Solution: 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?

Solution: 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 externa

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?

Solution

We're being asked to calculate the milliliters of N₂ gas 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:

$\boxed{{\mathbf{S}}_{\mathbf{gas}}\mathbf{ }\mathbf{=}\mathbf{ }{\mathbf{k}}_{\mathbf{H}}\mathbf{·}{\mathbf{P}}_{\mathbf{gas}}}$

where:

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

k_H = Henry’s law constant for the gas

P_gas = partial pressure of the gas

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