Question

Because the van der Waals constant \(b\) is the excluded volume per mole of a gas, we can use the value of \(b\) to estimate the radius of a molecule or atom. Consider a gas that consists of molecules, for which the van der Waals constant \(b\) is \(0.0315 \mathrm{~L} / \mathrm{mol}\). Estimate the molecular radius in pm. Assume that the molecules are spherical.

Short answer

The molecular radius is approximately 234 pm.

Step-by-step solution

Understanding van der Waals Constant

The van der Waals constant \(b\) represents the excluded volume per mole of molecules in a gas, essentially accounting for the volume occupied by the gas molecules themselves. Given \(b = 0.0315 \text{ L/mol}\), we need to convert this into cubic meters since later calculations involving radii will use standard metric units.

Convert Volume Units

First, convert the van der Waals constant from liters to cubic meters. We know that \(1 \text{ L} = 0.001 \text{ m}^3\). Therefore, \(b = 0.0315 \text{ L/mol} = 0.0315 \times 0.001 \text{ m}^3/\text{mol} = 3.15 \times 10^{-5} \text{ m}^3/\text{mol}\).

Volume per Molecule

Since \(b\) is given per mole, we need to divide it by Avogadro's number \(N_A = 6.022 \times 10^{23} \text{ molecules/mol}\) to find the excluded volume for a single molecule. Thus: \(V_{\text{molecule}} = \frac{3.15 \times 10^{-5} \text{ m}^3}{6.022 \times 10^{23}} \approx 5.23 \times 10^{-29} \text{ m}^3\).

Relating Volume to Radius

Assuming each molecule is spherical, the volume \(V\) of a sphere is \(\frac{4}{3}\pi r^3\). Set this equal to \(V_{\text{molecule}}\): \(\frac{4}{3}\pi r^3 = 5.23 \times 10^{-29} \text{ m}^3\). Solve for \(r\).

Solve for Molecular Radius

Rearrange to solve for \(r\): \[ r^3 = \frac{5.23 \times 10^{-29} \text{ m}^3 \times 3}{4\pi} \] \[ r^3 = \frac{1.569 \times 10^{-28} \text{ m}^3}{4\pi} \approx 1.25 \times 10^{-29} \text{ m}^3 \] Now, take the cube root: \(r \approx \left(1.25 \times 10^{-29}\right)^{\frac{1}{3}} \text{ m}\).

Convert Radius to Picometers

Calculate the cube root from Step 5: \(r \approx 2.34 \times 10^{-10} \text{ m}\). Finally, convert meters to picometers knowing \(1 \text{ pm} = 10^{-12} \text{ m}\): \(r \approx 234 \text{ pm}\).

Key concepts

Van der Waals Constant

The van der Waals constant, often denoted as \( b \), is an important parameter in the van der Waals equation. This equation is an attempt to describe the behavior of real gases more accurately than the ideal gas law. In essence, \( b \) represents the **excluded volume** {}
occupied by one mole of gas molecules.{}
It accounts for the finite volume of gas particles themselves.{}
When dealing with spherical molecules, \( b \) helps us understand just how much space is unavailable due to the physical presence of the molecules.{}
In the context of an exercise calculating the molecular radius, \( b \) is crucial because it helps us estimate the volume occupied by individual molecules. Knowing the van der Waals constant allows you to work backward to find the radius of a molecule, which can give insights into molecular size and structure. This constant is essential for students who want a realistic picture of gas behavior beyond the ideal gas assumption.

Excluded Volume

The concept of **excluded volume** becomes particularly pertinent when studying real gases in contrast to ideal gases. Ideal gases assume that molecules do not occupy any space and only interact through elastic collisions. However, the excluded volume reflects the actual volume that gas particles occupy and how they exclude each other from this space. While working with the van der Waals equation, this parameter \( b \) plays a pivotal role. The volume \( b \), provided for each mole, represents the cumulative space that a mole of gas molecules cannot occupy due to the presence of other molecules. When converting \( b \) from liters to cubic meters, you shift focus from a macroscopic scale to a more intimate molecular level, zooming into how a singular molecule fits into this broader space allocation.

Molecular Radius Estimation

Estimating the **molecular radius** involves cleverly navigating from the concept of excluded volume to the size of a single molecule. Here, Avogadro's number becomes a handy tool. It allows us to shift from mole-scale calculations to those applicable to individual molecules by assigning an average size to each particle.{}
This transformation provides insight into the microscopic world.{}
By assuming spherical molecules, the volume of a single molecule can be equated to the volume of a sphere using the formula \( \frac{4}{3}\pi r^3 \), which describes the relationship between volume and radius. Solving this equation grants us the radius of the molecules, which is essential when conceptualizing how molecules interact within a given space. When calculating this radius, remember to convert units appropriately to maintain consistency and accuracy in your results.

Spherical Molecules

In scientific calculations related to molecular size, **spherical molecules** are often assumed for simplicity. This assumption stems from the idea that a sphere is a simple shape that easily represents the three-dimensional space molecules might occupy. Although real molecules may not be perfect spheres, this approximation allows for straightforward mathematical calculations.{}
Using the sphere volume formula \( \frac{4}{3}\pi r^3 \), where \( r \) is the radius, simplifies the process of determining volumes and, consequently, dimensions like the radius.{}
This is particularly useful when we need to estimate the molecular radius from the volume information provided by a van der Waals constant like \( b \). By knowing the excluded volume per molecule and assuming spherical geometry, we translate the complex molecular structure into a more manageable calculation, which provides a clear visualization of molecular dimensions.