/*! 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 17 A gas mixture consists of \(7 \m... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 17

A gas mixture consists of \(7 \mathrm{lbm}\) of \(\mathrm{O}_{2}, 8 \mathrm{lbm}\) of \(\mathrm{N}_{2}\), and \(10 \mathrm{lbm}\) of \(\mathrm{CO}_{2}\). Determine \((a)\) the mass fraction of each component, \((b)\) the mole fraction of each component, and (c) the average molar mass of the mixture.

Short Answer

Expert verified
Question: Calculate the mass fractions, mole fractions, and average molar mass of a gas mixture containing 7 lbm of O鈧�, 8 lbm of N鈧�, and 10 lbm of CO鈧�. Answer: The mass fractions are as follows: - Mass fraction of O鈧� = 0.28 - Mass fraction of N鈧� = 0.32 - Mass fraction of CO鈧� = 0.4 The mole fractions are as follows: - Mole fraction of O鈧� = 0.2988 - Mole fraction of N鈧� = 0.3903 - Mole fraction of CO鈧� = 0.3104 The average molar mass of the mixture = 32.298 g/mol.

Step by step solution

01

Identify the molar mass of each gas component

Using periodic table we gather the molar mass of each gas component: (O) Oxygen: 16 g/mol (N) Nitrogen: 14 g/mol (C) Carbon: 12 g/mol For the given gases, the molar mass are: O鈧� = 2 脳 16 = 32 g/mol N鈧� = 2 脳 14 = 28 g/mol CO鈧� = 12 + (2 脳 16) = 44 g/mol
02

Calculate the total mass of the mixture

Sum the mass values of the individual components. Total mass = mass of O鈧� + mass of N鈧� + mass of CO鈧� Total mass = 7 lbm + 8 lbm + 10 lbm = 25 lbm
03

Calculate the mass fractions

(a) Determine the mass fraction of each component by dividing the mass of each component by the total mass of the mixture Mass fraction of O鈧� = mass of O鈧� / total mass Mass fraction of O鈧� = 7 lbm / 25 lbm = 0.28 Mass fraction of N鈧� = mass of N鈧� / total mass Mass fraction of N鈧� = 8 lbm / 25 lbm = 0.32 Mass fraction of CO鈧� = mass of CO鈧� / total mass Mass fraction of CO鈧� = 10 lbm / 25 lbm = 0.4
04

Calculate the moles of each component

(b) Determine the mole fraction of each component by dividing the mass of each component by its respective molar mass Moles of O鈧� = mass of O鈧� / molar mass of O鈧� Moles of O鈧� = 7 lbm / 32 g/mol = 0.21875 lbm-mol Moles of N鈧� = mass of N鈧� / molar mass of N鈧� Moles of N鈧� = 8 lbm / 28 g/mol = 0.28571 lbm-mol Moles of CO鈧� = mass of CO鈧� / molar mass of CO鈧� Moles of CO鈧� = 10 lbm / 44 g/mol = 0.22727 lbm-mol
05

Calculate the total moles of the mixture

Sum the moles of each individual gas component. Total moles = moles of O鈧� + moles of N鈧� + moles of CO鈧� Total moles = 0.21875 lbm-mol + 0.28571 lbm-mol + 0.22727 lbm-mol = 0.73173 lbm-mol
06

Calculate the mole fractions

Determine the mole fraction of each component by dividing the moles of each component by the total moles of the mixture Mole fraction of O鈧� = moles of O鈧� / total moles Mole fraction of O鈧� = 0.21875 lbm-mol / 0.73173 lbm-mol = 0.2988 Mole fraction of N鈧� = moles of N鈧� / total moles Mole fraction of N鈧� = 0.28571 lbm-mol / 0.73173 lbm-mol = 0.3903 Mole fraction of CO鈧� = moles of CO鈧� / total moles Mole fraction of CO鈧� = 0.22727 lbm-mol / 0.73173 lbm-mol = 0.3104
07

Calculate the average molar mass of the mixture

(c) Determine the average molar mass of the mixture by taking the sum of the product of the mole fraction and molar mass of each component Average molar mass = 危(mole fraction 脳 molar mass) Average molar mass = (0.2988 脳 32) + (0.3903 脳 28) + (0.3104 脳 44) Average molar mass = 32.298 g/mol

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Key Concepts

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

Mass Fraction
In the analysis of gas mixtures, it's important to understand the concept of **mass fraction**. This is a way of expressing the proportion of each gas present in the mixture relative to the total mass. The mass fraction is calculated by dividing the mass of a specific gas component by the total mass of all components combined. For example, in the case of our gas mixture:
  • Mass of \(\mathrm{O}_2\): 7 lbm
  • Mass of \(\mathrm{N}_2\): 8 lbm
  • Mass of \(\mathrm{CO}_2\): 10 lbm
  • Total mass: \(7 + 8 + 10 = 25\) lbm
To find the mass fraction of \(\mathrm{O}_2\), \(\mathrm{N}_2\), and \(\mathrm{CO}_2\), you divide:
  • Mass Fraction of \(\mathrm{O}_2\) = \( \frac{7}{25} \approx 0.28 \)
  • Mass Fraction of \(\mathrm{N}_2\) = \( \frac{8}{25} \approx 0.32 \)
  • Mass Fraction of \(\mathrm{CO}_2\) = \( \frac{10}{25} = 0.4 \)
These fractions will always sum to 1. This concept helps chemists and engineers understand composition and concentration in mixtures.
Mole Fraction
**Mole fraction** offers another way to describe the composition of a gas mixture. While mass fractions focus on weight, mole fractions use the number of molecules or moles. To calculate mole fractions, first determine the number of moles of each component using their respective molar masses.
  • Molar Mass of \(\mathrm{O}_2\) = 32 g/mol
  • Molar Mass of \(\mathrm{N}_2\) = 28 g/mol
  • Molar Mass of \(\mathrm{CO}_2\) = 44 g/mol
From the given mass:
  • Moles of \(\mathrm{O}_2\) = \( \frac{7 \text{ lbm}}{32 \text{ g/mol}} \approx 0.21875 \text{ lbm-mol} \)
  • Moles of \(\mathrm{N}_2\) = \( \frac{8 \text{ lbm}}{28 \text{ g/mol}} \approx 0.28571 \text{ lbm-mol} \)
  • Moles of \(\mathrm{CO}_2\) = \( \frac{10 \text{ lbm}}{44 \text{ g/mol}} \approx 0.22727 \text{ lbm-mol} \)
Next, you find the total moles and calculate the fraction:
  • Mole Fraction of \(\mathrm{O}_2\) = \( \frac{0.21875}{0.73173} \approx 0.2988 \)
  • Mole Fraction of \(\mathrm{N}_2\) = \( \frac{0.28571}{0.73173} \approx 0.3903 \)
  • Mole Fraction of \(\mathrm{CO}_2\) = \( \frac{0.22727}{0.73173} \approx 0.3104 \)
The mole fraction is crucial for understanding the chemical and physical behavior of gas mixtures in various applications.
Molar Mass
Understanding **molar mass** is essential in chemistry, particularly in the study of gas mixtures. Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). Each element has a specific molar mass based on its atomic mass. Here are the molar masses for the gases in our example:
  • Oxygen (\(\mathrm{O}_2\)): 32 g/mol
  • Nitrogen (\(\mathrm{N}_2\)): 28 g/mol
  • Carbon Dioxide (\(\mathrm{CO}_2\)): 44 g/mol
To find the average molar mass of a mixture, use a weighted average based on the mole fractions of each component:
  • Average Molar Mass = \( (0.2988 \times 32) + (0.3903 \times 28) + (0.3104 \times 44) \)
  • Average Molar Mass = 32.298 g/mol
This calculation is important in applications where properties like diffusion rates or reaction kinetics depend on the average molecular weight.

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Most popular questions from this chapter

Exposure to high concentration of gaseous ammonia can cause lung damage. The acceptable shortterm ammonia exposure level set by the Occupational Safety and Health Administration (OSHA) is 35 ppm for 15 minutes. Consider a vessel filled with gaseous ammonia at \(30 \mathrm{~mol} / \mathrm{L}\), and a 10 -cm- diameter circular plastic plug with a thickness of \(2 \mathrm{~mm}\) is used to contain the ammonia inside the vessel. The ventilation system is capable of keeping the room safe with fresh air, provided that the rate of ammonia being released is below \(0.2 \mathrm{mg} / \mathrm{s}\). If the diffusion coefficient of ammonia through the plug is \(1.3 \times 10^{-10} \mathrm{~m}^{2} / \mathrm{s}\), determine whether or not the plug can safely contain the ammonia inside the vessel.

Air at \(40^{\circ} \mathrm{C}\) and 1 atm flows over a \(5-\mathrm{m}\)-long wet plate with an average velocity of \(2.5 \mathrm{~m} / \mathrm{s}\) in order to dry the surface. Using the analogy between heat and mass transfer, determine the mass transfer coefficient on the plate.

Write down the relations for steady one-dimensional heat conduction and mass diffusion through a plane wall, and identify the quantities in the two equations that correspond to each other.

Consider a 30-cm-diameter pan filled with water at \(15^{\circ} \mathrm{C}\) in a room at \(20^{\circ} \mathrm{C}, 1 \mathrm{~atm}\), and 30 percent relative humidity. Determine \((a)\) the rate of heat transfer by convection, (b) the rate of evaporation of water, and \((c)\) the rate of heat transfer to the water needed to maintain its temperature at \(15^{\circ} \mathrm{C}\). Disregard any radiation effects.

The mass diffusivity of ethanol \(\left(\rho=789 \mathrm{~kg} / \mathrm{m}^{3}\right.\) and \(M=46 \mathrm{~kg} / \mathrm{kmol}\) ) through air was determined in a Stefan tube. The tube has a uniform cross-sectional area of \(0.8 \mathrm{~cm}^{2}\). Initially, the ethanol surface was \(10 \mathrm{~cm}\) from the top of the tube; and after 10 hours have elapsed, the ethanol surface was \(25 \mathrm{~cm}\) from the top of the tube, which corresponds to \(0.0445 \mathrm{~cm}^{3}\) of ethanol being evaporated. The ethanol vapor pressure is \(0.0684\) atm, and the concentration of ethanol is zero at the top of the tube. If the entire process was operated at \(24^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\), determine the mass diffusivity of ethanol in air.
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