Lesson 9.6 · Unit 9 · CED topic 9.4
PV diagrams, work, and thermodynamic processes
An automobile engine runs through the same four steps thousands of times a minute: gas is drawn in, compressed, ignited so that it expands and pushes the piston, and exhausted. Engineers track each step on a graph of pressure against volume. That graph, the PV diagram, lets you read the work done on a gas directly as an area, see at a glance how the temperature changes, and recognize the special cases that appear on every AP Physics 2 exam: isovolumetric, isobaric, isothermal and adiabatic processes. This lesson develops those tools and then follows a gas around a complete cycle.
Describe the behavior of a system using thermodynamic processes.
- Pressure–volume graphs (PV diagrams) are representations used to represent thermodynamic processes. Lines of constant temperature on a PV diagram are called isotherms.
- The absolute value of the work done on a gas when the gas expands or compresses is equal to the area underneath the curve of a plot of pressure versus volume for the gas.
- Special cases of thermal processes depend on the relationship between the configuration of the system, the nature of the work done on the system, and the system’s surroundings. These include constant volume (isovolumetric), constant temperature (isothermal), and constant pressure (isobaric), as well as processes where no energy is transferred to or from the system through thermal processes (adiabatic).
Reading a PV diagram
Each point on a PV diagram is one state of a fixed amount of gas: its pressure, volume, and (through \(PV = nRT\)) its temperature. A curve or line joining two points is the process that carries the gas from the first state to the second. An arrow shows the direction.
- Work is an area. For a process at constant pressure, \(|W| = P|\Delta V|\), which is a rectangle on the diagram. For any process, \(|W|\) is the area between the curve and the volume axis.
- The sign comes from the direction. Moving to the right (expansion) means the gas does work, so \(W \lt 0\) for the gas. Moving to the left (compression) means work is done on the gas, so \(W \gt 0\).
- Temperature from \(PV\). Because \(T \propto PV\), points farther from the origin are hotter. All points on one hyperbola \(PV = \) constant have the same temperature (an isotherm), and \(U\) is the same for all of them.
- Internal energy. \(\Delta U = \tfrac{3}{2}\Delta(PV)\) for a monatomic gas, so you can find \(\Delta U\) directly from the coordinates of the two end points.
Four special processes
The first law \(\Delta U = Q + W\) takes a simple form in each special process, and the shapes of the paths are worth memorizing.
| Process | What is constant or zero | On the PV diagram | First law consequence |
|---|---|---|---|
| Isovolumetric (isochoric) | \(\Delta V = 0\) | Vertical line | \(W = 0\), so \(\Delta U = Q\) |
| Isobaric | \(P\) constant | Horizontal line | \(W = -P\Delta V\); \(Q\) and \(W\) both matter |
| Isothermal | \(T\) constant | Hyperbola \(PV = \) constant | \(\Delta U = 0\), so \(Q = -W\) |
| Adiabatic | \(Q = 0\) | Curve steeper than an isotherm | \(\Delta U = W\) |
The adiabatic curve deserves a closer look. It is steeper than the isotherm through the same point. In an adiabatic expansion the gas does work, so \(\Delta U = W \lt 0\) and the gas cools, ending at a lower temperature than where an isothermal expansion would end, hence lower pressure. In an adiabatic compression the gas warms. Adiabatic processes are fast (little time for energy to leave) or occur in an insulated container. A fire syringe, in which a plunger is driven down suddenly and the air gets hot enough to ignite a bit of tissue, is an adiabatic compression.
Cycles
A cycle returns the gas to its starting state, so over a complete cycle \(\Delta U = 0\) and therefore \(Q_{net} = -W_{net}\). On the PV diagram a cycle is a closed loop, and the net work is the enclosed area: the work is negative (done by the gas) if the loop is clockwise, and positive (done on the gas) if it is counterclockwise. Clockwise cycles are engines; counterclockwise cycles are refrigerators and heat pumps.
A gas expands at a constant pressure of \(3.0\times10^5\) Pa from 0.010 m\(^3\) to 0.030 m\(^3\). Then it is cooled at constant volume until its pressure drops to \(1.0\times10^5\) Pa. Find the work done on the gas in each step, and \(\Delta U\) for the whole process if the initial temperature is \(T_0\) and \(n R T_0= 3000\) J.
Step 1 (isobaric). \(W_1 = -(3.0\times10^5)(0.020) = -6000\) J.
Step 2 (isovolumetric). \(W_2 = 0\).
Temperatures. Initial state \(PV = (3.0\times10^5)(0.010) = 3000\) J \(= nRT_0\). Final state \(PV = (1.0\times10^5)(0.030) = 3000\) J. The initial and final points have the same product \(PV\), so they are on the same isotherm and \(\Delta U = 0\) overall.
Net heating. \(Q = \Delta U - W = 0 - (-6000) = +6000\) J: the net energy entering the gas by heating equals the work it did.
An ideal monatomic gas goes clockwise through the rectangle A \(\to\) B \(\to\) C \(\to\) D \(\to\) A: A (\(1.0\times10^5\) Pa, 0.010 m\(^3\)), B (\(3.0\times10^5\) Pa, 0.010 m\(^3\)), C (\(3.0\times10^5\) Pa, 0.030 m\(^3\)), D (\(1.0\times10^5\) Pa, 0.030 m\(^3\)). Find the work, \(\Delta U\) and \(Q\) for each step and the net work.
Use \(W = -P\Delta V\) and \(\Delta U = \tfrac{3}{2}\Delta(PV)\), then \(Q = \Delta U - W\).
| Step | Process | \(W\) (J) | \(\Delta U\) (J) | \(Q\) (J) |
|---|---|---|---|---|
| A \(\to\) B | isovolumetric | 0 | +3000 | +3000 |
| B \(\to\) C | isobaric expansion | \(-6000\) | +9000 | +15000 |
| C \(\to\) D | isovolumetric | 0 | \(-9000\) | \(-9000\) |
| D \(\to\) A | isobaric compression | +2000 | \(-3000\) | \(-5000\) |
| Cycle | \(-4000\) | 0 | +4000 |
Check. The enclosed area is \((2.0\times10^5\ \text{Pa})(0.020\ \text{m}^3) = 4000\) J, equal to the net work done by the gas, and \(\Delta U\) sums to zero. ✓ Energy enters by heating in steps A\(\to\)B and B\(\to\)C (total 18000 J) and leaves in C\(\to\)D and D\(\to\)A (total 14000 J).
A gas is expanded from the same initial state along an isotherm and along an adiabat, to the same final volume. Compare the final pressures, the work done by the gas, and the heating.
Argument. In the adiabatic expansion \(Q = 0\) and the gas does work, so its internal energy and temperature fall; since \(P \propto T\) at a given volume, the final pressure is lower than in the isothermal case, where the temperature is unchanged. The adiabat lies below the isotherm, so the area under it, and the work done by the gas, is smaller. In the isothermal case \(\Delta U = 0\), so the gas must absorb energy by heating equal to the work it does. In the adiabatic case no energy is absorbed by heating.
- Work. Area under the curve; sign from the direction (right means \(W \lt 0\)).
- Special cases. Isovolumetric \(W = 0\); isothermal \(\Delta U = 0\); adiabatic \(Q = 0\); isobaric \(W = -P\Delta V\).
- Temperature. \(T \propto PV\); compare products, or compare positions relative to isotherms.
- Cycles. \(\Delta U = 0\), net \(Q = -\) net \(W\), net work equals the enclosed area (clockwise: done by the gas).
- Common errors. Reading the area as the work without checking direction; drawing an adiabat shallower than an isotherm.
- A gas expands at \(2.0\times10^5\) Pa from 0.020 m\(^3\) to 0.050 m\(^3\). Find the work done on the gas.
Show answer
\(W = -P\Delta V = -(2.0\times10^5)(0.030) = -6000\) J.
- An ideal gas undergoes an isothermal expansion and does 800 J of work. How much energy is transferred by heating?
Show answer
\(\Delta U = 0\), so \(Q = -W = +800\) J (800 J enters by heating).
- A monatomic gas is compressed adiabatically with 500 J of work. Find \(\Delta U\), \(Q\), and say whether the temperature rises.
Show answer
\(Q = 0\), \(\Delta U = W = +500\) J, so the temperature rises.
- On a PV diagram a gas moves from state 1 (\(2.0\times10^5\) Pa, 0.010 m\(^3\)) to state 2 (\(1.0\times10^5\) Pa, 0.040 m\(^3\)). Is the final temperature higher or lower than the initial?
Show answer
\(PV\) goes from 2000 J to 4000 J. Since \(T \propto PV\), the final temperature is twice as high.
- Describe a clockwise cycle on a PV diagram in terms of the signs of the net work done on the gas and the net heating.
Show answer
Net work done on the gas is negative (the gas does net work). Since \(\Delta U = 0\) for the cycle, the net energy transferred by heating is positive, equal in size to the net work done by the gas.
- In the cycle of the worked example, the gas absorbed 18000 J in total and did 4000 J of net work. What fraction of the energy absorbed became net work?
Show answer
\(4000/18000 = 0.22\), or about 22%.