4.2.1: Illustrations

Illustration 1: Maxwell-Boltzmann Distribution

In this animation \(N = nR\) (i.e., \(k_{B} = 1\)). This, then, gives the ideal gas law as \(PV = NT\). The average values shown, \(<\: >\), are calculated over intervals of one time unit. Restart.

The particles that make up a gas do not all have the same speed. The temperature of the gas is related to the average speed of the particles, but there is a distribution of particle speeds called the Maxwell-Boltzmann distribution. The smooth black curve on the graph is the Maxwell-Boltzmann distribution for a given temperature. What happens to the distribution as you increase the temperature? The distribution broadens and moves to the right (higher average speed). At a specific temperature, there is a set distribution of speeds. Thus, when we talk about a characteristic speed of a gas particle at a particular temperature we use one of the following (where \(M\) is the molar mass, \(m\) is the atomic mass):

• Average speed: \((8RT/\pi M)^{1/2}  =  (8k_{B}T/\pi m)^{1/2}\)
• Most probable speed: \((2RT/M)^{1/2}  =  (2k_{B}T/m)^{1/2}\)
• Root-mean-square (rms) speed: \((3RT/M)^{1/2}  =  (3k_{B}T/m)^{1/2}\)

There is not simply one way to describe the speed because it is a speed distribution. This means that as long as you are clear about which one you are using, you can characterize a gas by any of them. The different characteristic speeds are marked on the graph.

Illustration authored by Anne J. Cox.

Illustration 2: Kinetic Theory, Temperature, and Pressure

In this animation \(N = nR\) (i.e., \(k_{B} = 1\)). This, then, gives the ideal gas law as \(PV = NT\). The average values shown, \(<\: >\), are calculated over intervals of one time unit. Restart.

The equipartition theorem says that the temperature of a gas depends on the internal energy of the gas particles. For monatomic particles, the internal energy of a given particle is its kinetic energy (there is \(1/2 k_{B}T\) of energy per degree of freedom, and for a monatomic gas there are \(3\) degrees of freedom). Thus, for an ideal gas made up of particles of different masses, the average kinetic energy is the same for all the particles. In this animation the yellow particles are \(10\) times more massive than the light blue ones. How does the kinetic energy of the blue particle (representative of the smaller particles) compare with the kinetic energy of the orange particle (which is representative of the yellow particles)? What would you expect in a comparison of the speeds of the two particles? While the average kinetic energy of the particles should be identical, their average speeds should be different, since they have different masses.

Now triple the temperature.  What happens to the kinetic energy of both the blue particle (representative of the smaller particles) and the orange particle (representative of the yellow particles)? If you triple the temperature, what happens to the kinetic energy? What happens to the speeds of the particles? The average kinetic energy should increase by three, and the speed of the average particle should go up by \(1.73\), the square root of \(3\).

Finally, note that \(<d\mathbf{p}/dt>\), the average momentum delivered to the walls, is slightly higher than the value of the pressure calculated by the ideal gas law (\(P = NT/V\)). This is because the ideal gas law assumes that the particles are "point-like" (point particles), while the animation has particles with a definite radius. This means that they interact with the wall sooner (at the edge of a particle instead of the center) and with each other more often. Thus, the average time between collisions with the wall (\(\Delta t\)) is smaller, making \(<\Delta\mathbf{p}/ \Delta t>=<d\mathbf{p}/dt>\) bigger than predicted by the ideal gas law. In reality, of course, particles are not "point-like," but the size of the particles is typically much smaller in relation to the size of the container (think about the air molecules in a room), so the "point-like" approximation works well. Now increase the particle size to see what happens when the particles are quite large.

Illustration authored by Anne J. Cox.

Illustration 3: Thermodynamic Processes

\(\color{red}{\text{There is a time delay—since the system must be in equilibrium—before the change of state occurs.}}\)

In this animation \(N = nR\) (i.e., \(k_{B} = 1\)). This, then, gives the ideal gas law as \(PV = NT\). Restart.

There are many ways for a gas to go from one state (as described by its pressure, volume, number of atoms and temperature) to another. There can be heat flow from the environment to the gas or from the gas to the environment. Work can be done on the gas or by the gas. How the gas goes from one state to another is determined by the heat flow and the work done. However, the change in internal energy depends only on the temperature change (not on how the temperature change occurs). In other words, the change in internal energy is process independent, but the work and heat required are process dependent.

To make things easier, we categorize a set of processes that have names that describe the type of process. We illustrate them for the case when the number of atoms in the gas remains constant (there is no requirement that the number of atoms remains constant, but for purposes of this Illustration, we assume a sealed container).

• Isobaric: The pressure on the gas remains constant. This means that as the temperature of the gas changes, so will the volume. For example, a balloon that is put in the refrigerator shrinks in size.
• Isochoric: The volume of the gas remains constant. This means that any temperature change is accompanied by a change in pressure. For example, the steam in a pressure cooker increases in pressure as the temperature increases.
• Isothermal: The temperature of the gas remains constant. As the volume increases, the pressure decreases. For example, a balloon in a vacuum chamber increases in volume as the pressure in the chamber decreases.
• Adiabatic: The volume, temperature, and pressure all change. This is a rapid process in which no heat is exchanged with the environment, for example, compressing a piston on a bicycle pump or syringe quickly.

One way to describe the state of a gas is with a \(PV\) diagram. You could just as easily use a \(PT\) or \(VT\) diagram, but we use a \(PV\) diagram because it is easy to see the work done by the gas. The work is simply the area under the curve (the red region in the graphs). If you know some calculus, this is because work is given by \(W =\int P dV\) (and an integral is a calculation of the area under the curve).

A gas does not have to follow any of these special, "named" methods of changing state (Unknown Process). For an ideal gas, as long as \(PV = NT\) remains true through the process, everything is fine. It is simply a bit harder to mathematically describe the process and, therefore, harder to calculate the work and heat.

When you get a good-looking graph, right-click on it to clone the graph and resize it for a better view.

Illustration authored by Anne J. Cox.

Illustration 4: Evaporative Cooling

For a gas at a given temperature, there are particles with different speeds, and the particles' speed distribution follows the Maxwell-Boltzmann distribution. In this animation the two containers are separated by a "membrane" in the middle. Initially, no particles can cross the membrane. Once the particles are fairly evenly distributed in the left chamber, you are ready to allow for evaporation-that is, the fastest molecules can escape the left chamber and enter the right chamber. What is the approximate temperature of both sides initially? The light blue wall of the right chamber is at a constant chilly temperature of \(20\text{ K}\) so that as particles hit it, they cool (slow down). Restart.

Try letting particles through the membrane. This animation only allows particles that hit the membrane at a speed of \(25\) or higher to pass through. This threshold is shown on the speed histogram. After some time passes, particles are no longer passing through the membrane as much. Notice what has happened to the speed distribution in the left chamber. (There may still be particles with speeds greater than the threshold because the speed distribution still follows a Maxwell-Boltzmann distribution.) What has happened to the temperature in the left chamber? This is what happens with evaporation: The fastest particles leave and so the temperature of what remains behind is cooler. This is why sweating cools you off—as the sweat evaporates off your skin, you are cooled down. Thus, evaporation is a cooling process.

Illustration authored by Anne J. Cox.

Physlets were developed at Davidson College and converted from Java to JavaScript using the SwingJS system developed at St. Olaf College.