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Developing a Coherent Approach
A frequent question on the AP Physics B Exam deals with what is commonly referred to as a P-V diagram. An ideal gas undergoes changes of state as a result of work processes, heat processes, or both. The pressure of the gas is then plotted versus volume, and questions are asked about the state of the gas and how much work is done or heat is transferred during the different processes. As common as these questions are on the exam, they are not always a point of emphasis in the various textbooks used for the B-level course. The object of this article is to present a coherent approach to this material based on the ideal gas law and the first law of thermodynamics.
Students typically learn about mechanical energy during the first semester of the B-level course. Analysis of the work necessary to accelerate an object or to reposition the object leads to the concepts of kinetic and potential energy. Should only conservative forces act on an object, the sum of the kinetic and potential energies will not change. This sum, the total mechanical energy, is associated with macroscopic motions. However, the objects we use in class to demonstrate mechanical energy (balls, textbooks, weights, springs, etc.) are all complicated systems composed of many molecules or atoms. Even when the center of mass of these objects is stationary, there is considerable energy within the object due to molecular motions and positions. I keep a small demonstration piece consisting of a dozen small spheres interconnected by springs on my front desk. Using this Einstein model of a solid, I can demonstrate kinetic and potential energy at the molecular level. The sum total of the kinetic and potential energy at the molecular level is called the internal energy of the system:
U = KE+PE (molecular level).
Of course, the springs in the demo piece are representative of an electric interaction between the molecules, and in a solid this interaction is strong enough to keep each molecule close to its equilibrium position. In a liquid, the molecules are still very close to each other, but the interaction is not strong enough to keep the molecules at fixed lattice sites. In a gas, the interaction is relatively weak (more precisely, the interaction energy is small compared to the average kinetic energy of a molecule), and the molecules can readily move throughout the enclosed volume if the density of the gas is not too great. In an ideal gas, the interaction energy between molecules is neglected, that is, there is no internal potential energy. This means that the internal energy of an ideal gas is the sum total of the kinetic energy of the molecules. A standard derivation in most textbooks based on kinetic molecular theory relates the average kinetic energy of a molecule to the temperature:
.
In this equation, k is Boltzmann's constant,
, and temperature is expressed in kelvins. For an ideal gas containing N molecules, the internal energy will just be N times this since there is no potential energy:
.
Problem statements usually give the amount of substance in moles, so you can write:
.
In the last equation, R is the ideal gas constant with a value of 
.
Actually, the formula 
is only true for a monatomic ideal gas. For more complex molecules, rotational motion will also contribute significantly to the KE. However, this level of complexity has never been a point of emphasis on the AP Exam, so a teacher's emphasis should be on monatomic ideal gases. It is worth emphasizing here that the internal energy of an ideal gas depends only on the amount of substance and the temperature. To change the internal energy of a fixed amount of gas, the temperature must change.
Example: Calculate the internal energy of the air in a typical room with volume 40 m3. Treat the air as if it were a monatomic ideal gas at 1 atm = 1.01 105 Pa.
You can use the gas law PV=nRT to express the internal energy in terms of pressure and volume:
.
Energy Transfers: Work and Heat
The first law deals with changes in the internal energy of a system. In teaching the first law, it is important to emphasize that changes in the internal energy of a system can only occur if the system is not isolated. This means that the system is embedded in its surroundings in a way that there can be an energy transfer. There are two types of energy transfer, and the difference between the two is determined not by what occurs in the system but by what occurs in the surroundings.
Work is an energy transfer between a system and its surroundings that is a result of organized motion in the surroundings. You can increase the internal energy of a wood block by rubbing it vigorously. You can increase the internal energy of a glass of water by stirring it rapidly. You can decrease the internal energy of a gas by letting it expand against some external pressure applied by a piston. In each example, the surroundings (rubbing cloth, stirrer, piston) are systems consisting of many particles that are all undergoing an organized motion as a whole to facilitate the energy transfer.
The system consisting of an ideal gas enclosed in a container with a moveable piston is commonly used to illustrate many of the concepts of thermodynamics. Let's look at it in more detail. Suppose the gas is allowed to expand slowly in such a way that at any point in the process the pressure is well defined and the system can be thought of as being close to equilibrium every step of the way. Such a process is said to be quasi-static. Since the gas is always in equilibrium, the gas pressure is always equal to the external pressure exerted by the piston, and the work done by the gas on the piston is just the negative of the work done on the gas by the piston. Let us further assume that while the gas expands, the pressure P remains constant, an isobaric process. This could be achieved by immersing the gas container in consecutively warmer water baths. After a time, the piston with cross-sectional area A would have expanded a distance Δx.
Since 
, the force that the gas exerts on the piston is given by F = PA . Then the work done by the gas during the expansion is given by:
.
It is a simple matter to display this process on a graph that plots pressure versus volume with an arrow to show the sequence of the process.
Note that the work done by the gas is just the area under the "curve," i.e., between the graph and the V-axis. By convention, this area is positive for positive work done by the gas, as in this example. For our quasi-static process, this would correspond to negative work done on the gas. The distinction between work done on the gas and work done by the gas is one that is often made on the AP Exam, so a teacher should make this distinction clear at this point. As long as the process is quasi-static so that the pressure is well defined at each step of the way, the area under the P-V curve will always be the work done by the gas during the process. This follows from the fact that any curve can be approximated by a series of steps over which the pressure is constant. In the limit of infinitesimal steps, you get the exact answer, the sum of all the infinitesimal areas, the area under the curve.
A special case is that of a cycle in which the system is brought back to its original state after going through several processes.
Example: One mole of monatomic ideal gas is enclosed under a frictionless piston. A series of processes occur, and eventually the state of the gas returns to its initial state with a P-V diagram as shown below. Answer the following in terms of P0, V0, and R.
- Find the temperature at each vertex.
- Find the change in internal energy for each process.
- Find the work by the gas done for each process.
a. Use the gas law to find the temperatures at A, B, C:
.
Similarly, 
and 
.
b. Since the internal energy depends only on temperature, the change in internal energy for each process depends only on the temperature difference that occurs during the process:
.
Similarly,
c. To find the work done by the gas, find the area under each segment, remembering the sign convention.
There are two things to note about part (c) that are true in general:
- For a constant volume process like B -> C, no work is done by the gas.
- The total work done for the entire cycle is the area enclosed within the graph. In this example, the sum of the work is
, the same as the area of the enclosed triangle.
Heat is an energy transfer between a system and its surroundings that is the result of random motion in the surroundings. Note the difference between a work process and a heat process. In the former, there must be organized motion in the surroundings, but in the latter, the energy transfer is a result of random motion in the surroundings. When a glass of water is placed on a hot plate, energy will spontaneously leave the hot plate and cause the internal energy of the water to increase. Heat will always flow spontaneously from the system at higher temperature to the system at lower temperature, but heat can be made to flow in the opposite direction as well if work is done in the process. For example, a refrigerator relies on work done by its compressor to move heat from a cold freezer into a warm kitchen.
Heat is a term that is used all the time in everyday life. In this context, it's okay to talk about the "amount of heat in a hot object," but in the classroom it is important that a teacher emphasize the precise meaning of this term. A hot object may contain a lot of internal energy, but it does not contain heat. Unfortunately, physics textbooks are often guilty of sloppy usage. Phrases like "as friction slowed the block, heat was generated in the sliding surface" are not difficult to find. As the block slows, the organized motion of the block does work on the sliding surface and increases its internal energy. Such sloppy usage masks the essential difference between a work process and a heat process, and a solid grasp of the first law of thermodynamics cannot be achieved without understanding this distinction.
The First Law of Thermodynamics
Heat processes and work processes account for all possible energy transfers to a system. It therefore follows from conservation of energy that the total change in the internal energy of a system is the sum of the work done on the system and the heat transferred to the system. This is the first law of thermodynamics:
.
This equation is often written simply as 
, but you have to keep the sign conventions in mind. It is probably easiest to remember the conventions for W and Q by imagining processes where only one of them is present. If Q = 0, then ΔU is positive if positive work is done on the system. If W = 0, then ΔU is positive if heat flows into the system.
Let's go back to the P-V diagram example and calculate the heat transfer for each process. Since the work done by the gas and the change in the internal energy have already been calculated, it is a simple matter to calculate the heat transfer from the first law, but be careful of signs. Remember that for our quasi-static process,
.
The efficiency of a cycle is defined as the ratio of the work done by the gas to the heat Qin that flows into the system. Any heat that is expelled into the surroundings is not included in the calculation of Qin. From the point of view of efficiency, this expelled heat is lost and its energy is not used by the system:
.
In the example above, you get:
.
Besides the constant pressure (isobaric) and constant volume processes involved in this example, there are two other processes the student of AP Physics must be familiar with. An isothermal process is one that occurs at constant temperature. The gas-piston container in our example could expand isothermally if it were kept immersed in a large hot-water bath while the gas expanded. Since the temperature doesn't change during an isothermal process, there is no change in internal energy. The first law then tells you that the work done by the gas is just equal to the heat that flows into the system from the bath.
.
An adiabatic process is one that occurs without the exchange of heat with the surroundings. If the gas-piston system were insulated so that heat could not get in or out, any expansion or compression would occur adiabatically. Since Q = 0 for an adiabatic process, the first law tells you that the change in internal energy is just equal to the work done on the system.
.
When a gas expands adiabatically, the work done in the expansion comes at the expense of the internal energy of the gas, causing the temperature of the gas to drop. The figure below shows P-V diagrams for these two processes.
The figure compares two processes that begin with the same state and involve expansion to the same final volume. For the isothermal process, the product of P·V remains constant since T remains constant. Since the temperature must decrease for the adiabatic process, it follows that the final pressure must be less for this process. Thus the adiabat lies below the isotherm.
Let's look at one more example that incorporates many of the AP points of emphasis.
Example: One mole of ideal gas is at pressure P0 and volume V0 . The gas then undergoes three processes:
- The gas expands isothermally to 2V0 while heat Q flows into the gas.
- The gas is compressed at constant pressure back to the original volume.
- The pressure is increased while holding the volume constant until the gas returns to its initial state.
a. Draw a P-V diagram that depicts this cycle. Label relevant points on the axes. In terms of T0, the initial temperature, label each vertex with the temperature of the gas at that point.
For the remaining sections, answer in terms of T0, Q, and R.
b. Find the change in internal energy for each leg of the cycle.
c. Find the work done by the gas on each leg of the cycle.
d. Find the heat that flows into the gas on legs 2 and 3.
e. Find the efficiency of this cycle.
a.
The two ends of the isotherm will both be at T0 . Since the product of P·V is constant along an isotherm, an expansion to twice the volume implies a pressure reduction to half the original pressure. Applying the gas law to the lower-left vertex then yields:
b.
c.
- Since
, it follows from the first law that 
.
d. Use the first law since both ΔU and W are known for each process.
e. Heat flows into the system on both legs 1 and 3. Use the definition of efficiency:
.
Conclusion
Older versions of the AP Physics B Exam often gave CP and CV , the specific heats at constant pressure and volume, within the problem statement involving a P-V diagram. In fact, this information was never actually necessary to solve the problem. A solid grasp of the first law of thermodynamics and facility with the ideal gas law are the only tools a student needs to deal with these problems.
Jim Mooney has been teaching AP Physics B and C for 29 years, the last 20 at the Taft School in Connecticut. For the last 18 years he has led summer workshops on the AP Physics curriculum, and he also participated in the AP Reading for five years. His study guides for AP Physics, AP Advantage, AP Physics B and AP Advantage, AP Physics C were recently published by Peoples Publishing Group.
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