At a depth of 50 m, the pressure is 512 kPa (kilopascals).
A diver breathing air (78% N2, 21% O2) at this depth would have what partial pressure of oxygen in his lungs?
Question 18 options:
108kPa
5.1 kPa
404 kPa
512 kPa
To find the partial pressure of oxygen in the diver's lungs, we need to first calculate the total pressure of the air at 50 m depth.
Using the equation:
Ptot = Pnitrogen + Poxygen
Given:
Ptot = 512 kPa (total pressure at 50 m depth)
Pnitrogen = 78% of Ptot
Poxygen = 21% of Ptot
Firstly, we find Pnitrogen:
Pnitrogen = 78/100 * 512 kPa
Pnitrogen = 399.36 kPa
Next, we find Poxygen:
Poxygen = 21/100 * 512 kPa
Poxygen = 107.52 kPa
Therefore, the partial pressure of oxygen in the diver's lungs at a depth of 50 m is approximately 108 kPa.