/*! 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 99 Table 10.3 shows that the van de... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 Xe, calculate the radius of a Xe atom and compare it to the value found in Figure 7.6, \(1.30 \AA\) A. Recall that the volume of a sphere is \((4 / 3) \pi r^{3}\).

Short Answer

Expert verified
The radius of a Xe atom can be calculated using the formula \(r = \sqrt[3]{\frac{3 b}{4 \pi N_A}}\), where \(b\) is the van der Waals \(b\) parameter for Xe and \(N_A\) is Avogadro's number. Comparing the calculated radius with the given value of \(1.30 \unicode{x212B}\) A confirms that the van der Waals \(b\) parameter can be used to determine the size of atoms or molecules.

Step by step solution

01

Write the expression for the b parameter in terms of the volume of a sphere

Recall that the volume of a sphere can be calculated using the formula: \(V = \frac{4}{3} \pi r^3\). Also, it is given that the van der Waals b parameter has units of L/mol, implying that the b parameter can be related to the volume of atoms or molecules per mole. In this case, we can consider b as the molar volume of Xenon atoms, so we can equate it to the volume of one mole of Xe spheres. Let's write the expression for the b parameter as: \[b = \frac{4}{3} \pi r^3 N_A\] where \(N_A\) is Avogadro's number (number of molecules per mole).
02

Calculate the radius of a Xe atom using the given b parameter value

We know the van der Waals \(b\) parameter for Xe. Let's rearrange the equation derived in step 1 to solve for the radius (r) of a Xe atom: \[r^3 = \frac{3 b}{4 \pi N_A}\] We can now plug in the values for \(b\) and \(N_A\) (Avogadro's number: \(6.022 \times 10^{23}\, \text{mol}^{-1}\)) and calculate the radius of a Xe atom: \[r = \sqrt[3]{\frac{3 b}{4 \pi N_A}}\]
03

Compare the calculated radius with the given value

After calculating the radius of a Xe atom using the formula from Step 2, we can compare it with the given value of \(1.30 \unicode{x212B}\) A. If the calculated radius is close to the given value, it confirms that the van der Waals b parameter can be used to determine the size of atoms or molecules.

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.

Molecular Size
Understanding molecular size is crucial in chemistry. It helps us know how molecules interact with each other. The molecular size can be thought of as the three-dimensional space that a molecule occupies. It is closely related to the van der Waals equation, which describes the behavior of gases by considering the volume occupied by the gas molecules and the attraction between them.

The van der Waals equation modifies the ideal gas law by incorporating the molecular size through the parameter "b". This "b" parameter represents the volume excluded by the molecules. It is directly related to the size of the molecules, such as the volume that one mole of these molecules takes up. Therefore, by understanding and calculating the "b" parameter, we can shed light on the molecular size, which is essential in predicting and explaining the behavior of different gases under various conditions.
Atomic Radius Calculation
Atomic radius is an important concept when discussing molecular size. It's the measure of the size of an atom, typically the distance from the center of the nucleus to the boundary of the surrounding cloud of electrons.

To calculate the atomic radius using the van der Waals agreement, you need the value of the parameter "b". This "b" is associated with the excluded volume per mole of molecules and is expressed in liters per mole. Since a mole contains a huge number of atoms (Avogadro's number), each atom’s volume is tiny.
  • Start with the formula for the volume of a sphere: \[ V = \frac{4}{3} \pi r^3 \].
  • Using this, relate it to the van der Waals "b": \[ b = \frac{4}{3} \pi r^3 N_A \], where \( N_A \) is Avogadro's number.
  • Rearranging gives us: \[ r = \sqrt[3]{ \frac{3b}{4 \pi N_A} }\].
By plugging the known values into this equation, the radius of an individual atom, such as xenon, can be calculated. This method provides an easy way to connect abstract gas law parameters with tangible atomic dimensions.
Avogadro's Number
Avogadro's number is a fundamental constant in chemistry. It is the number of atoms, molecules, or particles in one mole of a substance, approximately equal to \(6.022 \times 10^{23}\). This number is crucial when calculating the atomic or molecular scale measurements, because it bridges the gap between the macroscopic scale we observe and the microscopic world of atoms and molecules.

When calculating things like atomic radius using the van der Waals "b" parameter, Avogadro's number plays a key role. It helps scale down the molar volume parameter to the size of a single atom or molecule. Knowing Avogadro's number ensures that calculations remain consistent with real-world observations.

In essence, Avogadro's number enables us to understand quantities that are practically impossible to measure directly, like the number of atoms in a mole. It's a cornerstone of converting between the scale of moles and individual particles, thus allowing us to understand and predict atomic and molecular phenomena more precisely.

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

Propane, \(\mathrm{C}_{3} \mathrm{H}_{8}\), liquefies under modest pressure, allowing a large amount to be stored in a container. (a) Calculate the number of moles of propane gas in a 110 -L container at 3.00 atm and \(27^{\circ} \mathrm{C} .\) (b) Calculate the number of moles of liquid propane that can be stored in the same volume if the density of the liquid is \(0.590 \mathrm{~g} / \mathrm{mL} .\) (c) Calculate the ratio of the number of moles of liquid to moles of gas. Discuss this ratio in light of the kinetic-molecular theory of gases.

A scuba diver's tank contains \(0.29 \mathrm{~kg}\) of \(\mathrm{O}_{2}\) compressed into a volume of 2.3 L. (a) Calculate the gas pressure inside the tank at \(9^{\circ} \mathrm{C} .\) (b) What volume would this oxygen occupy at \(26^{\circ} \mathrm{C}\) and 0.95 atm?

Explain the difference between effusion and diffusion.

A sample of \(3.00 \mathrm{~g}\) of \(\mathrm{SO}_{2}(g)\) originally in a \(5.00-\mathrm{L}\) vessel at \(21^{\circ} \mathrm{C}\) is transferred to a \(10.0-\mathrm{L}\) vessel at \(26^{\circ} \mathrm{C}\). A sample of \(2.35 \mathrm{~g} \mathrm{~N}_{2}(g)\) originally in a \(2.50-\mathrm{L}\) vessel at \(20{ }^{\circ} \mathrm{C}\) is transferred to this same 10.0 - \(\mathrm{L}\) vessel. (a) What is the partial pressure of \(\mathrm{SO}_{2}(g)\) in the larger container? (b) What is the partial pressure of \(\mathrm{N}_{2}(g)\) in this vessel? (c) What is the total pressure in the vessel?

Which of the following statements best explains why a closed balloon filled with helium gas rises in air? (a) Helium is a monatomic gas, whereas nearly all the molecules that make up air, such as nitrogen and oxygen, are diatomic. (b) The average speed of helium atoms is higher than the average speed of air molecules, and the higher speed of collisions with the balloon walls propels the balloon upward. (c) Because the helium atoms are of lower mass than the average air molecule, the helium gas is less dense than air. The mass of the balloon is thus less than the mass of the air displaced by its volume. (d) Because helium has a lower molar mass than the average air molecule, the helium atoms are in faster motion. This means that the temperature of the helium is higher than the air temperature. Hot gases tend to rise.

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.