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Chapter 14: Problem 2

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.

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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鈥檚 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.

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Most popular questions from this chapter

To enhance the effective surface, and hence the chemical reaction rate, catalytic surfaces often take the form of porous solids. One such solid may be visualized as consisting of a large number of cylindrical pores, each of diameter \(D\) and length \(L\). To enhance the effective surface, and hence the chemical reaction rate, catalytic surfaces often take the form of porous solids. One such solid may be visualized as consisting of a large number of cylindrical pores, each of diameter \(D\) and length \(L\). Consider conditions involving a gaseous mixture of \(\mathrm{A}\) and B for which species A is chemically consumed at the catalytic surface. The reaction is known to be first order, and the rate at which it occurs per unit area of the surface may be expressed as \(k_{1}^{\prime \prime} C_{\mathrm{A}}\), where \(k_{1}^{\prime \prime}(\mathrm{m} / \mathrm{s})\) is the reaction rate constant and \(C_{\mathrm{A}}\left(\mathrm{kmol} / \mathrm{m}^{3}\right)\) is the local molar concentration of species A. Under steadystate conditions, flow over the porous solid is known to maintain a fixed value of the molar concentration \(C_{\mathrm{A}, 0}\) at the pore mouth. Beginning from fundamentals, obtain the differential equation that governs the variation of \(C_{\mathrm{A}}\) with distance \(x\) along the pore. Applying appropriate boundary conditions, solve the equation to obtain an expression for \(C_{\mathrm{A}}(x)\).

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.

Pulverized coal pellets, which may be approximated as carbon spheres of radius \(r_{o}=1 \mathrm{~mm}\), are burned in a pure oxygen atmosphere at \(1450 \mathrm{~K}\) and 1 atm. Oxygen is transferred to the particle surface by diffusion, where it is consumed in the reaction \(\mathrm{C}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}\). The reaction rate is first order and of the form \(\dot{N}_{\mathrm{O}_{2}}^{\prime \prime}=\) \(-k_{1}^{\prime \prime} C_{\mathrm{O}_{2}}\left(r_{o}\right)\), where \(k_{1}^{\prime \prime}=0.1 \mathrm{~m} / \mathrm{s}\). Neglecting changes in \(r_{o}\), determine the steady-state \(\mathrm{O}_{2}\) molar consumption rate in \(\mathrm{kmol} / \mathrm{s}\). At \(1450 \mathrm{~K}\), the binary diffusion coefficient for \(\mathrm{O}_{2}\) and \(\mathrm{CO}_{2}\) is \(1.71 \times 10^{-4} \mathrm{~m}^{2} / \mathrm{s}\).

The presence of \(\mathrm{CO}_{2}\) in solution is essential to the growth of aquatic plant life, with \(\mathrm{CO}_{2}\) used as a reactant in the photosynthesis. Consider a stagnant body of water in which the concentration of \(\mathrm{CO}_{2}\left(\rho_{\mathrm{A}}\right)\) is everywhere zero. At time \(t=0\), the water is exposed to a source of \(\mathrm{CO}_{2}\), which maintains the surface \((x=0)\) concentration at a fixed value \(\rho_{\mathrm{A}, 0}\). For time \(t>0, \mathrm{CO}_{2}\) will begin to accumulate in the water, but the accumulation is inhibited by \(\mathrm{CO}_{2}\) consumption due to photosynthesis. The time rate at which this consumption occurs per unit volume is equal to the product of a reaction rate constant \(k_{1}\) and the local \(\mathrm{CO}_{2}\) concentration \(\rho_{\mathrm{A}}(x, t)\). (a) Write (do not derive) a differential equation that could be used to determine \(\rho_{\mathrm{A}}(x, t)\) in the water. What does each term in the equation represent physically? (b) Write appropriate boundary conditions that could be used to obtain a particular solution, assuming a "deep" body of water. What would be the form of this solution for the special case of negligible \(\mathrm{CO}_{2}\) consumption \(\left(k_{1} \approx 0\right)\) ?
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