Question

At ordinary body temperature \(\left(37^{\circ} \mathrm{C}\right)\) the solubility of \(\mathrm{N}_{2}\) in water in contact with air at ordinary atmospheric pressure \((1.0 \mathrm{~atm})\) is \(0.015 \mathrm{~g} / \mathrm{L}\). Air is approximately \(78 \mathrm{~mol} \% \mathrm{~N}_{2}\). Calculate the number of moles of \(\mathrm{N}_{2}\) dissolved per liter of blood, which is essentially an aqueous solution. At a depth of \(100 \mathrm{ft}\) in water, the pressure is \(4.0 \mathrm{~atm}\). What is the solubility of \(\mathrm{N}_{2}\) from air in blood at this pressure? If a scuba diver suddenly surfaces from this depth, how many milliliters of \(\mathrm{N}_{2}\) gas, in the form of tiny bubbles, are released into the bloodstream from each liter of blood?

Short answer

The number of moles of N₂ dissolved per liter of blood at 1 atm is \(\frac{0.015 \, \text{g}}{28 \, \text{g/mol}}\). At 4 atm, the solubility of N₂ in blood increases proportionally, resulting in a solubility of \(\frac{0.015 \, \text{g/L}}{28 \, \text{g/mol}}\) × \(\frac{4 \, \text{atm}}{1 \, \text{atm}}\). The solubility difference between these two pressures is \(\frac{0.015 \, \text{g/L}}{28 \, \text{g/mol}}\) × \(\frac{4 \, \text{atm}}{1 \, \text{atm}}\) - \(\frac{0.015 \, \text{g/L}}{28 \, \text{g/mol}}\). Converting this difference in moles to milliliters of N₂ gas, we find that when a scuba diver suddenly surfaces from 100 ft, approximately \(15.3 \, \text{mL}\) of N₂ gas is released into the bloodstream as tiny bubbles per liter of blood.

Step-by-step solution

Find the number of moles of N₂ per liter of blood at 1 atm

At ordinary body temperature and atmospheric pressure, the solubility of N₂ is given as 0.015 g/L. We need to convert the mass of N₂ into moles. To do that, we can use the molar mass of N₂, which is 28 g/mol. Number of moles = (Mass of N₂) / (Molar mass of N₂) So, Number of moles = \(\frac{0.015 \, \text{g}}{28 \, \text{g/mol}}\)

Calculate the number of moles of N₂ per liter of blood at 4 atm

According to Henry's law, the solubility of a gas is directly proportional to the partial pressure of the gas. Since the pressure increases from 1 atm to 4 atm, the solubility of N₂ will also increase proportionally. Solubility at 4 atm = Solubility at 1 atm × (Pressure at 4 atm / Pressure at 1 atm) So, Solubility at 4 atm = \(\frac{0.015 \, \text{g/L}}{28 \, \text{g/mol}}\) × \(\frac{4 \, \text{atm}}{1 \, \text{atm}}\)

Calculate the amount of N₂ released when the pressure decreases from 4 atm to 1 atm

Now we need to calculate the difference in solubility of N₂ at 4 atm and 1 atm to find the amount of N₂ released when the pressure drops. Solubility difference = Solubility at 4 atm - Solubility at 1 atm So, Solubility difference = \(\frac{0.015 \, \text{g/L}}{28 \, \text{g/mol}}\) × \(\frac{4 \, \text{atm}}{1 \, \text{atm}}\) - \(\frac{0.015 \, \text{g/L}}{28 \, \text{g/mol}}\)

Convert the difference in solubility (moles) into milliliters of N₂ gas

Finally, we'll convert the difference in solubility from moles to milliliters of N₂ gas. To do this, we'll use the relationship 1 mole of any gas occupies 22.4 L at standard temperature and pressure (STP). First, convert the difference in moles to millimoles: \[\text{Difference in millimoles of N₂} = \;\text{Solubility difference} \times 1000\] Now, using the relationship mentioned above, convert millimoles to milliliters of N₂ gas: \[\text{Difference in milliliters of N₂ gas} = \;\text{Difference in millimoles}\] × \(\frac{22.4 \, \text{L}}{1000 \, \text{mmol}}\) × 1000\] Once you've calculated the difference in milliliters of N₂ gas, you have the answer - the amount of N₂ released into the bloodstream as tiny bubbles per liter of blood when a scuba diver suddenly surfaces from a depth of 100 ft.