Question

At the top of Mt. Everest, assuming that atmospheric pressure declines from 760 mm Hg to 238 mm Hg but oxygen in air remains 21%. What will be the partial pressure of oxygen at mountain top?​

Answer 1

Answer:

159 mm hg

Explanation:

Answer 2

Answer:

The partial pressure of the oxygen at the mountain top is 49.98 mmHg which is nearly equal to 50 mmHg.

Explanation:

Data given,

The atmospheric pressure after declining from 760 mmHg = 238 mmHg.

The proportion of oxygen in air = 21%

The partial pressure of oxygen, $p_{O_{2} }$ =?

Now,

The oxygen mole fraction in the air, $X_{O_{2} }$  =?

• The mole fraction is defined as the number of constituents/moles divided by the total number of constituents/moles.
• $X_{O_{2} } = \frac{21}{100}$ = 0.21

Now,

The partial pressure of the oxygen can be calculated by using the Raoult's law equation:

• $p_{O_{2} } = p\times X_{O_{2} }$

Here,

• $p_{O_{2} }$ = The oxygen partial pressure
• p = The total pressure of air
• $X_{O_{2} }$ = The oxygen mole fraction

After putting all the values in the equation, we get:

• $p_{O_{2} } = 238\times0.21$
• $p_{O_{2} } = 49.98 mm Hg$ ≈ 50 mmHg

Therefore, the partial pressure of the oxygen = 49.98 mmHg ≈ 50 mmHg.