PV Diagrams for AP Physics B

Practice problems for these concepts can be found at:

Thermodynamics Practice Problems for AP Physics B

Imagine a gas in a cylinder with a moveable piston on top, as shown in Figure 18.1.

PV Diagrams

When the gas expands, the piston moves up, and work is done by the gas. When the piston compresses, work is done on the gas. The point of a heat engine is to add heat to a gas, making the gas expand—in other words, using heat to do some work by moving the piston up.

To visualize this process mathematically, we use a graph of pressure vs. volume, called a PV diagram. On a PV diagram, the x-axis shows the volume of the gas, and the y-axis shows the pressure of the gas at a particular volume. For the AP exam, you're expected to be familiar with the PV diagrams for four types of processes:

Isothermal: The gas is held at a constant temperature.
Adiabatic: No heat flows into or out of the gas.
Isobaric: The gas is held at a constant pressure.
Isochoric: The gas is held at a constant volume.

Not all processes on a PV diagram fall into one of these four categories, but many do. We look at each of these processes individually.

Isothermal Processes

Look at the ideal gas law: PV = Nk_BT. In virtually all of the processes you'll deal with, no gas will be allowed to escape, so N will be constant. Boltzmann's constant is, of course, always constant. And, in an isothermal process, T must be constant. So the entire right side of the equation remains constant during an isothermal process. What this means is that the product of P and V must also be constant. A PV diagram for an isothermal process is shown in Figure 18.2.

PV Diagrams

What this diagram shows is the set of all P–V values taken on by a gas as it goes from point "a" to point "b." At point "a," the gas is held at high pressure but low volume. We can imagine that there's a lot of gas squished into a tiny cylinder, and the gas is trying its hardest to push up on the piston—it is exerting a lot of pressure.

As the volume of the gas increases—that is, as the piston is pushed up—the pressure of the gas decreases rapidly at first, and then slower as the volume of the gas continues to increase. Eventually, after pushing up the piston enough, the gas reaches point "b," where it is contained in a large volume and at a low pressure.

Since the temperature of the gas does not change during this process, the internal energy of the gas remains the same, because U = (3/2)Nk_bT. For an isothermal process, ΔU = 0.

Adiabatic Processes

In an adiabatic process, no heat is added to or removed from the system. In other words, Q = 0. By applying the first law of thermodynamics, we conclude that ΔU = W. This means that the change in internal energy of the gas equals the work done ON the system. In an adiabatic process, you can increase the temperature of the gas by pushing down on the piston, or the temperature of the gas can decrease if the gas itself pushes up on the piston. The PV diagram for an adiabatic process looks similar, but not identical, to the diagram for an isothermal process—note that when a gas is expanded adiabatically, as from "a" to "b" in Figure 18.3, the final state is on a lower isotherm. (An isotherm is any curve on a PV diagram for which the temperature is constant.) In other words, in an adiabatic expansion, the temperature of the gas decreases.

PV Diagrams

Isobaric Processes

This one's pretty easy: the pressure stays constant. So a PV diagram for an isobaric process looks like the graph in Figure 18.4.

PV Diagrams

In an isobaric process, the pressure of the gas is the same no matter what the volume of the gas is. Using PV = Nk_BT, we can see that, if we increase V but not P, then the left side of the equation increases. So the right side must also: the temperature must increase. Therefore, in an isobaric process, the temperature of the gas increases as the volume increases, but the pressure doesn't change.

Isochoric Processes

Even easier: the volume remains constant, as shown in Figure 18.5.

PV Diagrams

As with an isobaric process, the temperature of a gas undergoing an isochoric process must change as the gas goes from point "a" to point "b." The temperature of the gas must increase as the pressure increases, and it must decrease as the pressure of the gas decreases.

Using PV Diagrams

Working with PV diagrams can seem rather complicated, but, fortunately, you'll rarely come upon a complicated PV diagram on the AP exam. We have two tips for solving problems using PV diagrams: if you follow them, you should be set for just about anything the AP could throw at you.

TIP #1: Work done on or by a gas = area under the curve on a PV diagram.

For an isobaric process, calculating area under the curve is easy: just draw a line down from point "a" to the horizontal axis and another line from point "b" to the horizontal axis, and the area of the rectangle formed is equal to PΔV (where ΔV is the change in the volume of the gas). The equation sheet for the AP exam adds a negative sign, indicating that work done by the gas (i.e., when the volume expands) is defined as negative: W = –PΔV. It's even easier for an isochoric process: –PΔV = 0.

This tip comes in handy for a problem such as the following.

PV Diagrams

We have two steps here. The first is isobaric, the second is isochoric. During the isobaric step, the work done by the gas is

W = –PΔV = –(2 atm)(2000–10,000)m³.

Careful—you always must be using standard units so your answer will be in joules!

W = –(2 × 10⁵ N/m²)(–8000 m³) = +1.6 × 10⁹ J

During the isochoric step, no work is done by or on the gas. So the total work done on the gas, taken from the results of the isobaric step, is 1.6 × 10⁹ J.

Calculating PΔV can be a bit harder for isothermal and adiabatic processes, because the area under these graphs is not a rectangle, and –PΔV does not apply. But if you want to show the AP graders that you understand that the work done by the gas in question is equal to the area under the curve, just shade in the area under the curve and label it accordingly. This is a great way to get partial—or even full—credit.

Internal energy is directly related to temperature. So, when a process starts and ends at the same point on a PV diagram, or even on the same isotherm, ΔU = 0.

This tip is particularly useful for adiabatic processes, because, in those cases, ΔU = W. So being able to calculate ΔU allows you to find the work done on, or by, the system easily. We'll provide an example problem.

PV Diagrams

We know that, for an adiabatic process, ΔU = W. So we can calculate the final and initial values for U using and then set that value equal to W. The variable n represents the number of moles, in this case 0.25 moles

U_a = (3/2)(0.25)(8.1)J/mol . K(1150K) = 3620 J.

U_b = (3/2)(0.25)(8.1 J/mol . K)(400 K) = 1260 J.

The answer is W = 2360 J, or equivalently stated, 2360 J of work are done by the gas.

Practice problems for these concepts can be found at:

Thermodynamics Practice Problems for AP Physics B

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