Thermal equilibrium and the zeroth law
A thermometer is kept in your mouth for a minute until its reading stops changing. Link to earlier classes : heat flowing from hot to cold of chapter 10 now gets a precise meaning.
Two bodies are in thermal equilibrium when no heat flows between them, i.e. they are at the same temperature. The zeroth law says that if A and B are each in thermal equilibrium with C, then A and B are in thermal equilibrium with each other.
Rule 1: Zeroth law — The zeroth law makes temperature a well-defined property and allows the use of a thermometer (the "third body").
- Easy : A and C show the same reading ⇒ A and C have the same temperature.
- Medium : If A and C are equal and B and C are equal, then A and B are equal.
- Exam : A mercury thermometer can be used to compare the temperature of two cups of tea because of the zeroth law.
Same temperature ⇔ thermal equilibrium; the zeroth law justifies thermometers.
Believing that bodies in equilibrium have the same heat content.
Memory clue : "Zeroth = Zero flow : same temperature, no heat moves."
Spot the difference
| Condition | Heat flow |
|---|---|
| T₁ > T₂ | from 1 to 2 |
| T₁ = T₂ | none |
Heat, internal energy and work
Rubbing your palms makes them warm (work becomes heat), and a steam engine turns steam heat into work. Link to earlier classes : energy transfer by work (chapter 5) and by heat (chapter 10).
Internal energy U is the total of the kinetic and potential energies of the molecules of a system. Heat and work are two ways of transferring energy; they are not stored in a body.
Rule 1: Work by a gas — W = P ΔV at constant pressure (the area under the P–V graph in general). W is positive when the gas expands and negative when it is compressed.
- Easy : A gas at 2 × 10⁵ Pa expands from 1 L to 3 L, so W = 2 × 10⁵ × 2 × 10⁻³ = 400 J.
- Medium : If the volume does not change, W = 0.
- Exam : For an ideal gas, U depends only on temperature; it does not depend on the pressure or the volume separately.
U is a state function; Q and W depend on the path.
Speaking of "the heat in a body" : only U is stored.
Memory clue : "State has U; the path has Q and W."
Spot the difference
| Quantity | State function? |
|---|---|
| U | yes |
| Q, W | no |
The first law of thermodynamics
A bicycle pump becomes warm when you compress the air quickly. Link to earlier classes : the conservation of energy.
The first law of thermodynamics is the law of conservation of energy for a system : the heat given to a system is used partly to raise its internal energy and partly to do work.
Rule 1: First law — Q = ΔU + W (Q = heat supplied to the system, ΔU = change in internal energy, W = work done by the system).
- Easy : Q = 500 J and W = 200 J ⇒ ΔU = 300 J.
- Medium : A system receives 100 W of heat for 60 s and does work at 20 W ⇒ ΔU = (100 − 20) × 60 = 4800 J.
- Exam : For a cycle, ΔU = 0, so the net heat absorbed equals the net work done.
Q = ΔU + W; signs : Q positive when absorbed, W positive when done by the gas.
Mixing the sign of W (work done on the gas is negative in this convention).
Memory clue : "Heat in = energy up + work out."
Spot the difference
| Quantity | Positive when |
|---|---|
| Q | heat enters the system |
| W | the system does work |
Specific heats of a gas : C_p and C_v
A gas in a rigid can heats up faster than a gas in a balloon for the same heat, because the balloon also does work while expanding. Link to earlier classes : specific heat of chapter 10, now for a gas.
The molar specific heat at constant volume C_v and at constant pressure C_p are the heat needed to raise the temperature of 1 mol by 1 K when the volume or the pressure is kept fixed.
Rule 1: Relation between them — C_p − C_v = R. For a monatomic ideal gas C_v = (3 ÷ 2)R and C_p = (5 ÷ 2)R. The ratio γ = C_p ÷ C_v is 5 ÷ 3 for a monatomic gas and 7 ÷ 5 for a diatomic gas.
- Easy : C_p − C_v = 8.314 J mol⁻¹ K⁻¹.
- Medium : Heating 2 mol of a monatomic gas by 10 K at constant volume needs Q = 2 × 1.5 × 8.314 × 10 = 249 J.
- Exam : At constant pressure the same heating needs Q = 2 × 2.5 × 8.314 × 10 = 416 J; the extra 166 J (= nRΔT) is the work done by the gas.
C_p > C_v by R; heating at constant pressure also pays for the expansion work.
Using C_v for a process at constant pressure.
Memory clue : "p is bigger than v by R : Cp = Cv + R."
Spot the difference
| Gas | C_v | C_p | γ |
|---|---|---|---|
| Monatomic | 3R ÷ 2 | 5R ÷ 2 | 5 ÷ 3 |
| Diatomic | 5R ÷ 2 | 7R ÷ 2 | 7 ÷ 5 |
Thermodynamic processes
Pumping up a tyre quickly (nearly adiabatic) warms the pump, while a slow expansion in a warm room is nearly isothermal. Link to earlier classes : the gas laws of chapter 10 are now followed along a path on a P–V diagram.
A thermodynamic process is a change from one equilibrium state to another. A process that is slow enough for the system to stay in equilibrium throughout is called quasi-static.
Rule 1: Special processes — Isothermal (T constant) : PV = constant, ΔU = 0, W = nRT ln(V₂ ÷ V₁). Adiabatic (Q = 0) : PV^γ = constant, TV^(γ−1) = constant. Isobaric (P constant) : W = PΔV. Isochoric (V constant) : W = 0, Q = ΔU. Cyclic : ΔU = 0, net Q = net W.
In the figure : The adiabatic curve is steeper than the isothermal curve through the same point, because in an adiabatic expansion the gas also cools.
- Easy : 1 mol at 300 K expands isothermally to twice its volume : W = 8.314 × 300 × ln 2 = 1729 J.
- Medium : A monatomic gas at 300 K is compressed adiabatically to half its volume : T₂ = 300 × 2^(2÷3) = 476 K.
- Exam : In an isochoric process all the heat supplied goes to raise the internal energy, because W = 0.
Isothermal : ΔU = 0; adiabatic : Q = 0; isochoric : W = 0; cyclic : ΔU = 0.
Using PV = constant for an adiabatic process (use PV^γ = constant).
Memory clue : "Iso-thermal : T stays; Adia-batic : heat stays in (no Q)."
Spot the difference
| Process | Fixed quantity | Result |
|---|---|---|
| Isothermal | T | ΔU = 0 |
| Adiabatic | Q = 0 | PV^γ = constant |
| Isobaric | P | W = PΔV |
| Isochoric | V | W = 0 |
The second law, heat engines and refrigerators
A hot cup of tea never gets hotter by itself in a cool room; a fridge needs electricity to move heat from inside to outside. Link to earlier classes : the first law only counts energy; the second law tells in which direction processes can go.
Kelvin–Planck statement : no process is possible whose only result is the absorption of heat from a reservoir and its complete conversion into work. Clausius statement : no process is possible whose only result is the transfer of heat from a cooler body to a hotter body.
Rule 1: Heat engine — An engine absorbs Q₁ from a hot source, does work W and rejects Q₂ to a cold sink. η = W ÷ Q₁ = 1 − Q₂ ÷ Q₁.
- Easy : Q₁ = 1000 J and Q₂ = 600 J ⇒ W = 400 J and η = 0.4.
- Medium : An engine of η = 25% takes 2000 J per cycle ⇒ W = 500 J and rejects 1500 J.
- Exam : No heat engine can have η = 1, because that would convert heat fully into work.
Rule 2: Refrigerator — It takes Q₂ from the cold space, uses work W and gives out Q₁ = Q₂ + W. Coefficient of performance : α = Q₂ ÷ W.
- Easy : Q₂ = 800 J and W = 200 J ⇒ α = 4.
- Medium : Q₁ = 1000 J given out and W = 200 J ⇒ Q₂ = 800 J.
- Exam : A refrigerator with no work cannot transfer heat from cold to hot (Clausius).
η = W ÷ Q₁; α = Q₂ ÷ W; efficiency is always below 1.
Taking Q₂ (heat rejected) as the work done.
Memory clue : "Engine : Work out of Q one. Fridge : Q two over Work."
Spot the difference
| Device | Purpose | Measure |
|---|---|---|
| Heat engine | heat to work | efficiency η |
| Refrigerator | move heat from cold to hot | coefficient α |
The Carnot engine and reversibility
A drop of ink spreads in water but never gathers back; broken glass never joins by itself. Link to earlier classes : the direction of natural processes.
A process is reversible if it can be retraced exactly in the opposite direction with no change left in the surroundings. A Carnot engine is an ideal reversible engine working between two temperatures, made of two isothermal and two adiabatic steps.
Rule 1: Carnot efficiency — η = 1 − T₂ ÷ T₁ (T₁ = temperature of the source and T₂ = temperature of the sink, both in kelvin). No engine working between the same two temperatures is more efficient than a Carnot engine.
- Easy : T₁ = 500 K and T₂ = 300 K ⇒ η = 1 − 0.6 = 0.4.
- Medium : For T₁ = 600 K and T₂ = 300 K, η = 0.5. To reach 60% with T₂ = 300 K the source must be at T₁ = 300 ÷ 0.4 = 750 K.
- Exam : A claim that an engine takes 1000 J at 600 K and rejects 400 J at 300 K gives η = 0.6, more than the Carnot limit of 0.5, so it is impossible.
Carnot η = 1 − T₂ ÷ T₁ in kelvin; a larger temperature gap means higher efficiency; real processes are irreversible.
Using °C in the Carnot formula.
Memory clue : "Cold over Hot, subtract from one."
Spot the difference
| Process | Reversible? | Example |
|---|---|---|
| Ideal | yes | Carnot cycle |
| Real | no | friction, diffusion, free expansion |