Question

Enter your answer in the provided box. 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 3.8 atm. Assume that the total volume of blood in the body is 4.8 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 N2 is 0.80 atm.

Answer 1

Ans:-Given solubility of N₂   = 5.6 x 10^-4 mol/L

partial pressure P_N2 = 0.80 atm

We Know ,

           = 5.6x10^-4 / 0.80

           =7 x10^-4 mol/L atm

Now let C be the concentration of N₂ in blood at 3.8 atm

We know C_N2 = K_H x P_N2

                      = 7x10^-4 x 3.8

                      = 0.0028mol/L

Now ,given

Total volume of blood V=4.8L

Therefore number of moles of N₂ in blood at 3.8 atm pressure n₁ = C xV

=0.0028x4.8

=0.01344 mol

Again, At 0.8 atm pressure and 4.8 L blood

Number of moles of N₂ ,n₂ = 5.6x10^-4 x 4.8

   = 0.002688mol

Now number of moles of N₂ when the diver returns to the surface = 0.01344-0.002688

= 0.010752mol

Let V be the volume of N₂ released using ideal gas equation

Temperature T= 37⁰C = 310 K

pressure p=1atm

Now

= 0.010752 x 0.082057x 310 / 1

   = 0.273L