Thermodynamics
Thermodynamics governs the macroscopic behavior of energy, heat, work, and entropy. Its four laws establish the framework for understanding everything from steam engines to black holes. Historically, thermodynamics emerged from practical engineering (Carnot, 1824) and was placed on statistical foundations by Boltzmann, Maxwell, and Gibbs in the late 19th century. This skill covers the four laws, thermodynamic processes, heat engines, kinetic theory, phase transitions, and free energy.
Agent affinity: boltzmann (thermodynamics, Sonnet)
Concept IDs: phys-thermodynamics, phys-work-power, phys-kinetic-potential-energy, phys-conservation-of-energy
Thermodynamics at a Glance
| # | Topic | Key equations | Core idea | |---|---|---|---| | 1 | Zeroth law & temperature | Thermal equilibrium is transitive | Defines temperature as a measurable quantity | | 2 | Heat transfer | Q = mc Delta T, Fourier/Stefan-Boltzmann | Three mechanisms: conduction, convection, radiation | | 3 | First law | Delta U = Q - W | Energy conservation for thermal systems | | 4 | PV diagrams | W = integral of P dV | Work = area under curve | | 5 | Processes | Isothermal, adiabatic, isobaric, isochoric | Each constrains one variable | | 6 | Second law | Delta S >= 0 (isolated), eta_Carnot = 1 - T_C/T_H | Entropy never decreases; efficiency is bounded | | 7 | Carnot cycle | eta = 1 - T_C/T_H | Theoretical maximum efficiency | | 8 | Statistical mechanics | S = k_B ln W | Entropy counts microstates | | 9 | Kinetic theory | PV = nRT, KE_avg = (3/2)k_B T | Gas behavior from molecular motion | | 10 | Third law | S -> 0 as T -> 0 | Absolute zero is unattainable | | 11 | Phase transitions | Q = mL, Clausius-Clapeyron | Energy exchange at constant temperature | | 12 | Free energy | G = H - TS, F = U - TS | Spontaneity criteria at constant T, P or T, V |
Topic 1 — The Zeroth Law and Temperature
Zeroth law. If system A is in thermal equilibrium with system B, and B is in thermal equilibrium with system C, then A is in thermal equilibrium with C. This seemingly trivial statement justifies the existence of temperature as a well-defined quantity and the use of thermometers.
Temperature scales:
- Kelvin: T_K = T_C + 273.15 (absolute scale, zero = no molecular motion)
- Celsius: T_C = T_K - 273.15
- Fahrenheit: T_F = (9/5)T_C + 32
Why Kelvin matters. All thermodynamic equations involving temperature ratios (Carnot efficiency, ideal gas law, entropy calculations) require absolute temperature in Kelvin. Using Celsius or Fahrenheit in these equations produces wrong answers.
Topic 2 — Heat Transfer
Conduction. Heat flows through a material due to a temperature gradient. Fourier's law: dQ/dt = -kA (dT/dx), where k is thermal conductivity, A is cross-sectional area.
Convection. Heat transfer via bulk fluid motion. Newton's law of cooling: dQ/dt = hA(T_surface - T_fluid), where h is the convective heat transfer coefficient.
Radiation. All objects emit electromagnetic radiation. Stefan-Boltzmann law: P = epsilon sigma A T^4, where sigma = 5.67 * 10^-8 W/(m^2 K^4) and epsilon is emissivity (0 to 1).
Worked example. A copper rod (k = 390 W/(m K)) of length 0.5 m and cross-sectional area 10^-4 m^2 connects a 100 C reservoir to a 20 C reservoir. Find the heat flow rate.
Solution. dQ/dt = kA (Delta T / L) = 390 10^-4 (80/0.5) = 390 10^-4 160 = 6.24 W.
Specific heat capacity: Q = mc Delta T, where c is specific heat (J/(kg K)). Water: c = 4186 J/(kg K). This is why water is such an effective coolant.
Topic 3 — The First Law of Thermodynamics
Statement: Delta U = Q - W, where Delta U is the change in internal energy, Q is heat added to the system, and W is work done by the system.
This is conservation of energy applied to thermal systems. Energy can enter as heat or leave as work (and vice versa), but the total is tracked by internal energy U.
Sign conventions matter. In the physics convention: Q > 0 means heat flows INTO the system; W > 0 means work is done BY the system. Some engineering texts reverse the sign of W. Be consistent.
Worked example. A gas absorbs 500 J of heat and does 200 J of work on its surroundings. What is the change in internal energy?
Solution. Delta U = Q - W = 500 - 200 = 300 J. The internal energy increases by 300 J.
Topic 4 — PV Diagrams and Work
Work done by a gas: W = integral from V_i to V_f of P dV. On a PV diagram, work is the area under the curve.
Key insight. Work depends on the path, not just the endpoints. Different processes connecting the same initial and final states yield different amounts of work. This is why thermodynamic processes must be specified.
Clockwise cycles on a PV diagram represent heat engines (net work done by the system). Counterclockwise cycles represent refrigerators or heat pumps (net work done on the system).
Topic 5 — Thermodynamic Processes
Isothermal (constant T)
For an ideal gas at constant temperature: PV = nRT = constant, so P = nRT/V. Work: W = nRT ln(V_f/V_i). Since T is constant, Delta U = 0, so Q = W — all absorbed heat becomes work.
Adiabatic (Q = 0)
No heat exchange: Delta U = -W. The gas cools when it expands, heats when compressed. PV^gamma = constant, where gamma = C_P/C_V (5/3 for monatomic, 7/5 for diatomic). TV^(gamma-1) = constant.
Isobaric (constant P)
Work: W = P Delta V. Heat: Q = nC_P Delta T. The specific heat at constant pressure C_P exceeds C_V because the gas must also do work expanding.
Isochoric (constant V)
No volume change, so W = 0. Delta U = Q = nC_V Delta T. All heat goes into internal energy.
Worked example. One mole of an ideal monatomic gas (gamma = 5/3) at 300 K and 1 atm expands adiabatically to twice its volume. Find the final temperature.
Solution. TV^(gamma-1) = constant. T_i V_i^(2/3) = T_f V_f^(2/3). So T_f = T_i (V_i/V_f)^(2/3) = 300 (1/2)^(2/3) = 300 * 0.63 = 189 K.
The gas cools by 111 K during the adiabatic expansion. No heat was exchanged — the internal energy was converted to work.
Topic 6 — The Second Law of Thermodynamics
The second law has several equivalent formulations:
Clausius statement: Heat cannot spontaneously flow from a colder body to a hotter body.
Kelvin-Planck statement: No cyclic process can convert heat entirely into work with no other effect.
Entropy statement: The entropy of an isolated system never decreases: Delta S_universe >= 0.
Entropy definition: dS = dQ_rev / T for a reversible process. For an irreversible process, Delta S > integral of dQ/T.
Worked example. 500 J of heat flows from a 600 K reservoir to a 300 K reservoir. Find the entropy change of the universe.
Solution. Delta S_hot = -Q/T_hot = -500/600 = -0.833 J/K. Delta S_cold = +Q/T_cold = +500/300 = +1.667 J/K. Delta S_universe = -0.833 + 1.667 = +0.833 J/K > 0. Entropy increased, confirming the process is irreversible.
Topic 7 — The Carnot Cycle
Carnot's theorem. No heat engine operating between two temperatures can be more efficient than a reversible (Carnot) engine between those temperatures.
Carnot efficiency: eta = 1 - T_C/T_H, where T_C and T_H are the cold and hot reservoir temperatures in Kelvin.
The Carnot cycle consists of four steps:
- Isothermal expansion at T_H (absorb Q_H)
- Adiabatic expansion (cool from T_H to T_C)
- Isothermal compression at T_C (reject Q_C)
- Adiabatic compression (heat from T_C to T_H)
Worked example. A heat engine operates between a 500 K source and a 300 K sink. It absorbs 1000 J per cycle. Find the maximum work output and the minimum heat rejected.
Solution. eta_Carnot = 1 - 300/500 = 0.4 = 40%. Maximum work: W = eta Q_H = 0.4 1000 = 400 J. Minimum heat rejected: Q_C = Q_H - W = 600 J.
Coefficient of performance (refrigerator): COP = Q_C / W = T_C / (T_H - T_C) for a Carnot refrigerator.
Why 100% efficiency is impossible.