PV Diagrams and their Relationship to Work Done on or by a Gas
A process by which a gas does work on a piston at constant pressure is called an isobaric process. Since the pressure is constant, the force exerted is constant and the work done is given as
15.10
15.11
See the symbols as shown in Figure 15.10. Now , and so
15.12
Because the volume of a cylinder is its cross-sectional area times its length , we see that , the change in volume; thus,
15.13
Note that if is positive, then is positive, meaning that work is done by the gas on the outside world.
(Note that the pressure involved in this work that we have called is the pressure of the gas inside the tank. If we call the pressure outside the tank , an expanding gas would be working against the external pressure; the work done would therefore be (isobaric process). Many texts use this definition of work, and not the definition based on internal pressure, as the basis of the First Law of Thermodynamics. This definition reverses the sign conventions for work, and results in a statement of the first law that becomes .)
It is not surprising that , since we have already noted in our treatment of fluids that pressure is a type of potential energy per unit volume and that pressure in fact has units of energy divided by volume. We also noted in our discussion of the ideal gas law that has units of energy. In this case, some of the energy associated with pressure becomes work.
Figure 15.11 shows a graph of pressure versus volume (that is, a diagram for an isobaric process. You can see in the figure that the work done is the area under the graph. This property of diagrams is very useful and broadly applicable: The work done on or by a system in going from one state to another equals the area under the curve on a diagram.
We can see where this leads by considering Figure 15.12(a), which shows a more general process in which both pressure and volume change. The area under the curve is closely approximated by dividing it into strips, each having an average constant pressure . The work done is for each strip, and the total work done is the sum of the . Thus the total work done is the total area under the curve. If the path is reversed, as in Figure 15.12(b), then work is done on the system. The area under the curve in that case is negative, because is negative.
diagrams clearly illustrate that the work done depends on the path taken and not just the endpoints. This path dependence is seen in Figure 15.13(a), where more work is done in going from A to C by the path via point B than by the path via point D. The vertical paths, where volume is constant, are called isochoric processes. Since volume is constant, , and no work is done in an isochoric process. Now, if the system follows the cyclical path ABCDA, as in Figure 15.13(b), then the total work done is the area inside the loop. The negative area below path CD subtracts, leaving only the area inside the rectangle. In fact, the work done in any cyclical process (one that returns to its starting point) is the area inside the loop it forms on a diagram, as Figure 15.13(c) illustrates for a general cyclical process. Note that the loop must be traversed in the clockwise direction for work to be positive—that is, for there to be a net work output.
Example 15.2 Total Work Done in a Cyclical Process Equals the Area Inside the Closed Loop on a PV Diagram
Calculate the total work done in the cyclical process ABCDA shown in Figure 15.13(b) by the following two methods to verify that work equals the area inside the closed loop on the diagram. Take the data in the figure to be precise to three significant figures. (a) Calculate the work done along each segment of the path and add these values to get the total work. (b) Calculate the area inside the rectangle ABCDA.
Strategy
To find the work along any path on a diagram, you use the fact that work is pressure times change in volume, or . So in part (a), this value is calculated for each leg of the path around the closed loop.
Solution for (a)
The work along path AB is
15.14
Since the path BC is isochoric, , and so . The work along path CD is negative, since is negative (the volume decreases). The work is
15.15
Again, since the path DA is isochoric, , and so . Now the total work is
15.16
Solution for (b)
The area inside the rectangle is its height times its width, or
15.17
Thus,
15.18 Discussion
The result, as anticipated, is that the area inside the closed loop equals the work done. The area is often easier to calculate than is the work done along each path. It is also convenient to visualize the area inside different curves on diagrams in order to see which processes might produce the most work. Recall that work can be done to the system, or by the system, depending on the sign of A positive is work that is done by the system on the outside environment; a negative represents work done by the environment on the system.
Figure 15.14(a) shows two other important processes on a diagram. For comparison, both are shown starting from the same point A. The upper curve ending at point B is an isothermal process—that is, one in which temperature is kept constant. If the gas behaves like an ideal gas, as is often the case, and if no phase change occurs, then . Since is constant, is a constant for an isothermal process. We ordinarily expect the temperature of a gas to decrease as it expands, and so we correctly suspect that heat transfer must occur from the surroundings to the gas to keep the temperature constant during an isothermal expansion. To show this more rigorously for the special case of a monatomic ideal gas, we note that the average kinetic energy of an atom in such a gas is given by
15.19
The kinetic energy of the atoms in a monatomic ideal gas is its only form of internal energy, and so its total internal energy is
15.20 where is the number of atoms in the gas. This relationship means that the internal energy of an ideal monatomic gas is constant during an isothermal process—that is, . If the internal energy does not change, then the net heat transfer into the gas must equal the net work done by the gas. That is, because here, . We must have just enough heat transfer to replace the work done. An isothermal process is inherently slow, because heat transfer occurs continuously to keep the gas temperature constant at all times and must be allowed to spread through the gas so that there are no hot or cold regions.
Also shown in Figure 15.14(a) is a curve AC for an adiabatic process, defined to be one in which there is no heat transfer—that is, . Processes that are nearly adiabatic can be achieved either by using very effective insulation or by performing the process so fast that there is little time for heat transfer. Temperature must decrease during an adiabatic process, since work is done at the expense of internal energy
15.21
(You might have noted that a gas released into atmospheric pressure from a pressurized cylinder is substantially colder than the gas in the cylinder.) In fact, because for an adiabatic process. Lower temperature results in lower pressure along the way, so that curve AC is lower than curve AB, and less work is done. If the path ABCA could be followed by cooling the gas from B to C at constant volume (isochorically), Figure 15.14(b), there would be a net work output.
Applying the Science Practices: Work in a Potato Cannon
Plan an experiment using a potato cannon, meter stick, and pressure gauge to measure the work done by a potato cannon. Your experiment should produce P–V diagrams to analyze and determine the work done on a gas or by a gas. What do you need to measure? How will you measure it? Can you modify the potato cannon to make your measurements easier? When you perform multiple trials, what variables do you need to keep fixed between each trial? Which variables will you change?
One class decides to use a heavy piston, capable of being latched in place, to replace the potato. They latch the piston in place so that the contained volume is 0.50 L, load the cannon with fuel, and close the cannon with their pressure gauge, which reads 101 kPa. Then they light the fuel, and the pressure jumps to 405 kPa. Next, they release the latch, and the piston moves out until the internal volume is 2.0 L. The pressure is measured at this point to be 101 kPa again. Finally, they release the pressure gauge, and move the piston back down to 0.50 L, still at atmospheric pressure. Draw a diagram of this process, and calculate the net work performed by this system. Can you think of any ways to improve the measurements?
You should find that you have a right triangle on a P–V diagram, the area of which is the net work done by the system. Using a pressure gauge that can take continuous measurements during the expansion phase might be useful, as it is unlikely that this would actually be a linear process.