What thermodynamics studies
Thermodynamics deals with heat, work and temperature, and how energy changes form in large systems such as a gas in a cylinder. It describes the system with macroscopic variables: pressure P, volume V and temperature T.
System: the part being studied (the gas in a cylinder).
Surroundings: everything else.
Thermal equilibrium and the zeroth law
Two systems in contact are in thermal equilibrium when there is no net flow of heat between them, that is, when they are at the same temperature.
Zeroth law: if systems A and B are each in thermal equilibrium with a third system C, then A and B are in thermal equilibrium with each other. This law is what makes temperature a meaningful quantity and lets a thermometer work.
Internal energy, heat and work
The internal energy U of a gas is the sum of the kinetic and potential energies of its molecules. It is a state variable: it depends only on the state (P, V, T), not on how the system reached it. For an ideal gas, U depends only on temperature.
U can change in two ways:
Heat (Q) flows because of a temperature difference.
Work (W) is done when, for example, a gas pushes a piston.
Heat and work are not state variables; they depend on the process.
Work done by a gas expanding against a piston: W = ∫P dV, the area under the P-V curve. At constant pressure, W = P(V₂ − V₁).
The first law of thermodynamics
ΔQ = ΔU + ΔW
Heat supplied to a system goes partly into increasing its internal energy and partly into work done by the system. It is the law of conservation of energy applied to heat.
Example. 500 J of heat is supplied to a gas, and it does 200 J of work in expanding. Its internal energy increases by 500 − 200 = 300 J.
Sign convention: Q is positive when heat is supplied to the system; W is positive when work is done by the system.
Specific heat of gases
A gas has two principal specific heats:
Cᵥ at constant volume: all the heat raises internal energy.
Cₚ at constant pressure: some heat also does work in expanding.
So Cₚ > Cᵥ. For one mole of an ideal gas, Cₚ − Cᵥ = R (Mayer's relation), where R = 8.31 J mol⁻¹ K⁻¹. Their ratio is γ = Cₚ/Cᵥ (about 1.67 for monatomic and 1.4 for diatomic gases).
Thermodynamic processes
A quasi-static process happens so slowly that the system stays in equilibrium at every step.
Process | What stays constant | Key relation | First law |
|---|---|---|---|
Isothermal | temperature | PV = constant; W = nRT ln(V₂/V₁) | ΔU = 0, so Q = W |
Adiabatic | no heat exchange (Q = 0) | PV^γ = constant; W = (P₁V₁ − P₂V₂)/(γ − 1) | ΔU = −W |
Isochoric | volume | W = 0 | Q = ΔU |
Isobaric | pressure | W = P(V₂ − V₁) | Q = ΔU + W |
Cyclic | returns to its initial state | ΔU = 0 | Q = W (net heat = net work) |
In an isothermal expansion, heat flows in and is all converted to work. It needs a cylinder with conducting walls in contact with a large reservoir.
In an adiabatic expansion, the gas does work at the expense of its internal energy, so it cools. Air pumped into a tyre quickly warms up (adiabatic compression); a sudden release of compressed gas cools it.
On a P-V diagram, an adiabatic curve is steeper than an isothermal one through the same point, since γ > 1.
The second law of thermodynamics
The first law allows any process that conserves energy, yet some never happen: heat never flows by itself from a cold body to a hot one, and no engine turns all heat into work. The second law sets these limits.
Kelvin-Planck statement: no process is possible whose sole result is the absorption of heat from a reservoir and its complete conversion into work. (No engine can be 100% efficient.)
Clausius statement: no process is possible whose sole result is the transfer of heat from a colder body to a hotter one. (A refrigerator needs work.)
Heat engines and refrigerators
A heat engine takes heat Q₁ from a hot reservoir, does work W and rejects heat Q₂ to a cold reservoir, working in a cycle.
Efficiency η = W/Q₁ = 1 − Q₂/Q₁
A refrigerator (or heat pump) runs the cycle in reverse: work W is used to extract heat Q₂ from a cold reservoir and deliver Q₁ to a hot one.
Coefficient of performance α = Q₂/W
The Carnot engine
A reversible process can be run backwards so that both system and surroundings return to their original states. Real processes involving friction or heat flow across a finite temperature difference are irreversible.
The Carnot cycle is an ideal reversible cycle of two isothermal and two adiabatic steps between a hot reservoir at T₁ and a cold one at T₂. Its efficiency is
η = 1 − T₂/T₁ (temperatures in kelvin)
Carnot's theorem: no engine working between two temperatures can be more efficient than a Carnot engine, and the efficiency of a Carnot engine does not depend on the working substance.
Example. An engine works between 500 K and 300 K. Its maximum possible efficiency is 1 − 300/500 = 0.4, or 40%. No engine, however well built, can do better.
Summary
Zeroth law: thermal equilibrium defines temperature.
First law: ΔQ = ΔU + ΔW; U is a state variable.
Isothermal: ΔU = 0; adiabatic: Q = 0, PV^γ = constant.
Second law: no engine is 100% efficient; heat doesn't flow from cold to hot by itself.
Carnot efficiency 1 − T₂/T₁ is the upper limit for any engine.