Thermal Processes — AP Physics 2 Study Guide

For: AP Physics 2 candidates sitting AP Physics 2.

Covers: First law of thermodynamics for closed systems, four standard thermal processes (isochoric, isobaric, isothermal, adiabatic), PV work calculation, entropy change for reversible processes, and process comparison for AP exam questions.

You should already know: Ideal gas law for closed systems, internal energy of an ideal gas depends only on absolute temperature, definition of mechanical work.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 2 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Thermal Processes?

A thermal process (also called a thermodynamic process) is any change in the macroscopic state of a thermodynamic system, defined by changes in state variables pressure $P$, volume $V$, and temperature $T$, driven by energy transfer as heat or work. For AP Physics 2, we almost exclusively analyze closed systems with a fixed number of moles of gas, so all processes follow the ideal gas law and first law of thermodynamics. This topic makes up roughly 40% of Unit 2 (Thermodynamics), which itself accounts for 12-18% of the total AP Physics 2 exam score. Thermal processes appear in both multiple-choice (MCQ) and free-response (FRQ) sections: MCQs typically test process identification, sign conventions, or work/heat comparisons, while FRQs often require full calculations across multiple processes on a PV diagram. This guide uses the standard AP CED convention: $W$ = work done on the system, $Q$ = heat added to the system, $\Delta U$ = change in system internal energy.

2. First Law of Thermodynamics and PV Work

The first law of thermodynamics is simply conservation of energy applied to thermodynamic systems. It is written per AP convention as: $$\Delta U=Q+W$$ Where $\Delta U$ is positive for an increase in internal energy, $Q$ is positive if heat enters the system, and $W$ is positive if the surroundings do work on the system.

A core skill for thermal process problems is calculating work from a pressure-volume (PV) diagram. Work done on the gas is related to the area under the process path on the diagram: if the gas expands ($\Delta V>0$), the gas does work on the surroundings, so $W$ is negative with magnitude equal to the area under the path. If the gas is compressed ($\Delta V<0$), $W$ is positive, equal to the area under the path. For constant pressure processes, the area forms a rectangle, so the formula simplifies to $W=-P\Delta V$, where $\Delta V=V_{final}-V_{initial}$.

Worked Example

A gas expands from $0.002{\text{ m}}^3$ to $0.006{\text{ m}}^3$ at a constant pressure of $1.5\times 10^5\text{ Pa}$. 800 J of heat is added to the gas. What is the change in internal energy of the gas?

1. List known values: $P=1.5\times 10^5\text{ Pa}$, $V_i=0.002{\text{ m}}^3$, $V_f=0.006{\text{ m}}^3$, $Q=+800\text{ J}$ (positive because heat is added to the system).
2. Calculate $\Delta V=V_f-V_i=0.006-0.002=0.004{\text{ m}}^3$. Expansion means $W$ will be negative.
3. Calculate work for constant pressure: $W=-P\Delta V=-(1.5\times 10^5)(0.004)=-600\text{ J}$.
4. Apply first law: $\Delta U=Q+W=800+(-600)=+200\text{ J}$.

Exam tip: Always confirm the sign of $W$ by checking the volume change first: expansion = negative $W$, compression = positive $W$. This rule works for any process, not just constant pressure, and will prevent sign errors on the first law.

3. Four Standard Thermal Processes

Every thermal process is defined by a constraint that keeps one state variable constant, simplifying calculations for ideal gases. The four processes tested on AP Physics 2 are:

1. Isochoric (isovolumetric): Volume $V$ is constant ($\Delta V=0$). This means $W=0$, so the first law reduces to $\Delta U=Q$: all heat transfer changes the internal energy (and thus temperature) of the gas.
2. Isobaric: Pressure $P$ is constant. Work is calculated directly as $W=-P\Delta V$, as shown in the previous section.
3. Isothermal: Temperature $T$ is constant ($\Delta T=0$). For ideal gases, internal energy depends only on temperature, so $\Delta U=0$. The first law reduces to $Q=-W$: all heat added to the gas is converted to work done by the gas (for expansion), and all work done on the gas is removed as heat (for compression).
4. Adiabatic: No heat transfer between the system and surroundings ($Q=0$). The first law reduces to $\Delta U=W$: all work done on the gas changes the internal energy (and temperature) of the gas. Expansion cools the gas, compression heats it.

Worked Example

A 0.2 mol sample of ideal helium gas undergoes isothermal compression at 290 K from $0.010{\text{ m}}^3$ to $0.004{\text{ m}}^3$. What is the heat transfer $Q$ during this process?

1. For an isothermal ideal gas process, $\Delta T=0$, so $\Delta U=0$. First law gives $0=Q+W⟹Q=-W$.
2. For isothermal processes, work done on the gas is calculated as $W=nRT\ln \left(\frac{V_i}{V_f}\right)$, derived from integrating $W=-\int PdV$ with $P=\frac{nRT}{V}$.
3. Plug in values: $n=0.2\text{ mol}$, $R=8.31\text{ J/(molcdotpK)}$, $T=290\text{ K}$, $\frac{V_i}{V_f}=2.5$.
4. Calculate $W=(0.2)(8.31)(290)\ln (2.5)\approx 442\text{ J}$.
5. $Q=-W\approx -442\text{ J}$. The negative sign confirms heat leaves the system to keep temperature constant during compression, as expected.

Exam tip: When comparing work between two processes from the same initial to final state on a PV diagram, the process with the larger area under its path has a larger magnitude of work. For expansion, this means a more negative $W$.

4. Entropy Change for Thermal Processes

Entropy $S$ is a state function that measures the disorder of a thermodynamic system, and the second law of thermodynamics states that the total entropy of an isolated system never decreases for any real process. For AP Physics 2, we only calculate entropy change for reversible processes occurring at constant temperature (such as phase changes or isothermal processes): $$\Delta S=\frac{Q}{T}$$ Where $Q$ is the heat added to the system, and $T$ is the constant absolute temperature of the system in Kelvin. Because entropy is a state function, its change depends only on the initial and final states, not the path taken between them. For any full cycle that returns a system to its initial state, the total entropy change of the system is zero.

Worked Example

1.0 kg of ice at 0°C melts completely into liquid water at 0°C. The latent heat of fusion for water is 334 kJ/kg. What is the entropy change of the water during melting?

1. Melting occurs at constant temperature, so we use $\Delta S=\frac{Q}{T}$. Convert temperature to Kelvin: $T=0+273=273\text{ K}$.
2. Calculate $Q$: heat added to melt the ice is $Q=m{L}_f=(1.0\text{ kg})(334000\text{ J/kg})=334000\text{ J}$.
3. Calculate $\Delta S=\frac{334000}{273}\approx 1220\text{ J/K}$.
4. The positive entropy change matches the physical expectation: liquid water has more molecular disorder than solid ice, so entropy increases.

Exam tip: Always convert temperature to Kelvin before plugging into entropy formulas. Using Celsius will give an incorrect magnitude and can lead to nonsensical negative temperature values.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Using $\Delta U=0$ for adiabatic processes, confusing it with isothermal processes. Why: Students mix up the "constant X" definitions of the two processes, remembering the $\Delta U=0$ rule and applying it to the wrong process. Correct move: Write the constraint of each process on your paper before starting calculations: isothermal $\to \Delta T=0\to \Delta U=0$ (ideal gas); adiabatic $\to Q=0\to \Delta U=W$.
• Wrong move: Calculating $W=+P\Delta V$ for expansion, getting the wrong sign. Why: Students learn the alternate convention where $W$ is work done by the system and forget AP uses work done on the system. Correct move: Before calculating $W$, check if volume increases or decreases: if $V$ increases, $W$ is negative; if $V$ decreases, $W$ is positive. This rule works for all processes.
• Wrong move: Calculating a non-zero $\Delta U$ for an isothermal ideal gas process. Why: Students associate internal energy with heat transfer, so if $Q$ is non-zero they assume $\Delta U$ must be non-zero. Correct move: For any ideal gas process, check if $\Delta T$ is zero first: if $\Delta T=0$, $\Delta U=0$, no exceptions for AP Physics 2.
• Wrong move: Calculating work as the area enclosed by the process and the axes, instead of the area between the process path and the volume axis. Why: Students confuse area for a single process with area enclosed by a full cycle. Correct move: For a single process, draw vertical lines from the start and end of the process down to the volume (x) axis, and calculate the area of the shape between the path and the axis.
• Wrong move: Treating entropy as a path function, calculating non-zero entropy change for a full cycle. Why: Students forget that entropy, like internal energy, temperature, pressure, and volume, is a state function that only depends on the current state. Correct move: If a process returns the system to its initial state, the change in all state functions (including entropy and internal energy) is zero.

6. Practice Questions (AP Physics 2 Style)

Question 1 (Multiple Choice)

An ideal gas undergoes an adiabatic expansion. Which of the following correctly describes the change in temperature $\Delta T$ and change in internal energy $\Delta U$ of the gas? A) $\Delta T<0$, $\Delta U<0$ B) $\Delta T<0$, $\Delta U>0$ C) $\Delta T>0$, $\Delta U<0$ D) $\Delta T>0$, $\Delta U>0$

Worked Solution: By definition, an adiabatic process has $Q=0$, no heat transfer. Expansion means volume increases, so the gas does work on the surroundings, so work done on the gas $W$ is negative. Apply the first law: $\Delta U=Q+W=0+W=W<0$, so $\Delta U$ is negative. For ideal gases, internal energy is proportional to absolute temperature, so $\Delta U<0$ implies $\Delta T<0$. This matches option A. Correct answer: A.

Question 2 (Free Response)

A 0.1 mol sample of ideal monatomic gas undergoes three processes forming a closed cycle:

• Process A→B: isochoric pressure decrease from $P_A=6.0\times 10^5\text{ Pa}$, $V_A=1.0\times 10^{-3}{\text{ m}}^3$ to $P_B=2.0\times 10^5\text{ Pa}$
• Process B→C: isobaric expansion to $V_C=3.0\times 10^{-3}{\text{ m}}^3$
• Process C→A: isothermal compression back to initial state A. (a) Calculate the temperature of the gas at state A. (b) For process A→B, calculate $W$, $Q$, and $\Delta U$. (c) What is the total change in entropy of the gas for the full cycle A→B→C→A? Justify your answer.

Worked Solution: (a) Use the ideal gas law $PV=nRT$ to solve for $T_A$: $$T_A=\frac{P_A{V}_A}{nR}=\frac{(6.0\times 10^5)(1.0\times 10^{-3})}{(0.1)(8.31)}=\frac{600}{0.831}\approx 722\text{ K}$$

(b) Process A→B is isochoric, so $\Delta V=0$, meaning $W=-P\Delta V=0\text{ J}$. Calculate $\Delta T=T_B-T_A=\frac{P_B{V}_B}{nR}-\frac{P_A{V}_A}{nR}=\frac{V}{nR}(P_B-P_A)\approx -481\text{ K}$. For monatomic ideal gas, $\Delta U=\frac{3}{2}nR\Delta T=1.5(0.1)(8.31)(-481)\approx -300\text{ J}$. Apply first law: $\Delta U=Q+W⟹Q=\Delta U=-300\text{ J}$.

(c) Entropy is a state function, meaning it depends only on the current state of the system, not the path taken. A full cycle returns the gas to its initial state, so the total change in entropy is $\Delta S_{total}=0\text{ J/K}$.

Question 3 (Application / Real-World Style)

A bicycle pump compresses air quickly enough that no significant heat is exchanged with the surroundings. The pump initially contains 0.01 mol of air (ideal diatomic gas) at 293 K (room temperature) that is compressed adiabatically to half its original volume. For adiabatic ideal gas processes, $T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}$, where $\gamma =1.4$ for air. What is the final temperature of the compressed air, and what does this imply about how the pump feels after pumping?

Worked Solution: We start with the adiabatic relation $T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}$. We know $V_2=0.5V_1$, so rearrange to solve for $T_2$: $$T_2=T_1{\left(\frac{V_1}{V_2}\right)}^{\gamma -1}=293(2{)}^{0.4}\approx 293\times 1.32\approx 387\text{ K}$$ Convert to Celsius: $387-273=114^{\circ }\text{C}$. This final temperature is well above room temperature, so the pump will feel noticeably warm to the touch after several quick compressions, matching everyday experience.

7. Quick Reference Cheatsheet

8. What's Next

Mastery of thermal processes is the foundation for all remaining topics in Unit 2 Thermodynamics. Heat engines, the most heavily tested FRQ topic in the unit, operate on repeated cycles of thermal processes, so you need to correctly identify process constraints and calculate work, heat, and internal energy change for each step to find engine efficiency. Without a solid understanding of thermal processes, you will not be able to analyze efficiency or entropy changes for heat engines, which are common high-weight FRQ questions. Thermal processes also connect to ideal gas behavior from Unit 1 and extend to topics in thermal energy transfer across the AP Physics 2 curriculum.

Next topics to study: Heat Engines and Efficiency Second Law of Thermodynamics Entropy and Irreversibility Ideal Gas Law