First Law of Thermodynamics — AP Physics 2 Study Guide

For: AP Physics 2 candidates sitting AP Physics 2.

Covers: The first law of thermodynamics formula, standard AP sign conventions, work done by/on an ideal gas, internal energy change calculations, PV diagram area interpretation, and applications to four core ideal gas thermodynamic processes.

You should already know: Internal energy of an ideal gas depends only on temperature. Work can be calculated from pressure and volume change for constant-pressure processes. Basic interpretation of pressure-volume (PV) diagrams.

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 First Law of Thermodynamics?

The first law of thermodynamics is a statement of conservation of energy adapted for thermal systems, explicitly accounting for the two primary ways energy transfers into or out of a defined system: heat transfer and macroscopic work. Per the AP Physics 2 Course and Exam Description (CED), this topic is foundational to all thermodynamics questions and makes up approximately 1-2% of the total exam score, though it is implicitly required for nearly all thermodynamics questions on both the MCQ and FRQ sections. The law states that the change in a system’s internal energy equals the net heat added to the system minus the net work done by the system on its surroundings. AP uses a consistent sign convention required by the CED that we will detail in this guide; unlike mechanical conservation of energy that often ignores thermal losses, the first law unifies mechanical and thermal energy to predict system behavior for any thermodynamic process. It is the starting point for all analysis of ideal gas processes, which make up the bulk of thermodynamics testing on the AP exam.

2. The First Law Formula and AP Sign Conventions

The core of the first law is a simple energy balance, but consistent sign conventions are the most common source of lost points on the AP exam, so mastering the AP-specific convention is non-negotiable. The standard AP CED convention is: $$\Delta U=Q-W$$ Where each variable’s sign follows strict rules:

• $\Delta U$ = change in the system’s internal energy: positive if internal energy increases, negative if it decreases
• $Q$ = net heat transferred to the system: positive if heat flows into the system from the surroundings, negative if heat flows out of the system to the surroundings
• $W$ = net work done by the system on the surroundings: positive if the system expands and pushes on the surroundings, negative if the surroundings compress the system and do work on the system

A common alternate convention found in some introductory textbooks defines $W$ as work done on the system, leading to $\Delta U=Q+W$, but the AP CED explicitly requires the convention above. Intuition for the formula: adding heat to a system increases its internal energy, so Q has a positive contribution. When the system does work on the surroundings, it uses energy to push outward, so that reduces internal energy, hence the minus sign. If work is done on the system (compression), W is negative, so subtracting a negative increases ΔU, which matches our physical expectation.

Worked Example

Problem: A piston containing an ideal gas is compressed by the surroundings. During compression, 120 J of work is done on the gas, and 40 J of heat flows out of the gas to the surroundings. What is the change in internal energy of the gas?

Solution:

1. Assign signs per AP convention: W is work done by the gas. Since the gas is compressed, it does negative work on the surroundings: $W=-120\text{ J}$.
2. Q is heat added to the gas. Since 40 J flows out, Q is negative: $Q=-40\text{ J}$.
3. Substitute into the first law: $\Delta U=Q-W=(-40\text{ J})-(-120\text{ J})=80\text{ J}$.
4. The internal energy of the gas increases by 80 J, which makes sense because work done on the gas adds more energy than is lost as heat.

Exam tip: Always write down the definition of each variable (Q, W, ΔU) with their signs before plugging into the first law. AP FRQ graders award points for correct sign reasoning, and MCQ distractors are explicitly designed to match the wrong sign convention.

3. Work Calculations and PV Diagrams

Work done by a gas during a quasi-static (slow, equilibrium) process, which all AP processes assume unless stated otherwise, is directly related to the area under the process path on a PV diagram. The general formula for work is: $$W={\int }_{V_1}^{V_2}PdV$$ This integral simplifies to the net area between the process curve and the volume (x) axis of the PV diagram. If volume increases (expansion, moving right on the diagram), dV is positive, so W is positive, matching our sign convention. If volume decreases (compression, moving left on the diagram), dV is negative, so W is negative. For a cyclic process (a closed loop on a PV diagram that starts and ends at the same state), the net work done by the gas equals the area enclosed by the loop. If the loop is traversed clockwise, net work is positive; if counterclockwise, net work is negative. For constant pressure processes, the integral simplifies to $W=P\Delta V$, which is just the area of a rectangle under the horizontal process line.

Worked Example

Problem: A gas undergoes a process from state 1 ($P_1=2\times 10^5\text{ Pa},V_1=0.01{\text{ m}}^3$) to state 2 ($P_2=2\times 10^5\text{ Pa},V_2=0.03{\text{ m}}^3$) along a straight isobaric line, then cools at constant volume back to state 1, forming a closed cycle. What is the net work done by the gas over the full cycle?

Solution:

1. The cycle forms a rectangle on the PV diagram: one isobaric expansion step, one isochoric (constant volume) step.
2. For the isochoric step, $\Delta V=0$, so work done in that step is $W_2=0$.
3. Work done during expansion is $W_1=P\Delta V=(2\times 10^5\text{ Pa})(0.03{\text{ m}}^3-0.01{\text{ m}}^3)=4000\text{ J}$.
4. Net work for the cycle is $W_{net}=W_1+W_2=4000\text{ J}+0=4000\text{ J}$, which matches the area of the rectangle enclosed by the cycle.

Exam tip: For any cyclic process, immediately note the direction of the loop to get the sign of work right. Clockwise loops always have positive net work, counterclockwise loops always have negative net work, so you do not need to recalculate the sign from scratch.

4. First Law Simplifications for Common Ideal Gas Processes

AP Physics 2 regularly tests application of the first law to four standard ideal gas processes, each with a defining constraint that simplifies the general first law formula. These simplifications come directly from the definition of the process and the property that ideal gas internal energy depends only on temperature:

1. Isochoric (constant volume): $\Delta V=0$, so $W=0$. First law simplifies to $\Delta U=Q$: all heat added goes into changing internal energy.
2. Isobaric (constant pressure): $P=\text{constant}$, so $W=P\Delta V$, use the full first law as normal.
3. Isothermal (constant temperature): For ideal gas, $\Delta U=0$ because internal energy depends only on temperature. First law simplifies to $Q=W$: all heat added becomes work done by the gas.
4. Adiabatic (no heat transfer): $Q=0$, so first law simplifies to $\Delta U=-W$: any change in internal energy comes from work done on or by the gas.

Worked Example

Problem: One mole of monatomic ideal gas undergoes an adiabatic expansion, doing 250 J of work on the surroundings. What is the change in temperature of the gas, given $R=8.31\text{ J/(molcdotpK)}$?

Solution:

1. Adiabatic process means $Q=0$ by definition, so apply the simplified first law: $\Delta U=-W$.
2. The gas does 250 J of work, so $W=+250\text{ J}$, so $\Delta U=-250\text{ J}$.
3. For monatomic ideal gas, $\Delta U=n\frac{3}{2}R\Delta T$. Rearrange to solve for $\Delta T$: $$\Delta T=\frac{2\Delta U}{3nR}=\frac{2(-250\text{ J})}{3(1\text{ mol})(8.31\text{ J/(molcdotpK)})}\approx -20\text{ K}$$
4. The temperature of the gas decreases by 20 K, which matches the physical expectation that adiabatic expansion cools a gas.

Exam tip: Memorize the simplified forms for each process, but always derive them from the full first law on FRQs. Showing the derivation from $\Delta U=Q-W$ will earn you full credit even if you misremember the simplified form.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Using the alternate textbook convention $\Delta U=Q+W$ (W = work done on the system) in an AP FRQ, leading to flipped signs on W. Why: Many introductory textbooks use the alternate convention, and students mix it up when approaching the exam. Correct move: Always write the AP convention $\Delta U=Q-W$ (W = work done by the system) at the top of your work for any thermodynamics problem to anchor your signs.
• Wrong move: Taking the area of a cyclic PV loop as positive regardless of the direction the loop is traversed. Why: Students remember work equals the area of the loop, so they only calculate the magnitude and ignore direction. Correct move: For any closed loop, immediately mark "clockwise = +W, counterclockwise = -W" before calculating area to get the sign right.
• Wrong move: Calculating work as $P\Delta V$ for a non-isobaric curved process on a PV diagram. Why: $P\Delta V$ works for constant pressure processes, and students overgeneralize it. Correct move: Always calculate work as the area under the process path; use $P\Delta V$ only if pressure is explicitly constant along the entire path.
• Wrong move: Assigning Q = +50 J when the problem states "the system releases 50 J of heat", because students only note the magnitude of heat transfer. Why: Students rush through the problem and forget that sign depends on direction of transfer. Correct move: After reading the problem, explicitly write $Q=+$ (heat in) / $Q=-$ (heat out) before plugging into the formula.
• Wrong move: Forgetting that ΔU = 0 for any full cyclic process, even if temperature changes along the path of the cycle. Why: Students only associate ΔU = 0 with isothermal processes. Correct move: If a process starts and ends at the same state, ΔU for the full process is always 0, regardless of the path taken.
• Wrong move: Assuming internal energy changes when a process returns to the initial pressure and volume. Why: Students confuse state functions (internal energy) with path functions (work and heat). Correct move: Always check if the start and end states are identical; if yes, ΔU = 0 by definition.

6. Practice Questions (AP Physics 2 Style)

Question 1 (Multiple Choice)

A rigid, sealed container holds an ideal gas. 300 J of heat is added to the container from the surroundings. What is the change in internal energy of the gas? A) $\Delta U=-300\text{ J}$ B) $\Delta U=0\text{ J}$ C) $\Delta U=+150\text{ J}$ D) $\Delta U=+300\text{ J}$

Worked Solution: A rigid container means constant volume, so this is an isochoric process. For constant volume, ΔV = 0, so work done by the gas W = PΔV = 0. Applying the first law: ΔU = Q - W = Q - 0 = Q. Q is +300 J because heat is added to the system. Distractors correspond to common errors: A uses the wrong sign for Q, B incorrectly assumes no change because work is zero, C is an incorrect average of input values. The correct answer is D.

Question 2 (Free Response)

A 1 mol sample of ideal gas goes through a three-step cyclic process:

1. Step 1→2: Isobaric expansion from $P=1\times 10^5\text{ Pa},V=0.01{\text{ m}}^3$ to $V=0.02{\text{ m}}^3$
2. Step 2→3: Isochoric pressure decrease to $P=5\times 10^4\text{ Pa},V=0.02{\text{ m}}^3$
3. Step 3→1: Isobaric compression back to the original state 1.

(a) Calculate the net work done by the gas over the full cycle. (b) What is the total change in internal energy over the full cycle? Justify your answer. (c) Calculate the total net heat added to the gas over the full cycle.

Worked Solution: (a) The cycle forms a rectangle on a PV diagram, so net work equals the area enclosed by the rectangle. The height of the rectangle is $\Delta P=1\times 10^5\text{ Pa}-5\times 10^4\text{ Pa}=5\times 10^4\text{ Pa}$. The width is $\Delta V=0.02{\text{ m}}^3-0.01{\text{ m}}^3=0.01{\text{ m}}^3$. Net work is $W_{net}=\Delta P\Delta V=(5\times 10^4)(0.01)=500\text{ J}$. The cycle is traversed clockwise, so net work is positive: $W_{net}=+500\text{ J}$. (b) The total change in internal energy is $\Delta U=0$. Internal energy is a state function, meaning it depends only on the current state of the system, not the path taken to reach that state. Since the process starts and ends at the same state 1, the net change in internal energy is zero. (c) Apply the first law to the full cycle: $\Delta U=Q_{net}-W_{net}$. Substitute the known values $\Delta U=0$ and $W_{net}=500\text{ J}$: $0=Q_{net}-500\text{ J}$, so $Q_{net}=500\text{ J}$. The net heat added to the gas over the full cycle is 500 J.

Question 3 (Application / Real-World Style)

When you push the plunger of a bicycle pump quickly to inflate a tire, the compression is approximately adiabatic (no heat transfer during the compression). The pump holds 0.3 mol of air (treated as an ideal diatomic gas) initially at room temperature (293 K). You do 1200 J of work on the air in the pump. What is the final temperature of the air after compression? Use the relation $\Delta U=n\frac{5}{2}R\Delta T$ for diatomic air, with $R=8.31\text{ J/(molcdotpK)}$.

Worked Solution:

1. Quick compression means no heat transfer, so Q = 0 (adiabatic process). You do 1200 J of work on the gas, so work done by the gas is W = -1200 J.
2. Apply the first law: $\Delta U=Q-W=0-(-1200\text{ J})=1200\text{ J}$.
3. Relate ΔU to temperature change: $\Delta T=\frac{2\Delta U}{5nR}=\frac{2(1200)}{5(0.3)(8.31)}\approx 192\text{ K}$.
4. Final temperature: $T_{final}=293\text{ K}+192\text{ K}=485\text{ K}\approx 212^{\circ }C$. In context: This result explains why bicycle pump handles get noticeably hot when inflating a tire quickly: the work done compressing air greatly increases its temperature, which conducts to the pump casing.

7. Quick Reference Cheatsheet

8. What's Next

Mastery of the first law of thermodynamics is a non-negotiable prerequisite for all remaining thermodynamics topics in AP Physics 2. Next you will apply the first law to analyze cyclic processes for heat engines and refrigerators, which rely on first law energy balances to calculate efficiency and coefficient of performance. Without solid command of first law sign conventions and work/energy calculations, you will not be able to correctly solve heat engine efficiency problems, which are common high-weight FRQ questions on the exam. The first law also builds on the ideal gas law you learned earlier in this unit, and it provides the energy foundation for understanding entropy and the second law of thermodynamics, which is the next core topic.

Follow-on topics: Ideal Gas Law Thermodynamic Processes and PV Diagrams Heat Engines and Efficiency Second Law of Thermodynamics