You are viewing a free preview of this lesson.
Subscribe to unlock all 7 lessons in this course and every other course on LearningBro.
The first law of thermodynamics is the statement of energy conservation applied to a thermodynamic system. For a gas, it relates the heat supplied to the gas, the work done on or by the gas, and the resulting change in the gas's internal energy. AQA option C — Engineering Physics builds on the AS-level thermal physics by classifying the four standard quasi-static processes by which a gas may change state — isothermal, isobaric, isochoric and adiabatic — and developing the algebra of each on a pressure-volume (pV) diagram. These four processes are the building blocks from which engine cycles (Lesson 3) and refrigeration cycles (Lesson 5) are constructed. Mastery of this lesson is non-negotiable: every subsequent lesson in this course assumes you can decompose a thermodynamic problem into the four processes, apply the first law to each, and read work and heat off a pV indicator diagram.
Spec mapping (AQA 7408 §3.11.2 — Thermodynamics, option C): This lesson covers the first law of thermodynamics ΔU = Q + W (with the sign convention that W is the work done on the gas); the internal energy of a monatomic ideal gas U = (3/2)nRT; and the four quasi-static ideal-gas processes — isothermal, isobaric, isochoric and adiabatic — with their algebraic descriptions, pV-diagram shapes, and worked applications. The adiabatic relation pV^γ = constant with γ = Cp/Cv is introduced. (Refer to the official AQA specification document for exact wording.)
Synoptic links: This lesson sits directly on top of AS thermal physics (§3.6) — the ideal gas equation pV = nRT, kinetic theory ½m〈c²〉 = (3/2)kT, and the molecular interpretation of internal energy. It feeds into the engine-cycle Lesson 3 (Otto, Diesel cycles) and into the Carnot analysis of Lesson 4, both of which are constructed by stitching together the four processes around a closed loop. The first law also reappears synoptically in nuclear physics (mass-energy conservation in fission and fusion is the same principle restated) and in particle physics (energy conservation in collisions). At university the four-process classification generalises to polytropic processes pV^n = constant, of which all four AQA processes are special cases.
There are two conventions in use for the first law, and AQA uses the engineering convention:
First law (AQA convention): ΔU = Q + W
- Q is the heat supplied to the gas (Q > 0 means heat in).
- W is the work done on the gas (W > 0 means work in, gas compressed).
- ΔU is the change in internal energy of the gas (positive means U increases).
If the gas expands, it does work on its surroundings, so the work done on the gas is negative. If the gas is compressed, work is done on it, so W is positive.
Exam Tip: AQA papers use ΔU = Q + W with W = work done on the gas. Some textbooks (and some physics traditions) use ΔU = Q − W with W = work done by the gas. Both are correct first-law statements but the sign of W flips. Always check which convention a question is using before substituting. AQA specimen and past papers consistently use the +W (on the gas) convention.
The physical content is the same in both conventions: heat in plus work in equals change in internal energy.
For an ideal monatomic gas (helium, argon, etc.), there are no intermolecular forces and no rotational or vibrational modes — only translational kinetic energy. The equipartition theorem gives 3 quadratic degrees of freedom (½kT each), so the internal energy of n moles of monatomic ideal gas is:
U = (3/2) n R T
(Equivalently U = (3/2) N k T for N molecules.) The crucial property is that U depends only on T, not on V or p. Heating a sealed gas at constant V does raise U; doing work on a gas without supplying heat (an adiabatic compression) also raises U and therefore T. This T-only dependence is the deepest property of an ideal gas and underlies the simplicity of the four-process analysis below.
For diatomic gases (N₂, O₂ — most of the atmosphere) at room temperature, two rotational degrees of freedom unfreeze and U = (5/2) nRT. Vibrational modes only unfreeze at high temperatures and add another (2/2) per mode. AQA examines only the monatomic case in detail but expects you to know that real gases have additional degrees of freedom.
When a gas changes volume from V_i to V_f at pressure p (in general varying with V):
Work done on the gas: W_on = − ∫_{V_i}^{V_f} p dV
The minus sign captures the convention: when V increases (expansion), the integral is positive, so W_on is negative — work flows out of the gas to the surroundings. When V decreases (compression), the integral is negative, so W_on is positive — work flows into the gas.
Geometrically, the magnitude of work done is the area under the curve on a pV diagram between V_i and V_f. The sign is determined by the direction of motion.
A quasi-static (reversible) process is one in which the gas remains arbitrarily close to thermodynamic equilibrium at every instant — pressure and temperature are well-defined throughout. The four standard cases are:
The gas exchanges heat with a large thermal reservoir at temperature T while undergoing slow expansion or compression. T = constant, so U = (3/2)nRT = constant and ΔU = 0.
On a pV diagram, an isothermal is a hyperbola pV = nRT = constant.
By the first law: 0 = Q + W ⇒ Q = −W.
The work done on the gas during isothermal change from V_i to V_f:
W_on = − ∫ p dV = − ∫ (nRT/V) dV = − nRT ln(V_f / V_i)
For an isothermal expansion (V_f > V_i): ln(V_f/V_i) > 0, so W_on < 0 (work out of gas), and Q = −W_on > 0 (heat in from reservoir). The gas absorbs heat and converts it all into work.
For an isothermal compression: V_f < V_i, W_on > 0 (work in), Q < 0 (heat out to reservoir).
The gas is held at constant pressure (e.g. against a piston loaded with a fixed weight) while heat is supplied or removed. The volume changes; the temperature changes; U changes.
On a pV diagram, an isobaric is a horizontal straight line.
Work done on gas: W_on = − p (V_f − V_i) = − p ΔV.
For expansion (ΔV > 0): W_on < 0 (work out). For compression: W_on > 0 (work in).
From the ideal-gas law at constant p: V/T = const, so V_f/V_i = T_f/T_i. Therefore ΔU = (3/2) n R ΔT and Q = ΔU − W_on = (3/2)nRΔT + pΔV = (5/2)nRΔT for a monatomic ideal gas. (This is why the molar heat capacity at constant pressure is C_p = (5/2)R for a monatomic gas.)
The gas is sealed in a rigid container; no expansion or compression is possible. V = constant means ΔV = 0 and so W_on = 0.
On a pV diagram, an isochoric is a vertical straight line.
By the first law: ΔU = Q. All heat supplied raises U (and therefore T). For monatomic ideal gas, Q = (3/2)nRΔT. The molar heat capacity at constant volume is C_v = (3/2)R.
From the ideal-gas law at constant V: p/T = const, so p_f/p_i = T_f/T_i. Heating an isochoric gas raises p in proportion to T.
The gas is thermally insulated from its surroundings, so Q = 0. By the first law: ΔU = W_on.
For a reversible adiabatic process on an ideal gas, the pressure and volume are related by:
pV^γ = constant
where γ = Cp/Cv is the adiabatic index (also called the ratio of principal specific heats). For monatomic ideal gas, γ = (5/2)R / (3/2)R = 5/3 ≈ 1.67. For diatomic gas at room T, γ = (7/2)R / (5/2)R = 7/5 = 1.4. (Air is approximately γ = 1.4 — the standard value used in engine and atmospheric calculations.)
Combining with the ideal gas law: TV^(γ−1) = constant also holds, and Tp^((1−γ)/γ) = constant.
On a pV diagram, an adiabatic is steeper than the isothermal through the same point (because γ > 1) — the gas cools faster on expansion and heats faster on compression than it would in isothermal contact with a reservoir.
For an adiabatic expansion: V increases, so by pV^γ = const, p decreases. By TV^(γ−1) = const, T decreases. The gas cools because it does work on the surroundings using its own internal energy (no heat enters to replenish it). This is why escaping pressurised gas (from a CO₂ cartridge, or air leaving a punctured tyre) feels cold — adiabatic expansion.
For an adiabatic compression: V decreases, p and T both increase. This is why a bicycle pump heats up rapidly when pumping fast (so quickly that there's no time for heat to escape, so it's approximately adiabatic). It is also the principle of the Diesel engine: air is compressed so violently in the cylinder that its temperature rises above the fuel's ignition point, igniting the fuel without a spark plug.
| Process | Constraint | ΔU | Q | W_on | pV-diagram shape |
|---|---|---|---|---|---|
| Isothermal | T = const | 0 | −W_on = nRT ln(V_f/V_i) | −nRT ln(V_f/V_i) | Hyperbola |
| Isobaric | p = const | (3/2)nRΔT | (5/2)nRΔT | −pΔV | Horizontal line |
| Isochoric | V = const | (3/2)nRΔT | (3/2)nRΔT | 0 | Vertical line |
| Adiabatic | Q = 0 | W_on | 0 | nC_v ΔT | pV^γ = const (steep) |
graph TD
A["Thermodynamic process"] --> B{"Heat exchange?"}
B -->|"No (Q = 0)"| C["Adiabatic<br/>pV^γ = const<br/>ΔU = W"]
B -->|"Yes"| D{"What is held constant?"}
D -->|"T"| E["Isothermal<br/>pV = nRT<br/>ΔU = 0"]
D -->|"p"| F["Isobaric<br/>V/T = const<br/>W = -pΔV"]
D -->|"V"| G["Isochoric<br/>p/T = const<br/>W = 0"]
style C fill:#27ae60,color:#fff
style E fill:#27ae60,color:#fff
style F fill:#27ae60,color:#fff
style G fill:#27ae60,color:#fff
Question: A sealed cylinder contains 0.50 mol of nitrogen at an initial volume of 1.0 L. The gas is allowed to expand isothermally at 300 K to a final volume of 5.0 L. Calculate (a) the initial and final pressures, (b) the work done on the gas, (c) the heat absorbed.
Solution:
(a) p_i = nRT/V_i = 0.50 × 8.314 × 300 / (1.0 × 10⁻³) = 1.247 × 10⁶ Pa ≈ 12.5 bar. p_f = nRT/V_f = 0.50 × 8.314 × 300 / (5.0 × 10⁻³) = 2.494 × 10⁵ Pa ≈ 2.5 bar. (Or simply p_f = p_i × V_i/V_f = 12.5/5 = 2.5 bar.)
(b) W_on = − nRT ln(V_f/V_i) = − 0.50 × 8.314 × 300 × ln 5 = − 1247.1 × 1.609 = − 2.01 × 10³ J ≈ − 2.01 kJ.
The negative sign means 2.01 kJ of work flows out of the gas to the surroundings.
(c) Isothermal ⇒ ΔU = 0 ⇒ Q = −W_on = + 2.01 kJ. Heat absorbed from the reservoir to maintain constant T.
Question: Air (treat as ideal gas with γ = 1.4) initially at 1.0 × 10⁵ Pa and 290 K is rapidly compressed adiabatically to one-tenth of its initial volume — a compression ratio of 10. Calculate (a) the final pressure, (b) the final temperature.
Solution:
(a) Adiabatic: p_i V_i^γ = p_f V_f^γ, so p_f = p_i (V_i/V_f)^γ = 1.0 × 10⁵ × 10^1.4 = 1.0 × 10⁵ × 25.12 = 2.51 × 10⁶ Pa ≈ 25 bar.
(b) T_i V_i^(γ−1) = T_f V_f^(γ−1), so T_f = T_i (V_i/V_f)^(γ−1) = 290 × 10^0.4 = 290 × 2.512 = 728 K ≈ 455 °C.
That final temperature of about 728 K is well above the autoignition point of Diesel fuel (~480 K). This is exactly why a Diesel engine can ignite fuel by adiabatic compression alone, no spark required.
Question: 1.0 mol of monatomic ideal gas at 300 K and 1.0 atm undergoes (Step A) isochoric heating to 600 K, then (Step B) isothermal expansion to twice its volume. Calculate ΔU, Q, W_on for each step and for the whole process.
Solution:
R = 8.314 J K⁻¹ mol⁻¹; (3/2)R = 12.47 J K⁻¹ mol⁻¹.
Subscribe to continue reading
Get full access to this lesson and all 7 lessons in this course.