Question

The solubility of N2 in blood at 37°C and a partial pressure of 0.80 atm is 5.6 ✕ 10−4 mol·L−1. A deep-sea diver breathes compressed air with a partial pressure of N2 equal to 4.3 atm. Assume that the total volume of blood in this diver's body is 6.0 L. Calculate the amount of N2 gas released (in liters) when the diver returns to the surface of water, where the partial pressure of N2 is 0.80 atm. (2 sig fig)

______ L **the answer IS NOT 0.37 or 0.38 my online program homework marks those as wrong**

Answer 1

According to Henry's law, solubility, C = K.P

Henry's law constant, k = C / p
                                    = 5.6x10^-4 / 0.80

                                    = 7.0 x 10^-4 mol / L.atm

When pressure = 4.3 atm

Solubility, C = 7.0 x 10^-4 x 4.3

                  = 3.01 x 10^-3 mol / L

So in 6.0 L of blood,

moles of nitrogen dissoved = 3.01 x 10^-3 x 6

                                            = 0.01806 mol                           

At the surface, solubility = 5.6x10^-4 mol/L

So moles of nitrogen dissolved = 6 x 5.6x10^-4

                                                = 3.36 x 10^-3 moles

So moles of nitrogen that will be released = 0.01806 - 0.00336

                                                                  = 0.0147 mol

Total pressure = 1 atm

temperature = 37 C = 310 K

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

So volume of nitrogen , V = nRT / P

                                       = 0.0147 x 0.0821 x 310 K / 1 atm

volume of nitrogen   = 0.37 L

this is the way of doing this problems . the answer is also correct. check it once , why it says wrong