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Chapter 9: Problem 3

Methane is burned with \(20 \%\) excess air. At \(1250 \mathrm{~K}\) the equilibrium composition of the product is: \(\begin{array}{rlcccc}\text { Species } i: & \mathrm{CO}_{2} & \mathrm{H}_{2} \mathrm{O} & \mathrm{O}_{2} & \mathrm{~N}_{2} & \mathrm{NO} \\ x_{i}: & 0.0803 & 0.160 & 0.0325 & 0.727 & 0.000118\end{array}\) Determine the mean molecular weight \(M\) and gas constant \(R\) of the mixture, and the mass fraction of the pollutant nitric oxide in parts per million.

Short Answer

Expert verified
Mean molecular weight is 28.57 g/mol, gas constant is 291.04 J/(kg K), and NO is 123.62 ppm.

Step by step solution

01

Calculating the Mean Molecular Weight

To calculate the mean molecular weight \(M\) of the mixture, use the formula \( M = \sum x_i M_i \), where \(x_i\) are the mole fractions and \(M_i\) are the molecular weights of each component.- \(M_{\mathrm{CO}_2} = 44 \text{ g/mol}\)- \(M_{\mathrm{H}_2\mathrm{O}} = 18 \text{ g/mol}\)- \(M_{\mathrm{O}_2} = 32 \text{ g/mol}\)- \(M_{\mathrm{N}_2} = 28 \text{ g/mol}\)- \(M_{\mathrm{NO}} = 30 \text{ g/mol}\)Thus, \(M = 0.0803 \times 44 + 0.160 \times 18 + 0.0325 \times 32 + 0.727 \times 28 + 0.000118 \times 30 = 28.57 \text{ g/mol}\).
02

Calculating the Gas Constant for the Mixture

The specific gas constant \(R\) for the mixture is obtained using the equation \( R = \frac{R_u}{M} \), where \(R_u = 8.314 \text{ J/(mol K)}\) is the universal gas constant and \(M\) is the mean molecular weight in kg/mol.Convert \(M\) to kg/mol: \(28.57 \text{ g/mol} = 0.02857 \text{ kg/mol}\).So, \(R = \frac{8.314}{0.02857} = 291.04 \text{ J/(kg K)}\).
03

Calculating the Mass Fraction of Nitric Oxide

To find the mass fraction of \(\text{NO}\) in parts per million (ppm), use:\[ \text{Mass fraction of NO (ppm)} = \left( \frac{x_{\text{NO}} \cdot M_{\text{NO}}}{M} \right) \times 10^6 \]Substituting the values:\[ \text{Mass fraction of NO (ppm)} = \left( \frac{0.000118 \times 30}{28.57} \right) \times 10^6 \approx 123.62 \text{ ppm}\].

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Key Concepts

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

Mean Molecular Weight
The mean molecular weight is an average weight of all the molecules present in a mixture. Unlike a pure substance, mixtures contain multiple species which contribute differently to the overall weight. In this exercise, each component has a specific mole fraction and molecular weight. These two factors are used to find the mean molecular weight using the formula:
  • \( M = \sum x_i M_i \)
Where \(x_i\) is the mole fraction, and \(M_i\) is the molecular weight of each component in the mixture. The given data includes various components such as carbon dioxide, water vapor, oxygen, nitrogen, and nitric oxide. Each has known molecular weights:
  • \(M_{\mathrm{CO}_2} = 44 \text{ g/mol}\)
  • \(M_{\mathrm{H}_2\mathrm{O}} = 18 \text{ g/mol}\)
  • \(M_{\mathrm{O}_2} = 32 \text{ g/mol}\)
  • \(M_{\mathrm{N}_2} = 28 \text{ g/mol}\)
  • \(M_{\mathrm{NO}} = 30 \text{ g/mol}\)
To find the mean molecular weight, these individual weights are multiplied by their respective mole fractions and summed up, resulting in a mean molecular weight of about \(28.57 \text{ g/mol}\). This value reflects the average mass of one mole of the gas mixture present in the combustion products.
Gas Constant
The specific gas constant \(R\) is a derivative of the universal gas constant \(R_u\), which applies to any gas. \(R_u\) has a value of \(8.314 \text{ J/(mol K)}\). However, mixtures require a calculated \(R\) value that's specific to them. The relationship to find this specific gas constant is:
  • \( R = \frac{R_u}{M} \)
Where \(M\) is the mean molecular weight in kilograms per mol. To accurately calculate \(R\), the molecular weight obtained earlier must be converted from g/mol to kg/mol. This conversion is crucial because it aligns with the units of the universal gas constant. Here, \(28.57 \text{ g/mol}\) is \(0.02857 \text{ kg/mol}\). Using this conversion, the specific gas constant for this mixture is found to be approximately \(291.04 \text{ J/(kg K)}\). This value indicates how the pressure and volume of the gas will respond to temperature changes.
Nitric Oxide Pollution
Nitric oxide (NO) is a significant byproduct of certain combustion processes, including the burning of methane. It is a component of pollution that contributes to smog and has harmful effects on human health and the environment. To quantify its presence in the combustion products, the mass fraction of NO is calculated in parts per million (ppm) as shown below:
  • \( \text{Mass fraction of NO (ppm)} = \left( \frac{x_{\text{NO}} \cdot M_{\text{NO}}}{M} \right) \times 10^6 \)
Given the mole fraction \(x_{\text{NO}} = 0.000118\) and its molecular weight \(M_{\text{NO}} = 30 \text{ g/mol}\), the mass fraction is calculated with respect to the mean molecular weight of the mixture. The result is approximately \(123.62 \text{ ppm}\), indicating the concentration of NO in the mixture. Understanding and controlling such concentrations is vital, as they help in determining compliance with environmental standards and for taking actions to reduce pollution.

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Most popular questions from this chapter

Insulations used in house wall cavities have high thermal resistances (high \(\mathrm{R}\) values; see Exercise 1-5) because of their porous structure: the air filling the pores has a low thermal conductivity. If water vapor diffuses into the insulation and condenses, the air is displaced and a smaller \(R\) value results. In winter, water vapor in a humidified room can diffuse through the inner drywall and condense in the insulation. Estimate the rate of water vapor diffusion across a \(3 \mathrm{~m} \times 6 \mathrm{~m}\) section of drywall when the room air is \(295 \mathrm{~K}\) and \(60 \% \mathrm{RH}\), and the partial pressure of the water vapor in the insulation is \(440 \mathrm{~Pa}\). The drywall is \(1 \mathrm{~cm}\) thick and has a permeability to water vapor of approximately \(3.4 \times 10^{-6} \mathrm{~m}^{3}(\mathrm{STP}) / \mathrm{m}^{2} \mathrm{~s}(\mathrm{~atm} / \mathrm{m})\).

An automobile catalytic converter has a ceramic matrix with \(1 \mathrm{~mm}\)-square passages and \(0.2 \mathrm{~mm}\)-thick walls. The walls are coated with a \(20 \mu \mathrm{m}\)-thick layer of porous alumina that is impregnated with a platinum alloy catalyst (the washcoat). The purpose of the porous washcoat can be viewed as giving an effective reaction rate constant \(k_{\text {eff }}^{\prime \prime}\) based on \(s\)-surface area that is much larger than if the catalyst were simply coated onto a solid wall. (i) If the actual rate constant for \(\mathrm{CO}\) oxidation is \(k^{\prime \prime}=7 \times 10^{-3} \mathrm{~m} / \mathrm{s}, \mathscr{D}_{1, \text { eff }}\) in the washcoat is \(1.6 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\), and the washcoat area-to-volume ratio is \(a_{p}=5 \times 10^{6} \mathrm{~cm}^{2} / \mathrm{cm}^{3}\), determine \(k_{\text {eff }}^{\prime \prime}\) (ii) If the exhaust is at \(800 \mathrm{~K}, 1.1\) bar, is the removal of \(\mathrm{CO}\) controlled by diffusion of CO from the gas stream or by reaction within the washcoat?

A 3 m-diameter spherical steel shell is raised out of a thermal soaking pit maintained at \(700^{\circ} \mathrm{C}\), and a worker is required to stand \(3 \mathrm{~m}\) in front of the shell. (i) Estimate the irradiation on the worker for a steel emittance of \(0.21 .\) (ii) If the worker's bare skin is wet with sweat, estimate the temperature the skin will attain in air at \(23^{\circ} \mathrm{C}\) and \(50 \%\) relative humidity. Use \(R=0.01\left[\mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\right]^{-1}\) as the thermal resistance of a skin and fat layer, and make appropriate simplifying assumptions.

Experimental oxidation data for a titanium alloy, used to sheath hypersonic vehicle nose cones, indicates that the total weight gain can be described by a parabolic law of the form \(w^{2}=C t\), where \(C=480 \exp \left(-E_{a} / \mathscr{R T}\right) \mathrm{g}^{2} / \mathrm{cm}^{4} \mathrm{~s}\), and \(E_{a}=61,800\) \(\mathrm{cal} / \mathrm{mol}\). Examination of oxidized samples shows that about \(20 \%\) of the oxygen goes to form a rutile \(\left(\mathrm{Ti} \mathrm{O}_{2}\right)\) surface scale. Assuming that the diffusion coefficient for oxygen in alpha titanium can be expressed as \(\mathscr{D}_{12}=\mathscr{D}_{0} \exp \left(-E_{a} / \mathscr{R} T\right)\), with \(E_{a}\) again \(61,800 \mathrm{cal} / \mathrm{mol}\), estimate \(\mathscr{D}_{0}\). Show also that the parabolic oxidation law implies diffusion across the rutile scale characterized by a linear concentration gradient. Use an alloy density of \(4400 \mathrm{~kg} / \mathrm{m}^{3}\).

Plot a graph of the mass fraction of water vapor in saturated water vapor-air mixtures at 1 atm pressure for \(273.15 \mathrm{~K}
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