Problem: At ordinary body temperature (37 oC), the solubility of N2 in water at ordinary atmospheric pressure (1.0 atm) is 0.015 g/L. Air is approximately 78 mol % N2.At a depth of 100 ft in water, the external pressure is 4.0 atm. If a scuba diver suddenly surfaces from this depth, how many milliliters of N2 gas, in the form of tiny bubbles, are released into the bloodstream from each liter of blood?

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FREE Expert Solution

We're being asked to calculate the milliliters of Ngas released into the blood stream.


For this problem, we will do the following steps:


Step 1: Calculate Henry’s Law constant for the gas


Recall that the solubility of a gas is given by Henry’s law:

Sgas = kH·Pgas

where:

Sgas = solubility of the gas (in mol/L or M)

kH = Henry’s law constant for the gas

Pgas = partial pressure of the gas

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Problem Details

At ordinary body temperature (37 oC), the solubility of N2 in water at ordinary atmospheric pressure (1.0 atm) is 0.015 g/L. Air is approximately 78 mol % N2.

At a depth of 100 ft in water, the external pressure is 4.0 atm. If a scuba diver suddenly surfaces from this depth, how many milliliters of N2 gas, in the form of tiny bubbles, are released into the bloodstream from each liter of blood?