A diver breathes compressed air with a partial pressure of N2 equal to 4.0 atm. Assume the total volume of blood in the body is 5.0 L. Calculate the amount of N2 gas released (in liters at 37°C and 1 atm) when the diver returns to the surface of the water, where the pressure partial N2 is 0.80 atmosphere?
The solubility of gas is proportional to the partial pressure of the gas.
Therefore, the solubility of nitrogen in blood at a partial pressure of 4 atm is
c(4atm) = (4atm/0.8atm) ∙ c(0.8atm)
= (4atm/0.8atm) ∙ 5.6×10⁻⁴mol/L = 2.8×10⁻³mol/L
Assuming the blood is still saturated with nitrogen, the diver’s blood nitrogen concentration at the surface decreases by:
Δc = 2.8×10⁻³mol/L – 5.6×10⁻⁴mol/L = 2.24×10⁻³mol/L
Thus, the amount of nitrogen released is
n = Δc ∙ V_blood = 2.24×10⁻³mol/L ∙ 5L = 1.12×10⁻²mol
Using the ideal gas law, you can calculate the volume corresponding to 1 atm and 37°C = 310.15K
p∙V = n∙R∙T
=>
V = n∙R∙T/p
= 1.12×10⁻²mol ∙ 0.082 057atmL/molK ∙ 310.15K / 1atm
= 0.285L
Nitrogen solubility in blood