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Chapter 13: Problem 107

At ordinary body temperature \(\left(37^{\circ} \mathrm{C}\right)\), the solubility of \(\mathrm{N}_{2}\) in water at ordinary atmospheric pressure ( \(1.0 \mathrm{~atm})\) 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 \(100 \mathrm{ft}\) in water, the external pressure is \(4.0 \mathrm{~atm}\). 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?

Short Answer

Expert verified
(a) There are 5.35 脳 10鈦烩伌 moles of N鈧� dissolved per liter of blood. (b) The solubility of N鈧� from air in blood at 100 ft depth and 4.0 atm pressure is 0.0468 g/L. (c) When the scuba diver suddenly surfaces from 100 ft depth, 28.7 milliliters of N鈧� gas in the form of tiny bubbles are released into the bloodstream from each liter of blood.

Step by step solution

01

(a) Calculate the number of moles of N鈧� dissolved in blood

We need to use the given solubility of N鈧� in water at body temperature and atmospheric pressure, which is 0.015 g/L. To find the number of moles, we'll need to convert the mass of N鈧� to moles using the molar mass of N鈧�, which is 28.02 g/mol. Moles of N鈧� = (solubility of N鈧�) / (molar mass of N鈧�) Moles of N鈧� = (0.015 g/L) / (28.02 g/mol) Moles of N鈧� = 5.35 脳 10鈦烩伌 mol/L So, there are 5.35 脳 10鈦烩伌 moles of N鈧� dissolved per liter of blood.
02

(b) Calculate the solubility of N鈧� in blood at 100 ft depth and 4 atm pressure

At 100 ft depth, the external pressure is 4.0 atm. According to Henry's Law, the solubility of a gas in a liquid is proportional to the partial pressure of that gas. So, we need to find the partial pressure of N鈧� and use that to calculate the solubility of N鈧� at the new pressure. Partial pressure of N鈧� = (mole fraction of N鈧�) 脳 (total pressure) Partial pressure of N鈧� = (0.78) 脳 (4.0 atm) Partial pressure of N鈧� = 3.12 atm Since the solubility is proportional to the pressure: Solubility of N鈧� at 4.0 atm = (original solubility) 脳 (pressure ratio) Solubility of N鈧� at 4.0 atm = (0.015 g/L) 脳 (3.12 atm / 1.0 atm) Solubility of N鈧� at 4.0 atm = 0.0468 g/L So, the solubility of N鈧� from air in blood at 100 ft depth and 4.0 atm pressure is 0.0468 g/L.
03

(c) Calculate the volume of N鈧� gas bubbles released after surfacing

When the scuba diver surfaces suddenly, the external pressure is reduced from 4.0 atm to 1.0 atm. So, we need to calculate the difference in solubilities of N鈧� at these pressures, as this will give us the amount of N鈧� that turns into gas bubbles. Solubility difference = Solubility at 4.0 atm - Solubility at 1.0 atm Solubility difference = 0.0468 g/L - 0.015 g/L Solubility difference = 0.0318 g/L This solubility difference is the mass of N鈧� that turns into gas bubbles per liter of blood. To find the volume of the gas, we need to convert it into moles and then use the ideal gas law at Ordinary Body Temperature (37掳C) and 1 atm. To find the moles released: Moles of N鈧� = (solubility difference) / (molar mass of N鈧�) Moles of N鈧� = (0.0318 g/L) / (28.02 g/mol) Moles of N鈧� = 1.13 脳 10鈦宦� mol/L Now, we can use the ideal gas law (PV=nRT) to find the volume: Volume = (moles of N鈧�) 脳 (R) 脳 (Temperature) / (Pressure) Volume = (1.13 脳 10鈦宦� mol/L) 脳 (0.0821 L鈰卆tm/mol鈰匥) 脳 (310.15 K) / (1 atm) Volume = 2.87 脳 10鈦宦� L or 28.7 mL So, when the scuba diver suddenly surfaces from 100 ft depth, 28.7 milliliters of N鈧� gas in the form of tiny bubbles are released into the bloodstream from each liter of blood.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solubility of Gases in Water
Solubility refers to the ability of a substance, like nitrogen gas (N鈧�), to dissolve in a liquid such as water or blood. At ordinary body temperature (37掳C) and atmospheric pressure, nitrogen's solubility is given as 0.015 g/L. This number indicates how much nitrogen can reasonably dissolve in every liter of the liquid under those specific conditions.
Remember, solubility can vary with changes in external conditions like temperature and pressure.

As a general rule, the solubility of gases in a liquid will typically decrease with increased temperature. However, pressure has a different effect. According to Henry鈥檚 Law, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. This means that as you increase the pressure on a gas, its solubility in a liquid increases as well. This plays a crucial role when considering depths at which scuba divers swim and the potential hazards related to decompression when ascending to the surface.
Understanding Moles in Chemical Solutions
Moles are a fundamental unit in chemistry that quantify the amount of a substance. One mole of any substance contains approximately 6.022 脳 10虏鲁 particles of that substance, whether atoms, molecules, etc. This concept helps chemists convert between the microscopic scale of atoms and molecules and the macroscopic scale we observe.
In the context of the given exercise, we determine the moles of nitrogen dissolved in a liter of blood using the provided solubility and the molar mass of nitrogen, which is 28.02 g/mol. Here鈥檚 how you can convert the mass-based solubility to moles:
  • Use the formula: Moles of N鈧� = Solubility of N鈧� (g/L) / Molar Mass of N鈧� (g/mol)
  • Substitute: 0.015 g/L 梅 28.02 g/mol 鈮� 5.35 脳 10鈦烩伌 mol/L
This tells us the precise number of nitrogen molecules present per liter under certain conditions, providing valuable insight into the chemical composition of the blood at that state.
Pressure Effects on Gas Solubility
Pressure, particularly when related to gases, is a measure of force exerted by gas particles colliding with the walls of its container. In the case of diving, when we discuss a depth of 100 ft, we're looking at an increased pressure environment of about 4 atm.
According to Henry's Law, the solubility of a gas in a solution is proportional to the pressure of that gas above the solution. This is crucial when examining the behavior of gases in the bloodstream during scenarios like diving and ascending:
  • The increase in pressure increases the amount of gas that can be dissolved. At 4 atm, nitrogen's solubility increases to 0.0468 g/L.
  • Upon rapidly decreasing pressure, such as a diver surfacing quickly, gases come out of solution, forming bubbles.
This phenomenon can lead to decompression sickness, known as "the bends," if the diver resurfaces too fast.
Hence, understanding how pressure affects solubility is not just an academic exercise; it's essential for the safety of activities involving pressure changes, such as scuba diving.

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Most popular questions from this chapter

The presence of the radioactive gas radon \((\mathrm{Rn})\) in well water presents a possible health hazard in parts of the United States. (a) Assuming that the solubility of radon in water with 1 atm pressure of the gas over the water at \(30^{\circ} \mathrm{C}\) is \(7.27 \times 10^{-3} \mathrm{M}\), what is the Henry's law constant for radon in water at this temperature? (b) A sample consisting of various gases contains \(3.5 \times 10^{-6}\) mole fraction of radon. This gas at a total pressure of 32 atm is shaken with water at \(30^{\circ} \mathrm{C}\). Calculate the molar concentration of radon in the water.

Suppose you had a balloon made of some highly flexible semipermeable membrane. The balloon is filled completely with a \(0.2 \mathrm{M}\) solution of some solute and is submerged in a \(0.1 \mathrm{M}\) solution of the same solute: Initially, the volume of solution in the balloon is \(0.25 \mathrm{~L}\). Assuming the volume outside the semipermeable membrane is large, as the illustration shows, what would you expect for the solution volume inside the balloon once the system has come to equilibrium through osmosis? [Section 13.5]

(a) What is the molality of a solution formed by dissolving \(1.12 \mathrm{~mol}\) of \(\mathrm{KCl}\) in \(16.0 \mathrm{~mol}\) of water? (b) How many grams of sulfur \(\left(\mathrm{S}_{8}\right)\) must be dissolved in \(100.0 \mathrm{~g}\) of naphthalene \(\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)\) to make a \(0.12 \mathrm{~m}\) solution?

At \(63.5^{\circ} \mathrm{C}\), the vapor pressure of \(\mathrm{H}_{2} \mathrm{O}\) is 175 torr, and that of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) is 400 torr. A solution is made by mixing equal masses of \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\). (a) What is the mole fraction of ethanol in the solution? (b) Assuming ideal-solution behavior, what is the vapor pressure of the solution at \(63.5^{\circ} \mathrm{C}\) ? (c) What is the mole fraction of ethanol in the vapor above the solution?

(a) Does a \(0.10 \mathrm{~m}\) aqueous solution of \(\mathrm{NaCl}\) have a higher boiling point, a lower boiling point, or the same boiling point as a \(0.10 \mathrm{~m}\) aqueous solution of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) ? (b) The experimental boiling point of the \(\mathrm{NaCl}\) solution is lower than that calculated assuming that \(\mathrm{NaCl}\) is completely dissociated in solution. Why is this the case?
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