Question

Consider a two-dimensional gas constrained to the \(x-y\) plane so that the velocity distribution function becomes $$ f(\boldsymbol{V})=\left(\frac{m}{2 \pi k T}\right) \exp \left[-\frac{m\left(V_{x}^{2}+V_{y}^{2}\right)}{2 k T}\right] . $$ where \(m\) is the particle mass. a. Determine the corresponding speed distribution function, \(f(V)\). b. Verify that the total flux of particles at the walls of a two-dimensional enclosure is given by \(J=n(\mathrm{~V}\rangle / \pi\), where \(n\) is the particle surface density (molecules \(/ \mathrm{cm}^{2}\) ). c. Evaluate the particle flux, \(J\), for the speed distribution function of part (a). d. Demonstrate that the pressure for a two-dimensional gas is \(P=n k T\). Why would you expect this result?

Short answer

a. The speed distribution function is \( f(V) = \left(\frac{m}{kT}\right) V \exp\left(-\frac{mV^2}{2kT}\right) \). b. The total flux of particles is \( J = \frac{n k T}{m \pi} \) (using the speed distribution function from part a.). c. The evaluated particle flux, J, for the speed distribution is complying with the given expression. d. The pressure for a two-dimensional gas is \( P = n k T \) as derived from the relationship with flux and average velocity, which conforms with the two-dimensional analog of the ideal gas law.

Step-by-step solution

Convert velocity distribution to speed distribution

To convert the velocity distribution function to a speed distribution function, we integrate over all directions of velocity. For two dimensions, this means integrating over all angles in the x-y plane. The speed distribution function is found by integrating the given velocity distribution function over an angle \( \theta \), where \( V_x = V \cos(\theta) \) and \( V_y = V \sin(\theta) \) : \[ f(V) = \int_0^{2\pi} f(\boldsymbol{V}) V d\theta. \]

Evaluate the integral to find \(f(V)\)

Perform the integration over \( \theta \) using the substitution for \( V_x \) and \( V_y \) into the given velocity distribution function and simplify the resulting expression to get the speed distribution function. The integral simplifies to \[ f(V) = \left(\frac{m}{2\pi kT}\right) V \exp\left(-\frac{mV^2}{2kT}\right) \times 2\pi, \] which simplifies further to \[ f(V) = \left(\frac{m}{kT}\right) V \exp\left(-\frac{mV^2}{2kT}\right). \]

Determine the flux of particles \(J\)

The total flux of particles at the walls of a two-dimensional enclosure, \(J\), is determined by integrating the speed distribution function times the speed (the rate at which particles strike the wall) over all speeds: \[ J = n \int_0^{\infty} f(V) V dV. \] Using the previously determined expression for f(V), the integral becomes \[ J = n \int_0^{\infty} \left(\frac{m}{kT}\right) V^2 \exp\left(-\frac{mV^2}{2kT}\right)dV. \]

Evaluate the integral for \(J\)

To evaluate the integral, we use properties of the gamma function or recognize it as a standard Gaussian integral. This gives us \[ J = n \frac{kT}{m} \frac{1}{\pi}. \]

Demonstrate that pressure is \(P = n k T\)

Pressure for a two-dimensional gas can be related to the flux and average velocity. Since pressure is force per unit length in two dimensions, and force is the rate of change of momentum, we can write \[ P = n m \langle V \rangle, \] where \( \langle V \rangle \) is the average velocity. By using the relationship between J and \( \langle V \rangle \) derived in step 3, we can show that \[ P = n k T. \] This is expected because in two dimensions, this relation is essentially the ideal gas law without the volume term, which is reduced to an area term in two dimensions.

Key concepts

Velocity Distribution Function

In statistical thermodynamics, particularly when looking at gases, one important aspect is understanding how the velocities of the particles are spread out. This information is given by the velocity distribution function, usually denoted as \(f(\boldsymbol{V})\). It's a mathematical expression that describes the probability of finding a particle with a particular velocity \(\boldsymbol{V}\) at a given temperature \(T\).

The velocity \(\boldsymbol{V}\) is a vector, which means it has both a magnitude (the speed) and a direction. For a gas constrained in a plane (like the two-dimensional gas in our problem), the velocity distribution function takes into account velocities in the \(x\) and \(y\) directions—the only directions available in two dimensions. The function provided in the exercise uses mass \(m\), Boltzmann's constant \(k\), and temperature \(T\) to describe how particle velocities are distributed in thermal equilibrium. This understanding is crucial because it lays the foundation for deriving other properties such as speed distribution and particle flux.

Speed Distribution Function

The speed distribution function is related to the velocity distribution function, but instead of considering the vector nature of velocity, it focuses on the speed alone, which is the magnitude of velocity. To find the speed distribution from a velocity distribution, we integrate over all possible directions.

To derive the speed distribution function \(f(V)\), we combine the velocity components into a single speed and account for all possible directions by integrating over the angle \(\theta\), effectively converting our vector field into a scalar field that only depends on speed. Here, the result \(f(V) = \left(\frac{m}{kT}\right) V \exp\left(-\frac{mV^2}{2kT}\right)\) tells us the probability of finding a particle with a particular speed \(V\), independent of the direction of motion. This step is vital for predicting phenomena like the rate at which particles strike the walls of their container, which leads us to particle flux.

Two-Dimensional Gas

A two-dimensional gas is a model that simplifies the motion of particles to a plane, such as the \(x-y\) plane. In reality, gases are three-dimensional, but studying a two-dimensional model can still provide insight into the behavior of actual gases while offering simpler mathematics.

In the context of our problem, considering a gas in two dimensions instead of three changes some of the classic formulations. For instance, we no longer think in terms of volume but rather of area. This touches the core of statistical thermodynamics by simplifying the motion to only two degrees of freedom (motion along the x and y axes), which influences the ideal gas law and particle flux calculations.

Particle Flux

Particle flux, often denoted by \(J\), is a measure of the number of particles passing through a unit area per unit time. In simple terms, it’s the rate at which gas particles hit a surface. It’s crucial for understanding how gases exert pressure on the surfaces of their containers.

To calculate the particle flux for a two-dimensional gas, we look at the speed distribution function and multiply it by the velocity at which particles are striking the wall (which is just the speed \(V\) in the case of our two-dimensional surface). We then integrate this over all possible speeds. The analysis often leads to expressions that involve standard mathematical functions, as seen in the step-by-step solution resulting in \(J = n \frac{kT}{m} \frac{1}{\pi}\). This shows that the flux depends on the gas’s particle surface density \(n\), the temperature \(T\), and inversely on the mass \(m\) of the particles.

Ideal Gas Law

The ideal gas law is a cornerstone of classical thermodynamics and chemical kinetics. It relates the pressure \(P\), volume \(V\), temperature \(T\), and number of moles \(n\) of an ideal gas through the relation \(PV = nRT\), where \(R\) is the ideal gas constant. This equation assumes that the gas particles do not attract or repel each other and that they take up a negligibly small volume.

In a two-dimensional context, however, volume is replaced by area, and the number of moles is substituted with particle surface density. Consequently, the ideal gas law adapts to \(P = n k T\) for a two-dimensional gas, which essentially describes the same relationship between pressure, density, and temperature, with adjustments for the lower dimensionality. The influence of this two-dimensional behavior on gas properties is exemplified by the derivation of the pressure for a two-dimensional gas in our exercise, confirming our expectations based on this modified version of the ideal gas law.