A person accidentally swallows a drop of liquid oxygen, \(\mathrm{O}_{2}(l)\), which has a density of \(1.149 \mathrm{~g} / \mathrm{mL}\). Assuming the drop has a volume of \(0.050 \mathrm{~mL}\), what volume of gas will be produced in the person's stomach at body temperature \(\left(37^{\circ} \mathrm{C}\right)\) and a pressure of \(1.0 \mathrm{~atm}\) ?

Short Answer

Expert verified
The volume of oxygen gas produced in the person's stomach after accidentally swallowing a drop of liquid oxygen with a volume of 0.050 mL is approximately 0.0455 L at body temperature (37°C) and a pressure of 1.0 atm.

Step by step solution

01

Calculate the mass of liquid oxygen swallowed

Given that the density of liquid oxygen is 1.149 g/mL and the volume of the drop is 0.050 mL, the mass of liquid oxygen swallowed by the person can be found using the formula: Mass = Density * Volume We can now plug in the given values to find the mass: \(Mass = 1.149 \frac{g}{mL} \times 0.050 mL\) Mass ≈ 0.05745 g
02

Find the moles of O₂ gas

To find the moles of O₂ gas, we need to use the molar mass of oxygen (O) which is approximately 16 g/mol. Oxygen gas has a molecular formula of O₂, so the molar mass of O₂ is: Molar mass of O₂ = 2 * Molar mass of O ≈ 2 * 16 g/mol ≈ 32 g/mol Now we can find the moles of O₂ using the mass and molar mass: Moles of O₂ = Mass / Molar mass ≈ 0.05745 g / 32 g/mol ≈ 0.001798 mol
03

Use the ideal gas law to find the volume

The ideal gas law is given by: PV = nRT Where P is the pressure in atm, V is the volume in L, n is the moles of gas, R is the ideal gas constant (0.0821 L∙atm/mol∙K), and T is the temperature in Kelvin. We are given the pressure (1.0 atm) and temperature (37 °C, which is equivalent to 310 K). We can solve for the volume V using the given values: V = nRT / P V ≈ (0.001798 mol)(0.0821 L∙atm/mol∙K)(310 K) / (1.0 atm) V ≈ 0.0455 L
04

Report the final answer

The volume of oxygen gas produced in the person's stomach is approximately 0.0455 L.

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Key Concepts

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

Molar Mass
The molar mass is a fundamental concept in chemistry that refers to the mass of one mole of a given substance. A mole is a unit of measurement that represents Avogadro's number, which is approximately 6.022 x 1023 particles - atoms, molecules, or ions. Knowing the molar mass allows us to convert between the mass of a substance and the amount in moles, which is crucial for chemical calculations.

For elements, the molar mass is the same as the atomic mass listed on the periodic table, but in grams per mole. For compounds, like oxygen gas (O2), the molar mass is the sum of the atomic masses of all the atoms in a single molecule of that compound. For example, since oxygen has an atomic mass of approximately 16 g/mol, the molar mass of O2 is 32 g/mol (16 g/mol for each oxygen atom multiplied by 2). This is a vital step in many calculations, including those involving the ideal gas law.

Using Molar Mass in Calculations

When calculating the moles from a given mass, divide the mass by the molar mass. Similarly, to find the mass from moles, multiply the moles by the molar mass. These conversions are essential when working with chemical reactions and stoichiometry, where precise measurements in moles are often needed to determine the amounts of reactants and products.
Gas Volume Calculation
Calculating the volume of gas is a practical aspect of chemistry, particularly when dealing with gases in chemical reactions, or when determining the quantity of a gas produced or consumed. The volume of a gas can be directly related to the amount in moles, conditions of pressure, and temperature. To perform these calculations, an understanding of the ideal gas law is necessary.

The volume of a gas can vary greatly with changes in temperature and pressure. For instance, at high temperatures or low pressures, gases expand to occupy a larger volume. Conversely, at low temperatures or high pressures, gases compress to occupy a smaller volume. This relationship is quantified by the ideal gas law and can be calculated if the number of moles, the pressure, and the temperature are known.
PV=nRT
The ideal gas law, represented by the equation PV=nRT, is a cornerstone in the field of chemistry for understanding the behavior of gases under various conditions. This equation relates the pressure (P), volume (V), amount of gas in moles (n), temperature (T), and includes the ideal gas constant (R).

  • P (Pressure): Exerted by the gas particles when they collide with the walls of their container. It is usually measured in atmospheres (atm) or Pascals (Pa).
  • V (Volume): The space that the gas occupies, typically measured in liters (L).
  • n (Amount of Substance): Refers to the quantity of gas measured in moles.
  • R (Ideal Gas Constant): A proportionality constant that relates the other variables. Its value depends on the units used for pressure and volume but often quoted as 0.0821 L∙atm/mol∙K.
  • T (Temperature): Must be in Kelvin (K) for use in this equation. To convert from Celsius to Kelvin, add 273.15 to the Celsius temperature.

By manipulating the ideal gas law, you can solve for any of the variables if the other three are known. It is a powerful tool for predicting and calculating the behavior of gases under different conditions. This law assumes ideal conditions where the gas particles do not interact with one another and occupy no volume themselves, which is not strictly true for real gases but serves as a good approximation under many conditions.
Liquid Oxygen Properties
Liquid oxygen, denoted as O2(l), stands out with distinct physical and chemical properties. It is a pale blue liquid with a density of about 1.149 grams per milliliter at its boiling point. One of its most notable properties is that it's a powerful oxidizing agent, making it important in applications such as rocket propulsion.

The transformation from liquid oxygen to gas is accompanied by a substantial increase in volume. This expansion is due to the significant difference in the density of liquids and gases; gases have much lower density than liquids. When a substance changes state, as in the case of liquid oxygen evaporating in the stomach, the increase in volume can be predicted using the ideal gas law if the temperature and pressure conditions are known.

It's vital to handle liquid oxygen with care because it can react vigorously with combustible materials. Safety protocols should always be followed when dealing with cryogenic liquids like liquid oxygen to prevent mishaps or injuries.

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Most popular questions from this chapter

At elevated temperatures, sodium chlorate decomposes to produce sodium chloride and oxygen gas. A \(0.8765\) -g sample of impure sodium chlorate was heated until the production of oxygen gas ceased. The oxygen gas collected over water occupied \(57.2 \mathrm{~mL}\) at a temperature of \(22^{\circ} \mathrm{C}\) and a pressure of 734 torr. Calculate the mass percent of \(\mathrm{NaClO}_{3}\) in the original sample. (At \(22^{\circ} \mathrm{C}\) the vapor pressure of water is \(19.8\) torr.)

One way of separating oxygen isotopes is by gaseous diffusion of carbon monoxide. The gaseous diffusion process behaves like an effusion process. Calculate the relative rates of effusion of \({ }^{12} \mathrm{C}^{16} \mathrm{O},{ }^{12} \mathrm{C}^{17} \mathrm{O}\), and \({ }^{12} \mathrm{C}^{18} \mathrm{O}\). Name some advan- tages and disadvantages of separating oxygen isotopes by gaseous diffusion of carbon dioxide instead of carbon monoxide.

A container is filled with an ideal gas to a pressure of \(11.0\) atm at \(0^{\circ} \mathrm{C}\). a. What will be the pressure in the container if it is heated to \(45^{\circ} \mathrm{C} ?\) b. At what temperature would the pressure be \(6.50 \mathrm{~atm}\) ? c. At what temperature would the pressure be \(25.0 \mathrm{~atm}\) ?

You have a helium balloon at \(1.00\) atm and \(25^{\circ} \mathrm{C}\). You want to make a hot-air balloon with the same volume and same lift as the helium balloon. Assume air is \(79.0 \%\) nitrogen and \(21.0 \%\) oxygen by volume. The "lift" of a balloon is given by the difference between the mass of air displaced by the balloon and the mass of gas inside the balloon. a. Will the temperature in the hot-air balloon have to be higher or lower than \(25^{\circ} \mathrm{C}\) ? Explain. b. Calculate the temperature of the air required for the hotair balloon to provide the same lift as the helium balloon at \(1.00\) atm and \(25^{\circ} \mathrm{C}\). Assume atmospheric conditions are \(1.00\) atm and \(25^{\circ} \mathrm{C}\).

A 2.50-L container is filled with \(175 \mathrm{~g}\) argon. a. If the pressure is \(10.0 \mathrm{~atm}\), what is the temperature? b. If the temperature is \(225 \mathrm{~K}\), what is the pressure?

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