Question

The Maxwell-Boltzmann probability distribution function. According to the kinetic theory of gases, the energies of molecules moving along the \(x\) direction are given by \(\varepsilon_{x}=(1 / 2) m v_{x}^{2}\), where \(m\) is mass and \(v_{x}\) is the velocity in the \(x\) direction. The distribution of particles over velocities is given by the Boltzmann law, \(p\left(v_{x}\right)=e^{-m v_{x}^{2} / 2 k T}\). This is the Maxwell-Boltzmann distribution (velocities may range from \(-\infty\) to \(+\infty\) ). (a) Write the probability distribution \(p\left(v_{x}\right)\), so that the Maxwell-Boltzmann distribution is correctly normalized. (b) Compute the average energy \(\left\langle\frac{1}{2} m v_{x}^{2}\right\rangle\). (c) What is the average velocity \(\left\langle v_{x}\right\rangle\) ? (d) What is the average momentum \(\left\langle m v_{x}\right\rangle\) ?

Short answer

Normalized distribution: \( p(v_x) = \sqrt{\frac{m}{2\pi kT}} e^{-\frac{mv_{x}^{2}}{2kT}} \). Average energy: \( \frac{1}{2}kT \). Average velocity: 0. Average momentum: 0.

Step-by-step solution

- Normalize the Maxwell-Boltzmann Distribution

The probability distribution must satisfy the normalization condition: \[ \ \int_{-\infty}^{+\infty} p(v_{x})\, dv_{x} = 1 \] Given: \[ \ p(v_{x}) = \, e^{-\frac{mv_{x}^{2}}{2kT}} \] We need a constant \( \text{A} \) so that \[ \ p(v_{x}) = \text{A} \, e^{-\frac{mv_{x}^{2}}{2kT}} \] Integrate this expression over all velocities:\[ \ A \int_{-\infty}^{+\infty} e^{-\frac{mv_{x}^{2}}{2kT}} dv_{x} = 1 \ \] Since the integral of a Gaussian function \( e^{-\frac{ax^{2}}{b^{2}}} \) over all space \( -\rightarrow \infty \rightarrow +\rightarrow \infty \) is \( \sqrt{\frac{\pi}{a}} \), we have: \[ \ A \sqrt{\frac{2\pi kT}{m}} = 1 \] Thus, \[ \text{A} = \sqrt{\frac{m}{2\pi kT}} \] So the normalized probability distribution is: \[ \ p(v_{x}) = \sqrt{\frac{m}{2\pi kT}} e^{-\frac{mv_{x}^{2}}{2kT}} \]

- Compute the Average Energy

Consider average energy \( \left\langle \frac{1}{2}m v_{x}^{2} \right\rangle \):\[ \left\langle \frac{1}{2}m v_{x}^{2} \right\rangle = \int_{-\infty}^{+\infty} \frac{1}{2}m v_{x}^{2} p(v_{x}) dv_{x} \] Substitute \( p(v_{x}) \):\[ \left\langle \frac{1}{2}m v_{x}^{2} \right\rangle = \frac{m}{2}\int_{-\infty}^{+\infty} v_{x}^{2} \sqrt{\frac{m}{2\pi kT}} e^{-\frac{mv_{x}^{2}}{2kT}} dv_{x} \] Use: \[ \int_{-\infty}^{+\infty} x^{2} e^{-ax^{2}} dx = \sqrt{\frac{\pi}{a^{3}}} /\ 2 \ \]where \( a = \frac{m}{2kT} \):\[ \ = \frac{m}{2}\sqrt{\frac{m}{2\pi kT}} \times \sqrt{\frac{\pi}{(\frac{m}{2kT})^{3}}} /\ 2 \] Calculate:\[ \frac{m}{2}\frac{\pi (\frac{2kT}{m})^{3/2}}{\pi (\frac{m}{2kT})^{1.5}} = \frac{kT}{2} \left\langle \frac{1}{2}m v_{x}^{2} \right\rangle = \frac{1}{2}kT \]

- Compute the Average Velocity

Given the symmetric nature, where the Maxwell-Boltzmann conditioned symmetry is considered, of the velocity distribution: \[ \left\langle v_{x} \right\rangle = 0 \] Since the function \( v_x \cdot \exp{(-\frac{mv_x^2}{2kT})} \) is odd, its integral over the symmetric interval is zero:\[\left\langle v_x\right\rangle = 0\]

- Compute the Average Momentum

The average momentum is: \[ \left\langle m v_x \right\rangle = m \cdot \left\langle v_x \right\rangle = 0 \] This is directly derived from the mean velocity.

Key concepts

kinetic theory of gases

The kinetic theory of gases explains how gas particles move and behave. It tells us that gas consists of many small particles (atoms or molecules) in constant, random motion. These particles constantly collide with each other and with the walls of their container. The theory is based on several key assumptions:

• Gas particles are in constant, random motion.
• Collisions between gas particles and the walls are elastic, meaning no energy is lost.
• The volume of the gas particles themselves is negligible compared to the volume of the container.
• There are no attractive or repulsive forces between the particles.

One important conclusion from this theory is that the average kinetic energy of gas particles is proportional to the temperature. This relationship is critical in understanding the Maxwell-Boltzmann distribution and how gas particles' velocities are distributed at any given temperature.

probability distribution function

A probability distribution function gives us the likelihood of different outcomes in a random process. For gas particles' velocities, the Maxwell-Boltzmann distribution function is used. The function is given by: \( p(v_x) = \sqrt{ \frac{m}{2 \pi k T} } e^{-(mv_x^2)/(2kT)} \)
This equation shows how likely it is to find a particle with a specific velocity in the x-direction. The term \( e^{-(mv_x^2)/(2kT)} \) decays as the velocity increases, meaning particles with higher speeds are less probable. To ensure that the total probability across all velocities sums to one, we normalize this function. By integrating the distribution over all possible velocities, we find the appropriate normalization constant.

The integral here is crucial as the sum of all probabilities must equal one. Without normalization, the distribution function would not correctly represent our physical system.

average energy calculation

The average energy of gas particles can be calculated using the Maxwell-Boltzmann distribution. Specifically, we are interested in the average kinetic energy in the x-direction, given by: \( \left\langle \frac{1}{2} m v_x^2 \right\rangle = \frac{1}{2} k T \)
To compute this, we use an integral where we multiply the kinetic energy \( \frac{1}{2} m v_x^2 \) by the probability distribution function \( p(v_x) \) and integrate over all velocities: \[ \left\langle \frac{1}{2} m v_x^2 \right\rangle = \int_{-\infty}^{+\infty} \frac{1}{2} m v_x^2 p(v_x) dv_x \] Substituting the distribution function and solving the integral yields the result: \[ \left\langle \frac{1}{2} m v_x^2 \right\rangle = \frac{1}{2} k T \] This result shows that the average kinetic energy in one direction (x) is proportional to the temperature. This is a fundamental result from the kinetic theory of gases, revealing how energy is distributed among particles.

average velocity calculation

The average velocity of particles, \( \left\langle v_x \right\rangle \), is another important quantity in the kinetic theory of gases. It represents the mean velocity of all particles in the x-direction. For a perfectly symmetric Maxwell-Boltzmann distribution centered around zero, the average velocity is: \[ \left\langle v_x \right\rangle = 0 \]
This result arises because the distribution is symmetric around zero, meaning there are equally as many particles moving in the positive direction as in the negative direction. When summed, these velocities cancel out, resulting in a net average of zero.
Therefore, even though individual particles can have varying speeds, the overall average velocity in any direction for a system at equilibrium is zero. This conclusion aligns with our expectations for a large, random distribution of particle velocities.

average momentum

The average momentum, \( \left\langle m v_x \right\rangle \), of gas particles is directly related to the average velocity. Since \( m \) (mass) is a constant for all particles, it scales with the velocity. Therefore, if the average velocity in the x-direction is zero, the average momentum is also zero: \[ \left\langle m v_x \right\rangle = m \cdot \left\langle v_x \right\rangle = 0 \]
This result means that for an ideal gas in equilibrium, there is no net movement of particles in any direction. Even though individual particles are in motion and have momentum, the overall momentum averages out to zero.
This makes sense within the context of the kinetic theory of gases, ensuring that there is no preferred direction of motion in a stable gas system. When considering forces and collisions, this property helps maintain the system's equilibrium over time.