Question

the solubility of N2 in blood at 37C and at a partial pressure of 0.80 atm is 5.6x10-4 mol/L a deep sea diver breathes compressed air with the partial pressure of N2 equal to 3.5atm assume that the total volume of blood is5.8L Calculate the amount of n2 gas released(in liters at 37 degree C and 1.00atm) when the diver returns to the surface of the water where the partial pressure of N2 is 0.80 atm

Answer #1

As the solubility of a gas ( in terms of concentration) is proportional to the partial pressure of the gas.

So under deep sea water the solubility of nitrogen in blood at 3.5 atm partial pressure would be --

( p_{1}
c_{1} , p_{2}
c_{2}
p_{1}/p_{2} = c_{1}/c_{2} ,
c_{2}= p_{2} / p_{1} . c_{1} )

Hence the solubilty undersea = 3.5atm/ 0.80atm . 5.6 x
10^{-4}mol/L

= 24.5 x 10^{-4}mol/L

Now we will calculate the difference in the concentrations of
nitrogen in the surfacing divers blood which decreases
by-- c = 24.5
x 10^{-4}mol/L - 5.6 x 10^{-4}mol/L

= 18.9 x 10^{-4} mol/L

So the amount of nitrogen gas released is --

n = c . v

= 18.9 x
10^{-4} mol/L .
5.8L
(as the total volume of blood is given)

= 109.62
x 10^{-4} = 1.09 x 10^{-2} mol

Now by using the Ideal Gas Law -

pV = nRT

V = nRT/p

V = 1.09 x 10^{-2}mol . 0.082057 Latm/Kmol
. (273+37)K / 1atm

= **0.277L**

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