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Answer the following questions involving a mixture of two gases: (a) When would the analysis of the mixture in terms of mass fractions be identical to the analysis in terms of mole fractions? (b) When would the apparent molecular weight of the mixture equal the average of the molecular weights of the two gases?

Short Answer

Expert verified
a) When the molecular weights of the gases are equal.b) When their mole fractions are equal (0.5 each).

Step by step solution

01

Title - Understand Mass and Mole Fractions

Mass fraction is the ratio of the mass of a particular gas to the total mass of the mixture. Mole fraction is the ratio of the number of moles of a particular gas to the total number of moles of the mixture.
02

Title - Identify the Conditions for Identical Analysis

The analysis in terms of mass fractions would be identical to the analysis in terms of mole fractions if the molecular weights of the two gases are equal. This is because the mass fraction will be directly proportional to the mole fraction in this case.
03

Title - Examine the Apparent Molecular Weight

The apparent molecular weight of the mixture (M_mix) is calculated as the sum of the products of the mole fractions and the molecular weights of the individual gases: \[ M_{mix} = X_A M_A + X_B M_B \] For the apparent molecular weight to equal the average of the molecular weights of the two gases, each gas must contribute equally in terms of mole fraction, i.e., \( X_A = X_B = 0.5 \), resulting in: \[ M_{mix} = 0.5 M_A + 0.5 M_B \] This simplifies to: \[ M_{mix} = \frac{M_A + M_B}{2} \].

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Key Concepts

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

mass fractions
Mass fractions are a way to express the concentration of a particular gas in a gas mixture in terms of mass. The mass fraction of a gas is the ratio of the mass of that gas to the total mass of the gas mixture.
This concept is particularly useful in chemical engineering and thermodynamics.
To calculate the mass fraction of a gas, use the formula:
\(\text{{mass fraction}} = \frac{{\text{{mass of the gas}}}}{{\text{{total mass of the mixture}}}}\).
This helps to understand how much of the mixture's total mass is made up of a specific gas. For example, in a mixture of oxygen and nitrogen, if oxygen has a mass of 40 grams and the total mass is 100 grams, the mass fraction of oxygen is 0.4 or 40%.
  • Easy to conceptualize because it relates directly to the weight
  • Independent of pressure and temperature
mole fractions
Mole fractions deal with counting the number of molecules of a gas in a mixture. It is the ratio of the number of moles of a particular gas to the total number of moles of all gases in the mixture.
The formula to calculate mole fraction is:
\(\text{{mole fraction}} = \frac{{\text{{number of moles of the gas}}}}{{\text{{total number of moles in the mixture}}}}\).
As an example, if in a mixture we have 2 moles of gas A and 3 moles of gas B, the total number of moles is 5. The mole fraction of gas A would be 2/5 or 0.4.
  • Useful in chemical reactions where the ratio of particles is needed
  • High accuracy independent of temperature and pressure
apparent molecular weight
The apparent molecular weight is a weighted average based on the mole fractions and molecular weights of the individual gases in a mixture.
It is a convenient way to calculate the overall molecular weight of a mixture.
The formula to determine apparent molecular weight:
\(\text{{M}}_{mix} = X_A M_A + X_B M_B\),
where \(X_A\) and \(X_B\) are the mole fractions, and \(M_A\) and \(M_B\) are the molecular weights of gases A and B respectively.
If gases contribute equally, meaning \(X_A = X_B = 0.5\), the formula simplifies to:
\(\text{{M}}_{mix} = \frac{{M_A + M_B}}{2}\).
  • Important in determining how gas mixtures will behave chemically and physically
  • Useful in industries where different gases are mixed, like air conditioning or chemical manufacturing

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Most popular questions from this chapter

Natural gas at \(23^{\circ} \mathrm{C}, 1\) bar enters a furnace with the following molar analysis: \(40 \%\) propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right), 40 \%\) ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right), 20 \%\) methane \(\left(\mathrm{CH}_{4}\right)\). Determine (a) the analysis in terms of mass fractions. (b) the partial pressure of each component, in bar. (c) the mass flow rate, in \(\mathrm{kg} / \mathrm{s}\), for a volumetric flow rate of \(20 \mathrm{~m}^{3} / \mathrm{s}\).

Air at \(77^{\circ} \mathrm{C}, 1\) bar, and a molar flow rate of \(0.1 \mathrm{kmol} / \mathrm{s}\) enters an insulated mixing chamber operating at steady state and mixes with water vapor entering at \(277^{\circ} \mathrm{C}, 1\) bar, and a molar flow rate of \(0.3 \mathrm{kmol} / \mathrm{s}\). The mixture exits at 1 bar. Kinetic and potential energy effects can be ignored. For the chamber, determine (a) the temperature of the exiting mixture, in \({ }^{\circ} \mathrm{C}\). (b) the rate of entropy production, in \(\mathrm{kW} / \mathrm{K}\).

A stream consisting of \(35 \mathrm{~m}^{3} / \mathrm{min}\) of moist air at \(14^{\circ} \mathrm{C}\), \(1 \mathrm{~atm}, 80 \%\) relative humidity mixes adiabatically with a stream consisting of \(80 \mathrm{~m}^{3} / \mathrm{min}\) of moist air at \(40^{\circ} \mathrm{C}, 1 \mathrm{~atm}\), \(40 \%\) relative humidity, giving a single mixed stream at \(1 \mathrm{~atm}\). Using the psychrometric chart together with the procedure of Prob. 12.58, determine the relative humidity and temperature, in \({ }^{\circ} \mathrm{C}\), of the exiting stream.

Air at \(35^{\circ} \mathrm{C}, 1\) bar, and \(10 \%\) relative humidity enters an evaporative cooler operating at steady state. The volumetric flow rate of the incoming air is \(50 \mathrm{~m}^{3} / \mathrm{min}\). Liquid water at \(20^{\circ} \mathrm{C}\) enters the cooler and fully evaporates. Moist air exits the cooler at \(25^{\circ} \mathrm{C}, 1\) bar. If there is no significant heat transfer between the device and its surroundings, determine (a) the rate at which liquid enters, in \(\mathrm{kg} / \mathrm{min}\). (b) the relative humidity at the exit. (c) the rate of exergy destruction, in \(\mathrm{kJ} / \mathrm{min}\), for \(T_{0}=20^{\circ} \mathrm{C}\). Neglect kinetic and potential energy effects.

Air at \(30^{\circ} \mathrm{C}, 1\) bar, \(50 \%\) relative humidity enters an insulated chamber operating at steady state with a mass flow rate of \(3 \mathrm{~kg} / \mathrm{min}\) and mixes with a saturated moist air stream entering at \(5^{\circ} \mathrm{C}, 1\) bar with a mass flow rate of \(5 \mathrm{~kg} / \mathrm{min}\). A single mixed stream exits at 1 bar. Determine (a) the relative humidity and temperature, in \({ }^{\circ} \mathrm{C}\), of the exiting stream. (b) the rate of exergy destruction, in \(\mathrm{kW}\), for \(T_{0}=20^{\circ} \mathrm{C}\). Neglect kinetic and potential energy effects.

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