Suppose that the air in the atmosphere has 2.79 mol N2, 0.656 mol O2, and 0.0225 mol Ar. What is the partial pressure of each gas when the barometric pressure is 754.1 mm Hg?
Answers
Answer:
Since inspired air is 21% oxygen and atmospheric pressure is 760 mmHg (at sea level), the partial pressure of oxygen is 0.21 x 760 mmHg = 160 mmHg. As air moves into the alveoli, water vapor and carbon dioxide are added, and that reduces the partial pressure of oxygen to about 100 mmHg in the alveolar gas.
The vapour pressure of air over water is 10 atm.
The partial pressure of nitrogen =P
N
2
=
100
79×10
=7.9atm=7.9×760mmHg=6004 mm Hg
The partial pressure of oxygen P
O
2
=
100
20×10
=2.0atm=2.0×760mmHg=1520 mm Hg
According to Henry's law,
P
N
2
=K
H
(N
2
)×X
N
2
X
N
2
=
K
H
(N
2
)
P
N
2
=
6.51×10
7
6004
=9.22×10
−5
X
O
2
=
K
H
O
2
P
O
2
=
3.30×10
7
1520
=4.6×10
−5
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