Question

The solubility of N₂ in blood at 37

°

C and at a partial pressure of 0.80 atm is

5.6

×

10⁻⁴ mol/L.

A deep-sea diver breathes compressed air with the partial pressure of N₂ equal to 5.3 atm. Assume that the total volume of blood in the body is 5.5 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

According to Henry's law, solubility, c = kp

Henry's law constant, k = c / p
                                    = 0.00056mol/L / 0.80 atm

                                    =0.0007 mol / (L.atm)

When pressure = 5.3 atm

Solubility, c = 0.0007 mol / (L.atm) * 5.3 atm

                  =0.00371 mol / L

So in 5.5 L of blood, moles of nitrogen dissoved = 0.00371 * 5.5 =0.0204 moles

At the surface, solubility = 0.00056 mol/L

So moles of nitrogen dissolved = 5 * 0.00056mol/L

                                                =0.00308 moles

So moles of nitrogen that will be released = 0.0204 - 0.00308 =0.01732 moles

Total pressure = 1 atm

temperature = 37 C = 310 K

R = 0.0821 L-atm/ (mol.K)

So volume of nitrogen , V = nRT / P

                                       = 0.01732 * 0.0821 L-atm / (mol. K) * 310 K / 1 atm

                                       = 0.44 L