From the Maxwell-Boltzmann speed distribution, show that the most probable speed of a gas molecu

From the Maxwell-Boltzmann speed distribution, show that the...

Key Concepts

Most Probable Speed

The most probable speed is the speed at which the Maxwell-Boltzmann distribution reaches its maximum value, representing the speed most frequently observed in the gas. Identifying this speed is important because it emphasizes the typical behavior of particles in an ensemble and reflects the balance between the increasing availability of high-speed states and the decreasing likelihood of high energy due to thermal constraints.

Optimization via Derivative

To find the most probable speed, one generally differentiates the speed distribution function with respect to speed and sets the derivative to zero. This mathematical technique of finding extrema by analyzing where the slope of the function becomes zero is a common method in calculus, ensuring that the calculated speed corresponds to a maximum in the distribution.

Maxwell-Boltzmann Distribution

This is a probability distribution that describes the statistics of speeds among the molecules in an ideal gas at thermal equilibrium. It provides the likelihood that a molecule will have a certain speed by accounting for both the density of states and the exponential factor arising from the Boltzmann factor in the context of kinetic theory.

Transcript

00:01 Hello and welcome to chapter 16, problem number 46 in principles of physics.

00:07 So this problem asks us to calculate the most probable velocity from the boltzmann distribution.

00:14 And as the problem notes, that's basically just asking us to calculate the derivative with respect to velocity of the boltzman distribution, set it equal to zero, and then solve for velocity.

00:27 So i have the boltzman distribution written out here, and this is basically just a result.

00:31 Calculus problem we're going to just take the derivatives that it equal to zero and solve for v so you have to bear with me through the math here all right so we start things off we can see that this first whole chunk in blue here of our boltzmann distribution 4 pi n times quantity m over 2 pi k t to the three halves all of this doesn't have any any v any velocity in it at all so when we take the derivative of our entire equation this can just kind of come out because it's a constant with respect to velocity.

01:03 So we're really just going to be taking a look at the derivative of v squared e to the negative m, v squared over 2kt.

01:11 And so we're going to pull that out of our entire equation here.

01:16 And i don't have the equal to zero here.

01:19 I have to apologize for the space.

01:21 But then when we do this equation here, the derivative with respect to velocity of v squared, e, all this, you can see that's the chain rule.

01:31 So if you remember chain rule, it has us take the derivative of our first term with times the second term plus the derivative of the second term times the first term.

01:44 So i have their written out here, d -d -v of v -squared times r -e, plus v -squared times d -d -v -r -e.

01:54 So that's pretty much what we're going to keep going with.

01:56 And i'm also, before we go further, i'm just going to call all.

01:59 All of this constant here, all of this, i'm going to call it a.

02:03 Just for notation and simplicity, i ran out of space writing our constant up front here, so i didn't have space to write equals zero.

02:12 But keep in mind, remember, as i have up here, it is equal to zero.

02:17 So we're going to keep going.

02:19 So we have a, all of that, times the quantity.

02:22 We're going to take the derivative.

02:23 I'll move scroll up so you can see a little bit more with our, oops, sorry, with our calculus here.

02:30 Derivative with respect to velocity of v squared is just 2v.

02:33 So that's easy enough.

02:34 And then we multiply that by our whole e.

02:36 We add that to v squared.

02:39 And remember, when you take the derivative with respect to a term, you take the derivative of e.

02:44 It's another chain rule, basically.

02:46 So you have the e term, the normal e term, e to the negative m, v squared over 2kt.

02:52 See that it follows right down here.

02:54 The same thing.

02:56 Times the derivative of what our exponent is.

03:00 So we have d -dv of negative m v squared over 2kt is equal to zero...

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