From the values in Table \(8-3\) for the van der Waals constant \(a\) for the gases \(\mathrm{H}_{2}, \mathrm{CO}_{2}, \mathrm{N}_{2},\) and \(\mathrm{CH}_{4},\) predict which of these gas molecules show the strongest intermolecular attractions.

Short Answer

Expert verified
Based on the van der Waals constant 'a' values from Table 8-3, CO₂ has the largest value (3.59 L² atm/mol²), indicating that its gas molecules exhibit the strongest intermolecular attractions among the gases H₂, CO₂, N₂, and CH₄.

Step by step solution

01

Identify the van der Waals constant 'a' values

In this step, we need to find or recall the values of the van der Waals constant 'a' from the given table 8-3 for the gases H₂, CO₂, N₂, and CH₄. Refer to Table 8-3 to find the following values of 'a' (L² atm/mol²) for the gases: H₂: \phantom{0}0.244 CO₂: 3.59 N₂: \phantom{0}1.36 CH₄: 2.25
02

Compare the values of 'a'

Now, we will compare the values of 'a' for the four gases to determine which one has the strongest intermolecular attractions. H₂: \phantom{0}0.244 CO₂: 3.59 N₂: \phantom{0}1.36 CH₄: 2.25 From the values listed above, CO₂ has the largest value of the constant 'a', which indicates that its gas molecules exhibit the strongest intermolecular attractions of the four given gases.

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Key Concepts

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

Intermolecular Attractions
Intermolecular attractions are the forces that act between stable molecules or between ions and molecules other than those in chemical bonds. These forces can influence the physical properties of substances, such as boiling points, melting points, and solubilities. Understanding these forces helps in predicting how substances behave in different states.

When looking at different gases, the strength of their intermolecular forces can vary greatly and are pivotal in determining the behavior of the gas under various conditions. For instance, a gas with stronger intermolecular forces will typically condense into a liquid at a higher temperature than one with weaker forces, as more energy is required to overcome the attractions between molecules.

There are several types of intermolecular forces, including dipole-dipole interactions in polar molecules, London dispersion forces, which are important for non-polar molecules, and hydrogen bonds, which are a special type of dipole-dipole attraction that occur between molecules containing hydrogen attached to electronegative atoms like oxygen, nitrogen, or fluorine.
Van der Waals Constant 'a'
The van der Waals constant 'a' is a term from the van der Waals equation, which corrects the ideal gas law for the volume occupied by gas particles and the intermolecular forces between them. Specifically, 'a' measures the magnitude of the attractive forces at the molecular level within a gas.

The constant 'a' provides valuable insight into the behavior of real gases, particularly at high pressures and low temperatures, where deviations from the ideal gas law are more pronounced. A higher 'a' value indicates stronger intermolecular attractions, which means that the gas molecules tend to stick together more. As a result, gases with higher 'a' values typically have lower compressibility and deviate more from ideal gas behavior because the attractive forces between molecules need to be overcome for them to move freely as an ideal gas would.
Gas Molecules
Gas molecules are in constant, random motion, colliding with each other and the walls of their container. The behavior of gas molecules can be described by the kinetic molecular theory, which assumes that these collisions are elastic and that the volume of the molecules themselves is negligible compared to the container's volume.

In the context of van der Waals forces and the constant 'a', however, we acknowledge that gas molecules do occupy space and their motion is influenced by mutual attractions or repulsions. This recognition is crucial when dealing with real gases instead of ideal gases, as ideal gas laws do not account for these intermolecular forces. Understanding the behavior of gas molecules in relation to their intermolecular forces is key for numerous practical applications, including the storage of gases, the design of pressurized containers, and the interpretation of gas behavior under different temperature and pressure conditions.
Comparing Intermolecular Forces
Comparing intermolecular forces among different substances is vital for predicting their physical and chemical properties. The van der Waals constants provide a means for such comparisons, especially for gases. For example, by looking at the 'a' values from van der Waals equation, we can get a relative measure of the strength of the intermolecular attractions.

In the exercise provided, the van der Waals constant 'a' allows us to predict that among the given gases, CO₂ exhibits the strongest intermolecular attractions. This greater force of attraction can lead to CO₂ condensing into a liquid at higher temperatures than the other gases. The comparisons made through the 'a' constant are directly linked to how similar substances might behave differently under the same conditions, for instance in phase changes or when mixed. It's also important to remember that other factors, such as molecular size and shape, can influence these forces and should be considered alongside 'a' when comparing intermolecular attractions.

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Most popular questions from this chapter

One of the chemical controversies of the nineteenth century concerned the element beryllium (Be). Berzelius originally claimed that beryllium was a trivalent element (forming \(\mathrm{Be}^{3+}\) ions) and that it gave an oxide with the formula \(\mathrm{Be}_{2} \mathrm{O}_{3}\). This resulted in a calculated atomic mass of \(13.5\) for beryllium. In formulating his periodic table, Mendeleev proposed that beryllium was divalent (forming \(\mathrm{Be}^{2+}\) ions) and that it gave an oxide with the formula BeO. This assumption gives an atomic mass of \(9.0 .\) In \(1894,\) A. Combes (Comptes Rendus 1894 p. 1221 ) reacted beryllium with the anion \(C_{5} \mathrm{H}_{7} \mathrm{O}_{2}^{-}\) and measured the density of the gaseous product. Combes's data for two different experiments are as follows:If beryllium is a divalent metal, the molecular formula of the product will be \(\mathrm{Be}\left(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{2}\right)_{2} ;\) if it is trivalent, the formula will be \(\mathrm{Be}\left(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{2}\right)_{3} .\) Show how Combes's data help to confirm that beryllium is a divalent metal.

A large flask with a volume of \(936 \mathrm{mL}\) is evacuated and found to have a mass of \(134.66 \mathrm{g}\). It is then filled to a pressure of 0.967 atm at \(31^{\circ} \mathrm{C}\) with a gas of unknown molar mass and then reweighed to give a new mass of 135.87 g. What is the molar mass of this gas?

Calculate the average kinetic energies of \(\mathrm{CH}_{4}(g)\) and \(\mathrm{N}_{2}(g)\) molecules at \(273 \mathrm{K}\) and \(546 \mathrm{K}\).

Concentrated hydrogen peroxide solutions are explosively decomposed by traces of transition metal ions (such as Mn or Fe): $$2 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(i)+\mathrm{O}_{2}(g)$$ What volume of pure \(\mathbf{O}_{2}(g),\) collected at \(27^{\circ} \mathrm{C}\) and 746 torr, would be generated by decomposition of \(125 \mathrm{g}\) of a \(50.0 \%\) by mass hydrogen peroxide solution? Ignore any water vapor that may be present.

A \(2.00-\mathrm{L}\) sample of \(\mathrm{O}_{2}(g)\) was collected over water at a total pressure of \(785\) torr and \(25^{\circ} \mathrm{C}\). When the \(\mathrm{O}_{2}(g)\) was dried (water vapor removed), the gas had a volume of \(1.94 \mathrm{L}\) at \(25^{\circ} \mathrm{C}\) and \(785\) torr. Calculate the vapor pressure of water at \(25^{\circ} \mathrm{C}\)

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