In this section we'll consider how the microscopic behavior of atoms and/or molecules (we'll lump those terms together into "particle.") explain the bulk behavior (temperature, pressure and other properties) of the gas. It all comes down to the distribution of speeds among the particles of the gas. James Clerk Maxwell and Ludwig Boltzmann, working separately between 1860 and 1880 [Think about that], developed the theory that we still used to model that distribution.
In any container of gas there are particles traveling faster than the average speed and particles traveling more slowly.
Particles continually shuffle their speeds by exchanging kinetic energy through collisions and through collisions with the container walls.
The large number of particles and that constant change from particle to particle allowed Maxwell and Boltzmann to model the speeds of gas particles with a probability distribution.
We work here under the same assumptions that we used to develop the ideal gas law, the most important of which is that our gas particles undergo elastic collisions (see momentum), in which momentum and kinetic energy are conserved.
Sometimes students wonder how they'd ever be able to come up with a difficult derivation or proof in math or science. It's worth taking a moment to realize that it took about 20 years to develop the kinetic-molecular theory of gases, and it took the work of many people. Hang in there.
The Maxwell-Boltzmann distribution of particle speeds in a sample of an ideal gas at temperatre T looks like this.
$$f(v) = \sqrt{\left( \frac{m}{2 \pi kT} \right)^3} 4\pi v^2 e^{-\frac{mv^2}{2kT}}$$
Don't panic, it's not as bad as it seems! It's a function of the velocities (we're using velocity here when we really mean speed, because we don't actually care about the particle directions, but that's OK). k = 1.381 × 10-23 J/K is the Boltzmann constant, m is the mass of a particle and v is the velocity, the independent variable. The term under the radical is called a normalization factor, and it's there to ensure that the total probability of any given speed in the distribution is one, i.e. that every particl has to have some speed.
We could also rewrite the radical expression like this:
$$f(v) = \left( \frac{m}{2\pi kT} \right)^{\frac{3}{2}} 4\pi v^2 e^{-\frac{mv^2}{2kT}}$$
Just as a side note here, when you encounter exponential expressions like this one with complicated exponents, they're sometimes written as exp( ...) instead of e ..., like this:
$$e^{-\frac{mv^2}{2kT}} \; \longrightarrow \; exp \left( -\frac{mv^2}{2kT} \right)$$
When you see that in other literature, you'll know what it means.
We'll explore how this distribution was discovered later, but for now, let's see what its graph looks like.
The Maxwell-Boltzmann distribution for a sample of gas at T = 200K is plotte below. For simplicity, k and m have been set to one (k = 1, m = 1), That means that the velocity axis doesn't have any absolute meaning – no big deal for our purpose here. We'll calculate some actual average velocities later.
One feature of this distribution of speeds that stands out if you're used to looking at left-right symmetric probability distributions like the Guassian distribution (bell-shaped curve) is that it's asymmetric. Notice that the "tail" on the right is longer than the tail on the left.
Here is a comparison of Maxwell-Boltzmann distributions for particles of the same mass at temperatures 100K - 500K.
There are a couple of important things to notice:
The peak of the distribution shifts to higher speeds as we heat the sample.
The peak of the distribution, which represents the most-probable speed, is higher at lower temperatures.
The area under each curve, which represents the total probability, is the same.
The figure below might help you see that the areas under these curves is always the same. We ensure that this is true for any probability distribution by including the proper normalization factor.
Finally, here are two M-B distributions at T = 10K and T = 1000K. Notice that at low temperature, the distribution is "tight." The fastest and slowest particles have speeds very close to the most probable speed, but at high temperature, those speeds are much more "smeared out." The area under these curves is also equivalent.
These graphs show that the average speed and temperature of a gas are connected: If we reduce the temperature, we expect that the average speed of the molecules will be lower, and that the outlying speeds will be closer to the average. If we heat up a gas, we expect that the average speed will be higher and that the distribution of speeds will be much broader.
Interestingly, this works in the other direction, too. If we can find a way to reduce the average velocity of the particles in a gas, or if we can make the distribution of speeds narrower, we can reduce its temperature. That is the basis for a number of modern experiments that have allowed us to uncover some interesting properties of matter – more on that below.
You can manipulate the graph to explore how the M-B distribution of speeds is affected by particle mass and temperature. Select either helium (He), argon(Ar) or krypton (Kr), and use the slider to change the temperature between 10K and 1000K.
Notice that increasing the mass slows the atoms down. That's because kinetic energy (KE) is conserved. At a given temperature, all particles have the same average KE, and $KE = \frac{1}{2}mv^2$, so for a given mass, the speed has to drop.
Also notice that as you turn the temperature up, the most probable speed (MPS), the speed at the peak of the distribution, increases while the distribution smears out a bit.
Note: the P(s) axis rescales for each atom, so you can't compare the heights of the curves.
The sections below on most-probable speed and average speed contain a bit of calculus. If you don't know anything about derivatives, don't worry. The results are recapped in the box below.
The Maxwell Boltzmann distribution is a probability distribution curve of a continuous variable, v (for velocity, but we really mean speed). It's different than the discrete probability distribution that models flipping of coins or rolling of dice. That presents us with some challenges. For one, we can't just evaluate the function and get the probability of a particle traveling at that speed. What we can do is integrate the function (calculus) to get the probability of a particle moving within a range of speeds.
There are some convenient speeds we can calculate using the M-B distribution. Three are particularly useful, the most-probable speed, the mean speed and the root-mean-squared (RMS) speed.
Here they are graphically, but not exactly to scale horizontally:
The most probable speed is the speed at the maximum of the distribution. If you reached into a sample of gas and picked out a particle, it would most likely be traveling at this speed. We find it by setting the derivative of the M-B distribution to zero:
$$\frac{d \, f(v)}{dv} = 0$$
The derivative is
$$\frac{d}{dv} \left( \frac{m}{2 \pi kT} \right)^{\frac{3}{2}} 4 \pi v^2 e^{-\frac{mv^2}{2kT}}$$
We can remove the constants to simplify it first,
$$4\pi \left( \frac{m}{2\pi kT} \right)^{\frac{3}{2}} \frac{d}{dv} v^2 e^{-\frac{mv^2}{2kT}}$$
The product rule and chain rule give us
Because this derivative is set to zero, we can divide away everything that's cancelled inside of the brackets. The constant is never zero (or equivalently, we could just divide both sides by the constant term), so all that's necessary is for this simple expression to be true:
$$2 - \frac{mv^2}{kT} = 0$$
Finding a common denominator (kT) and rearranging gives us
$$\frac{2kT - mv^2}{kT} = 0$$
multiplying by the denominator on both sides and rearranging gives
$$ \begin{align} 2kT - mv^2 &= 0 \\ 2kT &= mv^2 \end{align}$$
Finally, the most probable velocity is:
$$v_{mp} = \sqrt{\frac{2kT}{m}}$$
We can draw important conclusions from this result. First, the velocity of paricles scales with the temperature, but only as the square-root of T – it's not linear. Second, as particle size increases, the velocity also drops as the square root of the mass. That's because kinetic energy is what is conserved here: $KE = 1/2 mv^2$, so as the mass increases, the velocity must decrease.
Calculating average speed is a little trickier. Recall that the average value of a function over an interval [a, b] is
$$\frac{1}{b - a} \int_a^b f(x) \, dx$$
In the case were we have a probability distribution function for the velocity, the average velocity is
$$\bar{v} = \int_0^{\infty} v P(v) \, dv,$$
where P(v) is the Maxwell-Boltzmann distribution, so our mean speed is.
$$4\pi \left( \frac{m}{2\pi kT} \right)^{3/2} v^3 \, exp \left( -\frac{mv^2}{2kT}\right) \, dv$$
This integral is done by parts. The result is:
$$ \begin{align} \bar{v} &= 4\pi \left( \frac{m}{2\pi kT} \right)^{3/2} \frac{4 k^2 T^2}{2m^2} \\ \\ &= \sqrt{\frac{8 RT}{\pi m}} \end{align}$$
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This graph shows the most probable, mean and RMS speeds of N2 gas at temperatures from 10K to 500K. At 300K, the most-probable speed of an N2 is 322 m/s.
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