Apply Gas Laws — Apply B.3 Gas laws in problem-solving and explanations - IB Physics

SECTION 1

Historical Context & Motivation

For centuries, people observed that air had weight, that heated gases expanded, and that compressing a gas made it push back harder. Yet it was not until the seventeenth and eighteenth centuries that scientists began to express these observations as precise, quantitative laws. The journey from curious observation to mathematical relationship is one of the great success stories in physics, and it laid the groundwork for modern thermodynamics and even the kinetic molecular theory of matter.

1662 • Boyle's Law

Robert Boyle published experiments showing that, for a fixed amount of gas at constant temperature, pressure and volume are inversely proportional. He trapped air in a J-shaped tube and measured how adding mercury changed the trapped volume.

1787 • Charles's Law

Jacques Charles discovered that gas volume increases linearly with temperature when pressure is held constant. His unpublished work was later formalized by Joseph Gay-Lussac, establishing the concept of absolute zero.

1802 • Gay-Lussac's Law

Gay-Lussac showed that, at constant volume, the pressure of a gas is directly proportional to its absolute temperature. This law is essential for understanding sealed containers like aerosol cans.

1811 • Avogadro's Hypothesis

Amedeo Avogadro proposed that equal volumes of gases at the same temperature and pressure contain equal numbers of particles. This idea introduced the mole as a counting unit for gas problems.

1834 • The Ideal Gas Law

Émile Clapeyron combined Boyle's, Charles's, and Avogadro's results into a single equation, PV = nRT, unifying all the individual gas laws into one powerful relationship.

The central question these scientists addressed was deceptively simple: How exactly do the measurable properties of a gas—pressure, volume, temperature, and amount—depend on one another? Answering that question gave us the tools to design engines, predict weather, inflate airbags, and understand stellar atmospheres. In IB Physics topic B.3, you will apply these laws to solve quantitative problems and explain everyday phenomena.

SECTION 2

Core Principles & Definitions

Before diving into calculations, you need to be comfortable with the key ideas that underpin all gas law problems. An ideal gas is a theoretical model in which gas particles have negligible volume and exert no intermolecular forces on each other. Real gases approximate ideal behavior at low pressures and high temperatures—conditions where particles are far apart and moving fast. All IB B.3 problems treat gases as ideal unless stated otherwise.

Pressure (P)

The force exerted per unit area by gas particles colliding with the walls of their container. SI unit: pascal (Pa). 1 atm = 1.013 × 10⁵ Pa.

Volume (V)

The three-dimensional space occupied by the gas. SI unit: cubic metre (m³). 1 litre = 1 × 10⁻³ m³.

Temperature (T)

A measure of the average kinetic energy of gas particles. Gas laws require absolute temperature in kelvin (K). Convert from °C by adding 273.

Amount (n)

The quantity of gas measured in moles (mol). One mole contains 6.02 × 10²³ particles (Avogadro's number, N_A).

Ideal Gas Constant (R)

The proportionality constant that links all four state variables: R = 8.314 J mol⁻¹ K⁻¹. Given in the IB data booklet.

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KEY TAKEAWAY

Think of gas particles like a crowd of bouncy balls inside a closed room. If you heat the room (raise T), the balls bounce faster and hit the walls harder (raise P) or push the walls outward (raise V). If you stuff more balls inside (raise n), collisions increase too. The gas laws simply put numbers on these intuitive relationships.

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IB Exam Tip

Always convert temperature to kelvin before substituting into any gas law equation. Forgetting this is the single most common mistake on IB gas law questions. If a problem gives 27 °C, use T = 27 + 273 = 300 K.

SECTION 3

Visualizing the Gas Laws

Each individual gas law describes what happens when you change one variable while holding the others constant. The three classic laws—Boyle's, Charles's, and Gay-Lussac's—each produce a distinctive graph shape. Understanding these shapes helps you predict gas behavior and verify your calculations.

Left: Boyle's Law shows a hyperbolic curve because P and V are inversely proportional. Centre: Charles's Law produces a straight line through the origin (0 K). Right: Gay-Lussac's Law also yields a straight line through 0 K. The dashed extensions remind us that gases liquefy before reaching absolute zero.

Notice that both Charles's Law and Gay-Lussac's Law produce straight lines that, if extended backward, pass through absolute zero (0 K, or −273 °C). This is not a coincidence—it is a direct consequence of the fact that the average kinetic energy of particles is proportional to absolute temperature. At 0 K, particle motion would theoretically cease, meaning zero pressure and zero volume. Boyle's Law, on the other hand, produces a hyperbola: doubling the pressure halves the volume, tripling the pressure reduces it to one-third, and so on.

SECTION 4

Mathematical Framework

The individual gas laws can all be derived from one master equation—the ideal gas law. In the IB data booklet, you will find it along with the Boltzmann form. Both are essential for B.3 problems.

IDEAL GAS LAW (MOLAR FORM)

PV = nRT

P = pressure (Pa), V = volume (m³), n = amount of substance (mol), R = 8.314 J mol⁻¹ K⁻¹, T = absolute temperature (K).

IDEAL GAS LAW (BOLTZMANN FORM)

PV = NkᵦT

N = total number of particles, k_B = Boltzmann constant = 1.38 × 10⁻²³ J K⁻¹. Use this form when counting individual molecules rather than moles.

COMBINED GAS LAW (COMPARING TWO STATES)

P₁V₁ / T₁ = P₂V₂ / T₂

Valid when the amount of gas (n) stays constant. Subscript 1 = initial state; subscript 2 = final state. If one variable is constant, cross it out to recover the individual law (Boyle's, Charles's, or Gay-Lussac's).

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Recovering Individual Laws

Start with P₁V₁/T₁ = P₂V₂/T₂. If temperature is constant (T₁ = T₂), the T's cancel and you get P₁V₁ = P₂V₂ (Boyle). If pressure is constant (P₁ = P₂), you get V₁/T₁ = V₂/T₂ (Charles). If volume is constant, you get P₁/T₁ = P₂/T₂ (Gay-Lussac). One equation, three laws.

AVERAGE KINETIC ENERGY PER PARTICLE

Eₖ = (3/2) kᵦT

This equation links the microscopic world (particle energy) to the macroscopic variable (temperature). It explains why the gas laws work: higher T means faster particles, which means harder and more frequent collisions with container walls, which means higher P.

SECTION 5

Gas Processes & P–V Diagrams

In IB Physics, you are expected to interpret and sketch pressure–volume (P–V) diagrams that illustrate how a gas moves between states. Each type of process—isothermal, isobaric, and isovolumetric—appears as a distinct curve or line on the diagram. Being able to read these diagrams is crucial for both exam questions and real-world applications like heat engines.

On a P–V diagram, an isothermal process (constant T) follows a curved hyperbola from A to B. An isobaric process (constant P) is a horizontal line from A to C. An isovolumetric process (constant V) is a vertical line from A to D.

Process	Constant Variable	Law Applied	P–V Shape
Isothermal	Temperature (T)	Boyle's: P₁V₁ = P₂V₂	Hyperbola
Isobaric	Pressure (P)	Charles's: V₁/T₁ = V₂/T₂	Horizontal line
Isovolumetric	Volume (V)	Gay-Lussac's: P₁/T₁ = P₂/T₂	Vertical line

Summary of gas processes and their P–V diagram representations

SECTION 6

Worked Example

Let's work through a typical IB-style problem that requires you to apply the combined gas law and convert units carefully.

Compressed Gas in a Scuba Tank

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Step 1 — Read the Problem

A scuba diver's tank holds 12.0 litres of air at a pressure of 2.10 × 10⁷ Pa and a temperature of 27 °C. The diver descends into cold water, and the tank temperature drops to 7 °C while the volume remains constant. What is the new pressure inside the tank?
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Step 2 — Identify the Given Values and Process

V₁ = V₂ = 12.0 L (constant volume, so this is an isovolumetric process). P₁ = 2.10 × 10⁷ Pa. T₁ = 27 °C. T₂ = 7 °C. We need P₂.
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Step 3 — Convert Temperatures to Kelvin

T₁ = 27 + 273 = 300 K. T₂ = 7 + 273 = 280 K. Always convert before substituting.

T₁ = 300 K, T₂ = 280 K
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Step 4 — Choose the Correct Law

Since volume is constant, use Gay-Lussac's Law: P₁ / T₁ = P₂ / T₂. Rearranging for P₂: P₂ = P₁ × (T₂ / T₁).
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Step 5 — Substitute and Solve

P₂ = (2.10 × 10⁷ Pa) × (280 K / 300 K) = (2.10 × 10⁷) × 0.9333 = 1.96 × 10⁷ Pa.

P₂ ≈ 1.96 × 10⁷ Pa
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Step 6 — Check Reasonableness

The temperature dropped (from 300 K to 280 K), so the pressure should decrease. Our answer is lower than the original 2.10 × 10⁷ Pa, which makes physical sense. The ratio 280/300 ≈ 0.93 tells us the pressure dropped by about 7%, a modest change matching a modest temperature change.

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PROBLEM-SOLVING STRATEGY

For every gas law problem, follow these steps: (1) Identify which variables change and which stay constant. (2) Convert all temperatures to kelvin. (3) Choose the correct law or use the combined gas law. (4) Rearrange the equation for the unknown variable. (5) Substitute values and calculate. (6) Check that the direction of change makes physical sense.

SECTION 7

Ideal vs. Real Gases: Strengths & Limitations

The ideal gas model is remarkably useful, but it has boundaries. Understanding where the model breaks down is important both for IB exams and for real engineering applications. The table below compares the assumptions of the ideal gas model with the behavior of real gases.

Feature	Ideal Gas Assumption	Real Gas Behavior
Particle volume	Particles have zero volume (point masses).	Particles have finite volume; matters at high pressures when particles are packed closely.
Intermolecular forces	No attractive or repulsive forces between particles.	Van der Waals forces exist; significant at low temperatures when particles move slowly.
Collisions	All collisions are perfectly elastic (no energy loss).	Nearly elastic for most gases, but energy can be transferred to rotational/vibrational modes.
Best accuracy	Low pressure, high temperature.	Deviations grow at high pressure and/or low temperature (close to liquefaction).
Phase changes	Cannot predict condensation or boiling.	Real gases liquefy when cooled or compressed sufficiently.

Ideal gas assumptions versus real gas behavior

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WHEN DOES IDEAL WORK?

Think of the ideal gas model like a simplified city map that shows only main roads. For most trips (low pressure, high temperature), this map gets you where you need to go. But if you need to navigate narrow side streets (high pressure, low temperature, gases near liquefaction), you need a more detailed map, such as the van der Waals equation. For IB B.3, the simple map is all you need.

SECTION 8

Connection to Kinetic Molecular Theory & Beyond

The gas laws you have learned are empirical—they describe what gases do. The kinetic molecular theory (KMT) explains why gases behave this way, by modelling them as large numbers of tiny particles in constant random motion. In IB Physics, Topic B.3 connects macroscopic measurements (P, V, T) to microscopic quantities (particle speed, kinetic energy) through the Boltzmann constant and the equation E_k = (3/2)k_BT.

Concept	Gas Laws (Macroscopic)	Kinetic Theory (Microscopic)
Pressure	Force per unit area on container walls	Rate of momentum transfer from particle collisions with walls
Temperature	Measured by a thermometer in kelvin	Proportional to the average translational kinetic energy of particles
Volume increase at constant P	V increases as T increases (Charles's Law)	Faster particles push walls outward to maintain the same collision rate per unit area
Equation	PV = nRT	PV = NkᵦT and Eₖ = (3/2)kᵦT

Macroscopic gas laws versus microscopic kinetic theory

Looking ahead, if you study physics at university level, you will encounter the Maxwell–Boltzmann distribution, which describes the range of speeds particles actually have at a given temperature. You will also meet the van der Waals equation and the virial expansion, which correct the ideal gas equation for real-gas effects. For now, mastering PV = nRT and its component laws gives you a solid foundation for all of these more advanced treatments.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A sealed, rigid metal container holds a gas at room temperature. The container is then placed in an oven and heated. Explain, using the kinetic molecular theory and an appropriate gas law, what happens to the pressure inside the container and why.

Show answer

Because the container is rigid, the volume is constant (isovolumetric process). According to Gay-Lussac's Law, P₁/T₁ = P₂/T₂, so pressure is directly proportional to absolute temperature. As the temperature increases, the gas particles gain kinetic energy and move faster. They strike the container walls more frequently and with greater force, so the pressure rises. The key microscopic explanation is that temperature is proportional to average kinetic energy (Eₖ = 3/2 kᵦT), and faster-moving particles exert more force per collision.

Question 2: BASIC CALCULATION

A gas occupies 4.00 × 10⁻³ m³ at a pressure of 1.00 × 10⁵ Pa. If the gas is compressed isothermally to a volume of 1.00 × 10⁻³ m³, what is the new pressure?

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This is an isothermal process (constant T), so we use Boyle's Law: P₁V₁ = P₂V₂. Rearranging gives P₂ = P₁V₁/V₂ = (1.00 × 10⁵ Pa)(4.00 × 10⁻³ m³) / (1.00 × 10⁻³ m³) = 4.00 × 10⁵ Pa. The volume decreased by a factor of 4, so the pressure increased by a factor of 4. This confirms the inverse proportionality between P and V at constant temperature.

Question 3: INTERMEDIATE

A weather balloon is filled with 0.500 mol of helium at ground level where the temperature is 20.0 °C and the pressure is 1.01 × 10⁵ Pa. Calculate the volume of the balloon. Then determine the new volume when the balloon rises to an altitude where the temperature is −40.0 °C and the pressure is 3.50 × 10⁴ Pa.

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First, convert temperatures: T₁ = 20.0 + 273 = 293 K, T₂ = −40.0 + 273 = 233 K. Use PV = nRT for the initial volume: V₁ = nRT₁/P₁ = (0.500)(8.314)(293) / (1.01 × 10⁵) = 1218 / (1.01 × 10⁵) ≈ 0.01206 m³ ≈ 12.1 L. For the new volume, use the combined gas law (n constant): V₂ = V₁ × (P₁/P₂) × (T₂/T₁) = 0.01206 × (1.01 × 10⁵ / 3.50 × 10⁴) × (233/293) = 0.01206 × 2.886 × 0.7952 ≈ 0.0277 m³ ≈ 27.7 L. The volume more than doubles because the large drop in pressure outweighs the decrease in temperature.

Question 4: APPLIED

A car tyre has an internal volume of 0.0140 m³ and is inflated to a gauge pressure of 2.40 × 10⁵ Pa on a morning when the temperature is 10.0 °C. After a long highway drive, the tyre temperature rises to 55.0 °C. Assuming the volume changes negligibly, calculate the new gauge pressure. (Atmospheric pressure = 1.01 × 10⁵ Pa.)

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Gauge pressure does not include atmospheric pressure, so first convert to absolute pressure: P₁ = 2.40 × 10⁵ + 1.01 × 10⁵ = 3.41 × 10⁵ Pa. Convert temperatures: T₁ = 10.0 + 273 = 283 K, T₂ = 55.0 + 273 = 328 K. With constant volume, apply Gay-Lussac's Law: P₂ = P₁ × (T₂/T₁) = 3.41 × 10⁵ × (328/283) = 3.41 × 10⁵ × 1.159 ≈ 3.95 × 10⁵ Pa (absolute). Convert back to gauge: gauge P₂ = 3.95 × 10⁵ − 1.01 × 10⁵ = 2.94 × 10⁵ Pa. The gauge pressure rises from 240 kPa to about 294 kPa—a roughly 22% increase that explains why manufacturers recommend checking tyre pressure when tyres are cold.

Question 5: CRITICAL THINKING

Two identical sealed flasks, each of volume V, are connected by a thin tube of negligible volume. Initially the system contains an ideal gas at temperature T₀ and pressure P₀. Flask A is then heated to 2T₀ while flask B is maintained at T₀. The total amount of gas in the system remains constant. Derive an expression for the new equilibrium pressure P in terms of P₀.

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Initially the total number of moles is n = P₀(2V)/(RT₀) = 2P₀V/(RT₀). After heating, the system reaches a uniform pressure P (connected flasks). In flask A at temperature 2T₀, the number of moles is

n_A

= PV/(R × 2T₀). In flask B at temperature T₀,

n_B

= PV/(RT₀). Conservation of moles gives

n_A

n_B

= n: PV/(2RT₀) + PV/(RT₀) = 2P₀V/(RT₀). Factor out V/(RT₀): P/2 + P = 2P₀, so 3P/2 = 2P₀, giving P = 4P₀/3. The pressure increases by a factor of 4/3 because gas migrates from the hotter flask into the cooler one, but the average temperature of the gas has increased overall. This type of problem tests your ability to apply PV = nRT to multiple connected regions while conserving total moles.

SUMMARY

Lesson Summary

Summary

The ideal gas law PV = nRT unifies three classical relationships: Boyle's Law (P ∝ 1/V at constant T), Charles's Law (V ∝ T at constant P), and Gay-Lussac's Law (P ∝ T at constant V). The combined gas law P₁V₁/T₁ = P₂V₂/T₂ lets you compare two states of a fixed amount of gas. All temperatures must be in kelvin, and units must be consistent (Pa and m³ when using R = 8.314 J mol⁻¹ K⁻¹).

On the microscopic side, the kinetic molecular theory explains gas behavior by linking temperature to average kinetic energy via Eₖ = (3/2)kᵦT. The Boltzmann form PV = NkᵦT connects particle count to macroscopic state variables. When solving problems, always identify the constant variable, select the appropriate law, convert to kelvin, and verify that the direction of change makes physical sense. The ideal gas model works best at low pressure and high temperature; deviations occur near liquefaction.