Question

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 4.5 atm. Assume that the total volume of blood in the body is 5.2 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

Apply Henry law

PN2 = 0.8 atm

M = 5.6*!0^-4 M

then, solve for H

H = M/P = (5.6*10^-4)/(0.8) = 0.0007 M/atm

then

if P increases to = 4.5 atm

find the new concnetration

H = M/P

M = H*P = 0.0007*4.5 = 0.00315 mol per liter

assume V = 5.2 L so

mol = M*V = 0.00315*5.2 = 0.01638 mol of N2

so the "excess" when going back

M original = ( 5.6*10^-4)(5.2) = 0.002912 mol of N2

The net change = 0.00315 - 0.002912 = 0.000238 mol of N2

then

at such conditions

PV = nRT

V = nRT/P

V = (0.000238)(0.082)(37+273)/(1) = 0.00604996 L

V = 6 ml of N2

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