ThermodynamicsNEET Physics · Class 11 · NCERT Chapter 11

Introduction

Thermodynamics is the physics of heat, work and energy as they move between a system and its surroundings. It does not care about how a system is built at the atomic level. It just bookkeeps the flow of energy and tells us which directions are allowed.

For NEET 2027 you can expect 1 to 2 questions from this chapter. The repeated favourites are: the first law with sign conventions, isothermal versus adiabatic processes, the Mayer relation between $C_P$ and $C_V$, Carnot efficiency, and refrigerator COP. Master these five and you have most of the marks.

Two big laws run the chapter. The first law says energy is conserved when you account for heat. The second law says you cannot turn all of that heat into useful work. Together they explain everything from a steam engine to a household fridge.

System, surroundings and state variables

A system is the part of the universe you decide to study (a gas in a cylinder, a cup of coffee). Everything outside it is the surroundings. The boundary between them can be:

• Isolated: no exchange of matter or energy. A perfect thermos is the model.
• Closed: energy can cross but matter cannot. A sealed metal can.
• Open: both energy and matter cross. An open beaker.

A system is in a particular state when its macroscopic variables (P, V, T, n) all have definite values. Variables that depend only on the current state, not on the path used to reach it, are called state variables: pressure, volume, temperature, internal energy, entropy. Heat and work are not state variables; they depend on the path.

Zeroth law of thermodynamics

If two systems are each in thermal equilibrium with a third system, they are also in thermal equilibrium with each other. This sounds obvious but it is the law that lets us define temperature as a property: two systems at the same temperature stay at equilibrium when brought together.

Practical use: a thermometer is body C. When it reads the same value with body A and with body B, then A and B are at the same temperature.

Heat, work and internal energy

These three are the core quantities of the first law. Mix them up and the signs go wrong.

• Internal energy U: the total kinetic plus potential energy of all the molecules. For an ideal gas it depends only on temperature: $U=n{C}_{V}T$. State function.
• Heat Q: energy transferred because of a temperature difference. Not stored; it is energy in transit. Path function. NEET sign: Q is positive when heat enters the system.
• Work W: energy transferred by a force acting through a distance. For a gas pushing a piston: $W=\int PdV$. Path function. NEET sign: W is positive when the gas expands (does work on the surroundings).

First law of thermodynamics

Energy conservation, written for thermal systems:

Heat added to a system goes into changing its internal energy and into doing work. The same equation rearranged is also useful: $\Delta U=Q-W$.

Watch the sign convention. Some books use $\Delta U=Q+W$ with W as the work done on the system. NEET (and NCERT) use the IUPAC convention with W as the work done by the gas. We follow that throughout.

Pick which quantity to solve for. Enter the other two. Sign convention: Q positive when heat enters, W positive when gas expands.

Memory hook for first-law signs

• Heat added to gas → Q is positive.
• Heat lost by gas → Q is negative.
• Gas expands (pushes piston out) → W is positive.
• Gas is compressed (piston pushed in) → W is negative.
• Temperature rises (for ideal gas) → ΔU is positive. Falls → negative.

Specific heats of an ideal gas (Cp and Cv)

For a solid or a liquid one specific heat is enough. For a gas, the heat needed to change temperature depends on whether you keep volume fixed or pressure fixed:

• Cv (heat at constant volume): all the heat goes into raising T. $Q=n{C}_V\Delta T=\Delta U$. (No volume change ⇒ no work.)
• Cp (heat at constant pressure): heat raises T and the gas expands, doing work on the surroundings. So Cp must be greater than Cv.

Mayer's relation links them:

From kinetic theory, the internal energy of an ideal gas is $\frac{f}{2}{k}_{B}T$ per molecule, where f is the number of degrees of freedom. So:

Standard values to remember:

• Monoatomic (He, Ne, Ar): f = 3, Cv = (3/2)R, Cp = (5/2)R, γ = 5/3 ≈ 1.667.
• Diatomic (H₂, O₂, N₂) at room T: f = 5, Cv = (5/2)R, Cp = (7/2)R, γ = 7/5 = 1.4.
• Polyatomic (CO₂, NH₃) non-linear: f = 6, Cv = 3R, Cp = 4R, γ = 4/3 ≈ 1.333.

Degrees of freedom f set Cv. Mayer's relation gives Cp. Their ratio is gamma.

Examples: H₂, O₂, N₂

Work in a PV diagram

For a quasi-static process the gas exerts pressure P and pushes the piston by dV. The infinitesimal work is $dW=PdV$. For a finite change:

Geometrically, the work equals the area under the curveon a P versus V plot. If the gas expands (V₂ > V₁) the area is positive and W > 0. If the gas is compressed, W < 0.

Isothermal process

Temperature stays constant (T = const). For an ideal gas, U depends only on T, so $\Delta U=0$. The first law collapses to:

Work done in an isothermal expansion from V₁ to V₂:

Heat absorbed equals the work done, so every joule of heat that flows in becomes useful work. To keep T fixed, the process must happen slowly, in good thermal contact with a reservoir.

Isobaric and isochoric processes

Isobaric (P constant)

Easy to read on a PV diagram: a horizontal line, work = P × ΔV (a rectangle).

Isochoric (V constant)

A vertical line on the PV diagram. No volume change means no piston motion, so no work. All heat goes into raising temperature (or lowering it, if cooled).

Adiabatic process

No heat is exchanged with the surroundings (Q = 0). Either the system is well insulated, or the change is so fast that there is no time for heat to flow. The first law becomes:

Equation of state for a quasi-static adiabatic change:

Work done by the gas:

Adiabatic compression heats the gas (think of a bicycle pump getting hot). Adiabatic expansion cools it (a compressed gas escaping a CO₂ canister forms frost).

Same start point, same final volume, but different processes. The adiabat is steeper than the isotherm because temperature also drops (during expansion). For the same expansion the adiabat does less work.

Adiabatic vs isothermal: the slope

Slope of an adiabat at any point is steeper than the isotherm by a factor of γ:

That is why for the same expansion, the adiabat does less work than the isotherm: the pressure drops faster.

Cyclic processes

In a cyclic process the system returns to its starting state after one full cycle. So:

The net work done in one cycle equals the area enclosed on the PV diagram. Direction sets the sign:

• Clockwise cycle: W_net > 0, a heat engine. Gas does net work each cycle.
• Counter-clockwise cycle: W_net < 0, a refrigerator. Net work must be supplied to the gas.

A rectangular cycle on the PV diagram. Net work in one cycle = enclosed area. Direction matters: clockwise = engine (work done by gas, positive); counter-clockwise = refrigerator (work done on gas, negative).

Second law of thermodynamics

The first law allows many processes that we never see in real life, like heat flowing from cold to hot, or a glass un-shattering. The second law forbids them. Two equivalent statements:

• Kelvin Planck: no engine can convert heat completely into work in a cycle. There must be a cold reservoir to dump some heat into.
• Clausius: heat cannot flow on its own from a colder body to a hotter body. To reverse the flow you must do work (this is what your fridge does).

Equivalent way to state it: the entropy of an isolated system never decreases. Natural processes are irreversible because they increase total entropy.

Heat engine and the Carnot cycle

A heat engine takes in Q_h from a hot reservoir, converts part of it into work W, and rejects Q_c to a cold reservoir. By energy conservation:

The Carnot cycle is the most efficient cycle possible between two reservoirs. It uses two isotherms and two adiabats. For a Carnot engine the heat ratio equals the temperature ratio:

Both temperatures must be in kelvin. No real engine between the same two reservoirs can exceed this efficiency . That is Carnot's theorem.

Carnot is the most efficient possible engine between two reservoirs. Real engines fall below this limit.

Refrigerator and heat pump

Run a heat engine in reverse. Now you put work W in, pull heat Q_c from a cold space and dump Q_h = Q_c + W into the hot side. We measure how much heat is moved per joule of work using the coefficient of performance:

A heat pump is the same machine, but the goal is the heat dumped at the warm side (heating a room). Its COP is always one more than the refrigerator's:

A refrigerator pumps heat from cold to hot. You pay work W and you get Q_c removed from the inside. The same machine used in reverse becomes a heat pump.

Entropy (brief)

Entropy S is a state function that measures the "spread" of energy. For a reversible heat exchange:

Two takeaways for NEET:

• For an isolated system, ΔS ≥ 0. Equality holds only for reversible processes.
• Entropy of the universe (system plus surroundings) always increases for any real process.

Entropy gives the second law a quantitative form: natural processes go in the direction that raises total entropy.

Worked NEET problems

Summary cheat sheet

• First law: $Q=\Delta U+W$. Q in, W out (gas does work).
• Internal energy (ideal gas): $U=n{C}_{V}T$. Depends only on T.
• Mayer: $C_P-C_V=R$. $\gamma =1+2/f$.
• γ values: mono 5/3, di 7/5, polyatomic 4/3.
• Isothermal: $PV=\text{const}$, $\Delta U=0$, $W=nRT\ln (V_2/V_1)$, $Q=W$.
• Isobaric: $W=P\Delta V$, $Q=n{C}_P\Delta T$.
• Isochoric: $W=0$, $Q=\Delta U=n{C}_V\Delta T$.
• Adiabatic: $P{V}^{\gamma }=\text{const}$, $Q=0$, $W=(P_1{V}_1-P_2{V}_2)/(\gamma -1)$.
• Cyclic: $\Delta U=0$, $W_{\text{net}}=$ enclosed area.
• Heat engine: $\eta =W/Q_h=1-Q_c/Q_h$.
• Carnot: ${\eta }_{\text{C}}=1-T_c/T_h$. T in kelvin. Maximum possible.
• Refrigerator COP: $Q_c/W=T_c/(T_h-T_c)$.
• Heat pump COP: $Q_h/W=1+{\text{COP}}_{\text{fridge}}$.

Next: try the interactive widgets for first law, PV processes, Carnot and refrigerator COP, or work through the 32 NEET PYQs with full solutions. To time yourself, take the free 10-question mock test.