The solubility of N2 in blood at 37 degree and at a partial pressure of 0.8 atm is 5.6 x 10-4 mol/L?

Question

a deep sea diver breathes compressed air with the partial pressure of N2 equal to 4 atm. Assume that the total volume of blood in the body is 5 L. Calculate the amount of N2 gas released (in liters at 37 degree and 1 atm) when the diver returns to the surface of water, where the partial pressure of N2 is 0.8 atm

Answer 1

Total soluble N2 in blood at diving depth  

= 5.6 x 10-4 mol/L x 5 L x 4.0 atm / 0.8 atm

= 1.4 x 10-2 mol

Total soluble N2 in blood at surface of water  

= 5.6 x 10-4 mol/L x 5 L

= 2.8 x 10-3 mol

Release N2 when return to surface = (1.4 x 10-2) - (2.8 x 10-3) mol

= 1.12 x 10-2 mol  

= 1.12 x 10-2 mol x 22.4 Liter / mol x 310 deg / 273 deg

= 0.285 Liter