Gas Laws

The gas laws describe the relationships between pressure (p), volume (V), and temperature (T) for a fixed mass of gas. These laws were discovered experimentally before the kinetic theory provided a molecular explanation. They are fundamental to AQA A-Level Physics and appear regularly in both calculation and explanation questions.

Spec mapping (AQA 7408 §3.6.2.2 — Gas laws): This lesson covers the three empirical gas laws — Boyle's law (pV = constant at constant T and n), Charles's law (V/T = constant at constant p and n), and the pressure law (p/T = constant at constant V and n). Candidates must be able to interpret graphs of each law, perform calculations using the combined gas law p₁V₁/T₁ = p₂V₂/T₂, and explain why all three equations require absolute temperature in kelvin. The kinetic-molecular interpretation of each law is also assessed. (Refer to the official AQA specification document for exact wording.)

Synoptic links: The empirical gas laws are the macroscopic face of the molecular kinetic theory (§3.6.2.3): the same relationships emerge from the microscopic derivation pV = (1/3)Nm〈c²〉. They are also the precursors to the ideal gas equation pV = nRT (§3.6.2.2), which generalises all three. Experimentally they invoke uncertainty propagation (§3.1.3) when slopes are extracted from p-vs-V or p-vs-T graphs. Charles's law and the pressure law together define absolute zero — the kelvin scale's anchor point — which connects forward to §3.6.2.3 (mean kinetic energy → 0 as T → 0). Synoptically they reappear in atmospheric physics (lapse rates), thermodynamic cycles, and any A-Level question on gas heating or compression.

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Temperature Scales

Before studying the gas laws, it is essential to understand that all gas law equations require absolute temperature measured in kelvin (K).

T (K) = θ (°C) + 273.15

For most A-Level calculations, 273 is sufficiently accurate.

Absolute zero (0 K = −273.15 °C) is the lowest possible temperature. At absolute zero:

• Molecules have minimum internal energy.
• All thermal motion ceases (molecules have zero kinetic energy).
• The pressure of an ideal gas would be zero.
• The volume of an ideal gas would be zero.

Absolute zero cannot actually be reached in practice (Third Law of Thermodynamics), but temperatures within a fraction of a kelvin have been achieved in laboratories.

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Boyle's Law

Boyle's Law: For a fixed mass of gas at constant temperature, the pressure is inversely proportional to the volume.

pV = constant (at constant T) p₁V₁ = p₂V₂

Molecular Explanation

At constant temperature, the molecules have the same average kinetic energy. If the volume is halved:

• The molecules are in a smaller space, so they hit the walls more frequently.
• The molecules travel a shorter distance between collisions with the walls.
• The rate of change of momentum at the walls increases.
• Therefore the pressure doubles.

Graphical Representations

• p vs V: A rectangular hyperbola (curved line) — pressure decreases as volume increases. Each curve (isotherm) corresponds to a different temperature; higher-temperature isotherms are further from the origin.
• p vs 1/V: A straight line through the origin with gradient = pV = constant.
• pV vs p: A horizontal straight line (since pV = constant).

Worked Example — Boyle's Law

Question: A gas syringe contains 80 cm³ of gas at a pressure of 1.0 × 10⁵ Pa. The gas is slowly compressed at constant temperature to a volume of 20 cm³. Calculate the new pressure.

Solution:

Using p₁V₁ = p₂V₂:

p₂ = p₁V₁ / V₂ = (1.0 × 10⁵ × 80) / 20 = 4.0 × 10⁵ Pa

The pressure quadruples when the volume is reduced to one quarter, as expected from the inverse proportionality.

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Charles's Law

Charles's Law: For a fixed mass of gas at constant pressure, the volume is directly proportional to the absolute temperature.

V/T = constant (at constant p) V₁/T₁ = V₂/T₂

Molecular Explanation

If the temperature increases at constant pressure:

• The molecules gain kinetic energy and move faster.
• They hit the walls harder and more frequently.
• To maintain constant pressure, the volume must increase so that the molecules travel further between collisions with the walls, reducing the collision rate.
• The volume increases in direct proportion to the absolute temperature.

Graphical Representations

• V vs T (in kelvin): A straight line through the origin (V ∝ T).
• V vs θ (in °C): A straight line that does NOT pass through the origin. If extrapolated back, it crosses the temperature axis at −273 °C (absolute zero).

Worked Example — Charles's Law

Question: A balloon contains 2.0 litres of helium at 20 °C. The balloon is placed in a freezer at −18 °C. Calculate the new volume, assuming the pressure remains constant.

Solution:

Convert temperatures to kelvin: T₁ = 20 + 273 = 293 K T₂ = −18 + 273 = 255 K

Using V₁/T₁ = V₂/T₂:

V₂ = V₁ × T₂/T₁ = 2.0 × 255/293 = 2.0 × 0.8703 = 1.74 litres

The volume decreases because the temperature decreases.

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Pressure Law (Gay-Lussac's Law)

Pressure Law: For a fixed mass of gas at constant volume, the pressure is directly proportional to the absolute temperature.

p/T = constant (at constant V) p₁/T₁ = p₂/T₂

Molecular Explanation

If the temperature increases at constant volume:

• The molecules gain kinetic energy and move faster.
• They hit the walls harder (greater momentum change per collision).
• They hit the walls more frequently (less time between collisions).
• Both effects increase the pressure.

Worked Example — Pressure Law

Question: A sealed gas cylinder at 15 °C has a pressure of 2.5 × 10⁵ Pa. If the cylinder is left in sunlight and its temperature rises to 45 °C, what is the new pressure?

Solution:

T₁ = 15 + 273 = 288 K T₂ = 45 + 273 = 318 K

p₂ = p₁ × T₂/T₁ = 2.5 × 10⁵ × 318/288 = 2.5 × 10⁵ × 1.1042 = 2.76 × 10⁵ Pa

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Experimental Verification

Verifying Boyle's Law

Apparatus: A sealed gas syringe or Boyle's law apparatus (a column of trapped gas above oil, connected to a pressure gauge and pump).

Procedure:

1. Record the volume of the trapped gas and the pressure reading.
2. Increase the pressure in small steps and record V and p each time.
3. Allow time for the gas to return to room temperature after each compression (to ensure constant temperature).
4. Plot p against 1/V — a straight line through the origin confirms Boyle's law.

Sources of error:

• Gas may not be at constant temperature (compression heats the gas).
• Friction in the syringe or piston.
• Leaks from the system.

Estimating Absolute Zero

Apparatus: A capillary tube sealed at one end, containing a trapped column of air separated by a small plug of concentrated sulfuric acid (to keep the air dry). The tube is immersed in a water bath.

Procedure:

1. Heat the water bath to various temperatures, recording the temperature and the length of the air column each time.
2. The length of the air column is proportional to the volume (since the cross-sectional area of the tube is constant).
3. Plot length (or volume) against temperature in °C.
4. Extrapolate the straight line back to where the length would be zero.
5. The x-intercept gives an estimate of absolute zero.

A well-conducted experiment gives a value close to −273 °C.

Exam Tip: When plotting gas law graphs, always use kelvin for the temperature axis if you want a straight line through the origin. If you plot against °C, the line is still straight but does not pass through the origin — the x-intercept gives an estimate of absolute zero.

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Combined Gas Law

When none of the three variables is constant, the combined gas law applies:

p₁V₁/T₁ = p₂V₂/T₂

This is simply a combination of all three individual gas laws.

Worked Example — Combined Gas Law

Question: A weather balloon contains 500 litres of helium at ground level, where the temperature is 20 °C and the pressure is 1.01 × 10⁵ Pa. At an altitude of 18 km, the temperature is −55 °C and the pressure is 7.5 × 10³ Pa. Calculate the volume of the balloon at this altitude.

Solution:

T₁ = 20 + 273 = 293 K T₂ = −55 + 273 = 218 K p₁ = 1.01 × 10⁵ Pa p₂ = 7.5 × 10³ Pa V₁ = 500 litres

Using p₁V₁/T₁ = p₂V₂/T₂:

V₂ = V₁ × (p₁/p₂) × (T₂/T₁)

V₂ = 500 × (1.01 × 10⁵ / 7.5 × 10³) × (218/293)

V₂ = 500 × 13.47 × 0.744

V₂ = 5010 litres ≈ 5000 litres

The balloon expands by a factor of about 10, primarily because of the much lower atmospheric pressure at altitude.

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Common Misconceptions

1. "The gas laws work in °C." The individual proportionality laws (Charles's, pressure law) only work with kelvin. If you use °C, you get incorrect answers. Only Boyle's law does not involve temperature directly.

2. "At absolute zero, molecules stop completely." Classically, yes. But quantum mechanics shows that molecules retain a small residual energy called zero-point energy. For A-Level, stating that "thermal motion ceases" is acceptable.

3. "Boyle's law works for all gases at all conditions." It works well for ideal gases and approximately for real gases at moderate pressures and temperatures. At high pressures or low temperatures, real gases deviate significantly.

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Additional Worked Example — Charles's Law via Absolute-Zero Extrapolation

A student investigates Charles's law by measuring the volume of a fixed mass of dry air at constant atmospheric pressure as the temperature is varied. The data are:

θ (°C)	V (cm³)
20	23.5
40	25.1
60	26.7
80	28.4
100	30.0

(a) Plot V (cm³) on the y-axis against θ (°C) on the x-axis. Extrapolate the straight line back to the temperature axis intercept where V = 0. (b) Comment on the physical significance of the intercept.

Step 1 — confirm linearity and find gradient. The data should lie on a straight line. Take the first and last points: gradient = (30.0 − 23.5) / (100 − 20) = 6.5 / 80 = 0.0813 cm³ K⁻¹.

Step 2 — find the intercept (where V = 0). Using point (20 °C, 23.5 cm³): V = mθ + V₀, where V₀ is the V-intercept at θ = 0 °C. 23.5 = 0.0813 × 20 + V₀ ⇒ V₀ = 23.5 − 1.626 = 21.87 cm³ ≈ 21.9 cm³. At V = 0: 0 = 0.0813 × θ + 21.87 ⇒ θ = −21.87 / 0.0813 = −269 °C (3 s.f.).

Step 3 — compare with the accepted value. The accepted absolute zero is −273.15 °C. The student's result of −269 °C agrees to within about 4 °C, a ≈1.5% deviation — within typical RP-grade experimental uncertainty.

Step 4 — convert to V vs T(K). Define T(K) = θ + 273. At the extrapolated intercept V = 0, T_intercept = −269 + 273 = 4 K, not 0 K — but the difference reflects experimental error in the straight-line extrapolation, not a physical phenomenon. The student would quote: T_abs_zero = −269 ± 3 °C, consistent with the accepted value of −273 °C within experimental uncertainty.

Commentary. This graphical method is one of the most elegant experiments in A-Level Physics. Long before absolute zero is approached experimentally (the gas would liquefy and then solidify well before T = 0 K), the trend of gas-volume data above the boiling point allows the extrapolation. Charles, Gay-Lussac, and others in the 18th and 19th centuries arrived at consistent values from gases as different as air, hydrogen, and carbon dioxide — convergent evidence that absolute zero is a universal property of temperature, not a property of any particular gas.