Thermal Processes

AP Physics 2· AP Physics 2 CED — Thermodynamics· 14 min read

1. First Law of Thermodynamics and PV Work★★☆☆☆⏱ 4 min

A thermal (thermodynamic) process is any change in the macroscopic state of a system, defined by changes in pressure $P$ , volume $V$ , and temperature $T$ , driven by energy transfer as heat or work. For AP Physics 2, we 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 ~40% of Unit 2, which accounts for 12-18% of your total AP Physics 2 exam score.

📘 Definition

First Law of Thermodynamics

\Delta U = Q + W

Conservation of energy for thermodynamic systems, following AP CED convention: $Δ U$ is change in system internal energy (positive for an increase), $Q$ is heat added to the system (positive if heat enters), $W$ is work done on the system by the surroundings (positive if work is done on the system).

A core AP skill is calculating work from a pressure-volume (PV) diagram. The magnitude of work done on the gas equals the area under the process path on the diagram:

If the gas expands ( $Δ V > 0$ ), the gas does work on the surroundings, so $W$ is negative.
If the gas is compressed ( $Δ V < 0$ ), $W$ is positive. For constant pressure processes, the area forms a rectangle, so the formula simplifies to $W = - P Δ V$ , where $Δ V = V_{f ina l} - V_{ini t ia l}$ .

📐 Worked Example

A gas expands from $0.002 m^{3}$ to $0.006 m^{3}$ at a constant pressure of $1.5 \times 1 0^{5} Pa$ . 800 J of heat is added to the gas. What is the change in internal energy of the gas?

1
List known values: $P = 1.5 \times 1 0^{5} Pa$ , $V_{i} = 0.002 m^{3}$ , $V_{f} = 0.006 m^{3}$ , $Q = + 800 J$ (positive because heat is added to the system).
2
Calculate the change in volume:
3
$\Delta V = V_f - V_i = 0.006 - 0.002 = 0.004 \text{ m}^3$
4
Since the gas expands, $W$ will be negative, which aligns with the rule that expansion = negative work done on the system.
5
Calculate work for constant pressure:
6
$W = -P\Delta V = -(1.5 \times 10^5)(0.004) = -600 \text{ J}$
7
Apply the first law of thermodynamics to find $Δ U$ :
8
$\Delta U = Q + W = 800 + (-600) = +200 \text{ J}$

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 prevents sign errors on the first law.

2. Four Core Thermal Processes★★★☆☆⏱ 5 min

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

Isochoric (isovolumetric): Volume $V$ is constant ( $Δ V = 0$ ). This means $W = 0$ , so the first law reduces to $Δ U = Q$ : all heat transfer changes the internal energy (and thus temperature) of the gas.
Isobaric: Pressure $P$ is constant. Work is calculated directly as $W = - P Δ V$ , as shown in the previous section.
Isothermal: Temperature $T$ is constant ( $Δ T = 0$ ). For ideal gases, internal energy depends only on temperature, so $Δ U = 0$ . The first law reduces to $Q = - W$ : all heat added to the gas is converted to work done by the gas (for expansion).
Adiabatic: No heat transfer between the system and surroundings ( $Q = 0$ ). The first law reduces to $Δ U = W$ : 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 $0.010 m^{3}$ to $0.004 m^{3}$ . What is the heat transfer $Q$ during this process?

1
For an isothermal ideal gas process, $Δ T = 0$ , so $Δ U = 0$ . Apply the first law:
2
$0 = Q + W \implies Q = -W$
3
For isothermal processes, work done on the gas is calculated as (derived from integrating $W = - \int P d V$ with $P = \frac{n R T}{V}$ ):
4
$W = nRT \ln\left(\frac{V_i}{V_f}\right)$
5
Substitute values: $n = 0.2 mol$ , $R = 8.31 J/(mol \cdotp K)$ , $T = 290 K$ , $\frac{V _{i}}{V _{f}} = 2.5$ :
6
$W = (0.2)(8.31)(290)\ln(2.5) \approx 442 \text{ J}$
7
Solve for $Q$ :
8
$Q = -W \approx -442 \text{ J}$
9
The negative sign confirms heat leaves the system to keep temperature constant during compression, which matches physical expectations.

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 $W$ .

3. Entropy Change for Thermal Processes★★★☆☆⏱ 3 min

Entropy $S$ is a state function that measures the disorder of a thermodynamic system. The second law of thermodynamics states that the total entropy of an isolated system never decreases for any real process. For AP Physics 2, you only need to calculate entropy change for reversible processes occurring at constant temperature, such as phase changes or isothermal processes.

📘 Definition

Entropy Change (Constant Temperature Reversible Process)

\Delta S = \frac{Q}{T}

Change in entropy for a system at constant absolute temperature, where $Q$ is heat added to the system, and $T$ is temperature measured 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 $Δ S = \frac{Q}{T}$ . First convert temperature to Kelvin:
2
$T = 0 + 273 = 273 \text{ K}$
3
Calculate total heat added to melt the ice:
4
$Q = mL_f = (1.0 \text{ kg})(334000 \text{ J/kg}) = 334000 \text{ J}$
5
Calculate entropy change:
6
$\Delta S = \frac{334000}{273} \approx 1220 \text{ J/K}$
7
The positive entropy change matches physical expectation: liquid water has more molecular disorder than solid ice, so entropy increases during melting.

Exam tip:

Always convert temperature to Kelvin before plugging into entropy formulas. Using Celsius will give an incorrect magnitude and can lead to nonsensical results.

4. Concept Check★★☆☆☆⏱ 2 min

✓ Quick check

Test your understanding of adiabatic processes with this AP-style multiple choice question:

An ideal gas undergoes an adiabatic expansion. Which of the following correctly describes the change in temperature $Δ T$ and change in internal energy $Δ U$ of the gas?
- A) $Δ T < 0$ , $Δ U < 0$
- B) $Δ T < 0$ , $Δ U > 0$
- C) $Δ T > 0$ , $Δ U < 0$
- D) $Δ T > 0$ , $Δ U > 0$
Reveal answer
A) $\Delta T < 0$, $\Delta U < 0$ —
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: $Δ U = Q + W = 0 + W = W < 0$ , so $Δ U$ is negative. For ideal gases, internal energy is proportional to absolute temperature, so $Δ U < 0$ implies $Δ T < 0$ .

5. Common Pitfalls

Wrong move:

Using $Δ U = 0$ for adiabatic processes, confusing it with isothermal processes.

Why:

Students mix up the 'constant X' definitions of the two processes, remembering the $Δ U = 0$ rule and applying it to the wrong process.

Correct move:

Write the constraint of each process on your paper before starting calculations: isothermal $\to Δ T = 0 \to Δ U = 0$ (ideal gas); adiabatic $\to Q = 0 \to Δ U = W$ .

Wrong move:

Calculating $W = + P Δ V$ 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 $Δ U$ for an isothermal ideal gas process.

Why:

Students associate internal energy with heat transfer, so if $Q$ is non-zero they assume $Δ U$ must be non-zero.

Correct move:

For any ideal gas process, check if $Δ T$ is zero first: if $Δ T = 0$ , $Δ U = 0$ , 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. Quick Reference Cheatsheet

Category	Formula/Rule	Notes
First Law of Thermodynamics	$Δ U = Q + W$	W = work done on the system; Q = heat added to the system (AP CED convention)
PV Work (constant pressure)	$W = - P Δ V$	Only valid for constant pressure; W negative for expansion, positive for compression
PV Work (any process)	$∣ W ∣ =$ area under process path	Magnitude of W equals area between process path and volume axis
Isochoric Process	$Δ V = 0, W = 0, Δ U = Q$	Constant volume; all heat transfer changes internal energy
Isothermal Process (ideal gas)	$Δ T = 0, Δ U = 0, Q = - W$	Constant temperature; all heat transfer becomes work
Adiabatic Process	$Q = 0, Δ U = W$	No heat transfer; all work changes internal energy
Entropy Change (constant T reversible)	$Δ S = \frac{Q}{T}$	T must be in Kelvin; positive ΔS = increasing disorder
Adiabatic Temperature Relation	$T_{1} V_{1}^{γ - 1} = T_{2} V_{2}^{γ - 1}$	$γ = C_{p} / C_{v}$ , specific to the type of ideal gas

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Adiabatic process temperature change
2022 · FRQ
Cycle of thermal processes calculation

What's Next

Mastery of thermal processes is the foundation for all remaining topics in Unit 2 Thermodynamics. Heat engines, the most heavily tested free-response 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 process conventions and calculations, you will not be able to analyze efficiency or entropy changes for heat engines, which are common high-weight FRQ questions that make up a large portion of your unit score. Thermal processes also connect to ideal gas behavior from Unit 1 and extend to all topics involving thermal energy transfer across the AP Physics 2 curriculum.