Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Without looking at a table of values, which of the following gases would you expect to have the largest value of the van der Waals constant \(b: \mathrm{H}_{2}, \mathrm{N}_{2}, \mathrm{CH}_{4}, \mathrm{C}_{2} \mathrm{H}_{6},\) or \(\mathrm{C}_{3} \mathrm{H}_{8} ?\)

Short Answer

Expert verified
Based on the molecular structure and size, we expect that the gas with the largest value of the van der Waals constant \(b\) would be \(\mathrm{C}_{3} \mathrm{H}_{8}\).

Step by step solution

01

Understanding the molecular structure

With each molecule, the number of carbon (C) and hydrogen (H) atoms will determine the molecular size. More carbon and/or hydrogen atoms in the molecule increase the molecule's size.
02

Comparing the given molecules

Now, let's compare the number of carbon and hydrogen atoms for each molecule: 1. \(\mathrm{H}_{2}\): 0 carbon atoms, 2 hydrogen atoms 2. \(\mathrm{N}_{2}\): 0 carbon atoms, 0 hydrogen atoms (nitrogen gas) 3. \(\mathrm{CH}_{4}\): 1 carbon atom, 4 hydrogen atoms 4. \(\mathrm{C}_{2}\mathrm{H}_{6}\): 2 carbon atoms, 6 hydrogen atoms 5. \(\mathrm{C}_{3} \mathrm{H}_{8}\): 3 carbon atoms, 8 hydrogen atoms
03

Identifying the gas with the largest molecular size

Now that we have compared the number of carbon and hydrogen atoms for each molecule, we can see that \(\mathrm{C}_{3} \mathrm{H}_{8}\) has the most carbon (3) and hydrogen (8) atoms.
04

Conclusion

Based on the molecular structure and size, we expect that the gas with the largest value of the van der Waals constant \(b\) would be \(\mathrm{C}_{3} \mathrm{H}_{8}\).

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 Vaia!

Key Concepts

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

Molecular Structure
Molecular structure refers to the arrangement of atoms within a molecule. It determines a molecule's properties and how it interacts with other substances. In the context of van der Waals forces, which are weak attractions between molecules, the structure is crucial. These forces become more significant as molecules become larger and more complex in structure. This complexity often results in a larger molecular "surface area" for interaction. Understanding molecular structure helps predict how molecules will behave in various physical conditions, such as changes in temperature or pressure. For example, more complex structures can lead to higher boiling points. This is because it takes more energy to break the intermolecular forces in a large, complex molecule compared to a smaller, simpler one. In the original exercise, recognizing the differences in molecular structure helps predict which gas has the largest van der Waals constant, an indicator of molecular size and interaction capabilities.
Carbon and Hydrogen Atoms
Atoms are the building blocks of molecules, and both carbon and hydrogen play pivotal roles in organic chemistry. Carbon atoms can form four bonds due to their four valence electrons, allowing for a variety of configurations. Hydrogen atoms typically make one bond. Together, these atoms create diverse organic compounds, including hydrocarbons like methane (\(\mathrm{CH}_4\)), ethane (\(\mathrm{C}_2\mathrm{H}_6\)), and propane (\(\mathrm{C}_3\mathrm{H}_8\)).
  • Carbon: Tends to form long chains or rings, contributing significantly to a molecule's size and complexity.
  • Hydrogen: Often controls the saturation of a molecule, influencing its reactivity and boiling point.
In the exercise, by counting the carbon and hydrogen atoms, we determine the relative sizes of different molecules. More carbon and hydrogen atoms generally equate to a larger molecule. This has direct implications for the van der Waals constant, as larger molecules tend to have higher b values due to the increased area available for intermolecular interactions.
Molecular Size
Molecular size fundamentally affects a molecule's physical and chemical properties. Larger molecules exhibit stronger van der Waals forces due to increased surface areas available for interactions. This is why molecular size is a crucial consideration in the context of the van der Waals constant (denoted \(b\)). The van der Waals constant \(b\) is part of a modified ideal gas law, accounting for the volume occupied by gas particles themselves. Larger molecules have larger b values because they deplete more space. For instance, in our exercise with gases like hydrogen (\(\mathrm{H}_2\)) and propane (\(\mathrm{C}_3\mathrm{H}_8\)), propane's larger size results in a higher b constant. This indicates both a physically larger size and a greater propensity for intermolecular attraction; facts directly derived from comparing molecular sizes. Recognizing the connection between these properties is essential for understanding gas behavior under various conditions, providing insight into why molecular size matters in practical applications such as chemical reactions and industrial processes.

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

In the "Méthode Champenoise," grape juice is fermented in a wine bottle to produce sparkling wine. The reaction is $$\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+2 \mathrm{CO}_{2}(g)$$ Fermentation of \(750 .\) mL grape juice (density \(=1.0 \mathrm{g} / \mathrm{cm}^{3}\) ) is allowed to take place in a bottle with a total volume of \(825 \mathrm{mL}\) until \(12 \%\) by volume is ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right) .\) Assuming that the \(\mathrm{CO}_{2}\) is insoluble in \(\mathrm{H}_{2} \mathrm{O}\) (actually, a wrong assumption), what would be the pressure of \(\mathrm{CO}_{2}\) inside the wine bottle at \(25^{\circ} \mathrm{C} ?\) (The density of ethanol is \(0.79 \mathrm{g} / \mathrm{cm}^{3} .\) )

A bicycle tire is filled with air to a pressure of \(75\) psi at a temperature of \(19^{\circ} \mathrm{C}\). Riding the bike on asphalt on a hot day increases the temperature of the tire to \(58^{\circ} \mathrm{C}\). The volume of the tire increases by \(4.0 \% .\) What is the new pressure in the bicycle tire?

Natural gas is a mixture of hydrocarbons, primarily methane \(\left(\mathrm{CH}_{4}\right)\) and ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right) .\) A typical mixture might have \(\chi_{\text {mathane }}=\) 0.915 and \(\chi_{\text {ethane }}=0.085 .\) What are the partial pressures of the two gases in a \(15.00\) -\(\mathrm{L}\) container of natural gas at \(20 .^{\circ} \mathrm{C}\) and \(1.44\) atm? Assuming complete combustion of both gases in the natural gas sample, what is the total mass of water formed?

Calculate the pressure exerted by \(0.5000\) mole of \(\mathrm{N}_{2}\) in a \(1.0000-\mathrm{L}\) container at \(25.0^{\circ} \mathrm{C}\) a. using the ideal gas law. b. using the van der Waals equation. c. Compare the results.

You have a sealed, flexible balloon filled with argon gas. The atmospheric pressure is \(1.00\) atm and the temperature is \(25^{\circ} \mathrm{C}\). Assume that air has a mole fraction of nitrogen of \(0.790\), the rest being oxygen. a. Explain why the balloon would float when heated. Make sure to discuss which factors change and which remain constant, and why this matters. b. Above what temperature would you heat the balloon so that it would float?

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.

Sign-up for free