Write the Maxwell distribution, P(vx, vy, vz), for the velocities of molecules of mass M in a ga

Write the Maxwell distribution, P(vx, vy, vz), for the...

Write the Maxwell distribution, $P\left(v_x, v_y, v_z\right)$, for the velocities of molecules of mass $M$ in a gas at pressure $p$ and temperature $T$. (If you have forgotten the normalization constant, derive it from the Gaussian integral, $$ \int_{-\infty}^{\infty} \exp \left(-x^2 / 2 \sigma^2\right) d x=\sqrt{2 \pi} \sigma . $$ When a clean solid surface is exposed to this gas it begins to absorb molecules at a rate $W$ (molecules $/ \mathrm{s} \cdot \mathrm{cm}^2$ ). A molecule has absorption probability 0 for a normal velocity component less than a threshold $v_T$, and absorption probability 1 for a normal velocity greater than $v_T$. Derive an expression for $W$.

Write the Maxwell distribution, $P\left(v_x, v_y, v_z\right)$, for the velocities of molecules of mass $M$ in a gas at pressure $p$ and temperature $T$. (If you have forgotten the normalization constant, derive it from the Gaussian integral,
$$
\int_{-\infty}^{\infty} \exp \left(-x^2 / 2 \sigma^2\right) d x=\sqrt{2 \pi} \sigma .
$$

When a clean solid surface is exposed to this gas it begins to absorb molecules at a rate $W$ (molecules $/ \mathrm{s} \cdot \mathrm{cm}^2$ ).

A molecule has absorption probability 0 for a normal velocity component less than a threshold $v_T$, and absorption probability 1 for a normal velocity greater than $v_T$. Derive an expression for $W$.

Key Concepts

Kinetic Theory and Molecular Flux

Kinetic theory connects microscopic molecular motion to macroscopic observable quantities like pressure and absorption rates. In this framework, the flux of particles—i.e., the number of molecules striking a surface per unit time per unit area—is computed by integrating the product of the molecular velocity component normal to the surface and the Maxwell-Boltzmann probability density. This concept is essential in predicting how a gas interacts with surfaces, which is fundamental in areas ranging from material science to vacuum technology.

Threshold Effects in Molecular Absorption

When molecules interact with a surface, there can be a threshold condition for absorption, meaning only molecules with normal velocity components above a certain critical value contribute to the absorption process. This threshold effect introduces a conditional probability in the integration over the velocity distribution, affecting the overall absorption rate. It is important for modeling scenarios where energy barriers or kinetic constraints determine whether molecules stick to a surface or desorb.

Maxwell-Boltzmann Distribution

This distribution describes the probability of a gas molecule having a certain velocity in three-dimensional space when the system is in thermal equilibrium. It is derived from statistical mechanics and encapsulates the idea that molecular speeds have a spread governed by temperature, and that the velocities in different directions are statistically independent. The distribution is foundational in understanding the microscopic behavior of gases and provides the basis for calculating macroscopic properties such as pressure and temperature.

Gaussian Integrals and Normalization

Gaussian integrals arise when evaluating the normalization of the Maxwell-Boltzmann distribution, ensuring that the total probability over all possible velocities equals one. The integrals over the exponential terms are computed using standard Gaussian integral results. Normalization is a crucial step to validate that the probability distribution is physically meaningful and correctly scaled to account for all microstates accessible to the system.

Transcript

00:01 All right, so let's say we're given this normal density distribution equation, which has a lot of variables in it, which in which of the v is the speed of molecules in this equation of gas particles.

00:19 And we want to find the average v.

00:21 So we can integrate v times p of v, which equals and pulling out all the constants.

00:31 4 pi times m over 2 pi k t to the three halves times the integral of of v cube now since it's v squared e to the minus v squared over 2 kt times v dv and we can integrate this by parts and so if we let v equal or if we let u equal v squared and z prime equal v e equal v e to the minus v squared over 2k t and there's a reason for doing this it this way not u equals v cubed so we want it so that we want this z prime and i'm calling it z prime instead of v prime because we already have a v we want this z prime to have a v otherwise this expression is going to be really tough to integrate since we want the v since there's a v squared in the exponent here so u prime is going to be 2v and z is going to be just so e to the minus v squared over 2k t but then we have to add something to cancel out so if you took the derivative of e to the minus v squared over 2kt you would get a 2kt to come out a 1 over 2 kt so we need to cancel out that kt and you would have a minus m come out so we need an m to cancel this out and then the 2 from the v squared is going to cancel out the 2 in the denominator here so we don't have to worry about the two.

02:00 And actually these expressions, these expressions, we want zero to be the lower bounds since we want for v is greater than zero.

02:15 So this should be zero as well.

02:18 Okay, so that is going to transform our integral into four pi times m over two pi kt to the three halves.

02:33 Times, so i'm going to get uz.

02:37 So uz is going to be negative kt over m times v squared.

02:46 E to the minus mv squared over 2 kt from zero to infinity minus z u prime, but i already have a minus, so i'm just going to write plus and cancel out the minus, plus the integral from 0 to infinity of z prime.

03:10 So that's, i should pull out the constants again.

03:14 So i have kt over m and i have a 2.

03:18 So 2kt over m times the integral 0 to infinity of v times e to the minus v squared over 2kt, which is actually z prime.

03:39 So let's evaluate this first part.

03:42 So i might notice, so when i plug in infinity, i'm going to have infinity squared times e to the minus infinity.

03:52 And so that's infinity over infinity.

03:54 And you might wonder which one wins out.

03:57 And this e term is going to cause the whole term to go to zero, even though i'm multiplying by infinity.

04:04 And you can check that with lopitau's rule...