Question

A scuba diver is at a depth of 445 m, where the pressure is 45.5 atm.

What should be the mole fraction of O₂ in the gas mixture the diver breathes in order to have the same partial pressure of oxygen in his lungs as he would at sea level? Note that the mole fraction of oxygen at sea level is 0.209.

Answer 1

Let mole fraction of O2 = x at 445 m

At 445 meters;

P(o2) = x * P(total) = x * 45.5 ....(1)

At sea level;

Total pressure = 1 atm = atmospheric pressure

Mole fraction at sea level = 0.209

P(o2) = 0.209 * 1 = 0.209 ....(2)

From equation (1) and (2)

x * 45.5 = 0.209

x = 0.209 / 45.5 = 0.004593 OR 0.0046 (Approx) .....Answer