AP Physics 2 · Unit 9 · Topic 9.4
Energy Accounting for Thermodynamic Processes
Track energy across a clearly chosen system boundary. The first law connects heat transfer, mechanical work, and internal-energy change, while a pressure-volume diagram reveals how the route between two states affects work.
Learning Goals
- Choose a thermodynamic system and classify energy crossing its boundary as heat or work.
- Apply one stated sign convention consistently in words, equations, and diagrams.
- Derive boundary work and interpret signed area on a pressure-volume diagram.
- Compare isochoric, isobaric, isothermal, adiabatic, and cyclic processes.
- Distinguish the state function \(U\) from the path quantities \(Q\) and \(W\).
1. Begin with the System Boundary
A system is the matter or region chosen for analysis; everything else is the surroundings. In this lesson the system is usually a fixed amount of gas inside a piston, so it is a closed system: matter stays inside, but energy can cross the boundary.
Heat is not a substance stored in the gas. Heat and work describe energy crossing the boundary; internal energy describes a property of the state.
2. First Law and Sign Convention
We use the common physics convention: \(Q>0\) when heating adds energy to the system, and \(W_{\mathrm{by}}>0\) when the system does work on its surroundings.
The same relationship can be written using work done on the system:
Other textbooks may define \(W\) as work on the system. Either convention works; mixing them within one solution does not.
3. State Function versus Path Quantities
Internal energy is a state function. Once the initial and final equilibrium states are fixed, \(\Delta U=U_f-U_i\) is fixed, regardless of the path.
Heat \(Q\) and work \(W\) are path dependent. Two processes can reach the same final state with different amounts of work. The first law then requires different heat transfers so that the same \(\Delta U\) results.
4. Deriving Pressure-Volume Work
Consider a piston with cross-sectional area \(A\). The surroundings exert external pressure \(P_{\mathrm{ext}}\), producing force \(F=P_{\mathrm{ext}}A\). If the piston moves outward by a small distance \(dx\), then
Adding the work over the entire volume change gives
- Expansion has \(dV>0\), so the gas usually does positive work.
- Compression has \(dV<0\), so work by the gas is negative and work on the gas is positive.
- For a constant external pressure, \(W_{\mathrm{by}}=P_{\mathrm{ext}}(V_f-V_i)\).
On a \(P\)-versus-\(V\) diagram, the area under a process curve represents work. Replacing \(P_{\mathrm{ext}}\) with the gas pressure is justified for a quasi-static process, where the system passes through a sequence of near-equilibrium states.
5. Five Process Shortcuts
| Process | Defining condition | Energy accounting | Diagram clue |
|---|---|---|---|
| Isochoric | \(\Delta V=0\) | \(W_{\mathrm{by}}=0\), so \(\Delta U=Q\) | Vertical path on a \(PV\) graph |
| Isobaric | \(P=\text{constant}\) | \(W_{\mathrm{by}}=P\Delta V\) | Horizontal path on a \(PV\) graph |
| Isothermal ideal gas | \(\Delta T=0\) | \(\Delta U=0\), so \(Q=W_{\mathrm{by}}\) | Hyperbolic path with \(PV=\text{constant}\) |
| Adiabatic | \(Q=0\) | \(\Delta U=-W_{\mathrm{by}}\) | No energy crosses by heating |
| Complete cycle | Final state = initial state | \(\Delta U_{\mathrm{cycle}}=0\), so \(Q_{\mathrm{net}}=W_{\mathrm{net,by}}\) | Enclosed loop; area gives net work |
These are constraints, not separate laws. Always begin with the first law, then apply the condition given by the process.
6. Internal Energy of an Ideal Gas
For an ideal gas, internal energy depends only on temperature. Over a temperature change,
For a monatomic ideal gas, \(C_V=\frac{3}{2}R\), so
If the temperature rises, \(\Delta U>0\); if it falls, \(\Delta U<0\). This result does not depend on whether the change was produced by heating, compression, or both.
7. Reading a \(PV\) Diagram
- Area under one path gives the work for that path.
- Area enclosed by a cycle gives net work for the cycle.
- A clockwise loop gives positive \(W_{\mathrm{net,by}}\); a counterclockwise loop gives negative \(W_{\mathrm{net,by}}\).
- A steeper or higher path can produce more work even when its endpoints match another path.
Pressure and volume identify the temperature of a fixed amount of ideal gas through \(PV=nRT\), but the graph does not directly show \(Q\). Find work from area and then use the first law.
8. Worked Example — Direct First-Law Accounting
Problem: A gas absorbs \(600\,\mathrm J\) of energy by heating and does \(250\,\mathrm J\) of work on a piston. Find the change in internal energy.
- Heat enters, so \(Q=+600\,\mathrm J\).
- The gas does work, so \(W_{\mathrm{by}}=+250\,\mathrm J\).
- Apply the chosen convention:
Answer: the gas's internal energy increases by \(350\,\mathrm J\). Some incoming energy remains in the system; the rest leaves as mechanical work.
9. Worked Example — Compression with Heat Loss
Problem: The surroundings do \(400\,\mathrm J\) of work on a gas while \(100\,\mathrm J\) of energy leaves the gas by heating. Find \(\Delta U\).
Heat leaves, so \(Q=-100\,\mathrm J\). Work on the gas is \(W_{\mathrm{on}}=+400\,\mathrm J\), equivalent to \(W_{\mathrm{by}}=-400\,\mathrm J\).
Answer: \(\Delta U=+300\,\mathrm J\). Compression adds more energy than the gas loses by heating.
10. Worked Example — Constant-Pressure Expansion
Problem: A gas expands at \(2.0\times10^5\,\mathrm{Pa}\) from \(2.0\times10^{-3}\,\mathrm{m^3}\) to \(5.0\times10^{-3}\,\mathrm{m^3}\), while absorbing \(1000\,\mathrm J\) by heating.
Answer: the gas does \(600\,\mathrm J\) of work and gains \(400\,\mathrm J\) of internal energy. Notice that \(\mathrm{Pa\,m^3}=\mathrm J\).
11. Worked Example — Rectangular Cycle
Problem: A gas travels clockwise around a rectangle between \(P_{\mathrm{high}}=300\,\mathrm{kPa}\), \(P_{\mathrm{low}}=100\,\mathrm{kPa}\), and two volumes separated by \(2.0\times10^{-3}\,\mathrm{m^3}\). Find the net work and net heat transfer.
The vertical legs do zero work. The signed sum of the two horizontal legs equals the rectangle's area:
A complete cycle returns to its starting state, so \(\Delta U_{\mathrm{cycle}}=0\). Therefore,
Answer: the gas produces \(400\,\mathrm J\) of net work and must absorb \(400\,\mathrm J\) of net heat over the cycle.
12. A Reliable Solution Workflow
- Draw the system boundary and list the initial and final states.
- Write the sign convention before substituting numbers.
- Determine the signs of \(Q\), \(W_{\mathrm{by}}\), and \(\Delta U\) from the physical story.
- Use geometry or \(\int P_{\mathrm{ext}}dV\) to find work when needed.
- Apply the process condition and first law.
- Check units, signs, and whether the energy transfers balance.
13. Common Traps
- Switching between work by the gas and work on the gas without changing the sign.
- Treating heat or work as a quantity contained in the system.
- Using enclosed area for a single open path, or area under one leg for an entire cycle.
- Assuming \(\Delta U=0\) for every isothermal material; the shortcut is for an ideal gas model.
- Forgetting that \(\Delta U=0\) over a cycle does not require \(Q=0\) or \(W=0\).
- Using gas pressure in a rapid irreversible process without justifying \(P_{\mathrm{gas}}\approx P_{\mathrm{ext}}\).
- Using Celsius instead of kelvin in the ideal-gas equation.
1. A gas receives \(800\,\mathrm J\) by heating and its internal energy rises by \(500\,\mathrm J\). How much work does the gas do?
Show reasoning and answer
From \(\Delta U=Q-W_{\mathrm{by}}\), \(W_{\mathrm{by}}=Q-\Delta U=800-500=300\,\mathrm J\). The positive sign means the gas does work on the surroundings.
2. A gas is compressed at constant pressure \(1.5\times10^5\,\mathrm{Pa}\) from \(4.0\times10^{-3}\,\mathrm{m^3}\) to \(2.0\times10^{-3}\,\mathrm{m^3}\). Find \(W_{\mathrm{by}}\).
Show reasoning and answer
\(W_{\mathrm{by}}=P(V_f-V_i)=(1.5\times10^5)(-2.0\times10^{-3})=-300\,\mathrm J\). Equivalently, the surroundings do \(+300\,\mathrm J\) of work on the gas.
3. Two paths connect the same initial and final equilibrium states. Path A has more area under its \(PV\) curve than path B. Compare \(\Delta U\), \(W_{\mathrm{by}}\), and \(Q\).
Show reasoning and answer
Both paths have the same \(\Delta U\) because internal energy is a state function. Path A has greater work by the gas. Since \(Q=\Delta U+W_{\mathrm{by}}\), path A also requires greater heat transfer into the system.
4. A monatomic ideal gas undergoes an adiabatic expansion. State the signs of \(Q\), \(W_{\mathrm{by}}\), \(\Delta U\), and \(\Delta T\).
Show reasoning and answer
Adiabatic means \(Q=0\). Expansion gives \(W_{\mathrm{by}}>0\), so \(\Delta U=-W_{\mathrm{by}}<0\). For a monatomic ideal gas \(U\propto T\), therefore \(\Delta T<0\).
Investigation: Compare Two Paths
On graph paper, choose the same initial and final \(P,V\) states. Connect them first with a constant-pressure segment followed by a constant-volume segment, then reverse the order. Estimate \(W_{\mathrm{by}}\) from rectangular areas for both paths. Use the ideal-gas equation to show that the endpoints give the same temperature change, then use the first law to explain why the required heat transfers differ.
Official curriculum reference: College Board AP Physics 2 course page. The explanation and worked example are independently written for this study site.