Question

The solubility of N₂ in blood at 37°C and at a partial pressure of 0.80 atm is5.6 × 10⁻⁴ mol/L. A deep-sea diver breathes compressed air with the partial pressure of N₂ equal to 5.0 atm. Assume that the total volume of blood in the body is 5.2 L. Calculate the amount of N₂ gas released (in liters at 37°C and 1.00 atm) when the diver returns to the surface of the water, where the partial pressure of N₂ is 0.80 atm

Answer 1

First, use the normal atmospheric conditions to calculate the Henry's Law constant for N2, which relates the pressure of the gas to its solubility:
Solubility = kH P
5.6X10^-4 mol/L = kH (0.8 atm)
kH = 7.0X10^-4 mol/Latm

Now, calculate the solubility at high pressure:
solubility = 7X10^-4 mol/Latm X 5 atm = 3.5X10^-3 mol/L

Now, at the surface, the blood stream has a total of 5.6X10^-4 mol/L X 5.2 L = 2.912X10^-3 mol N2 dissolved in it

Before ascending, the total amount of N2 in the blood is 3.5X10^-3 mol/L X 5.2 L = 1.82X10^-2 mol N2

So, on ascending, 1.82X10^-2 - 2.912X10^-3 mol = 0.015288 mol N2 will be released.

Using the ideal gas law, this will occupy:
PV = n RT
1 atm V = 0.015288 mol (0.0821 Latm/molK) (310 K)
V = 0.389 L of Nitrogen