Thermodynamic Systems — AP Physics 2 Study Guide

For: AP Physics 2 candidates sitting AP Physics 2.

Covers: Definition and classification of open, closed, and isolated thermodynamic systems, distinction between state and path functions, first law of thermodynamics for closed systems, P-V work, and pressure-volume diagram analysis per AP Physics 2 CED.

You should already know: Ideal gas law and properties of ideal gases. Basic energy conservation principles. Definition of work in classical mechanics.

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 Thermodynamic Systems?

A thermodynamic system is any clearly defined region of matter or space that we choose to analyze for energy transfer and property changes; everything outside the system that can exchange energy or matter with it is called the surroundings. This topic is the foundational framework for all thermodynamics problems on the AP Physics 2 exam, and Unit 2 Thermodynamics accounts for 18-22% of the total exam score, with questions about thermodynamic systems appearing in both multiple-choice (MCQ) and free-response (FRQ) sections.

Standard notation defines system internal properties as unprimed variables, and surrounding properties as primed. Systems are classified by whether they can exchange energy and/or matter with the surroundings, a classification that dictates which thermodynamic laws apply to a given problem. Unlike many other topics in physics, thermodynamics analysis always starts with explicitly defining your system—this choice simplifies calculations and avoids sign errors in energy transfer problems.

2. Classification of Thermodynamic Systems

All thermodynamic systems are grouped into three categories based on what can cross the system boundary (the edge separating the system from the surroundings).

• Open system: Can exchange both energy and matter with the surroundings. Common examples include an open cup of hot tea (water vapor leaves as matter, heat leaves as energy) and a living plant exchanging carbon dioxide and oxygen with the air.
• Closed system: Can exchange energy, but not matter, with the surroundings. All calculation-based thermodynamics problems on the AP exam use closed systems, such as a sealed piston containing gas or an unopened soda can.
• Isolated system: Cannot exchange either energy or matter with the surroundings. Perfect isolation is an idealization, but highly insulated sealed containers (like a well-made thermos with a sealed lid) are approximated as isolated for problem-solving.

When you first approach any thermodynamics problem, drawing the system boundary and classifying the system is the critical first step, as energy accounting rules change based on the type. AP Physics 2 regularly tests classification in multiple-choice questions, so memorizing the rules for each category is essential.

Worked Example

A physiologist defines the system of interest as the human body during a 30-minute workout. The body sweats (loses water mass) and generates heat that dissipates to the environment. Classify this system.

1. First check for matter exchange: Sweating releases water molecules from the body (system) to the surroundings, so matter exchange occurs. This immediately rules out closed and isolated systems.
2. Next confirm energy exchange: Heat generated by the body is transferred to the environment, so energy exchange also occurs.
3. Match to the classification rules: Systems that exchange both energy and matter are open systems.
4. Conclusion: The human body during exercise is an open thermodynamic system.

Exam tip: When classifying systems, always check for matter exchange first—if any matter crosses the system boundary, it cannot be closed or isolated, so it is automatically open.

3. State vs Path Functions

A core distinction in thermodynamics that is heavily tested on the AP exam is between state functions and path functions.

• A state function is any property that depends only on the current equilibrium state of the system, not on the path taken to reach that state. The change in a state function between two states is always $\Delta X=X_{\text{final}}-X_{\text{initial}}$, and this value is the same no matter what process you use to go from the initial to final state. Common state functions on the AP exam: internal energy $U$, pressure $P$, volume $V$, temperature $T$, entropy $S$. For ideal gases, internal energy depends only on temperature, so $\Delta U=0$ for any process that starts and ends at the same temperature.
• A path function is a quantity whose value depends on the specific process (path) taken between two states. Two different paths between the same two states will give different values for path functions. Common path functions: heat $Q$ and work $W$.

This distinction is the basis for many common AP exam questions that test conceptual understanding of thermodynamics, beyond just calculation.

Worked Example

Two separate processes take a fixed amount of ideal gas from state 1 ($P_1=2\text{atm},V_1=1\text{L},T_1=300\text{K}$) to state 2 ($P_2=1\text{atm},V_2=2\text{L},T_2=300\text{K}$). Process A does 100 J of work on the gas, and Process B has the gas do 100 J of work on the surroundings. What is the difference in $\Delta U$ between Process A and Process B?

1. Recall that internal energy $U$ is a state function, and for an ideal gas $U$ depends only on temperature.
2. Both processes start at $T_1=300\text{K}$ and end at $T_2=300\text{K}$, so $\Delta U_A=U_2-U_1=0$ and $\Delta U_B=U_2-U_1=0$.
3. The work done is different between the two processes, but work is a path function, so it does not affect the change in internal energy.
4. The difference between $\Delta U_A$ and $\Delta U_B$ is $0\text{J}-0\text{J}=0\text{J}$.

Exam tip: Any question asking about change in internal energy between two states for an ideal gas only depends on the initial and final temperatures, not the work or heat transferred along the path.

4. First Law of Thermodynamics and P-V Work

For closed thermodynamic systems (the only type AP Physics 2 asks you to calculate for), energy conservation is written as the first law of thermodynamics. The AP exam uses the following standard sign convention:

• $Q$ is positive when heat is added to the system, negative when heat leaves the system.
• $W$ is the work done on the system by the surroundings, positive when work is done on the system, negative when the system does work on the surroundings.

With this convention, the first law is: $$\Delta U=Q+W$$

For a system that expands or compresses at constant pressure, the work done on the system is given by the P-V work formula: $$W=-P\Delta V$$ where $\Delta V=V_{\text{final}}-V_{\text{initial}}$. The negative sign accounts for the sign convention: if the system expands, $\Delta V>0$, so $W$ is negative (the system does work on the surroundings), which matches our convention. On a pressure-volume (P-V) diagram, the work done by the system is equal to the area under the process curve between the initial and final volume.

Worked Example

A closed system containing 0.5 mol of ideal gas undergoes an isobaric (constant pressure) expansion at $P=1.0\times 10^5\text{Pa}$, from an initial volume of $0.001{\text{m}}^3$ to a final volume of $0.003{\text{m}}^3$. The process adds 220 J of heat to the system. Calculate the change in internal energy $\Delta U$.

1. Calculate $\Delta V$: $\Delta V=V_f-V_i=0.003{\text{m}}^3-0.001{\text{m}}^3=0.002{\text{m}}^3$.
2. Calculate work done on the system: $W=-P\Delta V=-(1.0\times 10^5\text{Pa})(0.002{\text{m}}^3)=-200\text{J}$. The negative sign confirms the system does work on the surroundings as it expands.
3. Heat added to the system is $Q=+220\text{J}$, following the sign convention.
4. Apply the first law: $\Delta U=Q+W=220\text{J}+(-200\text{J})=+20\text{J}$.
5. Conclusion: The internal energy of the gas increases by 20 J.

Exam tip: Always confirm the sign convention for work in the problem—if the problem explicitly states it uses $W$ as work done by the system, adjust the first law to $\Delta U=Q-W$ to avoid sign errors.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Classifying a sealed, thermally insulated piston of gas as isolated. Why: Students confuse no matter exchange with no energy exchange—insulation prevents heat transfer, but you can still do work on the gas by compressing the piston, so energy can cross the boundary. Correct move: Always check for both matter and energy transfer: no matter exchange, energy exchange possible, so it is closed, not isolated.
• Wrong move: Calculating a different $\Delta U$ for two different paths between the same initial and final states. Why: Students confuse path functions (Q, W) with the state function (U), and assume the change in U depends on work or heat. Correct move: For any ideal gas, $\Delta U$ depends only on initial and final temperature, regardless of the process path.
• Wrong move: Getting a positive work $W$ (work done on the system) for an expanding gas at constant pressure. Why: Students forget the negative sign in $W=-P\Delta V$, and just multiply P and the magnitude of volume change. Correct move: Always write $\Delta V=V_f-V_i$ explicitly before plugging into the work formula, so the sign of $\Delta V$ gives the correct sign of $W$.
• Wrong move: Calling a puddle of water evaporating on a table a closed system, because the water doesn't move horizontally. Why: Students forget that water vapor is matter that leaves the system, so matter exchange does occur. Correct move: Any process involving evaporation, condensation, or gas exchange across the boundary is an open system, because water vapor is matter.
• Wrong move: Calculating work for a single process as the area enclosed by a cycle on a P-V diagram. Why: Students mix up work for a single process and net work for a cycle. Correct move: For a single process between two volumes, work done by the system is area under the P-V curve; net work for a full cycle is the area enclosed by the cycle.

6. Practice Questions (AP Physics 2 Style)

Question 1 (Multiple Choice)

A camper places an unopened, well-insulated soda can at room temperature into a campfire. The can does not change shape or volume, and remains sealed. The soda inside the can is defined as the system. What type of system is this, and what is the value of work done on the system? A) Open system, $W=0$ B) Closed system, $W>0$ C) Closed system, $W=0$ D) Isolated system, $W<0$

Worked Solution: First, check for matter exchange: the can is sealed, so no matter can enter or leave the system, so it cannot be open, eliminating option A. Next, heat from the campfire transfers into the can, so energy crosses the system boundary, meaning it cannot be isolated, eliminating option D. The can does not change volume, so $\Delta V=0$, so from the P-V work formula $W=-P\Delta V=0$, so work done on the system is zero, eliminating B. The correct answer is C.

Question 2 (Free Response)

A fixed quantity of ideal gas is contained in a frictionless, sealed piston. The gas undergoes three processes that form a clockwise cycle on a P-V diagram: Process 1→2 is isobaric expansion from $(V=1{\text{m}}^3,P=2\text{Pa})$ to $(V=3{\text{m}}^3,P=2\text{Pa})$; Process 2→3 is isochoric (constant volume) pressure reduction to $(V=3{\text{m}}^3,P=1\text{Pa})$; Process 3→1 is isobaric compression back to the initial state 1. (a) Calculate the total net work done by the gas for the full cycle. (b) If the heat added to the gas during Process 1→2 is 8 J, calculate the change in internal energy for Process 1→2. (c) Is the net change in internal energy for the full cycle positive, negative, or zero? Justify your answer.

Worked Solution: (a) Net work done by the gas for a clockwise cycle equals the area enclosed by the cycle on the P-V diagram. The cycle forms a rectangle with height $\Delta P=2\text{Pa}-1\text{Pa}=1\text{Pa}$ and width $\Delta V=3{\text{m}}^3-1{\text{m}}^3=2{\text{m}}^3$. Net work $W_{\text{net, by}}=\Delta P\times \Delta V=(1\text{Pa})(2{\text{m}}^3)=\boxed{2\text{J}}$.

(b) For Process 1→2, work done on the gas is $W=-P\Delta V=-(2\text{Pa})(2{\text{m}}^3)=-4\text{J}$. With $Q=+8\text{J}$, apply the first law: $\Delta U=Q+W=8\text{J}-4\text{J}=\boxed{+4\text{J}}$.

(c) The net change in internal energy is $\boxed{0}$. Internal energy is a state function, so after a full cycle the gas returns to its initial state, so $\Delta U=U_{\text{final}}-U_{\text{initial}}=0$, regardless of the path.

Question 3 (Application / Real-World Style)

A scuba diver breathes air from a regulator that supplies air at constant pressure equal to the surrounding water pressure at 10 m depth, $P=2.0\times 10^5\text{Pa}$. When the diver inhales, the volume of air in their lungs (the system) increases from 2.0 L to 6.0 L ($1\text{L}=0.001{\text{m}}^3$). Assume the process is isobaric. Calculate the work done by the lungs (the system) on the surrounding air when the diver inhales, and state the sign of work done on the lungs.

Worked Solution: First convert volumes to SI units: $V_i=2.0\text{L}=0.002{\text{m}}^3$, $V_f=6.0\text{L}=0.006{\text{m}}^3$. $\Delta V=V_f-V_i=0.004{\text{m}}^3$. Work done by the system at constant pressure is $W_{\text{by}}=P\Delta V=(2.0\times 10^5\text{Pa})(0.004{\text{m}}^3)=800\text{J}$. Work done on the system is the negative of work done by the system, so the work done on the lungs is $-800\text{J}$ (negative). In context, this result matches real breathing mechanics: the lungs do work on the surroundings to expand the chest cavity during inhalation.

7. Quick Reference Cheatsheet

8. What's Next

Thermodynamic systems are the foundational framework for all remaining topics in Unit 2 Thermodynamics. Next you will apply the first law of thermodynamics, system classification, and P-V work concepts to analyze specific ideal gas processes and then heat engines and refrigerators. Without mastering system boundaries, sign conventions for work and heat, and the distinction between state and path functions, you will not be able to correctly calculate efficiency or entropy change for thermodynamic cycles, which are heavily tested FRQ topics. This topic also connects to the broader theme of energy conservation across the entire AP Physics 2 course, linking thermal energy transfer to conservation principles in other domains. Follow-up topics to study next: Ideal Gas Processes, Heat Engines and Thermal Efficiency, Entropy and the Second Law of Thermodynamics