5. A deep-sea diver descends to a depth of about 40 m, while breathing an air mi

Question

5. A deep-sea diver descends to a depth of about 40 m, while breathing an air mixture (nitrogen and oxygen). After spending some time at this depth, the diver is pulled to the surface very rapidly (to avoid a shark). As a result, the partial pressure of nitrogen experienced by the diver decreases from 4.74 atm to 0.79 atm (a) What is the concentration of nitrogen in the blood plasma of the diver at these two pressures? The Henry's Law constant for nitrogen in blood plasma is k 5.0 x 10-4 mol L-1 atm-1. Use the following form of Henry's Law; CA-kPa, where CA and pA are the solution concentration and partial pressure, respectively, of gas A What volume of nitrogen bubbles will form in the diver's bloodstream during the ascent? Assume that the volume of blood in the body is 6.0 L and that the volume of one mole of N2 gas at surface pressure and body temperature is 25.51 (b)

Explanation / Answer

given depth, d = 40 m

as a result of ascending rapidly partial pressure of nitrogen changes form pi = 4.7 atm to Pf = 0.79 atm

total pressure at 40 m depth = (1.01*10^5 + rho*g*d)atm/1.01*10^5 [where rho is density of water and g is acceleration due to gravity]

Pt = (1.01*10^5 + 1000*9.81*40)/1.01*10^5 = 4.885 atm

a. concentration of nitrogen at pressure p , c = kpa

hence

for pf

cf = 5*10^-4*4.7 = 2.35*10^-3 mol/L

for pi

ci = 5*10^-4*0.79 = 3.95*10^-4 mol/L

b. now volume of blood in body, V = 6 L

volume of n moles of N2 at bodyt temp, surface temp , Vns = 25.51n L

now volume of nitrogen in blood at surface, Vs = Vns*ci*V = 25.51*3.95*10^-4*6 = 0.0604587 L

now from ideal gas law, let surface pressure be Ps and pressure at depth 40 m be P

PiVns = P*Vn

Vn =PiVns/P = 1*25.51/4.885 = 5.222 L per mol

hence

volume of nitroge in blood at depth d, V' = Vn*cf*V = 5.222*2.35*10^-3*6 = 0.07363 L

so volume of bubbles released = V' - Vs = 0.013 L

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