7.3 Gas Laws, Thermodynamic Laws & Expanding Gases

Gas Laws, Thermodynamic Laws & Expanding Gases

Gases in tyres, oleo struts, pneumatic systems, cabins, and engine working fluid obey simple ideal-gas relationships under Module 2 assumptions. This section formalises Boyle, Charles, and Gay-Lussac, the combined/ideal gas law, the first and second laws of thermodynamics, specific heats Cv and Cp, and work done by an expanding gas. The single most common calculation error is using Celsius instead of kelvin — fix that habit before the exam.

Ideal Gas Model (Module 2 Level)

An ideal gas is a model in which:

• Molecules have negligible volume compared with the container
• Intermolecular forces are neglected except during collisions
• Collisions are elastic; pressure arises from molecular impacts on walls

Real air and nitrogen approximate ideal behaviour well at ordinary temperatures and pressures used in many aircraft systems. Extreme cold, very high pressure, or near-condensation conditions deviate — exam problems usually stay ideal.

For a fixed mass of ideal gas:

PV / T = constant

or between two states:

P₁V₁ / T₁ = P₂V₂ / T₂

with T in kelvin. The full molar form is PV = nRT (n = amount of substance, R = universal gas constant) or PV = mRT with a specific gas constant; Module 2 calculations often use the ratio form without needing R explicitly.

Boyle’s Law — Constant Temperature

Boyle’s law: for a fixed mass of gas at constant temperature,

P ∝ 1/V or P₁V₁ = P₂V₂

Compress the volume → pressure rises; expand the volume → pressure falls, if temperature is held constant (isothermal compression/expansion).

Worked example — Boyle. A gas occupies 0.020 m³ at 200 kPa. It is compressed isothermally to 0.010 m³. Find the new pressure.

P₂ = P₁V₁ / V₂ = (200 × 0.020) / 0.010 = 400 kPa

Volume halved → pressure doubled at constant T.

Aviation link: Pumping air into a tyre reduces free volume available per mass of free air and raises pressure; temperature also rises during rapid pumping (not purely isothermal), so pressure can fall slightly as the tyre cools — real process mixes Boyle with temperature change.

Charles’s Law — Constant Pressure

Charles’s law: for a fixed mass of gas at constant pressure,

V ∝ T_abs or V₁ / T₁ = V₂ / T₂

Absolute temperature only. Heat a gas at constant pressure and it expands; cool it and it contracts.

Worked example — Charles (kelvin conversion). A balloon holds 1.50 m³ of gas at 27 °C and constant pressure. Temperature falls to −3 °C. Find the new volume.

T₁ = 27 + 273 = 300 K
T₂ = −3 + 273 = 270 K
V₂ = V₁ × (T₂ / T₁) = 1.50 × (270/300) = 1.50 × 0.90 = 1.35 m³

If you wrongly used 27 and −3 without converting, the ratio would be nonsense (negative temperature ratio).

Gay-Lussac’s Law — Constant Volume

Gay-Lussac’s law (pressure law): for a fixed mass of gas at constant volume,

P ∝ T_abs or P₁ / T₁ = P₂ / T₂

Worked example — Gay-Lussac. An oxygen bottle (fixed volume) reads 50 bar at 15 °C. After standing in the sun the gas reaches 45 °C. Estimate the new pressure (ideal gas, fixed mass).

T₁ = 15 + 273 = 288 K
T₂ = 45 + 273 = 318 K
P₂ = P₁ × (T₂ / T₁) = 50 × (318/288) ≈ 50 × 1.104 = ≈ 55.2 bar

Pressure rises with absolute temperature in a rigid bottle — never leave pressurised vessels where temperature can climb uncontrollably beyond design limits.

Combined and Ideal Gas Law

When P, V, and T all change:

P₁V₁ / T₁ = P₂V₂ / T₂

Worked example — combined. A mass of gas has P₁ = 100 kPa, V₁ = 0.030 m³, t₁ = 20 °C. It is taken to P₂ = 250 kPa, t₂ = 80 °C. Find V₂.

T₁ = 20 + 273 = 293 K
T₂ = 80 + 273 = 353 K
V₂ = V₁ × (P₁/P₂) × (T₂/T₁) = 0.030 × (100/250) × (353/293)
V₂ = 0.030 × 0.4 × 1.205 ≈ 0.0145 m³

Always: convert both temperatures → apply ratio → interpret volume (here compression and heating compete; net volume falls).

Work Done by an Expanding Gas

When a gas expands against a piston or atmosphere, it can do work on the surroundings.

At constant pressure:

W = P ΔV = P (V₂ − V₁)

(with consistent SI units: P in Pa, V in m³ → W in J).

If the gas is compressed, the surroundings do work on the gas (work done by the system is negative in the common sign convention below).

Worked example — expansion work. Gas expands at constant 200 kPa from 0.010 m³ to 0.025 m³.

W = 200 × 10³ Pa × (0.025 − 0.010) m³ = 200 000 × 0.015 = 3000 J

In engines, expanding combustion gases push pistons or drive turbines — that mechanical work is the goal of the thermodynamic cycle (section 7.4).

First Law of Thermodynamics

The first law is conservation of energy for a thermodynamic system:

ΔU = Q − W

(Common engineering convention: Q positive when heat enters the system; W positive when the system does work on the surroundings. Some physics texts write ΔU = Q + W with opposite work sign — know which convention a formula sheet uses. Module 2 conceptual point: heat and work both change internal energy; energy is conserved.)

• U = internal energy (microscopic energy of the system)
• For an ideal gas, U depends only on temperature (for fixed mass/composition): raise T → raise U

Implications:

• Add heat with no work (constant volume, rigid container): ΔU = Q → temperature rises
• System does work with no heat (adiabatic expansion idealisation): ΔU = −W → temperature falls
• Steady cruise machinery converts chemical energy ultimately into work and waste heat; first law accounts for the balance even when second law limits efficiency

Second Law of Thermodynamics

The second law introduces direction and limits:

1. Heat does not flow spontaneously from a colder body to a hotter body. Refrigerators and heat pumps move heat “uphill” only with work input.
2. You cannot convert heat completely into work in a cyclic engine with no other effect — some heat is always rejected to a colder sink in a real heat engine. Efficiency is limited (Carnot limit awareness: higher source temperature and lower sink temperature improve the ideal ceiling).
3. Real processes generate irreversibility (friction, unrestrained expansion, heat transfer across finite ΔT); not all first-law-allowed processes occur spontaneously.

Module 2 does not require full entropy calculus; it requires the ideas of spontaneous direction, need for work in refrigeration, and waste heat from engines.

Specific Heat Capacities Cv and Cp

For gases, heat capacity depends on the process constraints:

• Cv — specific heat capacity at constant volume. Heat added at constant volume goes entirely into internal energy: Q_v = m Cv ΔT (and for ideal gas ΔU = m Cv ΔT).
• Cp — specific heat capacity at constant pressure. At constant pressure the gas expands and does work, so more heat is needed for the same ΔT: Q_p = m Cp ΔT.

For an ideal gas:

Cp − Cv = R_specific (positive)

so Cp > Cv. The ratio γ = Cp / Cv appears in adiabatic relations (next section).

Solids and liquids expand little, so their single c is close to either process; gases need the Cv/Cp distinction in thermodynamic reasoning.

Process Labels Preview

Formula Recap

• Boyle: P₁V₁ = P₂V₂ (const T)
• Charles: V₁/T₁ = V₂/T₂ (const P, T in K)
• Gay-Lussac: P₁/T₁ = P₂/T₂ (const V, T in K)
• Combined: P₁V₁/T₁ = P₂V₂/T₂
• T_K = t_°C + 273 always before absolute-T ratios
• Work (const P): W = PΔV
• First law: ΔU = Q − W (stated convention)
• Second law: direction + efficiency/refrigeration limits
• Cp > Cv for gases; ΔU = m Cv ΔT (ideal gas)

Drill kelvin conversion until automatic; then Boyle/Charles/Gay-Lussac problems become routine arithmetic for Module 2.