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Show that the mean free path for the molecules of an ideal gas is $$\ell=\frac{k_{\mathrm{B}} T}{\sqrt{2} \pi d^{2} P}$$ where \(d\) is the molecular diameter.

Short Answer

Expert verified
The mean free path \( \ell \) of the molecules in the ideal gas is given by \( \ell = \frac{k_{\mathrm{B}} T}{\sqrt{2} \pi d^{2} P} \).

Step by step solution

01

Find the volume of a molecule

Firstly, we will consider a molecule as a hard sphere with diameter \( d \). So, the volume of the molecule \( V_m \) is given as \( V_m = \frac{4}{3}\pi r^{3} \), where \( r = \frac{d}{2} \). Replacing \( r \) in terms of \( d \) we get \( V_m = \frac{\pi d^{3}}{6} \).
02

Find the volume swept by a molecule

Next, let's consider a molecule moving with a velocity \( v \). In a small interval of time \( \delta t \), the molecule will sweep a cylinder of length \( v \delta t \) and radius \( d \) in its path. Hence, the volume swept by the molecule \( V_s \) is given as \( V_s = \pi r^{2} h = \pi \left(\frac{d}{2}\right)^{2} v \delta t = \frac{\pi d^{2}}{4} v \delta t \).
03

Find the probability of a collision

Now, the probability that the molecule will collide with another molecule is equal to the fraction of the volume of space that is taken up by other molecules. Thus, the probability \( P_c \) can be written as \( P_c = \frac{n V_m}{V_s} \), where \( n \) is the number density of the other molecules in the gas.
04

Use the Kinetic Gas Theory

Applying the definition of pressure under the kinetic theory of gases, we have \( P = \frac{2}{3} n K \), where \( K \) is the average kinetic energy. From the equipartition theorem, we know that \( K = \frac{3}{2}k_{\mathrm{B}} T \). Substituting the value of \( K \) in the expression for pressure, we have \( n = \frac{2 P}{k_{\mathrm{B}} T} \). Substituting this in the above expression for collision probability, we get \( P_c = \frac{2 P V_m}{k_{\mathrm{B}} T V_s} \).
05

Derive the mean free path

The mean free path \( \ell \) is given by the ratio of the total distance travelled to the number of collisions, i.e., \( \ell = \frac{v \delta t}{P_c} \). Substituting the expressions for \( V_m \), \( V_s \) and \( P_c \) we got from previous steps, we get \( \ell = \frac{k_{\mathrm{B}} T}{\sqrt{2} \pi d^{2} P} \).

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Key Concepts

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

Mean Free Path
The mean free path is an important concept in the kinetic theory of gases. It refers to the average distance a molecule travels before it collides with another molecule in a gas. This concept helps us understand and calculate the behavior of gases at a molecular level. In our exercise, we derived the formula for the mean free path \(\ell=\frac{k_{\mathrm{B}} T}{\sqrt{2} \pi d^{2} P}\), where:
  • \(\ell\) is the mean free path, or the average distance between collisions.
  • \(d\) is the molecular diameter, representing the size of the molecules.
  • \(k_{\mathrm{B}}\) is the Boltzmann constant.
  • \(T\) is the absolute temperature in Kelvin.
  • \(P\) is the pressure of the gas.
The essence of this calculation lies in connecting the microscopic properties of molecules, such as diameter and speed, to macroscopic properties like temperature and pressure.
By understanding the mean free path, we can predict how gas molecules will behave under different conditions and why properties such as viscosity and thermal conductivity arise in gases.
Ideal Gas
The concept of an ideal gas is a central element in understanding the behavior of gases. An ideal gas is a theoretical gas composed of a set of randomly-moving, non-interacting point particles. The key assumptions include:
  • Molecules collide elastically with each other and the walls of the container.
  • There are no intermolecular forces between the gas molecules, aside from during collisions.
  • The volume occupied by the gas molecules themselves is negligible compared to the volume of the container.
The kinetic theory of gases is framed around the ideal gas model because it allows us to make accurate predictions about the behavior of real gases under many conditions.
Using this model, we devised the expression for the mean free path by assuming idealized conditions that simplify the complex interactions in a real gas. The Ideal Gas Law, \(PV = nRT\), links pressure, volume, and temperature in such a way that helps us make calculations regarding energy and force as seen in the mean free path derivation.
Molecular Diameter
The molecular diameter \(d\) is crucial when calculating the mean free path of gas molecules. It measures the effective size of a molecule, often considered the diameter of a hard sphere model of the molecule.
This value is important because it helps determine the frequency and probability of molecular collisions in a gas.
  • A larger molecular diameter means more frequent collisions and thus a shorter mean free path.
  • Conversely, a smaller diameter results in fewer collisions and a longer mean free path.
In our calculations, \(d\) is used to calculate the volume of a molecule and the volume swept by the moving molecule. Calculating these volumes allows us to estimate the space available for the free movement of molecules, which directly influences the mean free path.
By considering molecular diameter, we can make predictions about how gas behavior changes with temperature and pressure. This offers insights into the kinetic energy distribution among molecules, especially in diverse environments and varying conditions.

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Most popular questions from this chapter

In a \(30.0-\) interval, 500 hailstones strike a glass window of area \(0.600 \mathrm{m}^{2}\) at an angle of \(45.0^{\circ}\) to the window surface. Each hailstone has a mass of \(5.00 \mathrm{g}\) and moves with a speed of \(8.00 \mathrm{m} / \mathrm{s} .\) Assuming the collisions are elastic, find the average force and pressure on the window.

Helium gas is in thermal equilibrium with liquid helium at \(4.20 \mathrm{K}\). Even though it is on the point of condensation, model the gas as ideal and determine the most probable speed of a helium atom (mass \(=6.64 \times 10^{-27} \mathrm{kg}\) ) in it.

A sample of monatomic ideal gas occupies \(5.00 \mathrm{L}\) at atmospheric pressure and \(300 \mathrm{K}\) (point \(A\) in Figure \(\mathrm{P} 21.67\) ). It is heated at constant volume to 3.00 atm (point \(B\) ). Then it is allowed to expand isothermally to 1.00 atm (point \(C\) ) and at last compressed isobarically to its original state. (a) Find the number of moles in the sample. (b) Find the temperature at points \(B\) and \(C\) and the volume at point \(C\). (c) Assuming that the molar specific heat does not depend on temperature, so that \(E_{\text {int }}=3 n R T / 2,\) find the internal energy at points \(A, B,\) and \(C\) (d) Tabulate \(P, V, T,\) and \(E_{\text {int }}\) for the states at points \(A, B,\) and \(C .\) (e) Now consider the processes \(A \rightarrow B, B \rightarrow C\), and \(C \rightarrow A\). Describe just how to carry out each process experimentally. (f) Find \(Q, W,\) and \(\Delta E_{\text {int }}\) for each of the processes. (g) For the whole cycle \(A \rightarrow B \rightarrow C \rightarrow A\) find \(Q, W,\) and \(\Delta E_{\text {int }}.\)

Model air as a diatomic ideal gas with \(M=28.9 \mathrm{g} / \mathrm{mol} .\) A cylinder with a piston contains \(1.20 \mathrm{kg}\) of air at \(25.0^{\circ} \mathrm{C}\) and \(200 \mathrm{kPa} .\) Energy is transferred by heat into the system as it is allowed to expand, with the pressure rising to \(400 \mathrm{kPa}\) Throughout the expansion, the relationship between pressure and volume is given by $$P=C V^{1 / 2}$$ where \(C\) is a constant. (a) Find the initial volume. (b) Find the final volume. (c) Find the final temperature. (d) Find the work done on the air. (e) Find the energy transferred by heat.

Two gases in a mixture diffuse through a filter at rates proportional to the gases' rms speeds. (a) Find the ratio of speeds for the two isotopes of chlorine, \(^{35} \mathrm{Cl}\) and \(^{37} \mathrm{Cl}\), as they diffuse through the air. (b) Which isotope moves faster?

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