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Question

1.00 points 2 attempts let Check my work Enter your answer in the provided box. The solubility of N, in blood at 370C and at a partial pressure of 0.50 atm is 5.,6 x 10 moVL. A deep sea diver breathes compressed air with the partial pressure of N2 equal to 4.5 atm. Assume that the total volume of blood in the body is 5.8 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 N is 0.80 atm. O Type here to search

Explanation / Answer

Henry's law

Henry's law constant Kh = Cocnetration / pressure

                                     = 0.00056mol/L / 0.80 atm

                                    = 0.0007 mol / (L.atm)

When pressure = 4.5 atm

Solubility, c = 0.0007 mol / (L.atm) x 4.5 atm

                  =0.00315 mol / L

So in 5.8 L of blood, moles of nitrogen dissoved = 0.00315 x 5.8

                                              = 0.01827 moles

At the surface, solubility = 0.00056 mol/L

So moles of nitrogen dissolved = 5 .8 x 0.00056mol/L

                                                = 0.00325 moles

So moles of nitrogen that will be released = 0.01827 - 0.00325 = 0.01502

Total pressure = 1 atm

temperature = 37 C = 310 K

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

volume of nitrogen , V = nRT / P

                                       = 0.01502 x 0.0821 x 310 / 1

volume of nitrogen  = 0.382 L

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