Thermodynamics: NEET notes

1. Heat as energy: what thermodynamics studies

NCERT §11.1

• Thermodynamics is about the laws that govern thermal energy: how work turns into heat and heat into work. Rubbing your palms together in winter warms them (work → heat); in a steam engine the steam's heat pushes the pistons that turn the wheels (heat → work).
• The old picture treated heat as an invisible fluid, 'caloric', held in the pores of a substance and flowing from one body to another until the 'caloric levels' (temperatures) matched, much as water levels equalise in two tanks joined by a pipe.
• Count Rumford (Benjamin Thomson), 1798: boring a brass cannon produced enough heat to boil water. The heat depended on the work done by the horses turning the drill, not on how sharp the drill was. A sharper drill should have released more caloric; it did not.
• The natural reading: heat is a form of energy, and the cannon showed work being converted into heat. The caloric idea was dropped in favour of this.
• Thermodynamics is a macroscopic science. It describes bulk systems through a few measurable variables such as pressure, volume, temperature, mass and composition, and never needs the positions and velocities of individual molecules. Its laws were set down in the nineteenth century, before the molecular picture was firmly established.
• Kinetic theory sits in between: it does not track each molecule but does use the distribution of molecular velocities.
• Mechanics studies the motion of a body as a whole under forces and torques; thermodynamics studies its internal state. A fired bullet gains kinetic energy, not temperature. When it stops in a block of wood, that kinetic energy becomes heat and warms the bullet and the wood around it.
• Temperature is tied to the disordered internal motion of a body's molecules, not to the motion of the body as a whole.

2. Thermal equilibrium and the zeroth law

NCERT §11.2, §11.3

• In mechanics, equilibrium means zero net external force and torque. In thermodynamics, a system is in equilibrium when its macroscopic variables stay constant in time, e.g. a gas in a closed, rigid, fully insulated container with fixed P, V, T, mass and composition.
• Whether a system reaches equilibrium depends on its surroundings and on the wall between them. An adiabatic wall is insulating: it lets no heat through. A diathermic wall is conducting: heat can flow across it.
• Two gases A and B separated by an adiabatic wall: any pair of values (P_A, V_A) can sit in equilibrium with any pair (P_B, V_B). Replace the wall with a diathermic one and the variables of both change on their own until each settles to a new steady state; then no more energy flows. A and B are now in thermal equilibrium.
• Not every variable has to change. If both gases are in fixed-volume containers, only their pressures change on the way to equilibrium.
• Zeroth law: if A and B are each in thermal equilibrium with a third system C, then A and B are in thermal equilibrium with each other. Test: let A and B each reach equilibrium with C through conducting walls, then put a conducting wall between A and B and insulate C; the states of A and B do not change further.
• R. H. Fowler put this law into words in 1931, long after the first and second laws had been stated and numbered, hence the name 'zeroth'.
• The zeroth law says there is a quantity that has the same value for any two systems in thermal equilibrium. That quantity is temperature T: if T_A = T_C and T_B = T_C, then T_A = T_B.
• So the zeroth law gives the formal basis of temperature; thermometry then deals with putting numbers on it.

3. Internal energy, heat and work

NCERT §11.4

• Temperature marks how hot a body is and fixes the direction of heat flow: from higher to lower temperature, stopping when the temperatures are equal.
• Internal energy U adds up the kinetic and the potential energies of all a system's molecules, measured in the frame in which the system's centre of mass is at rest. It counts only the disordered energy of random molecular motion.
• If a whole box of gas moves with some velocity, the kinetic energy of the box as a whole is not part of U. In a gas, the molecular kinetic energy includes translational, rotational and vibrational motion.
• For a gas, neglecting the small intermolecular forces, U is just the sum of the kinetic energies of the molecules' random motions.
• U is a state variable: its value depends only on the present state (given P, V, T), not on the path by which that state was reached.
• Two ways to change the internal energy of a gas in a cylinder with a piston: put it in contact with a hotter body so heat flows in, or push the piston in so work is done on it. Both can also run the other way: heat flows out to cooler surroundings, or the gas pushes the piston out and does work.
• Heat is energy transferred because of a temperature difference between the system and its surroundings. Work is energy transferred by other means, such as moving a piston by raising or lowering a weight attached to it, with no temperature difference involved.
• Heat is energy in transit. 'This gas holds so much heat' makes no more sense than 'this gas holds so much work'; 'this gas has so much internal energy' is meaningful, as is 'so much heat was supplied' or 'so much work was done'.
• Heat and work are not state variables: they are two ways of moving energy that change U, which is a state variable. Everyday speech often mixes up heat and internal energy; thermodynamics keeps them apart.

4. First law of thermodynamics

NCERT §11.5

• Let ΔQ be the heat supplied to the system by its surroundings, ΔW the work done by the system on its surroundings, and ΔU the change in its internal energy.
• First law: ΔQ = ΔU + ΔW. The heat put in partly raises the internal energy and partly goes out as work. It is conservation of energy for a system that exchanges energy with its surroundings.
• Written as ΔQ − ΔW = ΔU: a gas can go from (P₁, V₁) to (P₂, V₂) by many routes, e.g. first at constant pressure to (P₁, V₂) and then at constant volume, or the other way round. ΔQ and ΔW generally depend on the route, but ΔQ − ΔW = ΔU does not, because U is a state variable.
• If a process has ΔU = 0 (an ideal gas expanding isothermally, say), then ΔQ = ΔW: all the heat supplied is used as work on the surroundings.
• Work by a gas pushing a piston at constant pressure P: force = P × area and area × displacement = volume change, so ΔW = PΔV. Then ΔQ = ΔU + PΔV.
• Worked example, boiling 1 g of water: latent heat 2256 J/g, so ΔQ = 2256 J. Under atmospheric pressure this gram fills 1 cm³ as liquid and 1671 cm³ as vapour.
• Work against the atmosphere: ΔW = P(V_g − V_l) = 1.013 × 10⁵ × (1671 × 10⁻⁶) = 169.2 J. So ΔU = 2256 − 169.2 = 2086.8 J.
• Most of the heat supplied in boiling raises the water's internal energy; only a small part pushes back the atmosphere.
• Sign convention: Q > 0 when heat is added to the system, Q < 0 when heat is taken out; W > 0 when the system does work, W < 0 when work is done on it.

5. Specific heat capacity

NCERT §11.6

• Heat capacity S = ΔQ/ΔT for a temperature rise ΔT. It is proportional to the mass of the body and may vary with temperature.
• Specific heat capacity s = S/m = (1/m)(ΔQ/ΔT), in J kg⁻¹ K⁻¹, depends on the substance and its temperature but not on its amount.
• Molar specific heat capacity C = S/μ = (1/μ)(ΔQ/ΔT), in J mol⁻¹ K⁻¹, for μ moles. It also depends on the conditions under which the heat is supplied.
• Table 11.1, solids at room temperature and atmospheric pressure, specific heat (J kg⁻¹ K⁻¹) then molar specific heat (J mol⁻¹ K⁻¹): aluminium 900.0, 24.4; carbon 506.5, 6.1; copper 386.4, 24.5; lead 127.7, 26.5; silver 236.1, 25.5; tungsten 134.4, 24.9.
• Equipartition for a solid: each atom oscillates about its mean position; a one-dimensional oscillator has average energy 2 × ½k_BT = k_BT, so in three dimensions 3k_BT. For one mole, U = 3k_BT × N_A = 3RT.
• At constant pressure ΔQ = ΔU + PΔV ≈ ΔU for a solid, since its ΔV is tiny. So C = ΔU/ΔT = 3R ≈ 24.9 J mol⁻¹ K⁻¹, which matches the measured values at ordinary temperatures. Carbon (6.1) is the exception, and the agreement fails at low temperatures.
• The calorie was the old unit of heat. Since water's specific heat varies slightly with temperature (Fig. 11.5, 0–100 °C), 1 cal is defined as the heat that takes 1 g of water from 14.5 °C up to 15.5 °C. In SI units water's specific heat is 4186 J kg⁻¹ K⁻¹ = 4.186 J g⁻¹ K⁻¹, so 1 cal = 4.186 J.
• The 'mechanical equivalent of heat' (work needed to produce 1 cal) is only a conversion factor between calorie and joule; with joules used for every form of energy the term is no longer needed.
• For gases the conditions matter, so two specific heats are defined: at constant volume (C_v) and at constant pressure (C_p). For an ideal gas, C_p − C_v = R.
• Proof for 1 mole: at constant volume ΔV = 0, so C_v = ΔU/ΔT. At constant pressure C_p = ΔU/ΔT + P(ΔV/ΔT). U of an ideal gas depends only on T, so ΔU/ΔT is the same in both; and PV = RT gives P(ΔV/ΔT) at constant P = R. Hence C_p − C_v = R.

6. State variables and equation of state

NCERT §11.7

• An equilibrium state is fully described by the values of a few macroscopic state variables. For a gas: pressure, volume, temperature and mass (and composition, for a mixture).
• A system is not always in equilibrium. A gas rushing into vacuum after a partition is suddenly removed (free expansion) has non-uniform pressure; a petrol-vapour and air mixture exploding after a spark has non-uniform temperature and pressure. State variables cannot describe these states.
• Such a system eventually settles to uniform temperature and pressure and to thermal and mechanical equilibrium with its surroundings. State variables describe only equilibrium states.
• The relation connecting the state variables is the equation of state. For an ideal gas it is PV = μRT, so for a fixed amount of gas only two variables are independent, e.g. P and V, or T and V. Real gases can have more complicated equations of state.
• A pressure–volume curve drawn at fixed temperature is an isotherm.
• Extensive variables tell the 'size' of the system; intensive ones do not. Split a system in equilibrium into two equal halves: variables that are halved in each part are extensive, those that stay the same are intensive.
• Extensive: internal energy U, volume V, total mass M. Intensive: pressure P, temperature T, density ρ.
• Checking an equation with this idea: in ΔQ = ΔU + PΔV every term is extensive. P (intensive) × ΔV (extensive) is extensive, and ΔQ is proportional to the mass even though Q is not a state variable.

7. Quasi-static and isothermal processes

NCERT §11.8.1, §11.8.2

• If the external pressure on a gas is suddenly reduced (say a weight is lifted off the piston), the piston accelerates outward and the gas passes through non-equilibrium states without a defined pressure or temperature. A finite temperature difference with the surroundings likewise causes a rapid, non-equilibrium heat exchange.
• A quasi-static ('nearly static') process is an idealised, infinitely slow process in which the system stays in thermal and mechanical equilibrium with its surroundings at every stage. The pressure and temperature of the surroundings differ from the system's only infinitesimally.
• To reach (P′, T′) from (P, T) quasi-statically: change the external pressure by a tiny amount, let the gas equalise, and repeat; for temperature, use a series of reservoirs whose temperatures differ from the gas by tiny amounts, running from T to T′.
• It is a hypothetical construct, but slow processes without accelerating pistons or large temperature gradients come close to it. From here on, processes are taken to be quasi-static unless stated otherwise.
• Table 11.2 special processes: isothermal (temperature constant), isobaric (pressure constant), isochoric (volume constant), adiabatic (no heat flow between system and surroundings, ΔQ = 0).
• An isothermal process: a gas expanding in a metal cylinder placed in a large reservoir at fixed temperature. The reservoir's huge heat capacity keeps its temperature essentially unchanged as it gives or takes heat.
• For an ideal gas at fixed T, PV = constant (Boyle's law): the pressure of a given mass varies inversely as its volume.
• Work in an isothermal change from V₁ to V₂: W = ∫P dV = μRT ∫dV/V = μRT ln(V₂/V₁).
• U of an ideal gas depends only on T, so ΔU = 0 and Q = W. Expansion (V₂ > V₁): W > 0, the gas absorbs heat and does work. Compression (V₂ < V₁): W < 0, work is done on the gas and it gives out heat.
• Heat can flow in an isothermal quasi-static process even though the gas is always at the reservoir's temperature, because an infinitesimal temperature difference is enough to drive it.

8. Adiabatic process

NCERT §11.8.3

• In an adiabatic process the system is insulated, so no heat enters or leaves: ΔQ = 0. The first law then gives ΔW = −ΔU: work done by the gas comes out of its internal energy, so an ideal gas cools as it expands adiabatically.
• For an ideal gas undergoing a quasi-static adiabatic change, PV^γ = constant, where γ = C_p/C_v (ratio of the specific heats, ordinary or molar). The result is quoted without proof.
• So between two states: P₁V₁^γ = P₂V₂^γ. Since γ > 1, an adiabat is steeper than an isotherm through the same point: for the same compression the pressure rises more.
• Fig. 11.8: two adiabats connect two isotherms on a P–V diagram.
• Work done by the gas from (P₁, V₁, T₁) to (P₂, V₂, T₂): W = ∫P dV with P = constant/V^γ, giving W = [P₁V₁ − P₂V₂]/(γ − 1) = μR(T₁ − T₂)/(γ − 1).
• If the gas does work (W > 0), T₂ < T₁: it cools. If work is done on it (W < 0), T₂ > T₁: it warms up.
• Example: a gas with γ = 1.4 (such as hydrogen) compressed adiabatically to half its volume: P₂/P₁ = 2^1.4 ≈ 2.64, while an isothermal halving only doubles the pressure.
• For the same case, TV^(γ−1) = constant gives T₂ = T₁ × 2^0.4 ≈ 1.32 T₁: 300 K becomes about 396 K.

9. Isochoric, isobaric and cyclic processes

NCERT §11.8.4, §11.8.5, §11.8.6

• Isochoric (V constant): the gas neither does work nor has work done on it. All the heat absorbed goes into internal energy and raises the temperature; the rise for a given heat is set by the specific heat at constant volume.
• Isobaric (P constant): the gas does work W = P(V₂ − V₁) = μR(T₂ − T₁).
• In an isobaric process the temperature changes, so U changes too. The heat absorbed goes partly into internal energy and partly into work; the temperature rise for a given heat is set by the specific heat at constant pressure.
• That is why C_p > C_v: at constant pressure part of the heat leaves as work, so more heat is needed for the same temperature rise.
• Cyclic process: the system returns to its initial state. U is a state variable, so ΔU = 0 over a cycle, and the net heat absorbed equals the net work done by the system.
• On a P–V diagram the work in a process is the area under its curve; for a cycle, the net work is the area enclosed by the loop.
• Summary of the first law in the four special processes: isothermal ΔU = 0, Q = W; adiabatic Q = 0, W = −ΔU; isochoric W = 0, Q = ΔU; isobaric Q = ΔU + PΔV.

10. Second law of thermodynamics

NCERT §11.9

• The first law is energy conservation, but many processes that conserve energy are never seen. A book lying on a table never jumps up by itself, even though the table could, in principle, cool slightly and hand that energy to the book as mechanical energy.
• The principle that rules out such processes, while allowing everything else the first law allows, is the second law of thermodynamics.
• It sets a basic limit: the efficiency of a heat engine can never be 1, and the coefficient of performance of a refrigerator can never be infinite.
• Kelvin–Planck statement: no process can have, as its only outcome, taking heat from a reservoir and turning all of it into work. A perfect heat engine is impossible.
• Clausius statement: no process can have, as its only outcome, moving heat from a colder body to a hotter one. A perfect refrigerator or heat pump is impossible.
• The two statements can be shown to be completely equivalent: breaking either one would let you break the other.

11. Reversible and irreversible processes

NCERT §11.10

• A process from state i to state f is reversible if it can be run backwards so that both the system and its surroundings return to their original states, with no change anywhere else in the universe.
• The spontaneous processes of nature are irreversible. A pot taken off the stove has a hotter base; heat spreads until it is uniform, and the pot never cools on one side to reheat its base. That would violate the second law.
• More irreversible processes: free expansion of a gas; the burning of a petrol–air mixture ignited by a spark; cooking gas leaking from a cylinder and spreading through a kitchen, which never gathers itself back into the cylinder.
• Stirring a liquid in contact with a reservoir turns work into heat in the reservoir. Undoing it exactly would mean turning that heat wholly into work, against the second law.
• Two main causes of irreversibility: the process passes through non-equilibrium states (free expansion, explosive reactions), or it involves friction, viscosity and other dissipative effects, such as a sliding body coming to rest or a spinning blade in a liquid being stopped by viscosity.
• Dissipative effects can be reduced but never removed completely, so irreversibility is the rule in nature rather than the exception.
• Reversibility needs two things together: the process must be quasi-static and free of dissipation. Example: slow isothermal expansion of an ideal gas in a cylinder with a frictionless piston.
• Reversibility matters because an engine built only from reversible processes reaches the highest efficiency possible between two temperatures; any engine with irreversibility, as every real engine has, does worse.

12. Carnot engine

NCERT §11.11

• Question posed by Sadi Carnot, a French engineer, in 1824: with a hot reservoir at T₁ and a cold one at T₂, what is the largest possible efficiency of a heat engine, and what cycle gives it? He found the right answer before the basic ideas of heat were firmly settled.
• The ideal engine must be reversible. Heat exchange with a finite temperature difference is not quasi-static, so heat must be taken in isothermally at T₁ and given out isothermally at T₂.
• To move the working substance between T₁ and T₂ without other reservoirs, only reversible adiabatic steps will do; any other process, e.g. isochoric, would need a whole series of reservoirs between T₂ and T₁.
• A reversible engine working between two temperatures is a Carnot engine. The Carnot cycle, with an ideal gas: 1→2 isothermal expansion at T₁, absorbing Q₁ = W₁₂ = μRT₁ ln(V₂/V₁); 2→3 adiabatic expansion from T₁ to T₂, work by the gas μR(T₁ − T₂)/(γ − 1).
• 3→4 isothermal compression at T₂, releasing Q₂ = W₃₄ = μRT₂ ln(V₃/V₄) (work done on the gas); 4→1 adiabatic compression from T₂ back to T₁, work on the gas μR(T₁ − T₂)/(γ − 1).
• The two adiabatic works cancel. Net work W = μRT₁ ln(V₂/V₁) − μRT₂ ln(V₃/V₄), and the adiabatic steps (TV^(γ−1) = constant) give V₃/V₄ = V₂/V₁.
• Efficiency: η = W/Q₁ = 1 − Q₂/Q₁ = 1 − T₂/T₁, with temperatures in kelvin. For example, between 500 K and 300 K, η = 1 − 300/500 = 0.40 = 40%.
• Every step can be reversed. Run backwards, the cycle takes Q₂ from the cold reservoir, has work W done on it, and delivers Q₁ to the hot reservoir: a reversible refrigerator.
• Carnot's theorem: (a) no engine working between two given temperatures is more efficient than a Carnot engine; (b) the Carnot efficiency does not depend on the working substance.
• Proof of (a): couple an irreversible engine I, used as an engine, to a Carnot engine R, run as a refrigerator that returns the same Q₁ to the source. If I were more efficient (W′ > W), the pair would take W′ − W of heat from the cold reservoir and turn it wholly into work with no other change, which the Kelvin–Planck statement forbids.
• Since Q₂/Q₁ = T₂/T₁ holds for every Carnot engine, whatever the substance, it can define a universal thermodynamic temperature scale; with an ideal gas as the working substance this scale is the same as the ideal-gas temperature.

Must-know facts

1. Rumford's cannon boring showed heat is energy produced by work, not a fluid (caloric).
2. Adiabatic wall: no heat passes; diathermic wall: heat passes.
3. Zeroth law: two systems each in equilibrium with a third are in equilibrium with each other; it defines temperature.
4. Internal energy U excludes the kinetic energy of the system moving as a whole.
5. U is a state variable; heat and work are not, they are modes of energy transfer.
6. First law: ΔQ = ΔU + ΔW; ΔQ − ΔW is path-independent.
7. Boiling 1 g of water: ΔQ = 2256 J, ΔW = 169.2 J, ΔU = 2086.8 J.
8. Solids: molar heat ≈ 3R (Table 11.1), carbon excepted.
9. 1 cal = 4.186 J; water's specific heat 4186 J kg⁻¹ K⁻¹.
10. C_p − C_v = R for an ideal gas.
11. Extensive: U, V, M. Intensive: P, T, ρ.
12. Quasi-static: infinitely slow, always in equilibrium with the surroundings.
13. Isothermal: PV = constant, ΔU = 0, W = Q = μRT ln(V₂/V₁).
14. Adiabatic: Q = 0, PV^γ = constant, W = μR(T₁ − T₂)/(γ − 1); expansion cools, compression heats.
15. Adiabat is steeper than isotherm on a P–V diagram.
16. Isochoric: W = 0, Q = ΔU. Isobaric: W = PΔV = μRΔT.
17. Cyclic: ΔU = 0, net Q = net W = area of the loop.
18. Kelvin–Planck: no perfect heat engine. Clausius: no perfect refrigerator. They are equivalent.
19. Reversible needs quasi-static and no dissipation; natural processes are irreversible.
20. Carnot cycle: two isothermals joined by two adiabatics; η = 1 − T₂/T₁.
21. No engine between T₁ and T₂ beats Carnot, whatever the working substance.

Common traps

Formulas

Key terms

Caloric

The discarded idea of heat as an invisible fluid flowing from hot to cold bodies.

Thermal equilibrium

The state in which a system's macroscopic variables no longer change in time.

Adiabatic wall

An insulating wall that lets no heat through.

Diathermic wall

A conducting wall that lets heat flow between systems.

Zeroth law

Two systems each in thermal equilibrium with a third are in thermal equilibrium with each other.

Internal energy

Sum of the molecular kinetic and potential energies, measured in the centre-of-mass frame.

State variable

A quantity fixed by the present equilibrium state alone, not by the path taken, e.g. P, V, T, U.

Equation of state

The relation between the state variables, e.g. PV = μRT for an ideal gas.

Quasi-static process

An infinitely slow process in which the system stays in equilibrium with its surroundings.

Isothermal process

A process at constant temperature.

Adiabatic process

A process with no heat exchange between system and surroundings.

Isochoric process

A process at constant volume.

Isobaric process

A process at constant pressure.

Reversible process

A process that can be undone so that system and surroundings both return to their initial states with no other change.

Carnot engine

A reversible engine between two temperatures, working on two isothermal and two adiabatic steps.