AP Physics 2 • Thermodynamics

The Ideal Gas Law

A unifying equation that connects pressure, volume, temperature, and the number of particles in a gas.

An atom with wave and electric energy symbols.
Inside this lesson 9 sections 5 questions 9 figures

SECTION 1

Historical Context & Motivation

The behavior of gases captivated natural philosophers for centuries, driven by practical questions about steam engines, weather, and the very nature of matter. Before the ideal gas law emerged as a single unifying equation, a series of empirical gas laws were discovered independently, each isolating the relationship between two state variables while holding others constant. The synthesis of these separate laws into one elegant expression represents a triumph of scientific reasoning—combining observation, mathematical generalization, and the emerging kinetic theory of matter into a tool of remarkable predictive power.

1662 • Boyle's Law

Robert Boyle demonstrated that for a fixed amount of gas at constant temperature, pressure and volume are inversely proportional: PV = constant. This was one of the first quantitative laws in physical science.

1787 • Charles's Law

Jacques Charles found that the volume of a gas at constant pressure increases linearly with temperature. His unpublished work, later confirmed by Gay-Lussac, established the concept of absolute zero as the temperature at which volume extrapolates to zero.

1811 • Avogadro's Hypothesis

Amedeo Avogadro proposed that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. This revolutionary idea linked macroscopic volume to microscopic particle count.

1834 • Clapeyron's Synthesis

Émile Clapeyron combined Boyle's, Charles's, and Gay-Lussac's laws into a single equation of state: PV = nRT. This compact expression unified decades of empirical gas research into one framework.

1857 • Kinetic Theory Foundation

Rudolf Clausius provided a microscopic derivation of the ideal gas law from the kinetic theory of gases, showing that pressure arises from molecular collisions with container walls—connecting macroscopic thermodynamics to atomic-scale mechanics.

The central question these scientists addressed was deceptively simple: how do the macroscopic properties of a gas—pressure, volume, and temperature—relate to each other and to the amount of gas present? The ideal gas law provides the answer under conditions where intermolecular forces and molecular volumes are negligible, which serves as the starting point for all gas-phase thermodynamics in AP Physics 2.

SECTION 2

Core Principles & Definitions

An ideal gas is a theoretical model in which gas molecules are treated as point particles that undergo perfectly elastic collisions and exert no long-range forces on one another. While no real gas behaves ideally under all conditions, the model provides an excellent approximation at low pressures and high temperatures—precisely the regime explored in most AP Physics 2 problems. Understanding the assumptions and variables of this model is essential before applying the equation.

1

Point-Particle Assumption

Gas molecules occupy negligible volume compared to the container. The actual molecular volume is ignored, so the entire container volume is available for motion.
2

No Intermolecular Forces

Between collisions, molecules neither attract nor repel each other. The only interactions are perfectly elastic collisions with walls and other molecules, conserving total kinetic energy.
3

Thermal Equilibrium

Temperature reflects the average translational kinetic energy of molecules. At equilibrium, every molecule shares the same average KE regardless of mass, following Kavg = (3/2)kBT.
4

State Variables

The ideal gas law connects four macroscopic state variables: pressure (P), volume (V), absolute temperature (T), and number of moles (n) or number of molecules (N). Each must be in consistent SI units.

SECTION 3

Visualizing Gas Behavior

The diagram below illustrates how the three classic gas laws—Boyle's, Charles's, and Gay-Lussac's—are each special cases of the ideal gas law. Each panel shows a piston-cylinder system with the constrained variable indicated, demonstrating how fixing one quantity produces a predictable relationship between the other two.

Three Special Cases of PV = nRTBOYLE'S LAWConstant T & nV₁P₁→V₂P₂ > P₁P₁V₁ = P₂V₂↑P means ↓V(inverse relationship)CHARLES'S LAWConstant P & nV₁T₁→V₂T₂ > T₁V₁/T₁ = V₂/T₂↑T means ↑V(direct relationship)GAY-LUSSAC'S LAWConstant V & nP₁, T₁→P₂ > P₁, T₂ > T₁P₁/T₁ = P₂/T₂↑T means ↑P(direct relationship)All three are special cases of PV = nRT when one or more variables are held fixed.
Each panel shows a piston-cylinder system. Boyle's law (left) holds temperature constant and shows the inverse P–V relationship. Charles's law (center) holds pressure constant, revealing the direct V–T relationship. Gay-Lussac's law (right) holds volume constant, showing the direct P–T relationship. Faster-moving molecules (longer velocity arrows) indicate higher temperature.

Notice that in every case, one variable is held constant while the other two change. The ideal gas law, PV = nRT, encodes all three relationships simultaneously—fixing any one variable and the amount of gas recovers the corresponding historical law. This is why a single equation replaces three separate ones, and why recognizing which variables are held constant in a given AP problem is the essential first step in selecting the right approach.

SECTION 4

Mathematical Framework

The ideal gas law can be written in two equivalent forms depending on whether you count gas in moles or in individual molecules. Both forms appear on the AP Physics 2 exam, and you must be comfortable converting between them.

MOLAR FORM

PV=nRTPV = nRT

P = absolute pressure (Pa) • V = volume (m³) • n = number of moles (mol) • R = universal gas constant = 8.314 J/(mol·K) • T = absolute temperature (K)

MOLECULAR FORM

PV=NkBTPV = Nk_BT

N = number of molecules • kB = Boltzmann constant = 1.381 × 10⁻²³ J/K • Since N = nNA and R = NAkB, the two forms are algebraically equivalent.

COMBINED GAS LAW (FIXED n)

P1V1/T1=P2V2/T2P₁V₁ / T₁ = P₂V₂ / T₂

When the amount of gas is unchanged between two states, dividing PV = nRT for each state yields this ratio form. This is extremely useful for AP problems where a sealed gas undergoes a change in conditions.

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UNIT ALERT

The ideal gas law also connects to the microscopic picture through the average translational kinetic energy of a molecule. From kinetic theory, Kavg = (3/2)kBT. This means temperature is not merely a thermometer reading; it is a direct measure of the average kinetic energy per molecule. Doubling the absolute temperature doubles Kavg and increases the root-mean-square speed by a factor of √2.

KINETIC ENERGY – TEMPERATURE LINK

Kavg=(3/2)kBTK_avg = (3/2) k_B T

This equation holds for any ideal gas regardless of molecular mass. Heavier molecules move more slowly but carry the same average KE at the same temperature.

A transparent vertical cylinder contains a tinted gas region beneath a sealed metal piston and guided load. A copper heating jacket connects by two leads to a bench supply.For a fixed amount of ideal gas, compare equilibrium states using P₁V₁/T₁ = P₂V₂/T₂, with absolute pressure and temperature in kelvin. Endpoint states alone do not determine work.

Changing pressure and temperature together. For the same amount of gas, temperature falls from 293.15 K to 253.15 K while pressure falls from 101.3 to 50.7 kPa. Volume rises from about 48.1 to 83.0 L because the pressure decrease outweighs the temperature decrease.For the same amount of gas, temperature falls from 293.15 K to 253.15 K while pressure falls from 101.3 to 50.7 kPa. Volume rises from about 48.1 to 83.0 L because the pressure decrease outweighs the temperature decrease.

SECTION 5

PV Diagrams & Processes

A PV diagram (pressure vs. volume) is the standard graphical tool for analyzing gas processes in thermodynamics. Each point on a PV diagram represents a unique equilibrium state of the gas, and a curve connecting two states represents the path of a quasi-static process. For an ideal gas, the equation PV = nRT defines a family of isotherms—hyperbolic curves along which temperature remains constant. Different thermodynamic processes trace distinct paths on this diagram, and the area under any path equals the work done by the gas.

Ideal Gas Processes on a PV DiagramVolume (V)Pressure (P)Isobaric (const P)Isochoric(const V)Isothermal (const T)PV = constAdiabatic (Q = 0)PVᵞ = constABCDArea under any curve = Work done by gas (W = ∫P dV)
Four common thermodynamic processes on a PV diagram. The isobaric process is horizontal (constant P), the isochoric process is vertical (constant V), the isothermal process follows a hyperbola (PV = constant), and the adiabatic process (no heat exchange) curves more steeply than the isotherm. The area under each curve represents the work done by the gas.
ProcessHeld ConstantIdeal Gas ConsequenceWork Done by Gas
IsobaricPressure (P)V/T = constantW = PΔV
IsochoricVolume (V)P/T = constantW = 0
IsothermalTemperature (T)PV = constantW = nRT ln(V₂/V₁)
AdiabaticQ = 0 (no heat)PVγ = constantW = −ΔU

Summary of common ideal gas processes

SECTION 6

Worked Example

Worked example

  1. 1

    Step 1 — Read and Identify

    A sealed weather balloon contains 2.00 mol of helium at sea level where the temperature is 20.0 °C and the pressure is 1.013 × 10⁵ Pa. The balloon rises to an altitude where the pressure is 5.07 × 10⁴ Pa and the temperature is −20.0 °C. Find the volume of the balloon at altitude.
  2. 2

    Step 2 — List Known Quantities

    n = 2.00 mol (sealed, so constant). State 1: P₁ = 1.013 × 10⁵ Pa, T₁ = 20.0 + 273.15 = 293.15 K. State 2: P₂ = 5.07 × 10⁴ Pa, T₂ = −20.0 + 273.15 = 253.15 K. Unknown: V₂. We also need V₁ to use the combined gas law, or we can compute V₂ directly from PV = nRT.
    T₁ = 293.15 K, T₂ = 253.15 K
  3. 3

    Step 3 — Choose Equation

    Since we know n, R, P₂, and T₂, we can find V₂ directly from PV = nRT without needing V₁. Rearranging: V₂ = nRT₂ / P₂.
  4. 4

    Step 4 — Substitute and Calculate

    V₂ = (2.00 mol)(8.314 J/(mol·K))(253.15 K) / (5.07 × 10⁴ Pa). Numerator: 2.00 × 8.314 × 253.15 = 4209.7 J. Denominator: 5.07 × 10⁴ Pa. Since 1 J = 1 Pa·m³, we get V₂ = 4209.7 / 50700 = 0.08303 m³.
    V₂ ≈ 0.0830 m³ ≈ 83.0 L
  5. 5

    Step 5 — Verify with Combined Gas Law

    As a check, find V₁ = nRT₁/P₁ = (2.00)(8.314)(293.15)/(1.013 × 10⁵) = 0.04808 m³. Now apply P₁V₁/T₁ = P₂V₂/T₂, solving for V₂: V₂ = V₁ × (P₁/P₂) × (T₂/T₁) = 0.04808 × (1.013 × 10⁵ / 5.07 × 10⁴) × (253.15/293.15) = 0.04808 × 1.998 × 0.8636 = 0.0830 m³. ✓ Consistent result.
    Both methods confirm V₂ ≈ 0.0830 m³
  6. 6

    Step 6 — Physical Interpretation

    The balloon's volume at altitude is roughly 1.73 times its sea-level volume. The decrease in pressure tends to expand the balloon (Boyle's law effect), while the decrease in temperature tends to contract it (Charles's law effect). The net result is expansion because the pressure drops by about half, which dominates over the relatively modest temperature decrease of about 14%.

Flexible balloon expands when pressure drop outweighs temperature drop.With fixed gas amount, the pressure decrease outweighs cooling, increasing volume by about 1.725 and equivalent-sphere diameter by about 1.199.

2 mol:293.15 K/101.3 kPa→48.1 L;253.15 K/50.7 kPa→83.0 L; volume ratio 1.725 and equal-volume sphere diameter ratio 1.199.Computed equivalent-sphere size comparison; balloon scene is illustrative.

SECTION 7

Strengths & Limitations

The ideal gas law is remarkably powerful within its domain of validity, but it is a model with clear boundaries. Understanding when the approximation breaks down is just as important as knowing how to apply it—the AP exam frequently tests this conceptual understanding.

StrengthsLimitations
Accurate for most gases at atmospheric pressure and moderate-to-high temperatures.Fails at very high pressures where molecular volume becomes significant compared to container volume.
Simple algebra—no calculus required for most applications.Fails near the boiling point of a substance, where intermolecular attractions cause condensation.
Unifies Boyle's, Charles's, Gay-Lussac's, and Avogadro's laws into one equation.Cannot account for phase transitions (liquid ↔ gas ↔ solid).
Connects macroscopic variables (P, V, T) to microscopic energy via kinetic theory.Does not describe polar or highly interactive gases (e.g., H₂O, NH₃) well at moderate pressures.

When the ideal gas law works—and when it doesn't

★

Key takeaway

WHEN TO TRUST THE MODEL

SECTION 8

Connection to Real Gases & Advanced Theory

While AP Physics 2 focuses on the ideal gas law, it is instructive to see how the model extends to real gases. The most famous correction is the van der Waals equation, which modifies PV = nRT by adding two parameters: one that accounts for intermolecular attraction (reducing effective pressure) and one that accounts for finite molecular volume (reducing effective volume). This equation smoothly reduces to the ideal gas law when the corrections are small, confirming that PV = nRT is the correct low-density limit.

FeatureIdeal Gas (PV = nRT)van der Waals Gas
Molecular volumeZero (point particles)Finite (correction parameter b)
Intermolecular forcesNoneAttractive (correction parameter a)
Phase transitionsNot predictedPredicted (liquid–gas coexistence)
EquationPV = nRT(P + an²/V²)(V − nb) = nRT
AP Physics 2 relevanceCore equation—used in most problemsConceptual awareness only—not calculated on exam

Ideal vs. van der Waals gas models

The ideal gas law also connects forward to statistical mechanics, where PV = nRT is derived from first principles by computing the ensemble average of molecular momenta colliding with container walls. In AP Physics 2, you should be prepared to explain qualitatively why increasing temperature increases pressure (faster molecules strike walls harder and more frequently) and why decreasing volume increases pressure (same molecules hit walls more often in a smaller space). These microscopic explanations reinforce the macroscopic equation and are commonly assessed in free-response questions.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A sealed, rigid container holds an ideal gas at temperature T. If the absolute temperature is doubled, which of the following correctly describes the changes to the gas?
  • A. The pressure doubles and the average molecular kinetic energy doubles.
  • B. The pressure doubles and the average molecular speed doubles.
  • C. The pressure quadruples and the average molecular kinetic energy doubles.
  • D. The pressure doubles and the number of moles doubles.
Show answer (A)
A is correct. For a sealed rigid container (constant V and n), PV = nRT gives P ∝ T, so doubling T doubles P. The average translational KE is Kavg = (3/2)kBT, which also doubles. B is wrong because vrms ∝ √T, so speed increases by √2, not 2. C is wrong because pressure only doubles (P ∝ T, not T²). D is wrong because a sealed container cannot change the number of moles.

Question 2: BASIC CALCULATION

A 0.50 m³ tank contains nitrogen gas at a pressure of 2.0 × 10⁵ Pa and a temperature of 300 K. How many moles of nitrogen are in the tank? (R = 8.314 J/(mol·K))
  • A. 20 mol
  • B. 40 mol
  • C. 80 mol
  • D. 120 mol
Show answer (B)
B is correct. Using PV = nRT: n = PV/(RT) = (2.0 × 10⁵ Pa)(0.50 m³) / (8.314 J/(mol·K) × 300 K) = 1.0 × 10⁵ / 2494.2 ≈ 40.1 mol ≈ 40 mol. A (20) results from forgetting to use the given volume (using 0.25 m³ instead). C (80) results from a factor-of-two arithmetic error. D (120) is a distractor with no physical basis from the given values.

Question 3: INTERMEDIATE

An ideal gas initially at pressure P₁, volume V₁, and temperature T₁ undergoes an isothermal expansion to volume 3V₁, followed by an isochoric (constant-volume) heating that doubles its absolute temperature. What is the final pressure?
  • A. 2P₁/3
  • B. P₁/3
  • C. 6P₁
  • D. 2P₁
Show answer (A)
A is correct. Isothermal expansion: P₁V₁ = P₂(3V₁), so P₂ = P₁/3. Then isochoric heating doubles T at constant V = 3V₁: P/T = const, so P₃ = P₂ × (2T₁/T₁) = 2(P₁/3) = 2P₁/3. B (P₁/3) is the pressure after only the first step, ignoring the isochoric heating. C (6P₁) results from multiplying by 3 and 2 instead of dividing then multiplying. D (2P₁) ignores the volume change entirely.

Question 4: APPLIED

A student has access to a sealed syringe (whose volume can be measured by reading the graduations), a digital pressure sensor, a beaker of hot water, a beaker of ice water, and a thermometer. Design an experiment to verify that the pressure and temperature of a fixed amount of gas at constant volume obey Gay-Lussac's law (P/T = constant). (a) Describe a step-by-step procedure the student should follow. (b) State what quantities should be measured and how they should be plotted. (c) Explain how the resulting graph should look if Gay-Lussac's law is correct. (d) Identify one significant source of error and explain its effect on the results.
Show answer
(a) Procedure: Fix the syringe plunger in place with a clamp so the volume cannot change. Connect the pressure sensor to the syringe tip. Record the initial pressure and temperature at room conditions. Immerse the sealed syringe in ice water and wait 2–3 minutes for thermal equilibrium; record P and T. Repeat by immersing in progressively warmer water baths (e.g., 20°C, 40°C, 60°C, 80°C), recording P and T each time. (b) Measure pressure in Pa and temperature in K (convert from °C by adding 273.15). Plot P (vertical axis) vs. T (horizontal axis). (c) If Gay-Lussac's law holds, the graph of P vs. T should be a straight line through the origin (or extrapolating to zero pressure at 0 K), confirming P ∝ T. The slope equals nR/V. (d) A significant source of error is the gas inside the syringe not reaching thermal equilibrium with the water bath. If the gas is measured before equilibrium, the recorded temperature will be that of the water while the actual gas temperature is different, causing data points to deviate from the expected linear trend—typically underestimating pressure at higher bath temperatures and overestimating it at lower bath temperatures.

Question 5: CRITICAL THINKING

Two identical rigid containers, A and B, each hold an ideal gas at the same initial pressure P₀ and temperature T₀. Container A holds helium (monatomic, molar mass 4 g/mol) and container B holds nitrogen (diatomic, molar mass 28 g/mol). Both containers are heated so that each gas absorbs the same amount of thermal energy Q. (a) Determine which container reaches a higher final temperature. Justify your answer using the ideal gas law and the concept of degrees of freedom. (b) Determine which container reaches a higher final pressure. Justify your answer. (c) A student claims that since both gases started at the same pressure and temperature in identical containers, they must have the same number of moles. Evaluate this claim. (d) Compare the final rms speeds of helium and nitrogen molecules in their respective containers. Justify your reasoning.
Show answer
(a) For a constant-volume process, Q = nCvΔT. Monatomic helium has 3 translational degrees of freedom, giving Cv = (3/2)R. Diatomic nitrogen has 5 degrees of freedom (3 translational + 2 rotational), giving Cv = (5/2)R. From part (c), both containers contain the same number of moles n. Solving for the temperature change: ΔT = Q/(nCv). Since Cv is smaller for helium, the same energy Q produces a larger ΔT. Specifically, ΔT_He/ΔTNT_N₂ = (5/2)R / (3/2)R = 5/3. Therefore helium reaches a higher final temperature, with its temperature increase being 5/3 times that of nitrogen. (b) Both containers are rigid, so volume V is constant. From the ideal gas law, P = nRT/V. Since n and V are the same for both containers, pressure is directly proportional to temperature. Because helium reaches a higher final temperature (from part a), container A (helium) also reaches a higher final pressure. (c) The student's claim is correct. From PV = nRT, solving for n gives n = PV/(RT). Since both containers are identical (same V) and both gases start at the same pressure P₀ and temperature T₀, the number of moles n = P₀V/(RT₀) is the same for both gases. The ideal gas law depends only on the number of moles, not on the molecular identity or mass of the gas. (d) The rms speed is given by v_rms = √(3k_BT/m), where m is the mass of one molecule. Helium molecules have a smaller mass (4 u vs. 28 u for N₂), which alone would make helium's rms speed larger by a factor of √(28/4) = √7 ≈ 2.65 at the same temperature. Additionally, from part (a), helium reaches a higher final temperature than nitrogen. Both factors — lower molecular mass and higher final temperature — act in the same direction, so the final rms speed of helium molecules is substantially greater than that of nitrogen molecules. The combined ratio is v_rms,He/v_rms,N₂ = √(3k_BT_He/m_He) / √(3k_BT_N₂/mNm_N₂) = √[(T_He/TNT_N₂)(mNm_N₂/m_He)], which exceeds √7 because T_He > TNT_N₂.

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