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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

Referring to Problem 14.34, a more representative model of respiration in a spherical organism is one for which oxygen consumption is governed by a firstorder reaction of the form \(\dot{N}_{\mathrm{A}}=-k_{1} C_{\mathrm{A}}\). (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. Hint: To simplify solution of the species diffusion equation, invoke the transformation \(y \equiv r C_{\mathrm{A}}\). (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 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 \(k_{1}=20 \mathrm{~s}^{-1}\), estimate the corresponding value of the molar concentration at the center of the organism. What is the rate of oxygen consumption by the organism?

An open pan of diameter \(0.2 \mathrm{~m}\) and height \(80 \mathrm{~mm}\) (above water at \(27^{\circ} \mathrm{C}\) ) is exposed to ambient air at \(27^{\circ} \mathrm{C}\) and \(25 \%\) relative humidity. Determine the evaporation rate, assuming that only mass diffusion occurs. Determine the evaporation rate, considering bulk motion.

A large sheet of material \(40 \mathrm{~mm}\) thick contains dissolved hydrogen \(\left(\mathrm{H}_{2}\right)\) having a uniform concentration of \(3 \mathrm{kmol} / \mathrm{m}^{3}\). The sheet is exposed to a fluid stream that causes the concentration of the dissolved hydrogen to be reduced suddenly to zero at both surfaces. This surface condition is maintained constant thereafter. If the mass diffusivity of hydrogen is \(9 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\), how much time is required to bring the density of dissolved hydrogen to a value of \(1.2 \mathrm{~kg} / \mathrm{m}^{3}\) at the center of the sheet?

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?

A He-Xe mixture containing \(0.75\) mole fraction of helium is used for cooling of electronics in an avionics application. At a temperature of \(300 \mathrm{~K}\) and atmospheric pressure, calculate the mass fraction of helium and the mass density, molar concentration, and molecular weight of the mixture. If the cooling system capacity is \(10 \mathrm{~L}\), what is the mass of the coolant?
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