AP PHYSICS 2: ALGEBRA-BASED • THERMODYNAMICS

The First Law of Thermodynamics

Energy cannot be created or destroyed — only transferred between heat, work, and internal energy.

Historical Context & Motivation

The First Law of Thermodynamics emerged from a centuries-long effort to understand the relationship between heat and mechanical work. Before the mid-nineteenth century, scientists widely accepted the caloric theory, which treated heat as an invisible, weightless fluid that flowed between objects. This view seemed intuitive — hot objects 'poured' caloric into cold ones — but it could not explain why rubbing your hands together generates warmth without any apparent caloric source. The resolution required an entirely new framework: the recognition that heat and work are both forms of energy transfer, governed by a universal conservation principle.

1798
Rumford's Cannon-Boring Experiment
Count Rumford observed that boring cannon barrels produced seemingly inexhaustible heat, challenging the caloric theory by suggesting heat arises from mechanical motion.
1843
Joule's Paddle-Wheel Experiment
James Prescott Joule measured the temperature rise of water stirred by falling weights, establishing a precise mechanical equivalent of heat (≈ 4.18 J per calorie).
1847
Helmholtz Formalizes Energy Conservation
Hermann von Helmholtz published a mathematical formulation of energy conservation encompassing mechanical, thermal, and electrical phenomena.
1850
Clausius States the First Law
Rudolf Clausius formally stated the First Law, distinguishing clearly between internal energy, heat, and work — the framework still used in modern physics.

The central question these scientists addressed was deceptively simple: when energy enters or leaves a system, where does it go? The First Law provides the bookkeeping rule — the change in a system's internal energy equals the net energy added as heat minus the energy lost as work done by the system. This principle is the thermodynamic analog of conservation of energy, applied specifically to thermal processes, and it underpins every engine, refrigerator, and biological metabolism on Earth.

Core Principles & Definitions

To apply the First Law correctly, you must distinguish three quantities that are easily confused: internal energy, heat, and work. Each plays a distinct role in the energy budget of a thermodynamic system — the portion of the universe we choose to analyze. Everything outside the system constitutes the surroundings. The First Law constrains how energy crosses the boundary between them.

1

Internal Energy (U)

The total microscopic kinetic and potential energy of all particles in the system. It is a state function — it depends only on the current state (T, P, V), not on how the system reached that state.
2

Heat (Q)

Energy transferred between system and surroundings due to a temperature difference. Heat is a process quantity, not a property the system 'has.' Positive Q means energy flows into the system.
3

Work (W)

Energy transferred when the system expands or compresses against an external pressure. In the AP convention, positive W means work done by the system on the surroundings.
4

State vs. Process Quantities

Internal energy (U) is a state function: ΔU depends only on initial and final states. Heat (Q) and work (W) are path-dependent process quantities — their values depend on how the process occurs.
KEY TAKEAWAY
KEY TAKEAWAY

Visual Explanation

Energy Flow Diagram for a Thermodynamic System

The diagram shows heat Q entering the system from the surroundings (pink arrow) and work W done by the system on the environment (amber arrow). The net effect determines the change in internal energy ΔU. Note the sign convention box at the bottom — mastering signs is crucial for AP success.

The diagram above captures the essence of the First Law as an energy balance. Every joule of heat that enters the system either increases the system's internal energy or is expended as work done by the system on its surroundings — no energy vanishes and none appears from nothing. When the system is a gas in a cylinder with a movable piston, the work is expansion work: the gas pushes the piston outward, transferring energy mechanically. If more heat enters than work is done, the internal energy rises and the gas temperature typically increases. If the gas does more work than the heat it absorbs, internal energy decreases and the gas cools.

Mathematical Framework

FIRST LAW OF THERMODYNAMICS
ΔU = Q − W
ΔU = change in internal energy (J), Q = heat added to the system (J), W = work done by the system (J). Positive Q means energy flows in; positive W means the system expands against external pressure.

Be careful with sign conventions — some textbooks define W as work done on the system, which changes the equation to ΔU = Q + W. The AP Physics 2 exam uses the convention ΔU = Q − W, where W is work done by the system. Always check which convention a problem uses before substituting values.

WORK DONE BY A GAS AT CONSTANT PRESSURE
W = PΔV
P = external pressure (Pa), ΔV = Vfinal − Vinitial (m³). If the gas expands (ΔV > 0), the work is positive; if compressed (ΔV < 0), work is negative (work done on the system).
WORK FROM A PV DIAGRAM
W = area under the curve on a P-V diagram
For any process, the work equals the area beneath the process path on a pressure-versus-volume graph. This is critical because work is path-dependent: different paths between the same endpoints yield different W values.
AP Exam Tip

Thermodynamic Processes in Detail

The First Law applies universally, but its application simplifies dramatically under special constraints. AP Physics 2 focuses on four idealized processes for an ideal gas, each holding one thermodynamic variable constant. Understanding how Q, W, and ΔU behave in each case is essential for both multiple-choice and free-response success.

The PV diagram shows the four fundamental processes. The isochoric (red) process is a vertical line (no volume change, so no work). The isobaric (cyan) process is horizontal (constant pressure). The isothermal (amber) curve follows a hyperbola (PV = constant). The adiabatic (green) curve is steeper than isothermal because no heat enters to buffer the pressure drop.
Summary of the First Law applied to each process type
ProcessConstraintQWΔU
IsobaricConstant PQ = ΔU + PΔVW = PΔVΔU = Q − PΔV
IsochoricConstant VQ = ΔUW = 0ΔU = Q
IsothermalConstant TQ = WW = area under PV curveΔU = 0
AdiabaticQ = 0Q = 0W = −ΔUΔU = −W

For an ideal gas, internal energy depends only on temperature: U = (3/2)nRT for a monatomic gas and U = (5/2)nRT for a diatomic gas at moderate temperatures. This means that in any isothermal process involving an ideal gas, ΔU = 0 regardless of how pressure or volume change — a powerful simplification that frequently appears on the AP exam.

Worked Example

1
Step 1 — Identify Given InformationA monatomic ideal gas at constant pressure P = 1.5 × 105 Pa expands from Vi = 0.020 m³ to Vf = 0.050 m³. During the expansion, 8000 J of heat is added to the gas. Find ΔU and the work done by the gas.
2
Step 2 — Calculate Work Done by the GasSince the process is isobaric, W = PΔV = P(Vf − Vi) = (1.5 × 10⁵ Pa)(0.050 m³ − 0.020 m³) = (1.5 × 10⁵)(0.030) J.
W = 4500 J
3
Step 3 — Apply the First LawΔU = Q − W = 8000 J − 4500 J.
ΔU = 3500 J
4
Step 4 — Interpret the ResultOf the 8000 J of heat added, 4500 J went into doing expansion work against the external pressure, and the remaining 3500 J increased the gas's internal energy. Since the gas is monatomic ideal, this internal energy increase corresponds to a temperature rise: ΔU = (3/2)nRΔT, so the gas got hotter. This is consistent with the general behavior of an isobaric expansion — the gas both expands and warms when heat is added at constant pressure.

Common Pitfalls & Comparisons

Common First Law mistakes on the AP exam
Common MistakeWhy It's WrongCorrect Approach
Confusing Q and T — 'adding heat always raises temperature'In an isothermal expansion, Q > 0 but ΔT = 0. All heat converts to work.Heat (Q) is energy in transit; temperature change depends on whether ΔU ≠ 0.
Wrong sign on W — using ΔU = Q + W when the problem uses the 'by' conventionMixing sign conventions flips the sign of W, giving the wrong ΔU.Identify the convention first. AP uses W = work BY system → ΔU = Q − W.
Saying a system 'has' heat or 'contains' heatHeat is a process, not a state property. A system has internal energy, not heat.Say 'heat was transferred to the system' — Q describes energy in transit.
Assuming ΔU = 0 for every cyclic stepΔU = 0 only for the entire cycle, not for individual steps within it.Apply ΔU = Q − W to each step independently; sum over the cycle gives ΔU_net = 0.
KEY TAKEAWAY
SIGN CONVENTION REMINDER

Connection to Advanced Topics

The First Law is the foundation upon which the rest of thermodynamics is built. It tells us how much energy is conserved but says nothing about which direction processes spontaneously go. That question belongs to the Second Law of Thermodynamics, which introduces entropy — a measure of energy dispersal that always increases in an isolated system. Together, the first two laws determine both the energy budget and the spontaneity of any thermodynamic process.

First Law vs. Second Law comparison
First LawSecond Law
Energy is conserved: ΔU = Q − WEntropy of an isolated system never decreases: ΔS ≥ 0
Tells us the quantity of energy availableTells us the direction and quality of energy flow
Does not forbid a cold object spontaneously heating a hot oneForbids spontaneous heat flow from cold to hot
Cannot determine the efficiency limit of a heat engineSets the Carnot efficiency as the maximum: e = 1 − T_C / T_H

In AP Physics 2, you will use the First Law to analyze heat engines and refrigerators quantitatively. The efficiency of a heat engine, e = Wnet / QH, is derived directly from the First Law applied over a complete cycle (where ΔU = 0, so Qnet = Wnet). Understanding this connection will be essential when you encounter Carnot cycles and entropy in subsequent lessons.

Practice Problems

1
An ideal gas undergoes a free expansion into a vacuum (no external pressure). Which of the following correctly describes Q, W, and ΔU for this process?
2
A gas absorbs 5000 J of heat while doing 2000 J of work on its surroundings. What is the change in internal energy of the gas?
3
An ideal gas undergoes an isothermal compression at 400 K. During the process, 1200 J of work is done on the gas. How much heat is transferred, and in which direction?
PROBLEM 4APPLIED
A student wants to experimentally verify the First Law of Thermodynamics for an isobaric process. The student has an insulated cylinder with a frictionless, movable piston, a thermometer, a ruler, a known mass to place on the piston, and a hot plate. (a) Describe an experimental procedure the student could use to measure Q, W, and ΔU for an isobaric expansion. (b) What measurements should the student record? (c) Describe how the student would use the measurements to calculate Q, W, and ΔU, and verify the First Law. (d) Identify one source of experimental error and explain how it would affect the results.
PROBLEM 5CRITICAL THINKING
A monatomic ideal gas undergoes a two-step process from state A to state C. In step 1 (A → B), the gas expands isobarically at P = 2.0 × 10⁵ Pa from V_A = 0.010 m³ to V_B = 0.025 m³. In step 2 (B → C), the gas is cooled at constant volume until its pressure drops to 1.0 × 10⁵ Pa. (a) Calculate the work done by the gas for the entire process A → C. (b) If the gas starts at temperature T_A = 240 K, find T_B and T_C using the ideal gas law. (c) Calculate ΔU for the entire process A → C. (d) Determine the total heat Q transferred during the process A → C and state whether the net heat flow is into or out of the gas.
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