Pressure of nitrogen gas fall from 4000 mm to 2000mm in 30 minutes when allowed to effuse through a pinhole in the cylinder. If the same cylinder is filled with an equimolar mixture of N2 and H2 gas at 4000 mm of Hg what would be the molar ratio of gases N2 upon He in the cylinder after 1 hour ?
Answers
Answer:
The change of pressure of oxygen in 47 min is 500 mm Hg. The change of pressure of oxygen after 74 min is
47
500
×74=787.2mm
In the 1:1 molar ratio mixture of oxygen and another gas, each of them has an equal pressure of 2000 mm because the total pressure is given to be 4000 mm Hg.
The pressure of oxygen left after 74 min is
2000−787.2=1212.8mmHg
r
O
2
r
gas
=
M
gas
M
O
2
(Graham's law of diffusion)
V
O
2
(diffused)
V
gas
(diffused)
×
t
gas
t
O
2
=
M
gas
M
O
2
P
O
2
(diffused)
P
gas
(diffused)
×
t
gas
t
O
2
=
M
gas
M
O
2
Both diffuse for the same time, so
P
O
2
(diffused)
P
gas
(diffused)
=
M
gas
M
O
2
or
787.2
P
gas
(diffused)
=
79
32
P
gas
(diffused)=500.8mm
The pressure of gas left after 74 min is
2000−500.8=1499.2mmHg
Molar ratio of the gas and oxygen left=
P
(O
2
)
left
P
(gas)
left
=
M
O
2
M
gas
=
1212.8
1499.2
=1.236
Hence, the correct option is B
M
O
2
M
gas
=1.236