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Chapter 14: Problem 16

The molar analysis of a gas mixture at \(290 \mathrm{~K}\) and \(250 \mathrm{kPa}\) is 65 percent \(\mathrm{N}_{2}, 20\) percent \(\mathrm{O}_{2}\), and 15 percent \(\mathrm{CO}_{2}\). Determine the mass fraction and partial pressure of each gas.

Short Answer

Expert verified
Question: Calculate the mass fractions and partial pressures of N2, O2, and CO2 in a gas mixture with a total pressure of 250 kPa, given the following molar analysis: N2 - 65%, O2 - 20%, and CO2 - 15%. Answer: The mass fractions of N2, O2, and CO2 in the gas mixture are 0.60, 0.25, and 0.15, respectively. The partial pressures of N2, O2, and CO2 in the mixture are 162.50 kPa, 50 kPa, and 37.50 kPa, respectively.

Step by step solution

01

Calculate the molar fractions of each gas

We are given the molar analysis of the gas mixture, so we can directly use the given percentages to find the molar fractions. The molar fraction of a gas is the ratio of its molar ratio to the total molar ratio. Molar fractions: - Nitrogen (N2): 65% - Oxygen (O2): 20% - Carbon Dioxide (CO2): 15%
02

Calculate the mass fraction

The mass fraction of a gas is the ratio of its mass to the total mass of the mixture. To find the mass fraction, we first need the molar mass of each gas: - Molar mass of N2: \(28 \mathrm{g/mol}\) - Molar mass of O2: \(32 \mathrm{g/mol}\) - Molar mass of CO2: \(44 \mathrm{g/mol}\) Next, we'll multiply the molar fraction of each gas by its respective molar mass, and then divide the result by the sum of the weighted molar masses of all the gases: Mass fraction of N2: \(\frac{0.65 \times 28}{(0.65\times 28) + (0.2\times 32) + (0.15\times44)}\) Mass fraction of O2: \(\frac{0.2 \times 32}{(0.65\times 28) + (0.2\times 32) + (0.15\times44)}\) Mass fraction of CO2: \(\frac{0.15 \times 44}{(0.65\times 28) + (0.2\times 32) + (0.15\times44)}\)
03

Calculate the partial pressures

The partial pressure of each gas can be determined by multiplying the total pressure of the mixture by its respective molar fraction: Partial pressure of N2: \(250\ \mathrm{kPa} \times 0.65\) Partial pressure of O2: \(250\ \mathrm{kPa} \times 0.20\) Partial pressure of CO2: \(250\ \mathrm{kPa} \times 0.15\) Now we can calculate the mass fractions and partial pressures for each gas. Mass fraction of N2: \(\frac{0.65 \times 28}{(0.65\times 28) + (0.2\times 32) + (0.15\times44)} = 0.60\) Mass fraction of O2: \(\frac{0.2 \times 32}{(0.65\times 28) + (0.2\times 32) + (0.15\times44)} = 0.25\) Mass fraction of CO2: \(\frac{0.15 \times 44}{(0.65\times 28) + (0.2\times 32) + (0.15\times44)} = 0.15\) Partial pressure of N2: \(250\ \mathrm{kPa} \times 0.65 = 162.50\ \mathrm{kPa}\) Partial pressure of O2: \(250\ \mathrm{kPa} \times 0.20 = 50\ \mathrm{kPa}\) Partial pressure of CO2: \(250\ \mathrm{kPa} \times 0.15 = 37.50\ \mathrm{kPa}\) Therefore, the mass fractions of N2, O2, and CO2 are 0.60, 0.25, and 0.15, respectively, and their partial pressures are 162.50 kPa, 50 kPa, and 37.50 kPa, respectively.

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Key Concepts

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

Gas Mixture Composition
Understanding the composition of a gas mixture is key to solving a variety of chemical and physical problems. It involves determining the percentage of each gas present in a mixture. The molar analysis, often provided as percentages, indicates how much of each component there is relative to the total amount.

For instance, when a molar analysis says a mixture contains 65% \textbf{N}\(_2\), 20% \textbf{O}\(_2\), and 15% \textbf{CO}\(_2\), we interpret these percentages as molar fractions. This means that out of every 100 parts by mole of the gas mixture, 65 parts are nitrogen, 20 are oxygen, and 15 are carbon dioxide. Having this understanding of molar fractions sets the stage for calculating important derived properties such as mass fractions and partial pressures, which can then be applied in engineering calculations, environmental monitoring, and various scientific research.
Molar Mass
When dealing with gas mixtures, the molar mass is a vital concept. The molar mass tells us the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It is crucial for converting between moles and grams because it serves as a conversion factor.

Molar Mass in Calculations

As an example, nitrogen (\textbf{N}\(_2\)) has a molar mass of 28 g/mol, oxygen (\textbf{O}\(_2\)) has a molar mass of 32 g/mol, and carbon dioxide (\textbf{CO}\(_2\)) has a molar mass of 44 g/mol. These values are used to translate the molar fractions from our gas mixture composition into mass fractions. Knowing the molar masses allows us to how much actual mass each component contributes relative to the entire mixture, which is especially useful for chemical reactions and physical processes within various fields such as respiratory sciences, meteorology, and materials engineering.
Mass Fraction
Another fundamental concept in understanding gas mixtures is the mass fraction, which describes the proportion of a component's mass in relation to the total mass of the mixture. Unlike the molar fraction, which is based on the number of moles, the mass fraction accounts for the differing molar masses of gases in the mixture.

To calculate mass fractions, we multiply the molar fraction by the molar mass of each gas and then divide by the total mass of all gases, weighted by their molar fractions and molar masses. To simplify, the mass fraction is essentially the part of the total mass contributed by that specific gas. It's an important measure when considering the physical properties of the gas mixture, such as its density or how it will behave when subjected to different temperatures and pressures in thermodynamic processes.

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Most popular questions from this chapter

A recent attempt to circumnavigate the world in a balloon used a helium-filled balloon whose volume was \(7240 \mathrm{~m}^{3}\) and surface area was \(1800 \mathrm{~m}^{2}\). The skin of this balloon is \(2 \mathrm{~mm}\) thick and is made of a material whose helium diffusion coefficient is \(1 \times 10^{-9} \mathrm{~m}^{2} / \mathrm{s}\). The molar concentration of the helium at the inner surface of the balloon skin is \(0.2 \mathrm{kmol} / \mathrm{m}^{3}\) and the molar concentration at the outer surface is extremely small. The rate at which helium is lost from this balloon is (a) \(0.26 \mathrm{~kg} / \mathrm{h}\) (b) \(1.5 \mathrm{~kg} / \mathrm{h}\) (c) \(2.6 \mathrm{~kg} / \mathrm{h}\) (d) \(3.8 \mathrm{~kg} / \mathrm{h}\) (e) \(5.2 \mathrm{~kg} / \mathrm{h}\)

A glass bottle washing facility uses a well agi(Es) tated hot water bath at \(50^{\circ} \mathrm{C}\) with an open top that is placed on the ground. The bathtub is \(1 \mathrm{~m}\) high, \(2 \mathrm{~m}\) wide, and \(4 \mathrm{~m}\) long and is made of sheet metal so that the outer side surfaces are also at about \(50^{\circ} \mathrm{C}\). The bottles enter at a rate of 800 per minute at ambient temperature and leave at the water temperature. Each bottle has a mass of \(150 \mathrm{~g}\) and removes \(0.6 \mathrm{~g}\) of water as it leaves the bath wet. Makeup water is supplied at \(15^{\circ} \mathrm{C}\). If the average conditions in the plant are \(1 \mathrm{~atm}, 25^{\circ} \mathrm{C}\), and 50 percent relative humidity, and the average temperature of the surrounding surfaces is \(15^{\circ} \mathrm{C}\), determine (a) the amount of heat and water removed by the bottles themselves per second, \((b)\) the rate of heat loss from the top surface of the water bath by radiation, natural convection, and evaporation, \((c)\) the rate of heat loss from the side surfaces by natural convection and radiation, and \((d)\) the rate at which heat and water must be supplied to maintain steady operating conditions. Disregard heat loss through the bottom surface of the bath and take the emissivities of sheet metal and water to be \(0.61\) and \(0.95\), respectively.

Determine the mole fraction of the water vapor at the surface of a lake whose temperature is \(15^{\circ} \mathrm{C}\) and compare it to the mole fraction of water in the lake. Take the atmospheric pressure at lake level to be \(92 \mathrm{kPa}\).

The basic equation describing the diffusion of one medium through another stationary medium is (a) \(j_{A}=-C D_{A B} \frac{d\left(C_{A} / C\right)}{d x}\) (b) \(j_{A}=-D_{A B} \frac{d\left(C_{A} / C\right)}{d x}\) (c) \(j_{A}=-k \frac{d\left(C_{A} / C\right)}{d x}\) (d) \(j_{A}=-k \frac{d T}{d x}\) (e) none of them

A thick nickel wall is exposed to pure hydrogen gas at \(165^{\circ} \mathrm{C}\) on one side of its surface. The hydrogen concentration at the wall surface is constant. Determine the hydrogen concentration at the penetration depth in percentage of its concentration at the wall surface.
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