Question

When scuba diving, it is important to come back to the surface of the water from deep dives slowly. If this does not happen, then the diver can get the "bends."

a.) Using Henry's law constants, determine the moles of N2 and O2 dissolved in 7.5 L of blood (assume it is water) at 298 K at the surface of the water before the dive and again at a depth of 100.0 ft below the surface.

b.) If the diver were to suddenly return to the surface, how many moles of N2 and O2 would "undissolve" from the diver's blood?

c.) What volume would this "expelled" gas occupy (assume it behaves ideally) at 298 K and 1.00 atm?

Answer 1

a) Partial pressure of gas = mole fraction X total pressure

pN2= 0.79 x 1bar = 0.79 bar

Now henry's law constant for nitrogen in water = 9.04 x10⁴ bar

Mole fraction of nitrogen in blood = Partial pressure / Henry's constant = 0.79 / 9.04 X 10^4 = 8.73 X 10^-6

Moles of nitrogen in blood = Mole fraction of nitrogen in blood X total moles of water and nitrogen

Moles of water = Volume x density / molecular weight = 7.5 X 1 X 1000 / 18 = 420

Moles of nitrogen in blood = 8.73 X 10^-6 X 420 = 3.67 X 10^-3

b) pO2 = 0.21 X 1 bar = 0.21 bar

Now henry's law constant for oxygen in water = 4.3×10⁴ bar

Mole fraction of O2 in blood = Partial pressure / Henry's constant = 0.21/ 4.3 X 10^4 = 4.88 X 10^-6

Moles of O2 in blood = Mole fraction of O2 in blood X total moles of water and O2

Moles of water = Volume x density / molecular weight = 7.5 X 1 X 1000 / 18 = 420

Moles of O2 in blood = 4.88 X 10^-6 X 420 = 2.049 X 10^-3