/*! 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 47 Rank the following gases from le... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Rank the following gases from least dense to most dense at \(101.33 \mathrm{kPa}\) and \(298 \mathrm{~K}: \mathrm{O}_{2}, \mathrm{Ar}, \mathrm{NH}_{3}, \mathrm{HCl}\).

Short Answer

Expert verified
The ranking of gases from least dense to most dense at 101.33 kPa and 298 K is: NH₃ (0.706 kg/m³) < O₂ (1.331 kg/m³) < HCl (1.517 kg/m³) < Ar (1.661 kg/m³).

Step by step solution

01

Find the molar mass of each gas

To rank the gases based on their densities, we must first determine their molar masses. Use the periodic table to find the atomic masses of each element in a molecule. - For O₂: Molar mass = 2 × 16.00 g/mol = 32.00 g/mol - For Ar: Molar mass = 39.95 g/mol (since it's a noble gas) - For NH₃: Molar mass = 14.01 g/mol (N) + 3 × 1.01 g/mol (H) = 17.03 g/mol - For HCl: Molar mass = 1.01 g/mol (H) + 35.45 g/mol (Cl) = 36.46 g/mol
02

Use the ideal gas law to find density

Now we will use the ideal gas law equation to find the density for each gas under the given conditions (pressure = 101.33 kPa and temperature = 298 K). The ideal gas law is given by: \(PV = nRT\) Where: - P = pressure - V = volume - n = number of moles - R = ideal gas constant (8.314 J/mol·K) - T = temperature The density of a gas is given by: \(\rho = \frac{mass}{volume} = \frac{molar~mass~ ×~ number~of~ moles}{volume}\) Combining the ideal gas law and the density formula, we get: \(\rho = \frac{molar~mass~ ×~ P}{R ~×~ T}\) Now plug in the values for pressure (101.33 kPa), temperature (298 K), and molar mass for each gas to find their respective densities.
03

Calculate density for each gas

Use the formula from the previous step to calculate the densities: - For O₂: \(\rho = \frac{32.00~ g/mol × 101.33~ kPa}{8.314~ J/mol·K × 298~ K} = 1.331~ kg/m³\) - For Ar: \(\rho = \frac{39.95~ g/mol × 101.33~ kPa}{8.314~ J/mol·K × 298~ K} = 1.661~ kg/m³\) - For NH₃: \(\rho = \frac{17.03~ g/mol × 101.33~ kPa}{8.314~ J/mol·K × 298~ K} = 0.706~ kg/m³\) - For HCl: \(\rho = \frac{36.46~ g/mol × 101.33~ kPa}{8.314~ J/mol·K × 298~ K} = 1.517~ kg/m³\)
04

Rank the gases by density

Now that we have calculated the densities for each gas, we can rank them from least dense to most dense: 1. NH₃ (0.706 kg/m³) 2. O₂ (1.331 kg/m³) 3. HCl (1.517 kg/m³) 4. Ar (1.661 kg/m³) Thus, the ranking of gases from least dense to most dense at 101.33 kPa and 298 K is: NH₃ < O₂ < HCl < Ar.

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 91Ó°ÊÓ!

Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is pivotal in understanding the behavior of gases under different conditions. It is represented by the equation \(PV = nRT\), where:
  • \(P\) stands for pressure
  • \(V\) is the volume
  • \(n\) is the number of moles
  • \(R\) is the ideal gas constant, approximately 8.314 J/mol·K
  • \(T\) represents temperature in Kelvin
This formula combines several laws, including Boyle’s Law and Charles’s Law. The Ideal Gas Law allows us to predict how a gas will react when we change its pressure, volume, or temperature.
In density calculations, this law is particularly useful. Combining it with the density formula, \(\rho = \frac{mass}{volume}\), gives us a way to express gas density as \(\rho = \frac{Molar~Mass \times P}{RT}\). This highlights how gas density depends on molar mass, pressure, and temperature. Making it a central tool for comparing gases under consistent conditions.
Molar Mass
Molar mass is an essential parameter in chemistry, referring to the mass of one mole of a substance, commonly expressed in grams per mole \((g/mol)\). It is calculated by summing the atomic masses of the constituent atoms in the molecule.
To determine the molar mass of a compound like oxygen (\(O_2\)), you would multiply the atomic mass of oxygen by two because there are two oxygen atoms. Likewise, for ammonia \((NH_3)\), add the atomic mass of nitrogen to three times the atomic mass of hydrogen.
Knowing the molar mass is crucial, as it directly influences the density of a gas. The greater the molar mass, the denser the gas will be under the same conditions of pressure and temperature. It also ties back to the Ideal Gas Law in density calculations, where molar mass is a direct factor in the formula \(\rho = \frac{Molar~Mass \times P}{RT}\). This relationship emphasizes its role in ranking gases by density.
Density Calculation
Density computation for gases involves determining how closely packed the gas particles are within a given volume. It is expressed as the mass of the gas per unit volume \((kg/m^3)\).
The density \(\rho\) can be calculated through the formula \(\rho = \frac{mass}{volume}\). For gases, we modify this formula using the Ideal Gas Law: \(\rho = \frac{Molar~Mass \times P}{RT}\), where \(P\) is the pressure and \(T\) is the temperature.
  • Molar Mass: Directly impacts density; higher molar mass typically results in greater density.
  • Pressure: Increasing pressure can increase density as the gas molecules are pushed closer together.
  • Temperature: As temperature rises, density tends to decrease since gas molecules move apart.
This intuitive approach highlights how slight changes in conditions affect gas density and facilitates ranking gases by their densities under the same conditions, as seen in this exercise.
Periodic Table
The Periodic Table is a comprehensive chart of chemical elements arranged by atomic number, electron configuration, and recurring chemical properties. It is a fundamental tool for chemists.
Each element's square contains crucial information, including the atomic number and atomic mass, essential for calculating molar masses. This table informs the calculations of molecular compounds like \(O_2\), \(Ar\), \(NH_3\), and \(HCl\).
For example, when determining the molar mass of a compound, the atomic weight of each element (found on the Periodic Table) is used to sum up to form the compound's molar mass. The accurate determination of molar masses from the Periodic Table allows for precise density calculations.
  • Facilitates chemical calculations by providing necessary atomic weights.
  • Helps predict properties and behaviors of gases based on their chemical composition.
Overall, the Periodic Table serves as a starting point for any chemical calculation, ensuring accuracy and facilitating a deeper understanding of substances and their interactions.

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

A neon sign is made of glass tubing whose inside diameter is \(3.0 \mathrm{~cm}\) and length is \(10.0 \mathrm{~m}\). If the sign contains neon at a pressure of \(265 \mathrm{~Pa}\) at \(30^{\circ} \mathrm{C}\), how many grams of neon are in the sign? (The volume of a cylinder is \(\pi r^{2} h\).)

Table 10.3 shows that the van der Waals \(b\) parameter has units of \(\mathrm{L} / \mathrm{mol}\). This implies that we can calculate the size of atoms or molecules from \(b\). Using the value of \(b\) for \(\mathrm{Xe},\) calculate the radius of a Xe atom and compare it to the value found in Figure \(7.7,\) that is, \(140 \mathrm{pm}\). Recall that the volume of a sphere is \((4 / 3) \pi r^{3}\).

Which statement concerning the van der Waals constants \(a\) and \(b\) is true? (a) The magnitude of \(a\) relates to molecular volume, whereas \(b\) relates to attractions between molecules. (b) The magnitude of \(a\) relates to attractions between molecules, whereas \(b\) relates to molecular volume. (c) The magnitudes of \(a\) and \(b\) depend on pressure. (d) The magnitudes of \(a\) and \(b\) depend on temperature.

Natural gas is very abundant in many Middle Eastern oil fields. However, the costs of shipping the gas to markets in other parts of the world are high because it is necessary to liquefy the gas, which is mainly methane and has a boiling point at atmospheric pressure of \(-164^{\circ} \mathrm{C}\). One possible strategy is to oxidize the methane to methanol, \(\mathrm{CH}_{3} \mathrm{OH},\) which has a boiling point of \(65^{\circ} \mathrm{C}\) and can therefore be shipped more readily. Suppose that \(3.03 \times 10^{8} \mathrm{~m}^{3}\) of methane at atmospheric pressure and \(25^{\circ} \mathrm{C}\) is oxidized to methanol. (a) What volume of methanol is formed if the density of \(\mathrm{CH}_{3} \mathrm{OH}\) is \(0.791 \mathrm{~g} / \mathrm{mL} ?(\mathbf{b})\) Write balanced chemical equations for the oxidations of methane and methanol to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l) .\) Calculate the total enthalpy change for complete combustion of the \(3.03 \times 10^{8} \mathrm{~m}^{3}\) of methane just described and for complete combustion of the equivalent amount of methanol, as calculated in part (a). (c) Methane, when liquefied, has a density of \(0.466 \mathrm{~g} / \mathrm{mL} ;\) the density of methanol at \(25^{\circ} \mathrm{C}\) is \(0.791 \mathrm{~g} / \mathrm{mL}\). Compare the enthalpy change upon combustion of a unit volume of liquid methane and liquid methanol. From the standpoint of energy production, which substance has the higher enthalpy of combustion per unit volume?

Both Jacques Charles and Joseph Louis Guy-Lussac were avid balloonists. In his original flight in 1783 , Jacques Charles used a balloon that contained approximately \(31,150 \mathrm{~L}\) of \(\mathrm{H}_{2}\). He generated the \(\mathrm{H}_{2}\) using the reaction between iron and hydrochloric acid: $$\mathrm{Fe}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{FeCl}_{2}(a q)+\mathrm{H}_{2}(g)$$ How many kilograms of iron were needed to produce this volume of \(\mathrm{H}_{2}\) if the temperature was \(22{ }^{\circ} \mathrm{C} ?\)

See all solutions

Recommended explanations on Chemistry Textbooks

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.