/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 Consider an ideal gas mixture of... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider an ideal gas mixture of \(n\) species. (a) Derive an equation for determining the mass fraction of species \(i\) from knowledge of the mole fraction and the molecular weight of each of the \(n\) species. Derive an equation for determining the mole fraction of species \(i\) from knowledge of the mass fraction and the molecular weight of each of the \(n\) species. (b) In a mixture containing equal mole fractions of \(\mathrm{O}_{2}\), \(\mathrm{N}_{2}\), and \(\mathrm{CO}_{2}\), what is the mass fraction of each species? In a mixture containing equal mass fractions of \(\mathrm{O}_{2}, \mathrm{~N}_{2}\), and \(\mathrm{CO}_{2}\), what is the mole fraction of each species?

Short Answer

Expert verified
In summary, for the mixture with equal mole fractions of \(O_2\), \(N_2\), and \(CO_2\), the mass fractions are approximately 0.3077, 0.2692, and 0.4231, respectively. For the mixture with equal mass fractions of \(O_2\), \(N_2\), and \(CO_2\), the mole fractions are approximately 0.3927, 0.4239, and 0.1834, respectively.

Step by step solution

01

Define mass fraction and mole fraction

Mass fraction is the fraction of the mass of a particular species in the mixture, denoted by \(Y_i\), in comparison to the total mass of the mixture. Similarly, mole fraction is the fraction of moles of a particular species, denoted by \(X_i\), in comparison to the total moles in the mixture.
02

Relate mass fraction and mole fraction using molecular weights

To derive the equations, we need to relate mass and mole fractions using molecular weights of the species. The molecular weight, denoted by (\(M_i\)), is the ratio of the mass of a molecule of species i to one twelfth the mass of a carbon-12 atom. For species i, we have mass fraction (\(Y_i\)) equal to: \[Y_i = \frac{m_i}{m_t}\] And mole fraction (\(X_i\)) equal to: \[X_i = \frac{n_i}{n_t}\] where \(m_i\) is the mass and \(n_i\) is the mole number of species i, whereas \(m_t\) and \(n_t\) are the total mass and total mole number of the mixture, respectively. Now, mass (\(m_i\)) can be expressed as: \[m_i = n_i M_i\]
03

Deriving mass fraction equation

To derive an equation for mass fraction, substitute the expression for mass from step 2 into the mass fraction equation: \[Y_i = \frac{n_i M_i}{m_t}\] Next, express the total mass of the mixture (\(m_t\)) as the sum of the mass of all n species: \[m_t = \sum_{j=1}^{n} n_j M_j\] Now, substitute this expression of \(m_t\) into the equation for \(Y_i\): \[Y_i = \frac{n_i M_i}{\sum_{j=1}^{n} n_j M_j}\] Finally, divide both the numerator and denominator by the total number of moles (\(n_t\)): \[Y_i = \frac{X_i M_i}{\sum_{j=1}^{n} X_j M_j}\]
04

Deriving mole fraction equation

To derive the equation for mole fraction, start with the mass fraction equation and solve for the mole number of species i (\(n_i\)): \[n_i = \frac{m_i}{M_i}\] Substitute this expression into the mole fraction equation: \[X_i = \frac{Y_i/M_i}{n_t}\] Since the total mole number (\(n_t\)) can be expressed as the sum of mole numbers of all n species: \[n_t = \sum_{j=1}^{n} \frac{m_j}{M_j}\] Substitute this expression for \(n_t\) into the equation for \(X_i\): \[X_i = \frac{Y_i/M_i}{\sum_{j=1}^{n} Y_j/M_j}\] #b) Calculation of mass and mole fractions for given mixtures#
05

Define molecular weights of the gases

We need to find the mass fractions and mole fractions of \(O_2\), \(N_2\), and \(CO_2\). The molar masses of these gases are as follows: 1. \(O_2\) - 32 g/mol 2. \(N_2\) - 28 g/mol 3. \(CO_2\) - 44 g/mol
06

Calculate mass fractions for equal mole fractions

Given that the mixture has equal mole fractions, \(X_{O_2} = X_{N_2} = X_{CO_2} = \frac{1}{3}\). Using the derived equation for mass fraction from part (a), we can calculate the mass fractions of these three gases. \[Y_{O_2} = \frac{\frac{1}{3} \times 32}{\frac{1}{3} \times (32 + 28 + 44)} = \frac{32}{104} \approx 0.3077\] \[Y_{N_2} = \frac{\frac{1}{3} \times 28}{\frac{1}{3} \times (32 + 28 + 44)} = \frac{28}{104} \approx 0.2692\] \[Y_{CO_2} = \frac{\frac{1}{3} \times 44}{\frac{1}{3} \times (32 + 28 + 44)} = \frac{44}{104} \approx 0.4231\]
07

Calculate mole fractions for equal mass fractions

Given that the mixture has equal mass fractions, \(Y_{O_2} = Y_{N_2} = Y_{CO_2} = \frac{1}{3}\). Using the derived equation for mole fraction from part (a), we can calculate the mole fractions of these three gases. \[X_{O_2} = \frac{\frac{1}{3}/32}{\frac{1}{3}/32 + \frac{1}{3}/28 + \frac{1}{3}/44} \approx \frac{1}{0.03125 + 0.03571 + 0.02273} \approx 0.3927\] \[X_{N_2} = \frac{\frac{1}{3}/28}{\frac{1}{3}/32 + \frac{1}{3}/28 + \frac{1}{3}/44} \approx \frac{1}{0.03125 + 0.03571 + 0.02273} \approx 0.4239\] \[X_{CO_2} = \frac{\frac{1}{3}/44}{\frac{1}{3}/32 + \frac{1}{3}/28 + \frac{1}{3}/44} \approx \frac{1}{0.03125 + 0.03571 + 0.02273} \approx 0.1834\] In summary, for the mixture with equal mole fractions of \(O_2\), \(N_2\), and \(CO_2\), the mass fractions are approximately 0.3077, 0.2692, and 0.4231, respectively. For the mixture with equal mass fractions of \(O_2\), \(N_2\), and \(CO_2\), the mole fractions are approximately 0.3927, 0.4239, and 0.1834, respectively.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Mass Fraction
Understanding the mass fraction of a gas species within a mixture is essential in thermodynamics, particularly when dealing with ideal gas interactions. The mass fraction (\(Y_i\)) is simply the ratio of the mass of a specific component to the total mass of the mixture.

Here's how it works:- For any gas species \(i\) in a mixture, the mass fraction is calculated using \(Y_i = \frac{m_i}{m_t}\), where \(m_i\) is the mass of species \(i\), and \(m_t\) is the total mass of the mixture.
- This fraction shows how much of the entire mixture's weight is made up by that particular gas.

To express mass fraction in terms of the mole fraction and molecular weights, use the formula:\[Y_i = \frac{X_i M_i}{\sum_{j=1}^{n} X_j M_j}\]

Here:- \(X_i\) is the mole fraction.- \(M_i\) is the molecular weight of the species \(i\).- \(\sum_{j=1}^{n} X_j M_j\) is the weighted sum of the molecular weights of all species in the mixture.

Using mass fractions, engineers and scientists can quickly deduce the composition and potentially the behavior of gas mixtures under various conditions.
Mole Fraction
The mole fraction is a fundamental concept for analyzing gas mixtures. It represents the proportion of one component's moles relative to the total moles in the entire mixture. This is crucial for calculating properties like pressure and concentration in gas systems.

Here's a straightforward breakdown:- The mole fraction (\(X_i\)) is defined by \(X_i = \frac{n_i}{n_t}\), where \(n_i\) is the number of moles of species \(i\), and \(n_t\) represents the total moles in the mixture.
- The sum of all mole fractions in a mixture always equals 1, indicating the complete accounting of all moles in the system.

To determine the mole fraction from mass fractions and molecular weights, the formula is:\[X_i = \frac{Y_i/M_i}{\sum_{j=1}^{n} Y_j/M_j}\]

In this formula:- \(Y_i\) represents the mass fraction.- \(M_i\) denotes the molecular weight of the species \(i\).
- \(\sum_{j=1}^{n} Y_j/M_j\) is the sum of ratios of mass fractions to molecular weights.

Knowing the mole fraction provides key insights into the mixture's properties and allows for assessments of each component's potential chemical reactivity or physical behavior.
Molecular Weight
Molecular weight is an intrinsic property of a substance that describes its mass per mole. It helps in bridging the gap between understanding mass and mole ratios in chemistry.

Let’s simplify:- Molecular weight (\(M_i\)) is essentially the sum of the atomic weights of all atoms in a molecule, conventionally expressed in grams per mole (g/mol).- It's calculated using the periodic table by adding the atomic weights of all the atoms present in the molecule.

Importance in ideal gas mixtures:- Molecular weight allows conversion between moles and mass, which is pivotal in stoichiometry and calculation of proportions.- It impacts the density and behavior of gases in a mixture since lighter gases will generally diffuse faster than heavier ones.

Understanding molecular weight is essential for physical and chemical properties calculation. This includes gas law calculations, such as those involving pressure and volume, based on amounts determined in terms of moles with known molecular weight.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

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?

Hydrogen gas is used in a process to manufacture a sheet material of \(6-\mathrm{mm}\) thickness. At the end of the process, \(\mathrm{H}_{2}\) remains in solution in the material with a uniform concentration of \(320 \mathrm{kmol} / \mathrm{m}^{3}\). To remove \(\mathrm{H}_{2}\) from the material, both surfaces of the sheet are exposed to an airstream at \(500 \mathrm{~K}\) and a total pressure of \(3 \mathrm{~atm}\). Due to contamination, the hydrogen partial pressure is \(0.1 \mathrm{~atm}\) in the airstream, which provides a convection mass transfer coefficient of \(1.5 \mathrm{~m} / \mathrm{h}\). The mass diffusivity and solubility of hydrogen (A) in the sheet material (B) are \(D_{\mathrm{AB}}=2.6 \times 10^{-8} \mathrm{~m}^{2} / \mathrm{s}\) and \(S_{\mathrm{AB}}=160 \mathrm{kmol} / \mathrm{m}^{3} \cdot\) atm, respectively. (a) If the sheet material is left exposed to the airstream for a long time, determine the final content of hydrogen in the material \(\left(\mathrm{kg} / \mathrm{m}^{3}\right)\). (b) Identify and evaluate the parameter that can be used to determine whether the transient mass diffusion process in the sheet can be assumed to be characterized by a uniform concentration at any time during the process. Hint: This situation is analogous to that used to determine the validity of the lumped-capacitance method for a transient heat transfer analysis. (c) Determine the time required to reduce the hydrogen mass density at the center of the sheet to twice the limiting value calculated in part (a).

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.

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?

Gaseous hydrogen at 10 bars and \(27^{\circ} \mathrm{C}\) is stored in a 100 -mm-diameter spherical tank having a steel wall \(2 \mathrm{~mm}\) thick. The molar concentration of hydrogen in the steel is \(1.50 \mathrm{kmol} / \mathrm{m}^{3}\) at the inner surface and negligible at the outer surface, while the diffusion coefficient of hydrogen in steel is approximately \(0.3 \times 10^{-12} \mathrm{~m}^{2} / \mathrm{s}\). What is the initial rate of mass loss of hrogen by diffusion through the tank wall? What is the initial rate of pressure drop within the tank?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.