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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.

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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.

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Most popular questions from this chapter

Cyclopropane, a gas used with oxygen as a general anesthetic, is composed of \(85.7 \% \mathrm{C}\) and \(14.3 \% \mathrm{H}\) by mass. \((\mathrm{a})\) If \(1.56 \mathrm{~g}\) of cyclopropane has a volume of \(1.00 \mathrm{~L}\) at 0.984 atm and \(50.0^{\circ} \mathrm{C}\), what is the molecular formula of cyclopropane? (b) Judging from its molecular formula, would you expect cyclopropane to deviate more or less than Ar from ideal-gas behavior at moderately high pressures and room temperature? Explain. (c) Would cyclopropane effuse through a pinhole faster or more slowly than methane, \(\mathrm{CH}_{4} ?\)

(a) What is an ideal gas? (b) Show how Boyle's law, Charles's law, and Avogadro's law can be combined to give the ideal-gas equation. (c) Write the ideal-gas equation, and give the units used for each term when \(R=0.08206 \mathrm{~L}-\mathrm{atm} / \mathrm{mol}-\mathrm{K}\). (d) If you measure pressure in bars instead of atmospheres, calculate the corresponding value of \(R\) in \(\mathrm{L}-\mathrm{bar} / \mathrm{mol}-\mathrm{K}\).

A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required 105 s for \(1.0 \mathrm{~L}\) of the gas to effuse. Under identical experimental conditions it required \(31 \mathrm{~s}\) for \(1.0 \mathrm{~L}\) of \(\mathrm{O}_{2}\) gas to effuse. Calculate the molar mass of the unknown gas. (Remember that the faster the rate of effusion, the shorter the time required for effusion of \(1.0 \mathrm{~L} ;\) that is, rate and time are inversely proportional.)

Calculate each of the following quantities for an ideal gas: (a) the volume of the gas, in liters, if \(1.50 \mathrm{~mol}\) has a pressure of 1.25 atm at a temperature of \(-6^{\circ} \mathrm{C} ;(\mathbf{b})\) the absolute temperature of the gas at which \(3.33 \times 10^{-3}\) mol occupies \(478 \mathrm{~mL}\) at 750 torr; \((\mathbf{c})\) the pressure, in atmospheres, if \(0.00245 \mathrm{~mol}\) occupies \(413 \mathrm{~mL}\) at \(138{ }^{\circ} \mathrm{C} ;(\mathbf{d})\) the quantity of gas, in moles, if \(126.5 \mathrm{~L}\) at \(54^{\circ} \mathrm{C}\) has a pressure of \(11.25 \mathrm{kPa}\).

An open-end manometer containing mercury is connected to a container of gas, as depicted in Sample Exercise \(10.2 .\) What is the pressure of the enclosed gas in torr in each of the following situations? (a) The mercury in the arm attached to the gas is \(15.4 \mathrm{~mm}\) higher than in the one open to the atmosphere; atmospheric pressure is 0.985 atm. (b) The mercury in the arm attached to the gas is \(12.3 \mathrm{~mm}\) lower than in the one open to the atmosphere; atmospheric pressure is \(0.99 \mathrm{~atm} .\)

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