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Chapter 7: Problem 3

A gas is a mixture of \(22 \% \mathrm{O}_{2}, 33 \% \mathrm{~N}_{2}\), and \(45 \% \mathrm{CO}_{2}\) by volume. Calculate (a) the mole fraction of the constituents in the mixture (b) the mixture molecular weight \(\mathrm{MW}_{\mathrm{m}}\)

Short Answer

Expert verified
The mole fractions of \(\mathrm{O}_{2}, \mathrm{~N}_{2}\), and \(\mathrm{CO}_{2}\) are 0.22, 0.33, and 0.45 respectively. And the mixture molecular weight, \(\mathrm{MW}_{\mathrm{m}}\), is \(32 \times 0.22 + 28 \times 0.33 + 44 \times 0.45\) g/mol.

Step by step solution

01

Mole Fraction Calculation

For a gas, under the same physical conditions, the volume percentage is identical to the mole fraction. Therefore, to calculate the mole fraction, simply convert the given volume percentages into decimal form. For \(\mathrm{O}_{2}\), 22 percent becomes 0.22. For \(\mathrm{N}_{2}\), 33 percent becomes 0.33. For \(\mathrm{CO}_{2}\), 45 percent becomes 0.45.
02

Molecular Weight of Individual Gases

Determining the individual molecular weights of the gases. This can be done using a periodic table or similar resource. For \(\mathrm{O}_{2}\), the double amount of Oxygen atoms will be \(16 \times 2 = 32 \mathrm{~g/mol}\). For \(\mathrm{N}_{2}\), this will be \(14 \times 2 = 28 \mathrm{~g/mol}\). And for \(\mathrm{CO}_{2}\), it is one Carbon atom and 2 Oxygen atoms, hence, \(12 + 16 \times 2 = 44 \mathrm{~g/mol}\).
03

Mixture Molecular Weight Calculation

The Mixture Molecular Weight, \(\mathrm{MW}_{\mathrm{m}}\), is found by the sum of the product of each individual molecular weight and its corresponding mole fraction. Therefore, \(\mathrm{MW}_{\mathrm{m}} = \mathrm{MW}_{\mathrm{O}_{2}} \times X_{\mathrm{O}_{2}} + \mathrm{MW}_{\mathrm{N}_{2}} \times X_{\mathrm{N}_{2}} + \mathrm{MW}_{\mathrm{CO}_{2}} \times X_{\mathrm{CO}_{2}} = 32 \times 0.22 + 28 \times 0.33 + 44 \times 0.45\)

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Key Concepts

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

Mole Fraction
Mole Fraction is a way to express the concentration of a component in a mixture. In the context of gases, it helps us understand the proportion of each individual gas in the mix. The mole fraction (\[ X_i \]) of a component i is calculated by taking the moles of the component and dividing it by the total moles in the mixture. However, for gases under the same conditions of temperature and pressure, the volume percentages provided directly translate to mole fractions due to the ideal gas law.
  • To translate a percentage to a mole fraction, you divide the percent by 100. For example, if the volume percentage of Oxygen (\(\mathrm{O}_2\)) in a gas mixture is 22%, its mole fraction will be 0.22.
  • The same applies to Nitrogen (\(\mathrm{N}_2\)) and Carbon Dioxide (\(\mathrm{CO}_2\)), with mole fractions of 0.33 and 0.45, respectively.
This understanding simplifies the process of analyzing gas mixtures and calculating associated properties.
Gas Mixture
A Gas Mixture is comprised of more than one component, allowing us to model the behavior of real gases in different environments. It is significant in fields like chemistry, environmental science, and engineering. When combining gases, each retains its properties within the mixture, and the overall behavior is determined by the blend of these components. An important concept when dealing with gas mixtures is the partial pressure of each gas. Each gas in a mixture exerts pressure independently, and these individual pressures sum up to the total pressure of the mixture.
  • Each gas contributes to the total pressure in proportion to its mole fraction.
  • The behavior of gases in a mixture is often predicted using Dalton's Law of Partial Pressures, which states that the total pressure of a gas mixture is equal to the sum of the partial pressures of its components.
Gas mixtures provide insights into how gases interact and influence conditions around them, such as in atmospheric studies or industrial processes.
Molecular Weight of Gases
The Molecular Weight of Gases is a critical concept for understanding the mass and composition of a gas. The molecular weight (MW) of a gas is calculated by summing the atomic weights of all atoms present in a molecule and is expressed in grams per mole (g/mol).To determine the molecular weight of a gas mixture, calculate the average based on individual molecular weights and mole fractions of each gas:
  • First, find the molecular weight of each gas. For example, \( \mathrm{O}_2 = 32 \text{ g/mol} \), \( \mathrm{N}_2 = 28 \text{ g/mol} \), and \( \mathrm{CO}_2 = 44 \text{ g/mol} \).
  • Then, use the formula: \[ \mathrm{MW}_m = (\mathrm{MW}_{\mathrm{O}_2} \times X_{\mathrm{O}_2}) + (\mathrm{MW}_{\mathrm{N}_2} \times X_{\mathrm{N}_2}) + (\mathrm{MW}_{\mathrm{CO}_2} \times X_{\mathrm{CO}_2}) \]
  • Substitute the values: \[ \mathrm{MW}_m = 32 \times 0.22 + 28 \times 0.33 + 44 \times 0.45 \]
Using these calculations provides the mixture's molecular weight, enabling further analysis like reaction predictions, transportation calculations, and more.

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Most popular questions from this chapter

Consider the combustion of butane and air according to $$ \begin{aligned} \mathrm{C}_{4} \mathrm{H}_{10}+8\left(\mathrm{O}_{2}+3.76 \mathrm{~N}_{2}\right) \rightarrow & 4 \mathrm{CO}_{2}+5 \mathrm{H}_{2} \mathrm{O} \\ &+1.5 \mathrm{O}_{2}+8(3.76) \mathrm{N}_{2} \end{aligned} $$ Calculate the (a) fuel-to-air ratio, \(f\) (b) equivalence ratio, \(\varphi\), for this reaction

A combustion chamber uses a prediffuser with a sudden area expansion (known as a dump diffuser) to decelerate the flow of air \((\gamma=1.4)\) before entering the combustor. Assuming the inlet Mach number to the dump diffuser is \(M_{1}=0.5\), the area ratio of the dump diffuser is \(A_{2} / A_{I}=2.0\), calculate (a) exit Mach number \(M_{2}\) (b) the ratio of total pressures, i.e., \(p_{\mathrm{t} 2} / p_{\mathrm{t} 1}\)

Write the stoichiometric combustion of methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) in (dry) theoretical air. If the lower heating value, LHV, of methanol is \(21.2 \mathrm{MJ} / \mathrm{kg}\), calculate its higher heating value, HHV.

Calculate the adiabatic flame temperature in the following reaction of octane in dry air at reference temperature \(\left(T_{\mathrm{f}}=298.16 \mathrm{~K}\right)\) and pressure (1 bar), $$ \begin{aligned} \mathrm{C}_{8} \mathrm{H}_{18(1)}+14\left(\mathrm{O}_{2}+3.76 \mathrm{~N}_{2}\right) \rightarrow & 8 \mathrm{CO}_{2(\mathrm{~g})}+9 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})} \\ &+\frac{3}{2} \mathrm{O}_{2(\mathrm{~g})}+14(3.76) N_{2(\mathrm{~g})} \end{aligned} $$ Assume the average molar specific heats of the chemical compounds at constant pressure are: $$ \begin{aligned} &\bar{c}_{p_{\mathrm{CO}_{2}}}=54.31 \mathrm{~kJ} / \mathrm{kmol} \cdot \mathrm{K} \quad \bar{c}_{p_{\mathrm{O}_{2}}}=34.88 \mathrm{~kJ} / \mathrm{kmol} \cdot \mathrm{K} \\ &\bar{c}_{p_{\mathrm{N}_{2}}}=32.7 \mathrm{~kJ} / \mathrm{kmol} \cdot \mathrm{K} \quad \bar{c}_{p_{\mathrm{H}_{2} \mathrm{O}}}=41.22 \mathrm{~kJ} / \mathrm{kmol} \cdot \mathrm{K} \end{aligned} $$ The standard heats of formation in this reaction are: $$ \begin{aligned} \left.\Delta \bar{h}_{f}^{o}\right|_{C_{8} H_{18(1)}} &=-249,930 \mathrm{~kJ} / \mathrm{kmol}, \\ \left.\Delta \bar{h}_{f}^{o}\right|_{\mathrm{H}_{2} \mathrm{O}(\mathrm{g})} &=-241,827 \mathrm{~kJ} / \mathrm{kmol}, \\ \left.\Delta \bar{h}_{f}^{o}\right|_{\mathrm{CO} 2(\mathrm{~g})} &=-393,522 \mathrm{~kJ} / \mathrm{kmol} \end{aligned} $$

Consider the flow in an afterburner, in dry mode, as shown. The flow upstream of the flameholder, i.e., the afterburner inlet, is characterized by: $$ \begin{aligned} &p_{1}=33 \mathrm{kPa}, \quad T_{1}=675 \mathrm{~K}, \quad M_{1}=0.3, \quad \dot{m}_{1}=100 \mathrm{~kg} / \mathrm{s} \\ &\gamma_{1}=1.33, \quad R_{1}=287 \mathrm{~J} / \mathrm{kgK} \end{aligned} $$ Assuming the flameholder drag coefficient is \(C_{\mathrm{D}}=1.0\), which is based on the duct cross-sectional area, and the flow is \(a d i\) abatic, calculate (a) the duct cross sectional area, \(A_{1}\), in \(m^{2}\) (b) flameholder drag, \(D_{\text {flameholder }}\), in \(\mathrm{kN}\) (c) static pressure in station 2, where \(M_{2}=0.371\), \(\gamma_{2}=\gamma_{1}=1.33\) and wall friction drag coefficient is neglected (i.e., \(C_{\mathrm{fw}}=0\) ) as shown (d) static temperature in station \(2, T_{2}\), in \(\mathrm{K}\)
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