Question

The solubility of N2in blood at 37°C and at a partial pressure of 0.80 atm is 5.6 × 10−4mol/L. A deep-sea diver breathes compressed air with the partial pressure of N2equal to 4.0 atm. Assume that the total volume of blood in the body is 5.0 L. Calculate the amount of N2gas released (in liters at 37°C and 1 atm) when the diver returns to the surface of the water, where the partial pressure ofN2is 0.80 atm.

Answer 1

Solution :

According to Henry's law of solubility, we have c = kp

Henry's law constant, $k=\frac{c}{p}$

                                      $=\frac{0.00056}{0.80}$

                                      = 0.0007 mol/L.atm

When the pressure is = 4 atm

The solubility is c = 0.0007 mol/(L.atm) x 4 atm

                             = 0.0028 mol/L

Therefore, in a 5 liter of blood, the moles of nitrogen dissolved

= 0.0028 x 5

= 0.014 moles

At the surface, the solubility is = 0.00056 mol/L

So the moles of the nitrogen dissolved = 5 x 0.00056

                                                                  = 0.0028 moles

Therefore, the number of moles of nitrogen released = 0.014 - 0.0028

                                                                                         = 0.0112 moles

Given total pressure = 1 atm                                                                        

Temperature = 37 degree C

                     = 37 + 273

                      = 310 K

R = $0.0821\ L -atm/ mol.K$

Therefore, volume of the nitrogen is

$V=\frac{nRT}{P}$

$V=\frac{0.0112 \times 0.0821 \times 310}{1}$

  = 0.285 L