Chapter 15: Problem 2

It is shown in the kinetic theory of gases that the probability for the distance a molecule travels between collisions to be between \(x\) and \(x+d x\), is proportional to \(e^{-x / \lambda} d x,\) where \(\lambda\) is a constant. Show that the average distance between collisions (called the "mean free path") is \(\lambda\). Find the probability of a free path of length \(\geq 2 \lambda\).

Short Answer

Expert verified

The mean free path is \lambda. The probability of a free path greater than \(2 \lambda\) is \(e^{-2}\).

Step by step solution

Understanding the Probability Density Function

Given the probability for the distance a molecule travels between collisions to be between \(x\) and \(x + dx\) is proportional to \(e^{-x / \lambda} dx\). This implies the probability density function can be written as:\[ P(x) dx = C e^{-x/\lambda} dx \] where \(C\) is a normalization constant.

Determining the Normalization Constant

To find the constant \(C\), we use the fact that the total probability must be 1:\[ \int_{0}^{\infty} C e^{-x/\lambda} dx = 1 \] Solve for \(C\) by computing the integral.

Solving the Integral for Normalization

Compute: \[ C \int_{0}^{\infty} e^{-x/\lambda} dx = 1 \] Use substitution \(u = x/\lambda\), then \(du = dx/\lambda\). Thus, the integral becomes:\[ C \lambda \int_{0}^{\infty} e^{-u} du = 1 \] \[ C \lambda [ -e^{-u} ]_{0}^{\infty} = 1 \] Evaluating the limits, \[ C \lambda (1 - 0) = 1 \] \[ C = \frac{1}{\lambda} \]

Finding the Average Distance (Mean Free Path)

The average distance (mean free path) is given by the expectation value of \(x\):\[ \langle x \rangle = \int_{0}^{\infty} x P(x) dx \] Substitute \(P(x) = \frac{1}{\lambda} e^{-x/\lambda}\):\[ \langle x \rangle = \int_{0}^{\infty} x \frac{1}{\lambda} e^{-x/\lambda} dx \] Use integration by parts to solve this.

Using Integration by Parts

Let \(u = x\), \(dv = \frac{1}{\lambda} e^{-x/\lambda}dx\). Then, \(du = dx\) and \(v = - e^{-x/\lambda}\). Applying integration by parts: \[ \langle x \rangle = \left[ -\frac{x}{\lambda} e^{-x/\lambda} \right]_{0}^{\infty} + \int_{0}^{\infty} \frac{1}{\lambda} e^{-x/\lambda} dx \] The first term evaluates to zero, and the second term is:\[ \lambda \left[1 - 0\right] = \lambda \]

Calculating the Probability of Free Path Greater than \(2 \lambda\)

The probability \(P(x \geq 2 \lambda)\) is found by integrating the probability density function from \(2 \lambda\) to infinity:\[ P(x \geq 2 \lambda) = \int_{2 \lambda}^{\infty} \frac{1}{\lambda} e^{-x/\lambda} dx \] Solve the integral: \[ P(x \geq 2 \lambda) = \left[ -e^{-x/\lambda} \right]_{2 \lambda}^{\infty} = 0 - (-e^{-2}) = e^{-2} \]

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Key Concepts

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

Mean Free Path

In the kinetic theory of gases, the mean free path is the average distance a molecule travels between collisions. This distance helps us understand how molecules move in a gas and how frequently they interact.

The mean free path formula can be derived from the probability density function. The problem definition suggests that the distance a molecule travels between collisions follows an exponential probability distribution. This probability density function is given by: \[ P(x) dx = C e^{-x/\textbackslash lambda} dx \] where \( C \) is a normalization constant, and \( \textbackslash lambda \) is a characteristic distance related to the mean free path.

To make sure the probability is correctly normalized (i.e., the total probability equals 1), we solve: \[ \textbackslash int_{0}^{\textbackslash infty} C e^{-x/\textbackslash lambda} dx = 1 \] After finding the normalization constant \( C = \textbackslash frac{1}{\textbackslash lambda} \), we can calculate the average distance a molecule travel, or mean free path, as: \[ \textbackslash langle x \textbackslash rangle = \textbackslash int_{0}^{\textbackslash infty} x P(x) dx \] Substituting \( P(x) \) into the integral and solving it, we find that: \[ \textbackslash langle x \textbackslash rangle = \textbackslash lambda \] This indicates that the mean free path is \( \textbackslash lambda \), as expected.

Probability Density Function

In statistics and probability theory, a probability density function (PDF) describes the likelihood of a continuous random variable to take on a particular value. For the kinetic theory of gases, the PDF determines how distances between molecular collisions are distributed.

Given our specific problem, the PDF is defined as: \[ P(x) dx = C e^{-x/\textbackslash lambda} dx \] Here’s a breakdown of its components:

C: Normalization constant ensuring the total probability is 1.
\( e^{-x/\lambda} \): Exponential decay function describing how the probability decreases as the distance (x) increases.
\( \lambda \): Characteristic distance parameter representing the mean free path.

To solve this integral: \[ \textbackslash int_{0}^{\textbackslash infty} C e^{-x/\textbackslash lambda} dx = 1 \] We recognize a common form of integrals involving exponential functions. Solving this with integration, we find that the normalization constant \( C \) is: \[ C = \frac{1}{\lambda} \] This allows us to correctly define our PDF as: \[ P(x) = \frac{1}{\lambda} e^{-x/\textbackslash lambda} \] This PDF helps us calculate probabilities and expected values, such as the mean free path and the probability for a path length greater than a certain value.

Integration by Parts

Integration by parts is a technique used in calculus to integrate products of functions. It's particularly helpful in this problem to find expected values or mean values by integrating functions that are products.

The formula for integration by parts is: \[ \textbackslash int u dv = uv - \textbackslash int v du \] In our problem, we use integration by parts to solve for the mean free path where:

\( u = x \)
\( dv = \frac{1}{\lambda} e^{-x/\textbackslash lambda}dx \)

Then, we find the derivatives and integrals:

\( du = dx \)
\( v = - e^{-x/\textbackslash lambda} \)

Applying integration by parts: \[ \textbackslash langle x \textbackslash rangle = \left[ - \frac{x}{\lambda} e^{-x/\textbackslash lambda} \right]_{0}^{\textbackslash infty} + \textbackslash int_{0}^{\textbackslash infty} \frac{1}{\lambda} e^{-x/\textbackslash lambda} dx \] Evaluating this, the first term vanishes due to its exponential factor, and the integral simplifies to: \[ \lambda \left[ 1 \right] = \lambda \] This confirms that the mean free path is \( \lambda \), demonstrating how integration by parts can simplify the calculation of expected values in exponential distributions.