/*! 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 34 An ideal-gas mixture contains \(... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 34

An ideal-gas mixture contains \(35 \%\) helium, \(20 \%\) methane, and \(45 \%\) nitrogen by volume at 2.00 atm absolute and \(90^{\circ} \mathrm{C}\). Calculate (a) the partial pressure of each component, (b) the mass fraction of methane, (c) the average molecular weight of the gas, and (d) the density of the gas in \(\mathrm{kg} / \mathrm{m}^{3}\).

Short Answer

Expert verified
The results are: (a) Partial pressures: Helium: 0.70 atm, Methane: 0.40 atm, Nitrogen: 0.90 atm, (b) Mass fraction of Methane: 0.19, (c) Average molecular weight: 17.04 g/mol, (d) Density of gas: 1.38 kg/m^3.

Step by step solution

01

Calculation of partial pressures

The partial pressure of a gas is given by the total pressure times the volume fraction of that gas. Thus, the partial pressures will be: - For helium: \(0.35 脳 2.00 atm = 0.70 atm\), - For methane: \(0.20 脳 2.00 atm = 0.40 atm\), - For nitrogen: \(0.45 脳 2.00 atm = 0.90 atm\).
02

Mass fraction calculation for methane

The mass of each component is given by the partial percentage multiplied by its molecular weight, and then the mass fraction is found by dividing the mass by the total mass. The molar masses are: Helium (He) = 4 \(g/mol\), Methane (CH4) = 16 \(g/mol\), Nitrogen (N2) = 28 \(g/mol\). Therefore, the mass of each component will be: - For He: \(0.35 脳 4 = 1.4 g\), - For CH4: \(0.20 脳 16 = 3.2 g\), - For N2: \(0.45 脳 28 = 12.6 g\). The sum of these gives the total mass, which is \(1.4 g + 3.2 g + 12.6 g = 17.2 g\). The mass fraction of methane is then the mass of methane divided by the total mass, which is \(3.2 g/17.2 g = 0.19\).
03

Average molecular weight calculation

The average molecular weight is found by weighting the molecular mass by the mass fraction of each component. This is given by: Average molecular weight = \(0.35 脳 4 + 0.19 脳 16 + 0.45 脳 28 = 1.4 g/mol + 3.04 g/mol +12.6 g/mol = 17.04 g/mol\).
04

Calculation of gas density

For this, we will use the equation of state for ideal gases: \(P = 蟻RT/M\), where P is the total pressure, 蟻 is the density, R is the universal gas constant = 0.0821 \(atm.L/mol.K\), T is the temperature in Kelvins and M is the average molar mass in kg per mole. First, convert the temperature to Kelvins: \(90^{\circ}C = 90+273.15 = 363.15 K\). The density is given by 蟻 = \(PM/RT = 2.00 atm 脳 17.04 脳 10^{-3} kg/mol / (0.0821 atm.L/mol.K 脳 363.15 K) = 1.38 kg/m^3\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Partial Pressure Calculation
Understanding the concept of partial pressure in an ideal gas mixture is important. Each gas in a mixture contributes to the total pressure based on its volume percentage.

The formula used to calculate the partial pressure of each component is simple. It's a product of the total pressure and the volume fraction of that gas.
  • For helium: Given volume fraction is 35%. So, its partial pressure is calculated as \(0.35 \times 2.00 \text{ atm} = 0.70 \text{ atm}\).
  • Methane, with a 20% volume fraction, has a partial pressure of \(0.20 \times 2.00 \text{ atm} = 0.40 \text{ atm}\).
  • Nitrogen holds 45% of the volume, leading to a partial pressure of \(0.45 \times 2.00 \text{ atm} = 0.90 \text{ atm}\).
Partial pressures are crucial as they directly affect reactions and dynamics in the gas mixture.
Mass Fraction Calculation
Mass fraction helps to measure the contribution of a single gas component in terms of mass within a mixture.

Calculating mass fractions involves a few steps, using molecular weights and given percentages.
  • The molar mass of helium is 4 \(\text{g/mol}\), methane is 16 \(\text{g/mol}\), and nitrogen is 28 \(\text{g/mol}\).
  • You first determine the mass of each gas:
    • Helium: \(0.35 \times 4 = 1.4 \text{g}\)
    • Methane: \(0.20 \times 16 = 3.2 \text{g}\)
    • Nitrogen: \(0.45 \times 28 = 12.6 \text{g}\)
  • Sum of these masses provides the total mass of the mixture: \(1.4 + 3.2 + 12.6 = 17.2 \text{g}\).
  • The mass fraction of methane is calculated by dividing its mass by the total mass: \(3.2 \text{g} / 17.2 \text{g} = 0.19\).
This fraction illustrates the weight contribution each gas partakes in the mixture.
Average Molecular Weight
The average molecular weight tells us about the typical mass of a molecule in a gas mixture. It's derived by considering both the mass fractions and the molecular weights.

The calculation mixes together the contribution of each gas component.
  • Use mass fractions: Helium contributes \(0.35 \), Methane \(0.19\), and Nitrogen \(0.45\).
  • Calculate average molecular weight:
    • Helium's contribution: \(0.35 \times 4 = 1.4 \text{ g/mol}\)
    • Methane's contribution: \(0.19 \times 16 = 3.04 \text{ g/mol}\)
    • Nitrogen's contribution: \(0.45 \times 28 = 12.6 \text{ g/mol}\)
  • The sum gives an average molecular weight of the mixture: \(1.4 + 3.04 + 12.6 = 17.04 \text{ g/mol}\).
This average molecular weight is vital for understanding the gas behavior under various conditions.
Gas Density Calculation
Calculating the density of a gas involves understanding the equation of state for ideal gases. It's influenced by pressure, temperature, and the gas's molecular weight.

Here's how you compute the density:
  • The equation used is \( P = \rho \frac{RT}{M} \), where:
    • \( P \) is total pressure.
    • \( \rho \) is density.
    • \( R \) is the universal gas constant \(0.0821 \text{ atm.L/mol.K}\).
    • \( T \) is the temperature in Kelvin. Convert Celsius to Kelvin: \(90^{\circ}C = 363.15 \text{ K}\).
    • \( M \) is the average molar mass \(17.04 \text{ g/mol} = 17.04 \times 10^{-3} \text{ kg/mol}\).
  • Rearrange the formula to find density: \[ \rho = \frac{PM}{RT} \]
  • Plug in the values to get: \( \rho = \frac{2.00 \times 17.04 \times 10^{-3}}{0.0821 \times 363.15} = 1.38 \text{ kg/m}^3\).
This final density value tells us about the mass of the gas per unit volume.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

When a liquid or a gas occupies a volume, it may be assumed to fill the volume completely. On the other hand, when solid particles occupy a volume, there are always spaces (voids) among the particles. The porosity or void fraction of a bed of particles is the ratio (void volume)/(total bed volume). The bulk density of the solids is the ratio (mass of solids)/(total bed volume), and the absolute density of the solids has the usual definition (mass of solids)/(volume of solids). Suppose \(600.0 \mathrm{g}\) of a crushed ore is placed in a graduated cylinder, filling it to the \(184 \mathrm{cm}^{3}\) level. One hundred \(\mathrm{cm}^{3}\) of water is then added to the cylinder, whereupon the water level is observed to be at the \(233.5 \mathrm{cm}^{3}\) mark. Calculate the porosity of the dry particle bed, the bulk density of the ore in this bed, and the absolute density of the ore.

The van der Waals equation of state (Equation \(5.3-7\) ) is to be used to estimate the specific molar volume \(\hat{V}(\mathrm{L} / \mathrm{mol})\) of air at specified values of \(T(\mathrm{K})\) and \(P(\mathrm{atm}) .\) The van der Waals constants for air are \(a=1.33 \mathrm{atm} \cdot \mathrm{L}^{2} / \mathrm{mol}^{2}\) and \(b=0.0366 \mathrm{L} / \mathrm{mol}\) (a) Show why the van der Waals equation is classified as a cubic equation of state by expressing it in the form $$f(\hat{V})=c_{3} \hat{V}^{3}+c_{2} \hat{V}^{2}+c_{1} \hat{V}+c_{0}=0$$ where the coefficients \(c_{3}, c_{2}, c_{1},\) and \(c_{0}\) involve \(P, R, T, a,\) and \(b .\) Calculate the values of these coefficients for air at \(223 \mathrm{K}\) and 50.0 atm. (Include the units when giving the values.) (b) What would the value of \(\hat{V}\) be if the ideal-gas equation of state were used for the calculation? Use this value as an initial estimate of \(\tilde{V}\) for air at \(223 \mathrm{K}\) and 50.0 atm and solve the van der Waals equation using Goal Seek or Solver in Excel. What percentage error results from the use of the ideal-gas equation of state, taking the van der Waals estimate to be correct? (c) Set up a spreadsheet to carry out the calculations of Part (b) for air at \(223 \mathrm{K}\) and several pressures. The spreadsheet should appear as follows: The polynomial expression for \(\hat{V}\left(f=c_{3} \hat{V}^{3}+c_{2} \hat{V}^{2}+\cdots\right)\) should be entered in the \(f(V)\) column, and the value in the \(V\) column should be determined using Goal Seek or Solver in Excel.

The concentration of oxygen in a 5000 -liter tank containing air at 1 atm is to be reduced by pressure purging prior to charging a fuel into the tank. The tank is charged with nitrogen up to a high pressure and then vented back down to atmospheric pressure. The process is repeated as many times as required to bring the oxygen concentration below 10 ppm (i.c., to bring the mole fraction of \(\mathrm{O}_{2}\) below \(10.0 \times 10^{-6}\) ). Assume that the temperature is \(25^{\circ} \mathrm{C}\) at the beginning and end of each charging cycle. When doing \(P V T\) calculations in Parts (b) and (c), use the generalized compressibility chart if possible for the fully charged tank and assume that the tank contains pure nitrogen. (a) Speculate on why the tank is being purged. (b) Estimate the gauge pressure (atm) to which the tank must be charged if the purge is to be done in one charge-vent cycle. Then estimate the mass of nitrogen (kg) used in the process. (For this part, if you can't find the tank condition on the compressibility chart, assume ideal-gas behavior and state whether the resulting estimate of the pressure is too high or too low.) (c) Suppose nitrogen at 700 kPa gauge is used for the charging. Calculate the number of charge-vent cycles required and the total mass of nitrogen used. (d) Use your results to explain why multiple cycles at a lower gas pressure are preferable to a single cycle. What is a probable disadvantage of multiple cycles?

The quantity of sulfuric acid used globally places it among the most plentiful of all commodity chemicals. In the modern chemical industry, synthesis of most sulfuric acid utilizes elemental sulfur as a feedstock. However, an alternative and historically important source of sulfuric acid was the conversion of an ore containing iron pyrites (FeS_) to sulfur oxides by roasting (burning) the ore with air. The following reactions occurred in an oven: $$\begin{array}{c} 2 \mathrm{FeS}_{2}(\mathrm{s})+\frac{11}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+4 \mathrm{SO}_{2}(\mathrm{g}) \\ \mathrm{SO}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{SO}_{3}(\mathrm{g}) \end{array}$$ The gases leaving the oven were fed to a catalytic converter in which most of the remaining \(\mathrm{SO}_{2}\) produced was oxidized to \(\mathrm{SO}_{3}\). Finally, the gas leaving the converter was sent to an absorption column where the \(S O_{3}\) was taken up by water to produce sulfuric acid \(\left(H_{2} S O_{4}\right)\) (a) The ore fed to the oven was 90.0 wt\% \(\mathrm{FeS}_{2}\), and the remaining material may be considered inert. Dry air was fed to the oven in \(30.0 \%\) excess of the amount required to oxidize all of the sulfur in the ore to \(S O_{3}\). Eighty-five percent of the \(\mathrm{FeS}_{2}\) was oxidized, and \(60 \%\) of the \(\mathrm{SO}_{2}\) produced was oxidized to \(S O_{3}\). Leaving the roaster were (i) a gas stream containing \(S O_{2}, S O_{3}, O_{2},\) and \(N_{2}\) and (ii) a solid stream containing unconverted pyrites, ferric oxide \(\left(\mathrm{Fe}_{2} \mathrm{O}_{3}\right),\) and the inert material. Calculate the required feed rate of air in standard cubic meters per \(100 \mathrm{kg}\) of ore fed to the process. Also determine the molar composition and volume (SCM/100 kg ore) of the gas leaving the oven. (b) The gas leaving the oven entered the catalytic converter, which operated at 1.0 atm. Reaction (2) proceeded to equilibrium, at which point the component partial pressures are related by the expression $$K_{\mathrm{P}}(T)=\frac{p_{\mathrm{SO}_{3}}}{p_{\mathrm{SO}_{3}} p_{\mathrm{O}_{2}}^{0.5}}$$ The gases were first heated to \(600^{\circ} \mathrm{C}\) to accelerate the rate of reaction, and then cooled to \(400^{\circ} \mathrm{C}\) to enhance \(S O_{2}\) conversion. The equilibrium constant \(K_{\mathrm{P}}\) at these two temperatures is 9.53 atm \(^{0.5}\) and 397 atm \(^{0.5}\), respectively. Calculate the equilibrium fractional conversions of \(S O_{2}\) at these two temperatures. (c) Estimate the production rate of sulfuric acid in \(\mathrm{kg} / \mathrm{kg}\) ore if all of the \(\mathrm{SO}_{3}\) leaving the converter was transformed to sulfuric acid. What would this value be if all the sulfur in the ore had been converted?

Lewis \(^{12}\) describes the hazards of breathing air containing appreciable amounts of an asphyxiant (a gas that has no specific toxicity but, when inhaled, excludes oxygen from the lungs). When the mole percent of the asphyxiant in the air reaches \(50 \%,\) marked symptoms of distress appear, and at \(75 \%\) death occurs in a matter of minutes. A small storage room whose dimensions are \(2 \mathrm{m} \times 1.5 \mathrm{m} \times 3 \mathrm{m}\) contains a number of expensive and dangerous chemicals. To prevent unauthorized entry, the door to the room is always locked and can be opened with a key from either side. A cylinder of liquid carbon dioxide is stored in the room. The valve on the cylinder is faulty and some of the contents have escaped over the weekend. The room temperature is \(25^{\circ} \mathrm{C}\). (a) If the concentration of \(\mathrm{CO}_{2}\) reaches the lethal 75 mole \(\%\) level, what would be the mole percent of \(\mathrm{O}_{2} ?\) (b) How much \(\mathrm{CO}_{2}(\mathrm{kg})\) is present in the room when the lethal concentration is reached? Why would more than that amount have to escape from the cylinder for this concentration to be reached? (c) Describe a set of events that could result in a fatality in the given situation. Suggest at least two measures that would reduce the hazards associated with storage of this scemingly harmless substance.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Chemical Analysis

Read Explanation

Making Measurements

Read Explanation

Nuclear Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Inorganic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.