Problem 107 - Chemistry The Central Science In Si Units

Question

At ordinary body temperature \(\left(37^{\circ} \mathrm{C}\right),\) the solubility of \(\mathrm{N}_{2}\) in water at ordinary atmospheric pressure is \(0.015 \mathrm{~g} / \mathrm{L}\). Air is approximately \(78 \mathrm{~mol} \% \mathrm{~N}_{2} .\) (a) Calculate the number of moles of \(\mathrm{N}_{2}\) dissolved per liter of blood, assuming blood is a simple aqueous solution. (b) At a depth of \(30.5 \mathrm{~m}\) in water, the external pressure is \(405 \mathrm{kPa}\). What is the solubility of \(\mathrm{N}_{2}\) from air in blood at this pressure? (c) 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?

Step-by-Step Solution

Answer

(a) 5.35 × 10⁻⁴ mol/L; (b) 0.00215 mol/L; (c) 41 mL of \( \mathrm{N}_2 \).

Step 1: Calculate moles of dissolved \( \mathrm{N}_2 \) at normal atmospheric pressure

First, we need to calculate the moles of nitrogen \( \mathrm{N}_2 \) dissolved in water (or blood, treated as a simple aqueous solution) at normal atmospheric pressure with a solubility of \( 0.015 \, \mathrm{g/L} \). The molar mass of \( \mathrm{N}_2 \) is \( 28.02 \, \mathrm{g/mol} \).Use the formula: \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \)Thus, the moles of \( \mathrm{N}_2 \) are given by:\[\text{moles of } \mathrm{N}_2 = \frac{0.015 \, \mathrm{g}}{28.02 \, \mathrm{g/mol}} = 5.35 \times 10^{-4} \, \mathrm{mol/L}\]

Step 2: Calculate solubility under increased pressure using Henry's Law

Next, we calculate the solubility of \( \mathrm{N}_2 \) at a depth of \( 30.5 \, \mathrm{m} \), where the pressure is \( 405 \, \mathrm{kPa} \), using Henry's Law. Henry's Law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid, expressed as: \[ S_1/P_1 = S_2/P_2 \]From Step 1, \( S_1 = 0.015 \, \mathrm{g/L} \) at \( P_1 = 101.3 \, \mathrm{kPa} \). We need to find \( S_2 \) at \( P_2 = 405 \, \mathrm{kPa} \):\[ 0.015 \, \mathrm{g/L} / 101.3 \, \mathrm{kPa} = S_2 / 405 \, \mathrm{kPa} \]\[ S_2 = \frac{0.015 \, \mathrm{g/L} \times 405 \, \mathrm{kPa}}{101.3 \, \mathrm{kPa}} = 0.0601 \, \mathrm{g/L} \]To convert this to moles, use the molar mass of \( \mathrm{N}_2 \):\[ \text{moles of } \mathrm{N}_2 = \frac{0.0601 \, \mathrm{g}}{28.02 \, \mathrm{g/mol}} = 0.00215 \, \mathrm{mol/L} \]

Step 3: Calculate volume of \( \mathrm{N}_2 \) released when surfacing

When the diver surfaces, the pressure drops back to \( 101.3 \, \mathrm{kPa} \), and the solubility returns to \( 0.015 \, \mathrm{g/L} \) or \( 5.35 \times 10^{-4} \, \mathrm{mol/L} \). The released \( \mathrm{N}_2 \) gas is the difference between the moles dissolved at depth and at surface pressure:\[ 0.00215 \, \mathrm{mol/L} - 5.35 \times 10^{-4} \, \mathrm{mol/L} = 0.00162 \, \mathrm{mol/L} \]Convert moles to volume using the ideal gas law: \( PV = nRT \), where \( R = 8.314 \, \mathrm{J/(mol \cdot K)} \), \( T = 310 \, \mathrm{K} \) (temperature of the human body), and \( P = 101.3 \, \mathrm{kPa} \).\[ V = \frac{nRT}{P} = \frac{0.00162 \, \mathrm{mol} \times 8.314 \, \mathrm{J/(mol \cdot K)} \times 310 \, \mathrm{K}}{101.3 \times 10^3 \, \mathrm{Pa}} \]\[ V = 0.041 \, \mathrm{L} = 41 \, \mathrm{mL} \]

Step 4: Summary and Final Answer

To summarize, the calculations show: (a) At 1 atm, the moles of \( \mathrm{N}_2 \) dissolved per liter of blood are approximately \( 5.35 \times 10^{-4} \, \mathrm{mol/L} \). (b) At 405 kPa, the moles of \( \mathrm{N}_2 \) dissolved increase to \( 0.00215 \, \mathrm{mol/L} \). (c) Upon sudden surfacing, \( 41 \, \mathrm{mL} \) of \( \mathrm{N}_2 \) gas is released per liter of blood.

Key Concepts

Henry's Law

Henry's Law is an essential principle that relates the solubility of gases in liquids to the partial pressure of the gas above the liquid. According to this law, the solubility of a gas is directly proportional to the gas's partial pressure. Mathematically, it can be expressed as:

• \[ \frac{S_1}{P_1} = \frac{S_2}{P_2} \]

Here, \( S_1 \) and \( S_2 \) represent the solubility of the gas (often in g/L), while \( P_1 \) and \( P_2 \) denote the initial and final partial pressures, respectively.
In the context of the given exercise, when the scuba diver is at depth, the external pressure increases, thus increasing the solubility of \( \text{N}_2 \) in blood. Using Henry's Law, we calculate how much more nitrogen gas is absorbed at an increased depth. Conversely, when the diver returns to surface pressure, the solubility decreases, leading to the release of nitrogen bubbles, which can be dangerous if not managed properly by a diver.

Molar Mass

Molar mass is a measure of the mass of one mole of a given chemical substance. It's typically represented in units of grams per mole \( (\text{g/mol}) \). For nitrogen \( (\text{N}_2) \), the molar mass is calculated by adding up the atomic masses of the constituent nitrogen atoms:

• Nitrogen atomic mass: approximately 14.01 g/mol
• Molar mass of \( \text{N}_2 \): \( 2 \times 14.01 \text{ g/mol} = 28.02 \text{ g/mol} \)

Understanding molar mass is crucial for converting between grams and moles, as used in the exercise. For example, to calculate the number of moles of \( \text{N}_2 \) in a specific solubility \( (0.015 \text{ g/L}) \), you would divide the mass by the molar mass:

• \[ \text{moles of } \text{N}_2 = \frac{0.015 \text{ g}}{28.02 \text{ g/mol}} \approx 5.35 \times 10^{-4} \text{ mol/L} \]

Molar mass thus acts as a bridge in chemical calculations, allowing us to move between the tangible mass and the abstract idea of moles.

Partial Pressure

Partial pressure refers to the pressure exerted by a single type of gas in a mixture of gases, as if it occupied the entire volume by itself. It contributes to the total pressure in the system. This concept is particularly important when discussing gases dissolved in liquids.
In the context of atmospheric \( \text{N}_2 \), while air is composed of multiple gases, each gas has its own partial pressure that combines to form total atmospheric pressure (about 101.3 kPa at sea level). Air is about 78% nitrogen, so nitrogen's partial pressure is 78% of the total atmospheric pressure. As depth increases and overall pressure rises, the partial pressure of nitrogen also increases proportionally.

• Surface atmospheric pressure: \( 101.3 \text{ kPa} \)
• Partial pressure of \( \text{N}_2 \): \( 78\% \times 101.3 \text{ kPa} \)

These pressures determine gas solubility in water or blood. When calculating solubility changes with depth, considering changes in partial pressure using Henry's Law becomes crucial. As the diver ascends, the partial pressure and solubility reduce, affecting the physiology of someone experiencing changes in pressures underwater.