# First Law of Thermodynamics

> AP Physics 2 · Unit 2: Thermodynamics
> Source: https://www.owlsprep.com/study/ap-physics-2-u2-first-law-of-thermodynamics/

This guide covers the first law of thermodynamics, AP-specific sign conventions, work calculations from PV diagrams, and applications to common ideal gas processes for AP Physics 2 exam preparation.

**Prerequisites:** Internal energy of an ideal gas depends only on temperature; [Basic interpretation of pressure-volume (PV) diagrams](https://www.owlsprep.com/study/ap-physics-2-u2-thermodynamic-processes-pv-diagrams/); Work calculation for constant-pressure processes

## Learning objectives

- State the first law of thermodynamics using AP sign conventions
- Calculate change in internal energy, heat, and work for ideal gas processes
- Interpret work as the area under a PV diagram path
- Apply simplified first law to common thermodynamic processes
- Avoid common sign and convention errors on the AP exam

## Core Definition of the First Law

The first law of thermodynamics is an application of the conservation of energy principle to thermal systems, explicitly accounting for two primary energy transfer methods: heat transfer between the system and surroundings, and macroscopic work done by or on the system. This law is foundational for all thermodynamics questions on the AP Physics 2 exam, making up 1-2% of total score directly and required for nearly all thermodynamics problems.

**First Law of Thermodynamics** — 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.

*Notation:* \Delta U = Q - W

## AP-Specific Sign Conventions

Consistent sign conventions are the most common source of lost points on the AP exam. The College Board explicitly requires the convention outlined below, which differs from some introductory textbooks.

- $\Delta U$: Positive when system internal energy increases, negative when it decreases
- $Q$: Positive when net heat flows *into* the system from surroundings, negative when heat flows out
- $W$: Positive when net work is done *by* the system on surroundings, negative when work is done *on* the system by surroundings

> **warning**
>
> Many textbooks use an alternate convention where $W$ is work done *on* the system, leading to $\Delta U = Q + W$. This is *not* accepted for AP Physics 2; always use the AP convention $\Delta U = Q - W$.

**Worked example:** 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?

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:
2. $$W = -120 \text{ J}$$
3. Q is heat added *to* the gas. Since 40 J flows out, Q is negative:
4. $$Q = -40 \text{ J}$$
5. Substitute into the first law formula:
6. $$\Delta U = Q - W = (-40 \text{ J}) - (-120 \text{ J}) = 80 \text{ J}$$
7. The internal energy of the gas increases by 80 J, which matches physical intuition: work done on the gas adds more energy than is lost as heat.

> **Exam tip:** Always write down each variable with its sign before substituting. AP FRQ graders award points for correct sign reasoning, and MCQ distractors are designed for wrong convention errors.

## Work Calculations and PV Diagrams

For any quasi-static (slow, equilibrium) process (the only type tested on AP Physics 2), work done by the gas equals the area under the process path on a pressure-volume (PV) diagram. The general formula for work is:

$$W = \int_{V_1}^{V_2} P dV$$

If volume increases (expansion, moving right on the diagram), work is positive. If volume decreases (compression, moving left), work is negative. For constant pressure processes, the integral simplifies to $W = P\Delta V$. For a cyclic process (closed loop on a PV diagram that starts and ends at the same state), net work equals the area enclosed by the loop: clockwise loops give positive net work, counterclockwise loops give negative net work.

**Worked example:** 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?

1. The cycle forms a rectangle on the PV diagram, with one isobaric expansion step and one isochoric (constant volume) step.
2. For the isochoric step, $\Delta V = 0$, so work done in this step is:
3. $$W_2 = 0$$
4. Work done during the isobaric expansion is calculated as $W_1 = P\//Delta V$:
5. $$W_1 = (2 \times 10^5 \text{ Pa})(0.03 \text{ m}^3 - 0.01 \text{ m}^3) = 4000 \text{ J}$$
6. Net work for the full cycle is the sum of work from each step:
7. $$W_{net} = W_1 + W_2 = 4000 \text{ J} + 0 = 4000 \text{ J}$$
8. This matches the area of the rectangle enclosed by the cycle, as expected.

> **Exam tip:** For any cyclic process, immediately note the direction of the loop to get the sign of work right. You do not need to recalculate the sign from scratch once you know direction.

## 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 formula, using the fact that ideal gas internal energy depends only on temperature.

| Process | Defining Constraint | Simplified First Law |
| --- | --- | --- |
| Isochoric (constant volume) | $\Delta V = 0$ | $\Delta U = Q$ |
| Isobaric (constant pressure) | $P = \text{constant}$ | $\Delta U = Q - P\Delta V$ |
| Isothermal (constant temperature) | $\Delta T = 0 \implies \Delta U = 0$ | $Q = W$ |
| Adiabatic (no heat transfer) | $Q = 0$ | $\Delta U = -W$ |

**Worked example:** 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 $R = 8.31 \text{ J/(mol·K)}$?

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

**Check your understanding**

Test your understanding of the key concepts above:

1. 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?

   - $\Delta U = -300 \text{ J}$
   - $\Delta U = 0 \text{ J}$
   - $\Delta U = +150 \text{ J}$
   - $\Delta U = +300 \text{ J}$

   *Answer:* $\Delta U = +300 \text{ J}$

   *Why:* A rigid container means constant volume (isochoric), so $W = 0$. By first law: $\Delta U = Q - 0 = +300 \text{ J}$.

> **Exam tip:** Memorize the simplified forms, but always derive them from the full first law $\Delta U = Q - W$ on FRQs to earn full credit even if you misremember the simplification.

## Common pitfalls

- **Wrong:** Using the alternate textbook convention $\Delta U = Q + W$ (W = work done on the system)
  - Why it fails: Many introductory textbooks use this alternate convention, leading to flipped signs that lose points.
  - Correct: Always write the AP convention $\Delta U = Q - W$ (W = work done *by* the system) at the top of your work.
- **Wrong:** Taking the area of a cyclic PV loop as positive regardless of traversal direction
  - Why it fails: Students remember work equals loop area but ignore the effect of direction on sign.
  - Correct: Immediately note: clockwise loops = positive net work, counterclockwise loops = negative net work before calculating area.
- **Wrong:** Calculating work as $P\Delta V$ for a non-isobaric curved process on a PV diagram
  - Why it fails: $P\Delta V$ only works for constant pressure processes, and students overgeneralize the formula.
  - Correct: Always calculate work as the area under the process path; only use $P\Delta V$ if pressure is constant along the entire path.
- **Wrong:** Assigning a positive Q when the problem states the system releases heat, only noting magnitude
  - Why it fails: Students rush through problems and forget that sign depends on the direction of heat transfer.
  - Correct: Explicitly mark $Q = +$ for heat into the system, $Q = -$ for heat out of the system before plugging into the formula.
- **Wrong:** Forgetting that $\Delta U = 0$ for any full cyclic process, even if temperature changes along the path
  - Why it fails: Students only associate $\Delta U = 0$ with isothermal processes.
  - Correct: If a process starts and ends at the same state, $\Delta U$ for the full process is always 0, regardless of path.
- **Wrong:** Assuming internal energy changes when a process returns to initial pressure and volume
  - Why it fails: Students confuse state functions (internal energy) with path functions (work and heat).
  - Correct: If start and end states are identical, $\Delta U = 0$ by definition, since internal energy is a state function.

## Cheatsheet

| Category | Formula/Rule | AP Exam Notes |
| --- | --- | --- |
| Core First Law (AP Convention) | $\Delta U = Q - W$ | W = work done *by* system, Q = heat added *to* system. Always use this. |
| General Work (PV Diagram) | W = area under process curve | Positive W for expansion, negative W for compression. |
| Constant Pressure Work | $W = P\Delta V$ | Only for isobaric (constant P) processes. |
| Isochoric Process | $W = 0, \Delta U = Q$ | Constant volume, all heat changes internal energy. |
| Isothermal Process (ideal gas) | $\Delta U = 0, Q = W$ | Constant T, ideal gas U depends only on T. |
| Adiabatic Process | $Q = 0, \Delta U = -W$ | No heat transfer between system and surroundings. |
| Cyclic Process | $\Delta U = 0, Q_{net} = W_{net}$ | Net work = enclosed loop area; clockwise = +W, counterclockwise = -W. |
| Ideal Gas Internal Energy Change | $\Delta U = n c_v \Delta T$ | $c_v = 3R/2$ monatomic, $5R/2$ diatomic. |

## 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 first law energy balances to analyze cyclic processes for heat engines and refrigerators, calculating efficiency and coefficient of performance, a common high-weight FRQ topic. Without solid command of first law sign conventions and work/energy calculations, you will not be able to correctly solve these problems. The first law builds on the ideal gas law you learned earlier in this unit, and provides the energy foundation for understanding entropy and the second law of thermodynamics, the next core topic in thermodynamics for AP Physics 2.

- [Second Law of Thermodynamics](https://www.owlsprep.com/study/ap-physics-2-u2-second-law-of-thermodynamics/)
- [Entropy](https://www.owlsprep.com/study/ap-physics-2-u2-entropy/)
- [Electric Force, Field, and Potential](https://www.owlsprep.com/study/ap-physics-2-u3-overview/)

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