Thermodynamics

Thermodynamics is the branch of physics that keeps track of energy as it moves between a system and its surroundings in the form of heat and work. For NEET it is a high-yield chapter because almost every question reduces to applying one equation (the first law) with the correct signs, or plugging temperatures into the Carnot efficiency formula.

The System, Its Surroundings, and the Zeroth Law

A thermodynamic system is whatever portion of matter we choose to study — typically a fixed mass of gas inside a cylinder. Everything outside it that can exchange energy with it is the surroundings, and the real or imaginary boundary between them decides what can cross. If the boundary blocks heat flow it is adiabatic (an insulated wall); if it permits heat flow it is diathermic (a thin metal wall). A system sealed off from all exchange of energy and matter is called isolated.

The condition of a system is specified by state variables — pressure P, volume V, temperature T, internal energy U, number of moles n. These have definite values only when the system is in equilibrium, meaning P and T are uniform throughout and not changing with time. For an equilibrium state of an ideal gas the state variables are tied together by the equation of state, PV = nRT, so only two of P, V, T can be chosen independently.

The Zeroth Law states that if system A is in thermal equilibrium with system C, and B is also in thermal equilibrium with C, then A and B are in thermal equilibrium with each other. This apparently obvious statement is what allows temperature to exist as a meaningful physical quantity: temperature is that property which is common to all bodies in mutual thermal equilibrium. It is also the logical basis of thermometry — a thermometer (C) placed successively in two bodies reads the same value only if those bodies would exchange no net heat with each other.

A useful classification of variables:

• Extensive — depend on the amount of matter: mass, volume, total internal energy, entropy.
• Intensive — independent of amount: pressure, temperature, density, specific heat capacity.

Internal Energy, Heat, Work and the First Law

Internal energy (U) is the total energy stored inside the system, measured in the frame in which the system as a whole is at rest — the sum of the kinetic energies of random molecular motion plus the potential energies of intermolecular interaction. It is a state function: its value depends only on the present state, not on how the system got there. For an ideal gas there are no intermolecular forces, so U depends on temperature alone: U = (f/2)nRT, where f is the number of degrees of freedom. Hence for an ideal gas, ΔT = 0 implies ΔU = 0, whatever else happens.

Heat (Q) and work (W), in contrast, are not properties of a system. They are two different modes of energy transfer across the boundary, and each depends on the path followed. Heat flows because of a temperature difference; work is done when the boundary moves against a pressure, or through stirring, electrical dissipation, etc. It is meaningless to say "the gas contains 200 J of heat" — heat exists only while it is crossing the boundary.

The First Law of Thermodynamics is conservation of energy for a thermodynamic system:

ΔQ = ΔU + ΔW

with the standard convention used in NCERT and NEET:

• Q is positive when heat is absorbed by the system, negative when released.
• W is positive when work is done by the system (expansion), negative when work is done on it (compression).
• ΔU is positive when internal energy (and hence, for an ideal gas, temperature) increases.

For a quasi-static process — one carried out so slowly that the system passes through a continuous succession of equilibrium states — the work done by a gas when its volume changes by dV is dW = P dV, so

W = ∫P dV = area under the P–V curve.

The area is counted positive for expansion (rightward path) and negative for compression (leftward path). Two consequences worth memorising: for a cyclic process the system returns to its initial state, so ΔU = 0 and Q = W, the work equalling the area enclosed by the loop (positive for a clockwise cycle, negative for anticlockwise). For a free expansion into vacuum, which is adiabatic (Q = 0) and involves no opposing pressure (W = 0), we get ΔU = 0 — an ideal gas undergoing free expansion does not change temperature, even though the process is violently non-quasi-static.

Molar Specific Heats and the Relation C_p − C_v = R

The heat needed to raise the temperature of a substance depends on how the heating is done. The molar specific heat capacity is C = (1/n)(ΔQ/ΔT). Since Q is path-dependent, gases need two values:

• C_v — molar specific heat at constant volume. Here ΔW = 0, so ΔQ = ΔU, giving C_v = (1/n)(dU/dT).
• C_p — molar specific heat at constant pressure. Part of the heat now leaves as expansion work, so C_p > C_v always.

For one mole at constant pressure: ΔQ = ΔU + PΔV, i.e. C_p ΔT = C_v ΔT + RΔT (using PV = RT), which gives the Mayer relation

C_p − C_v = R (per mole)

Defining the ratio γ = C_p/C_v, the kinetic-theory results follow from U = (f/2)nRT:

Useful working forms: ΔU = nC_vΔT for any ideal-gas process (not just constant volume), and C_v = R/(γ − 1), C_p = γR/(γ − 1).

The Four Standard Processes

1. Isothermal (T constant). The gas is kept in contact with a large reservoir and changed slowly, so PV = constant (Boyle's law). Since ΔU = 0,

W = Q = nRT ln(V₂/V₁) = 2.303 nRT log₁₀(P₁/P₂)

All the heat absorbed converts entirely into work. The P–V graph is a hyperbola.

2. Adiabatic (Q = 0). No heat is exchanged — either the walls are insulating or the change is very rapid (sudden compression of a gas, propagation of sound). Then ΔU = −W: the gas cools when it expands and heats when compressed. The governing relations are

PV^γ = constant, TV^(γ−1) = constant, P^(1−γ)T^γ = constant

and the work done is

W = (P₁V₁ − P₂V₂)/(γ − 1) = nR(T₁ − T₂)/(γ − 1)

On a P–V diagram an adiabatic curve is steeper than an isothermal through the same point, since its slope is γ times as large in magnitude.

3. Isobaric (P constant). W = PΔV = nRΔT, Q = nC_pΔT, ΔU = nC_vΔT. The P–V graph is a horizontal line.

4. Isochoric / isovolumetric (V constant). W = 0, so Q = ΔU = nC_vΔT. All heat supplied goes into raising the temperature. The graph is a vertical line.

Related terms: a reversible process can be run backwards through the same equilibrium states, returning both system and surroundings to their original condition; it requires infinite slowness and zero dissipation, so it is an idealisation. All real processes involving friction, viscosity, finite temperature differences or turbulence are irreversible.

Second Law, Heat Engines, Refrigerators and the Carnot Cycle

The first law permits many processes that never happen: heat spontaneously leaving a cold body for a hot one, or a gas absorbing heat from a single reservoir in a cycle and converting all of it to work. The Second Law rules these out. Two equivalent statements:

• Kelvin–Planck: no cyclic engine can take heat from a single reservoir and convert it completely into work; some heat must be rejected to a colder body.
• Clausius: no self-acting device can transfer heat from a colder body to a hotter one without external work.

A heat engine works in cycles, absorbing Q₁ from a hot source at T₁, doing work W, and rejecting Q₂ to a sink at T₂. Since ΔU = 0 over a cycle, W = Q₁ − Q₂, and the efficiency is

η = W/Q₁ = 1 − Q₂/Q₁

η = 1 (i.e. Q₂ = 0) is forbidden by the second law.

A refrigerator or heat pump is an engine run in reverse: work W is supplied to extract Q₂ from a cold body and dump Q₁ = Q₂ + W into the surroundings. Its coefficient of performance is

α = Q₂/W = Q₂/(Q₁ − Q₂), and for a Carnot refrigerator α = T₂/(T₁ − T₂)

Note that α can be much larger than 1, which is why it is never called "efficiency".

The Carnot engine is the ideal reversible engine, using an ideal gas taken through four quasi-static steps:

1. Isothermal expansion at T₁, absorbing Q₁ from the source.
2. Adiabatic expansion, cooling the gas from T₁ to T₂.
3. Isothermal compression at T₂, rejecting Q₂ to the sink.
4. Adiabatic compression, restoring the gas to its initial state at T₁.

Applying the isothermal and adiabatic relations to the four legs gives Q₁/Q₂ = T₁/T₂, hence

η_Carnot = 1 − T₂/T₁ (T in kelvin)

Carnot's theorem: no engine working between two given temperatures can be more efficient than a reversible (Carnot) engine, and the efficiency of a Carnot engine is independent of the working substance. Practical consequences: efficiency rises when the source is hotter or the sink colder, and 100% efficiency would demand a sink at absolute zero. Conversely, an efficient engine is a poor refrigerator between the same temperatures, since α = (1 − η)/η — as η approaches 1, α approaches 0, so an engine tuned for very high efficiency between two temperatures would make an extremely poor refrigerator working between those same two temperatures, and vice versa.

Common Mistakes and Exam Traps

• Mixing up the sign convention. The NCERT form ΔQ = ΔU + ΔW takes W as positive when done by the gas (expansion); other textbooks write ΔU = Q − W with the same physical meaning, or define W as positive when done on the gas. Fix which convention a problem is using before assigning any signs — this single confusion causes more wrong answers in this chapter than any calculation error.
• Assuming ΔU = 0 whenever a process is called "isothermal," for any substance. This shortcut is valid only for an ideal gas, where internal energy depends on temperature alone; it fails for real gases and for phase changes at constant temperature, where melting or boiling absorbs latent heat while ΔU still changes.
• Treating free expansion as identical to a slow isothermal expansion. Both leave an ideal gas's temperature unchanged, but free expansion is sudden, irreversible, and does zero work (there is nothing to push against), whereas isothermal expansion is quasi-static and does positive work equal to the heat absorbed — wrongly assigning W = nRT ln(V₂/V₁) to a free expansion is a common mistake.
• Forgetting that an adiabatic curve is steeper than an isothermal curve through the same point. On a P–V diagram, misjudging which of two intersecting curves is adiabatic — and hence which process does more work for a given volume change — is a frequent graph-reading error.