4.2.2: Explorations

Exploration 1: Kinetic Theory, Microscopic and Macroscopic Connections

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. Using the ideal gas law, we can make a connection between the macroscopic quantities of temperature (\(T\)) and pressure (\(P\)) and the individual microscopic properties of a particle of momentum (\(\mathbf{p} = m\mathbf{v}\)) and kinetic energy (\(1/2 mv^{2}\)). Restart.

Let's begin with one particle in an enclosed box and bouncing between two walls.

1. What is the change in momentum of the particle (use the velocity vs. time graph) as it hits the wall?
2. What is the average force that the right wall experiences over time? As a reminder, \(\mathbf{F}_{\text{avg}}=\Delta\mathbf{p}/\Delta t\), so pick a time frame (like \(20\) or so), multiply the change in momentum each time by the number of collisions with the right wall, and then divide by the total time for this number of collisions.
3. What is the average force on the left wall? The top and bottom walls?
4. Find the pressure on the surface of the box, Force/(area of all the walls). The dimension of the wall into the screen is \(1\).
5. Compare the pressure from (d) with the pressure on the box calculated from the ideal gas law and recorded in the table.

Increase the speed of the particle. The momentum delivered to the wall (and the force it experiences) will increase, thereby increasing the pressure. If the pressure of the gas increases (and if the volume is constant), the temperature of the gas will also increase.

6. What is the new speed of the particle?
7. What is the new pressure?
8. What is the new temperature?

By the same reasoning, increasing the mass of the particle will also increase the pressure, so the temperature should be connected to the mass of the particle as well. Increase the mass of the particle.

9. What is the new mass?
10. What is the new pressure?
11. What is the new temperature? The relation between the temperature and the increased speed and mass is that the temperature is proportional to the kinetic energy.

One particle in an enclosed box is not very realistic, so let's add a second particle (of the same mass) with a different speed. This time, however, we'll plot the kinetic energy of each particle as a function of time and the change in momentum at any wall (an average of this will give us the pressure). The table now shows the average momentum change at the walls. How does this compare with the pressure you calculate using the ideal gas law?

12. The collision between the particles is an elastic one. How can you tell?
13. What is the connection between the temperature and the total kinetic energy?

Now let's add some more particles of the same mass and different speeds. The table gives the momentum delivered to the wall as particles collide with the wall (\(<d\mathbf{p}/dt>\)), as well as the pressure calculated from the ideal gas law. This time we plot a histogram of the speeds of the particles. Stop the animation at some time and calculate the total kinetic energy of the ensemble of particles. This, divided by the number of particles, should be the same as the temperature of the system. This is the equipartition of energy theorem: The internal energy of a gas (the sum of the energy of all particles) is equal to \((f/2)k_{B}NT\), where \(f\) is the number of degrees of freedom for the atoms or molecules in a gas. In this case, the particles have \(2\) degrees of freedom; they can move in the \(x\) direction and \(y\) direction and thus \(f = 2\). Because we are treating the gas particles as "hard spheres" (one of our assumptions in the ideal gas model), the internal energy of the gas is due to the kinetic energy of the particles and is equal to \(k_{B}NT\) and for this animation, \(k_{B} = 1\).

Exploration authored by Anne J. Cox.

Exploration 2: Partial Pressure of Gases

In this animation \(N = nR\) (i.e., \(k_{B} = 1\)). This, then, gives the ideal gas law as \(PV = NT\). Two particles of different masses are in the same container. The total pressure on the container is due to the collisions of both types of particles on the walls. The blue particles are more massive than the red ones (\(10\) times more massive). How can you tell this by watching the animation? (Hint: Temperature is proportional to the average kinetic energy.) Restart.

1. What is the average pressure on the walls? Note: You need to watch the \(<P>\) number and wait until it stays around the same number (is not increasing or decreasing, but oscillating about the same number) and then estimate an approximate value. From this pressure and the temperature, use the ideal gas law (in the form \(PV = NT\)) and calculate the volume of the box the particles are in. If the dimension of the box into the computer screen is \(1\), what is the length of one wall? Measure the size of the container to verify your calculation.

Run the same animation with only the red particles.

2. What is the pressure on the walls? This is the partial pressure of the red particles.

Run the animation again, but this time with only the blue particles.

3. What is the pressure on the walls? This is the partial pressure of blue particles.
4. Compare the total pressure with the sum of the partial pressures.
5. The sum of the partial pressures and the total pressure should be equal. Why?

Now, start the second animation when the red and blue particles are in a container with a movable piston between them. The piston generally stays in a position where the pressures on both sides are essentially equal.

6. Where does the piston generally stay (right, left, or in the middle)?
7. Why?
8. Remembering that the blue particles have a mass \(10\) times the red particles, predict what the partial pressure of the red and the blue will be (in the first animation) if there are the same number of red and blue particles (\(25\) each for a total of \(50\) particles).

Try running the first simulation with an equal number of red and blue particles (after clicking, scroll back up to the first animation). Run the second simulation with only red particles. Run with only blue particles.

9. Was your prediction right? Explain.
10. Predict where you expect the piston will be in the second animation if there are \(25\) particles on either side of the partition.
11. Try the second animation with 25 particles on either side of the partition. Was your prediction right? Specifically, when the temperature is the same on both sides, where is the partition on average?

Exploration authored by Anne J. Cox.

Exploration 3: Ideal Gas Law

The relationship between the number of particles in a gas, the volume of the container holding the gas, the pressure of the gas, and the temperature of the gas is described by the ideal gas law: \(PV = nRT\). In this animation \(N = nR\) (i.e., \(k_{B} = 1\)). This, then, gives the ideal gas law as \(PV = NT\). Restart.

Notice what happens as you change the number of particles, the temperature, and the volume. The pressure is due to collisions with the walls of the container. The graph shows the instantaneous "pressure" (the change in momentum of the particles as they hit the wall and thus exert a force) as a function of time, while the table shows both \(NT/V\) (equal to the pressure for an ideal gas) and the average of the instantaneous pressure.

1. Keep the number of particles and the volume constant. What happens to the speed of the particles as the temperature changes? What happens to the pressure (\(N\ast T/V\)) if you increase the temperature? (This is known as Gay-Lussac's Law: \(P/T =\text{ constant}\).)
2. If you double the volume (while keeping the same number of particles and the same temperature), what happens to the pressure (and force on the wall)? Why? (This is known as Boyle's law: \(PV =\text{ constant}\).)
3. If the number of particles is increased (and the temperature and volume stay the same), what happens to the pressure (and the force on the wall)? Why?
4. If you double the volume and halve the temperature (while keeping the number of particles constant), what happens to the pressure? (This is known as Charles's Law: \(V/T =\text{ constant}\).)

Note that all of these "named" gas laws are included in the ideal gas law: \(PV = nRT\).

You can also drag the top of the piston to change the volume. In this process both the temperature and pressure can change.

5. Start with a volume of \(100\). Drag the piston up. What changes and why?
6. Once the particles have spread out into the entire volume available, drag the piston back down. Notice that the particles move fast and the temperature and pressure change dramatically. This is because as the piston comes down and particles hit it, the downward moving piston transfers some of its momentum to these particles and so they speed up. Faster moving particles mean a higher temperature. In a real system, you would not normally see this effect because the particles are moving much faster than any piston being compressed (but if they moved that fast in the animation, you wouldn't see individual particles).
7. How would you need to drag the piston to minimize the change in temperature? Start with a volume of \(100\) and temperature of \(100\) again and try to minimize the increase in temperature as you compress the piston.

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

Exploration authored by Anne J. Cox.

Exploration 4: Equipartition Theorem

The kinetic energy of a particle can be due to motion in the \(x,\: y,\) and \(z\) directions, as well as to rotations. The equiparition of energy theorem says that the kinetic energy of an atom or particle is, on average, equally distributed between the different modes (different degrees of freedom) available. In a monatomic gas, an individual atom has three degrees of freedom because it can move in the \(x,\: y\) and \(z\) directions. The energy per particle has an average value of \((f/2)k_{B}T\), where \(f\) is the number of degrees of freedom, \(k_{B}\) is the Boltzmann constant, and \(T\) is the temperature. Restart.

1. In this animation of a monatomic gas in a box, why do the particles only have \(2\) degrees of freedom? The table shows the total kinetic energy of all particles in the box, as well as the average kinetic energies of particles in the box (the animation averages over a \(10\text{-s}\) period, so you need to wait \(10\text{ s}\) to read the averages).
2. Record the total energy.
3. What is the energy per particle?
4. If the energy is given in \(\text{joules/}k_{B}\), what is the temperature inside the box?

Try this animation of a diatomic gas with 20 particles. Notice that the graph shows the total kinetic energy of the diatomic particles and the kinetic energies of translation (motion in \(x\) and \(y\) directions) and rotation.

5. Why is the translational kinetic energy, on average, about two times the rotational kinetic energy? (The animation averages over a \(10\text{-s}\) interval, so you need to wait for the animation to run for at least \(10\text{ s}\) to read the average values of kinetic energy).
6. From the total energy, what is the energy per particle?
7. If the energy is given in \(\text{joules/}k_{B}\), what is the temperature in the box? (Remember that \(\text{<energy>/particle }= (f/2)k_{B}T\) and in this case, \(f = 3\) (Why?).)

Now, try a mixture of 20 monatomic particles and 20 diatomic particles.

8. Why is the temperature of the gas in the box a single value (not one value for atoms and another for molecules)? Hint: Think about the air surrounding you at essentially a constant temperature, unless the heater or air conditioner just turned on and made one section of the air a different temperature. Air is made up of monatomic particles (helium) and diatomic particles (water, oxygen, and nitrogen).
9. After waiting at least \(10\text{ s}\), compare the average values of the kinetic energies. What value is the average monatomic kinetic energy close to?
10. Why should those two values (the two averages that you found in part (i)), averaged over a long period of time, be equal to each other and greater than the rotational kinetic energy of the diatomic particles?
11. Explain why the total energy should be equal to \((2/2)20k_{B}T + (3/2)20k_{B}T\).
12. From the total energy (given in \(\text{joules/}k_{B}\)), what is the temperature?
13. In this animation, if a mixture has \(15\) atoms, how many diatomic particles should it have so that the average kinetic energies of both particles are the same? Try setting the number of monatomic particles and diatomic particles to check your answer.

Exploration authored by Anne J. Cox.
Script authored by Wolfgang Christian modified by Anne J. Cox.
Applet authored by Ernesto Martin and modified by Wolfgang Christian.

Exploration 5: PV Diagrams and Work

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

Ideal gas law: \(PV = nRT\). In this animation \(N = nR\) (i.e., \(k_{B} = 1\)). This, then, gives the ideal gas law as \(PV = NT\). The work done during a thermodynamic process depends on the type of process (and can be positive, negative, or zero). Restart. Work is given by the equation

\[W=\int PdV\nonumber\]

so that on a pressure-volume diagram, the area under the curve is the work done by the gas during the expansion. In order to analytically solve for the work, you need to know how pressure depends on volume (is pressure constant, changing linearly with volume, etc.?). How pressure varies with volume depends on the type of process (isothermal, isobaric, isochoric, adiabatic).

The three animations show three different processes that start at a common temperature and end at a common temperature.

1. What is the change in internal energy (\(\Delta U\)) for these processes (remember that \(\Delta U = (3/2)nR\Delta T = (3/2)N\Delta T\) for an ideal monatomic gas)?
2. Estimate the area under the curve (count the blocks on the graph) when the system goes from one temperature to another (from one isotherm on the graph to another). This is the value of the work done since work is \(W = \int PdV\). Which process does positive work? Which process does negative work? Which process does zero work?
3. The first law of thermodynamics, \(\Delta U = Q - W\), when written as \(Q = W + \Delta U\), says that the heat into a system can be used to do work and/or increase the internal energy. Therefore, which process requires the most heat?
4. Compare the area under the curve that you estimated in (b) with the value you calculate using the equations below (found by using calculus and solving the integral):
   • Constant pressure: \(W = P(V_{f} - V_{i})\)
   • Adiabatic: \(W = (P_{f}V_{f} - P_{i}V_{i})/(1 - \gamma )\), where \(\gamma\) (the ratio of \(C_{P}/C_{V}\), specific heat at a constant pressure divided by the specific heat at a constant volume) for an ideal monatomic gas is \(5/3\).

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

Exploration authored by Anne J. Cox.

Exploration 6: Specific Heat at Constant Pressure and Constant Volume

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

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

For an ideal monatomic gas, the change in internal energy depends only on temperature, \(\Delta U = (3/2)nR\Delta T = (3/2)N\Delta T\).

1. Calculate the change in internal energy for the three cases.
2. What is the work done in each case? As a reminder, \(W =\int P dV\), and pressure can (and does in many instances) depend on volume. Calculate the work done in each case using the following two methods, then compare your answers.

• Graphically: To find the work done is to determine the area under the curve ("area" of the red region on the graph). After estimating the area by counting the grid blocks, click the checkbox above to show a calculation of the area (by numerical integration) on the simulation. Explain any significant differences between your estimation and the numerical integration.
• Analytic solution (a little bit of calculus required): When heat is added at constant pressure (isobaric process), then \(P\), pressure, in the above equation for work is simply a constant of integration. When heat is added at a constant volume (isochoric process), the work done is zero. Why? When heat is added at a constant temperature (isothermal), use the ideal gas law (\(PV = NT\)) and write the pressure as a function of volume: \(NT/V\) (where \(N\) and \(T\) are constant) and then you can integrate (the answer involves a natural logarithm).

3. Using the first law of thermodynamics, \(Q = W + \Delta U\), calculate the heat input and show that it is the same for all three cases.

The specific heat capacity of a material is a measure of the quantity of heat needed to raise a gram (or given quantity) of a material \(1^{\circ}\text{C}\). For a gas, it requires a different amount of heat to raise the same amount of gas to the same temperature depending on the circumstances under which the heat is added. If the same amount of heat is added, the final temperatures of the constant pressure and constant volume expansions are quite different (and, for a constant temperature, heat is added but the temperature does not change!).

4. In which case does the heat input raise the temperature the most? Why?

So, if the specific heat capacity of an ideal gas is to have any meaning at all, it must be defined in terms of the process: specific heat at a constant volume or specific heat at a constant pressure.

5. Go back to your calculation of heat in (c). Calculate the constant of proportionality between heat input and the change in temperature for the constant volume and constant pressure cases:  \(Q = (\text{Constant})N\Delta T\).
6. What is the constant in each case? Why is the constant for an expansion at constant pressure greater? (Hint: Think about whether the heat is used only to change temperature or to change temperature and do work.)

Generally, we write the heat capacity as a molar heat capacity (where n is the number of moles) and find that for constant pressure \(Q = C_{P}n\Delta T\) and \(C_{P} = (5/2)R\), and for constant volume \(Q = C_{V}n\Delta T\) and \(C_{V} = (3/2)R\).

We began this discussion by noting that for an ideal monatomic gas, the average internal energy is \((3/2)T\). This comes from kinetic theory and the equiparititon of energy, where the \(3\) comes from \(3\) degrees of freedom. For a diatomic gas, the average internal energy is \((5/2)T\) because there are two more degrees of freedom (rotation).

7. How are the heat capacity at a constant pressure and heat capacity at a constant volume different for a diatomic gas?

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

Exploration 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.