Workers in underwater diving suits necessarily breathe air at greater than norma

Question

Workers in underwater diving suits necessarily breathe air at greater than normal pressure. If they are returned to the surface too rapidly, N2, dissolved in their blood at the previously higher pressure, comes out of solution and may cause emboli (gas bubbles in the bloodstream), bends, and decompression sickness. Blood at 37C and 1 bar pressure dissolves 1.3 mL of N2 gas (measured for pure N2 at 37C and 1 bar) in 100 mL of blood.

Part A

Calculate the volume of N2 likely to be liberated from the blood of a diver returned to 1 bar pressure after prolonged exposure to air pressure at 300 m of water (below the surface). The total blood volume of the average adult is 4.7 L; air contains 78 vol% N2.

Correct answer from the back of the txtbk is 1800 mL .....i just cant figure out how they got it. Thanks

Explanation / Answer

Whne P = 1 bar (= 1 atm) and T = 37 DegC = 37 + 273 = 310 K :>

1.3 mL of N2 is present in 100 mL of blood.

Hence the volume of N2 that would be present in 1L(= 1000 mL) of blood

=(1.3 mLN2 / 100 mL blood) x (1000 mL blood / 1L blood) = 13 mL N2 / 1 L blood = 0.013 L N2 / 1L blood

Hence moles of N2 present in 1L blood, n = PV/RT =(1 atm x 0.013 L) / (0.0821L.atm.mol-1K-1x310K)

= 5.11x10-4 mol / 1L blood

Hence concentration of N2 in blood at 1 bar pressure, Caq= 5.11x10-4 mol /1L blood

Now we can caclculate the value of Henry constant, Hcp by applying Henry's law which is

Hcp = Caq / P = 5.11x10-4 mol /1L blood / 1 atm = 5.11x10-4 mol.atm-1L-1  

Whne P = 300 m water and T = 310 :>

P = 300 m Water = 29.06 atm

Now applying Henry's law

Hcp = Caq / P

or Caq = P x Hcp = 29.06 atm x 5.11x10-4 mol.atm-1L-1 = 0.0149 mol N2/L blood

i.e 0.0149 mol N2 is present in 1L of blood.

Hence moles of N2 that would present in 4.7 L of blood = 0.0149 mol N2/1 L blood x (4.7 L blood)

= 0.07003 mol N2

Now when the diver comes out of water the pressure agains becomes 1 bar (1 atm)

Hence volume occupied by 0.07003 mol N2 at 1 atm pressure,  V = nRT / P

=> V = (0.07003 mol x 0.0821L.atm.mol-1K-1x310K) / 1 atm = 1.80 L = 1800 mL (answer)

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