[hokychik]

"The solubility of N₂ in blood at 37^oC and at a partial pressure of 0.8 atm is 5.6*10^-4 mol/L. A deep-sea diver breathes compressed air with the partial pressure of N₂ equal to 4.0 atm. Assume that the total volume of blood in the body is 4.3 L. Calculate the amount of N₂ gas released (in liters at 37^oC and 1.00 atm) when the diver returns to the surface of the water, where the partial pressure of N₂ is 0.8 atm"

I used Henry's Law (S=kP) using 5.6*10^-4 mol/L as S and 0.8 atm as the P to get k=7*10^-4. I used that k with the 4.0atm as the P to get a S=.0028 mol/L. I then converted both of these S numbers into moles by multiplying by 4.3 L to get 2.408*10^-3 mol (at 0.8 atm) and 1.204*10^-2 mol (at 4.0 atm). I need to get the volume of gas released so I used the gas constant law (PV=nRT rewritten as nRT/P). I used the respective n's, R=8.314, T=310.15 and the respective P values. In the end I got the same volume for both, which obviously subtracting you get 0, which isn't the answer. Would someone please help me out?

[fledarmus]

"The solubility of N₂ in blood at 37^oC and at a partial pressure of 0.8 atm is 5.6*10^-4 mol/L. A deep-sea diver breathes compressed air with the partial pressure of N₂ equal to 4.0 atm. Assume that the total volume of blood in the body is 4.3 L. Calculate the amount of N₂ gas released (in liters at 37^oC and 1.00 atm) when the diver returns to the surface of the water, where the partial pressure of N₂ is 0.8 atm"

I used Henry's Law (S=kP) using 5.6*10^-4 mol/L as S and 0.8 atm as the P to get k=7*10^-4. I used that k with the 4.0atm as the P to get a S=.0028 mol/L. I then converted both of these S numbers into moles by multiplying by 4.3 L to get 2.408*10^-3 mol (at 0.8 atm) and 1.204*10^-2 mol (at 4.0 atm). I need to get the volume of gas released so I used the gas constant law (PV=nRT rewritten as nRT/P). I used the respective n's, R=8.314, T=310.15 and the respective P values. In the end I got the same volume for both, which obviously subtracting you get 0, which isn't the answer. Would someone please help me out?

The highlighted parts are your problem - once your diver surfaces he is not at his "respective P" anymore... what pressure should you be using?

[hokychik]

i might be confused there then, because i used 0.8 atm for when he is at the surface because it says the partial pressure of N₂ is 0.8 atm at the surface.

also with that i got the same answer of zero

[Borek]

I used the respective n's

No wonder you got zero.

When the diver starts to go up his blood contains 1.204*10^-2 moles of N₂, of which 2.408*10^-3 moles will be left in his blood on the surface. How many moles of nitrogen have to be released? (that is assuming both numbers listed are correct; I have not checked them).

[fledarmus]

Yes, you used 0.8 when you calculated the moles absorbed at the surface. However, you've also calculated the moles absorbed underwater (a much larger number), which when you go to the surface is ALSO at 0.8. R, T, and P for both of your final calculations should be the same - only n and V will be different. You've calculated the n's; what are the V's?