Question

The solubility (i.e. the maximum molarity at equilibrium) of N₂ gas in blood at 37^oC under a nitrogen partial pressure of 0.80 atm is 5.6 x 10^-4 mol.L^-1. When swimming at large depth in the ocean, a diver breathes compressed air with a partial pressure of N₂ equal to 4.0 atm. Henry’s law indicates that the solubility of N₂ gas increases with pressure. Assume that the total volume of blood in the body is 5.0 L. Calculate the amount of N₂ gas in the diver’s blood before diving. Calculate the amount of N₂ gas in the diver’s blood as he is breathing compressed air with a partial pressure of N₂ equal to 4.0 atm. Calculate the amount of nitrogen gas (in liters) that must be released when the diver returns to the surface of water, where the partial pressure of N₂ is 0.80 atm.

Answer 1

Solution :-

Lets first calculate the constant of the N2 gas using the given pressure and solubility

kH = C/p

kH= 5.6*10^-4 mol / L / 0.80 atm

KH = 7.0*10^-4 mol per L atm

Now using this constant lets calculate the concentration of the N2 at 4.0 atm pressure

C= KH*P

= 7.0*10^-4 mol per L atm * 4.0 atm

   = 2.8*10^3 mol per L

So the total amount of the gas dissolved in the divers blood before diving = 5.6*10^-4 mol per L * 5.0 L = 0.0028 mol

Now lets calculate the amount of the N2 dissolved after diving

Moles of N2 dissolved after diving = 2.8*10^-3 mol per L * 5.0 L = 0.014 mol N2

Now lets find the difference when the diver come back to the surface

Moles of N2 that be released = 0.014 mol – 0.0028 mol = 0.0112 mol N2

Now lets calculate the volume of the N2 using the ideal gas law

PV= nRT

V= nRT/P

T= 37 C + 273 = 310 K

P = 0.80 atm

V= 0.0112 mol * 0.08206 L atm per mol K * 310 K / 0.80 atm

V= 0.356 L

So the volume of the N2 released = 0.356 L