Hi. I'm having some trouble solving this problem. Can you check my work ?

The solubility of N

_{2} in blood at 37 ºC and at a partial pressure of 0.8 atm is 5.6 x 10

^{-4} mol/L. A diver breathes compressed air with a partial pressure of N

_{2} of 4 atm. Say the volume of blood in his body is 5 L. Determine the quantity of N

_{2} released in litres when the diver comes back to the surface where the partial pressure of N

_{2} is 0.8 atm.

The solution for this problem is 0.28 dm

^{3}. I got a different result.

First, I calculated Henry's constant for N

_{2} at 37 ºC. I got K = 7 x 10

^{-4} mol dm

^{-3} atm

^{-1}. Then, I determined the solubility of N

_{2} at a partial pressure of 4 atm using the value I got earlier for K. The solubility I got was 2,8 x 10

^{-3} mol/L. After that, I calculated the nº of moles for each solubility. For a solubility of 5.6 x 10

^{-4} mol/L the nº of moles is 2,8 x 10

^{-3} mol and for a solubility of 2,8 x 10 -3 the nº of moles is 0,014 mol. Then I subtracted 0,014 mol by 2,8 x 10

^{-3} mol and got 0,0112 mol. Finally, I applied the equation pV = nRT to determine the value of V using 0,8 atm as the value for pressure. I got V = 0,38 dm

^{3}.