CHEM20018 · Chemistry: Reactions And Synthesis
Physical Chemistry & Thermodynamics: Gases, Laws & the Carnot Cycle
This section builds classical thermodynamics from the kinetic model of gases up to free energy. The kinetic theory connects molecular speeds (mean, root-mean-square, most-probable) and the Maxwell–Boltzmann distribution to temperature and molar mass. The First Law (dU = dq + dw) handles work and heat, especially reversible isothermal expansion; the Second Law introduces entropy as a state function and the Carnot cycle as the efficiency limit of any heat engine; the Third Law fixes an absolute entropy scale. It all converges on the Gibbs free energy G = H − TS and the relations ΔG° = −RT ln K = −nFE°.
What this chapter covers
- 01Kinetic model of gases: pV = nRT and pressure from molecular momentum transfer at the walls
- 02Molecular speeds: vmp = √(2RT/M) < v̄ = √(8RT/πM) < vrms = √(3RT/M)
- 03Maxwell–Boltzmann distribution broadens with T↑ and narrows with M↑; average translational KE = (3/2)kBT
- 04First Law dU = dq + dw; reversible isothermal ideal gas: w = −nRT ln(Vf/Vi), q = −w, ΔU = 0
- 05Adiabatic processes: q = 0, ΔU = w, pVᵞ = constant
- 06Second Law dS ≥ dq/T; entropy as a state function; statistical S = kB ln W
- 07Carnot cycle: two isothermal + two adiabatic reversible steps; efficiency η = 1 − TC/TH
- 08Third Law and absolute entropies; Gibbs free energy G = H − TS and ΔG° = −RT ln K = −nFE°
Reversible isothermal expansion of an ideal gas (First/Second Law)
- 1 mark — correct formula and volume ratio(a) For a reversible isothermal expansion of an ideal gas, w = −nRT ln(Vf/Vi). Here Vf/Vi = 20.0/2.0 = 10, so ln 10 = 2.303.
- 2 marks — value and correct sign interpretationw = −(1)(8.314)(320)(2.303) = −(2660)(2.303) = −6.13×10³ J. The negative sign means work is done BY the gas on the surroundings.
- 2 marks — ΔU = 0 reasoning and q value(b) The process is isothermal and the gas is ideal, so ΔU = 0. The First Law dU = dq + dw then gives q = −w = +6.13×10³ J (heat is absorbed by the gas).
- 2 marks — ΔS value + correct adiabatic-step behaviour(c) ΔSsys = nR ln(Vf/Vi) = (1)(8.314)(2.303) = +19.1 J K⁻¹ (entropy rises on expansion). In a reversible adiabatic step of a Carnot cycle q = 0, so ΔS = 0 (isentropic) while the temperature changes; over the full cycle ΔU = 0 and ΔS = 0.
Key terms
- Root-mean-square speed
- vrms = √(3RT/M), the speed whose square equals the mean square molecular speed; it sets the average translational kinetic energy (3/2)kBT per molecule.
- First Law of thermodynamics
- Energy is conserved: dU = dq + dw, where dq is heat added to the system and dw is work done on it.
- Entropy
- A state function (statistically S = kB ln W) measuring the number of accessible microstates; the Second Law requires dS ≥ dq/T for any process.
- Carnot cycle
- A reversible four-step cycle (two isothermal, two adiabatic) defining the maximum efficiency η = 1 − TC/TH of a heat engine working between two temperatures.
- Gibbs free energy
- G = H − TS; its change ΔG predicts spontaneity at constant T and p, with ΔG° = −RT ln K = −nFE° linking it to equilibrium and electrochemistry.
Physical Chemistry & Thermodynamics: Gases, Laws & the Carnot Cycle FAQ
Why is ΔU = 0 for an isothermal ideal-gas process?
The internal energy of an ideal gas depends only on temperature. If the temperature is held constant (isothermal), U does not change, so ΔU = 0. The First Law then forces q = −w: any work done by the gas is exactly balanced by heat absorbed.
How do the three molecular speeds compare?
They always rank vmp < v̄ < vrms, with vmp = √(2RT/M), v̄ = √(8RT/πM) and vrms = √(3RT/M). All increase with temperature and decrease with molar mass, which is why light gases at high temperature move fastest.
Why can no real engine beat the Carnot efficiency?
The Carnot cycle is fully reversible, so it generates no entropy and converts the maximum possible fraction of heat into work for given hot and cold reservoir temperatures: η = 1 − TC/TH. Any real (irreversible) engine generates entropy and so must be less efficient — a direct consequence of the Second Law.
Exam move
Treat this as a section of clean, repeatable calculation pipelines and learn each with its sign convention. The kinetic-theory speeds, the isothermal w/q/ΔS trio, adiabatic q = 0, the Carnot efficiency, and ΔS for heating (nC ln(Tf/Ti)) or phase change (ΔHtrs/Ttrs) cover most marks. Practise carrying units and signs through every step, because the appendix gives you the relations but not the discipline. Build intuition for the laws (why ΔU = 0 isothermally, why entropy rises on expansion, why the Carnot bound is unbeatable) so you can write the one-sentence justifications examiners want alongside the numbers. Do the Smith tutorial problems and the mock-exam Section D under a 30-minute clock to lock in speed.
