and B is?

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Partial pressure of A  P
A
= 1 atm
Partial pressure of B  P
B
= (1.5 – 1) atm = 0.5 atm
Then,
P
A
= P
total
A
…(1)
and P
B
= P
total
B
…(2)
where, A
is mole fraction of A
and B
is mole fraction of B
Dividing (2) by (1), we get
B
B B A B B
A A A A
A B
n
P n n n
P n n
n n
 
  


n
B
=
B
B
W
M
; n
A
=
A
A
W
M
[where, W
A
, W
B
= Weight of A, B and M
A
, M
B
= Molecular mass ]

B
B B B A B
A A A B A
A
W
n M W M P
n W W M P
M
   
Putting the values, we get
A B A
B A B
M P W 0.5 atm 2 g 1
M P W 1atm 3g 3

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Easy method

Physical significance of ?a?:??a? is a measure of the magnitude of intermolecular attractive forces between the particles

Physical significance of ?b?:

?b? is the volume excluded by a mole of particles.

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Exercise 1.3 Solution

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