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Chapter 14: Problem 1

Assuming air to be composed exclusively of \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\), with their partial pressures in the ratio \(0.21: 0.79\), what are their mass fractions?

Short Answer

Expert verified
The mass fractions of O鈧� and N鈧� in air can be determined using the provided partial pressures and molar masses. First, we find the mass of O鈧� and N鈧�: O鈧� mass = 0.21 * 32 g/mol = 6.72 g/mol N鈧� mass = 0.79 * 28 g/mol = 22.12 g/mol Next, we calculate the total mass of air: Total mass of air = 6.72 + 22.12 = 28.84 g/mol Finally, we determine the mass fractions for O鈧� and N鈧�: O鈧� mass fraction = 6.72 / 28.84 鈮� 0.233 N鈧� mass fraction = 22.12 / 28.84 鈮� 0.767 Thus, the mass fractions of O鈧� and N鈧� in air are approximately 0.233 and 0.767, respectively.

Step by step solution

01

Mass of O2

The mass of O2 can be found by multiplying its partial pressure (0.21) by its molar mass (32 g/mol): O2 mass = 0.21 * 32 g/mol
02

Mass of N2

The mass of N2 can be found by multiplying its partial pressure (0.79) by its molar mass (28 g/mol): N2 mass = 0.79 * 28 g/mol #Step 2: Calculate the total mass of air#
03

Total mass of air

The total mass of air can be found by adding the mass of O2 and N2: Total mass of air = O2 mass + N2 mass #Step 3: Calculate the mass fractions of O2 and N2#
04

Mass fraction of O2

The mass fraction of O2 can be found by dividing its mass by the total mass of air: O2 mass fraction = O2 mass / Total mass of air
05

Mass fraction of N2

The mass fraction of N2 can be found by dividing its mass by the total mass of air: N2 mass fraction = N2 mass / Total mass of air By calculating the mass fractions of O2 and N2, we can now find their respective values.

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Key Concepts

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

Mass Fraction
The mass fraction of a component in a mixture represents how much that specific component contributes to the total mass of the mixture. It is expressed as a ratio or a percentage. Calculating the mass fraction helps in understanding the composition of mixtures, allowing us to see the weight contribution of each element.
To find the mass fraction, you first need to determine the mass of each component in the mixture. This requires multiplying the partial pressure of each gas by its molar mass, as seen in the exercise where oxygen (O鈧�) and nitrogen (N鈧�) have specified partial pressures.
Let's walk through the process:
  • Calculate the mass of O鈧�: Multiply the partial pressure of O鈧� (0.21) by its molar mass (32 g/mol).
  • Calculate the mass of N鈧�: Multiply the partial pressure of N鈧� (0.79) by its molar mass (28 g/mol).
  • Add the masses of O鈧� and N鈧� to find the total mass of air.
  • Divide each component's mass by the total mass of air to find the mass fraction.
This process highlights the importance of mass fraction in proportioning substances within a mixture, crucial for fields like chemistry and engineering.
Molar Mass
Molar mass is a fundamental concept in chemistry referring to the mass of one mole of a substance, expressed in grams per mole (g/mol). It provides a bridge between the microscopic world of atoms and molecules and the macroscopic quantities we can measure.
For any chemical element, the molar mass can be found on the periodic table. For compounds, it's the sum of the molar masses of the individual elements that make up the compound. In the exercise, we deal with diatomic gases, oxygen (O鈧�) and nitrogen (N鈧�), which have molar masses of 32 g/mol and 28 g/mol, respectively.
Molar mass plays a crucial role in computing other quantities:
  • Converting moles to grams: By multiplying the number of moles by the molar mass, you obtain the mass in grams.
  • Finding mass fractions: As seen in the exercise, we multiply the molar mass by the partial pressure to get the component's mass inside a mixture.
Understanding molar mass is essential for performing calculations related to chemical reactions and determining the amounts of reactants and products.
Air Composition
Air composition refers to the types and amounts of gases present in the atmosphere. In broad terms, dry air is composed predominantly of nitrogen ( N鈧�), oxygen ( O鈧�), with smaller amounts of argon, carbon dioxide, and other trace gases.
For simplicity, exercises often model air as a mixture of just O鈧� and N鈧�, focusing on their typical atmospheric proportions. These proportions can be expressed by volume percentages or partial pressures. In the exercise, the partial pressures are given as a ratio: 21% O鈧� and 79% N鈧�.
Air composition is crucial:
  • For studying environmental and atmospheric sciences, where the concentration of gases affects climate and weather patterns.
  • In engineering applications, where knowing the composition of gases is important for processes like combustion or air filtration.
  • Understanding these proportions also facilitates calculations, such as finding mass fractions necessary for diverse scientific explorations and applications.
So, comprehending air composition, especially when simplified, forms a critical foundation for analyzing atmospheric conditions and implementing related technologies.

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Most popular questions from this chapter

An old-fashioned glass apothecary jar contains a patent medicine. The neck is closed with a rubber stopper that is \(20 \mathrm{~mm}\) tall, with a diameter of \(10 \mathrm{~mm}\) at the bottom end, widening to \(20 \mathrm{~mm}\) at the top end. The molar concentration of medicine vapor in the stopper is \(2 \times 10^{-3} \mathrm{kmol} / \mathrm{m}^{3}\) at the bottom surface and is negligible at the top surface. If the mass diffusivity of medicine vapor in rubber is \(0.2 \times 10^{-9} \mathrm{~m}^{2} / \mathrm{s}\), find the rate \((\mathrm{kmol} / \mathrm{s})\) at which vapor exits through the stopper.

Insulation degrades (experiences an increase in thermal conductivity) if it is subjected to water vapor condensation. The problem may occur in home insulation during cold periods, when vapor in a humidified room diffuses through the drywall (plaster board) and condenses in the adjoining insulation. Estimate the mass diffusion rate for a \(3 \mathrm{~m} \times 5 \mathrm{~m}\) wall, under conditions for which the vapor pressure is \(0.03\) bar in the room air and \(0.0\) bar in the insulation. The drywall is \(10 \mathrm{~mm}\) thick, and the solubility of water vapor in the wall material is approximately \(5 \times 10^{-3} \mathrm{kmol} / \mathrm{m}^{3}\) - bar. The binary diffusion coefficient for water vapor in the drywall is approximately \(10^{-9} \mathrm{~m}^{2} / \mathrm{s}\).

Consider a spherical organism of radius \(r_{o}\) within which respiration occurs at a uniform volumetric rate of \(\dot{N}_{\mathrm{A}}=-k_{0}\). That is, oxygen (species A) consumption is governed by a zero-order, homogeneous chemical reaction. (a) If a molar concentration of \(C_{\mathrm{A}}\left(r_{o}\right)=C_{\mathrm{A}, o}\) is maintained at the surface of the organism, obtain an expression for the radial distribution of oxygen, \(C_{\mathrm{A}}(r)\), within the organism. From your solution, can you discern any limits on applicability of the result? (b) Obtain an expression for the rate of oxygen consumption within the organism. (c) Consider an organism of radius \(r_{o}=0.10 \mathrm{~mm}\) and a diffusion coefficient for oxygen transfer of \(D_{\mathrm{AB}}=10^{-8} \mathrm{~m}^{2} / \mathrm{s}\). If \(C_{\mathrm{A}, o}=5 \times 10^{-5} \mathrm{kmol} / \mathrm{m}^{3}\)and \(k_{0}=1.2 \times 10^{-4} \mathrm{kmol} / \mathrm{s}^{*} \mathrm{~m}^{3}\), what is the molar concentration of \(\mathrm{O}_{2}\) at the center of the organism?

A solar pond operates on the principle that heat losses from a shallow layer of water, which acts as a solar absorber, may be minimized by establishing a stable vertical salinity gradient in the water. In practice such a condition may be achieved by applying a layer of pure salt to the bottom and adding an overlying layer of pure water. The salt enters into solution at the bottom and is transferred through the water layer by diffusion, thereby establishing salt-stratified conditions. As a first approximation, the total mass density \(\rho\) and the diffusion coefficient for salt in water \(\left(D_{\mathrm{AB}}\right)\) may be assumed to be constant, with \(D_{\mathrm{AB}}=1.2 \times 10^{-9} \mathrm{~m}^{2} / \mathrm{s}\). (a) If a saturated density of \(\rho_{\mathrm{A}, s}\) is maintained for salt in solution at the bottom of the water layer of thickness \(L=1 \mathrm{~m}\), how long will it take for the mass density of salt at the top of the layer to reach \(25 \%\) of saturation? (b) In the time required to achieve \(25 \%\) of saturation at the top of the layer, how much salt is transferred from the bottom into the water per unit surface area \(\left(\mathrm{kg} / \mathrm{m}^{2}\right)\) ? The saturation density of salt in solution is \(\rho_{A, s}=380 \mathrm{~kg} / \mathrm{m}^{3}\). (c) If the bottom is depleted of salt at the time that the salt density reaches \(25 \%\) of saturation at the top, what is the final (steady-state) density of the salt at the bottom? What is the final density of the salt at the top?

Consider the DVD of Problem 14.49, except now the reacting polymer is blended uniformly with the polycarbonate to reduce manufacturing costs. Assume that a first-order homogeneous chemical reaction takes place between the polymer and oxygen; the reaction rate is proportional to the oxygen molar concentration. (a) Write the governing equation, boundary conditions, and initial condition for the oxygen molar concentration after the DVD is removed from the oxygen- proof pouch, for a DVD of thickness \(2 L\). Do not solve. (b) The DVD will gradually become more opaque over time as the reaction proceeds. The ability to read the DVD will depend on how well the laser light can penetrate through the thickness of the DVD. Therefore, it is important to know the volume-averaged molar concentration of product, \(\bar{C}_{\text {prod }}\), as a function of time. Write an expression for \(\bar{C}_{\text {prod }}\) in terms of the oxygen molar concentration, assuming that every mole of oxygen that reacts with the polymer results in \(p\) moles of product.
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