Question

At ordinary body temperature (37 C) the solubility of N2 in water in contact with air at ordinary atmospheric pressure (1.0 atm) is 0.015 g/l. Air is approximately 78 mol % N2.

a)Calculate the number of moles of N_2 dissolved per liter of blood, which is essentially an aqueous solution.
b)t a depth of 100ft in water, the pressure is 4.0 atm. What is the solubility of N_2 from air in blood at this pressure?
c)If a scuba diver suddenly surfaces from this depth, how many milliliters of N_2 gas, in the form of tiny bubbles, are released into the bloodstream from each liter of blood?

Answer 1

a. 0.015g/L x (1 molN2/28 gN2) = about 0.5.36E-4 M
b. calculate k. k = p/c
XN2 = 0.78 atm from the problem.
XN2 = pN2/Ptotal
0.78 x 1 atm = pN2 = 0.78 atm
k = p/c = 0.78/5.36E-4 = about 1450 or so. You need to do it more accurately.

At 100 ft the Ptotal = 4.0 atm.
Then pN2 = XN2 x Ptotal = 0.78*4.0 atm = 3.12 atm.
c = pN2/k = 3.12/1450 = about 2E-3 M.

c. Take the difference in moles in a liter at the two parts of the problem.
5.36E-4 - 2E-3 = about 0.0016 or so moles N2 for each liter.
The problem asks for mL, I would use PV = nRT, plug in n, R, T(37 + 273), and P (1 atm) and solve for volume, then convert to mL. I came out with about 41 mL N2 per liter of blood.

Answer 2

Thank you!!!!

Answer 3

But isn't the K (Henry's constant) suppose to be mol/(L x atm)???