NCERT Solutions for Class 11 Physics Chapter 11: Thermodynamics

Chapter at a glance — complete revision notes

11.1 Introduction

In Chapter 10 we studied temperature, heat, and how heat flows by conduction, convection and radiation. Thermodynamics goes a step further: it deals with the relation between heat, work and the internal energy of a system, and sets limits on how heat can be converted into work. The subject was born in the steam-engine age. Today the same laws explain refrigerators, air-conditioners, car engines and power stations, the weather, and even the behaviour of stars.

Thermodynamics is a macroscopic science. It describes a system by a few measurable large-scale variables such as pressure P, volume V, temperature T, mass and composition, without asking what individual molecules are doing. (The molecular picture comes in the next chapter, kinetic theory.) In thermodynamics the kinetic energy of the system as a whole (for example, a box of gas moving on a trolley) and the potential energy of the system in an external field are not considered: only the internal energy of the random molecular motion matters.

Key terms at a glance

11.2 Thermal Equilibrium

A gas enclosed in a rigid, completely insulated container with fixed P, V, T, mass and composition that do not change with time is in a state of thermodynamic equilibrium. Whether a system is in equilibrium depends on its surroundings and on the nature of the wall that separates them.

Take two gases A and B, with pressure and volume (P_A, V_A) and (P_B, V_B). For a fixed mass of gas, P and V are two independent variables.

• Adiabatic wall (Fig. 11.1a). A and B are separated by an insulating wall. Then any pair of values (P_A, V_A) is in equilibrium with any pair (P_B, V_B). No heat flows, so the two gases do not influence each other.
• Diathermic wall (Fig. 11.1b). The wall conducts heat. The variables of A and B change spontaneously, say to (P_A′, V_A′) and (P_B′, V_B′), until the two gases are in equilibrium with each other and no more energy flows. We then say A is in thermal equilibrium with B.

If the gases are in rigid containers of fixed volume, only the pressures change during thermal equilibration; both variables need not change. It depends on the constraints. The question is: what is common to A and B once they are in thermal equilibrium? Experience tells us the answer: the temperatures are equal. The Zeroth Law gives this idea a firm logical footing.

11.3 Zeroth Law of Thermodynamics

Place systems A and B, separated by an adiabatic wall, each in contact with a third system C through a conducting wall (Fig. 11.2a). The states of A and B change until both are in thermal equilibrium with C. Now insulate C from both with an adiabatic wall and replace the wall between A and B by a conducting one (Fig. 11.2b). It is found that the states of A and B do not change any further: they are in thermal equilibrium with each other.

Zeroth Law of Thermodynamics: two systems in thermal equilibrium with a third system separately are in thermal equilibrium with each other. R. H. Fowler formulated it in 1931, long after the first and second laws had been stated and numbered, which is why it carries the number zero.

What the Zeroth Law gives us

• There exists a physical quantity with the same value for all systems that are in thermal equilibrium: this is temperature T. If T_A = T_C and T_B = T_C, then T_A = T_B.
• Temperature determines the direction of heat flow: heat flows from the body at higher temperature to the body at lower temperature, and flow stops when temperatures are equal.
• It justifies the use of a thermometer: the thermometer (system C) is brought into contact with a body and gives its temperature once it reaches thermal equilibrium.

11.4 Heat, Internal Energy and Work

The Zeroth Law gave us temperature, the marker of “hotness”. We now describe three related ideas: internal energy, heat and work.

Internal energy U

Every bulk system consists of a very large number of molecules. Internal energy is the sum of the kinetic and potential energies of its molecules, measured in the frame of reference in which the centre of mass of the system is at rest. It therefore includes only the disordered energy of random molecular motion. If the intermolecular forces of a gas are neglected, U of the gas is just the sum of the kinetic energies of the molecules: translational, rotational and vibrational (Fig. 11.3).

The key point: U is a state variable. Its value depends only on the state of the system (its P, V, T), not on how that state was reached. Pressure, volume, temperature and internal energy are state variables.

Heat and work: two modes of energy transfer

Take a gas in a cylinder with a movable piston (Fig. 11.4). Its internal energy can be changed in two ways: (a) put the cylinder in contact with a body at a higher temperature, so energy (heat) flows in because of the temperature difference; (b) push the piston down, i.e. do work on the gas, by means that involve no temperature difference. Either can occur in the reverse direction: heat can flow out, or the gas can push the piston up and do work on the surroundings.

11.5 First Law of Thermodynamics

Let ΔQ be the heat supplied to the system by the surroundings, ΔW the work done by the system on the surroundings and ΔU the change in its internal energy. Conservation of energy gives

ΔQ = ΔU + ΔW (11.1) i.e. ΔU = ΔQ − ΔW (11.2)

This is the First Law of Thermodynamics: the energy ΔQ supplied to the system goes partly to increase its internal energy and the rest does work on the environment. It is simply the law of conservation of energy applied to a system with heat and work exchange taken into account.

Path dependence

A system can go from state (P₁, V₁) to (P₂, V₂) by many paths, e.g. first at constant pressure and then at constant volume, or the other way round. ΔU depends only on the end states, so it is the same for all paths; ΔQ and ΔW are in general path-dependent; but their difference ΔQ − ΔW is path-independent.

Sign convention (NCERT)

Work done by a gas

For a gas pushing a piston of area A through a distance Δx against constant pressure P, ΔW = P A Δx = P ΔV. Equation (11.1) becomes

ΔQ = ΔU + P ΔV (11.3)

If the process has ΔU = 0 (isothermal expansion of an ideal gas), ΔQ = ΔW: all the heat supplied is used up in doing work.

Worked illustration: boiling 1 g of water at 1 atm

Latent heat of vaporisation: ΔQ = 2256 J for 1 g. Volume of 1 g of water = 1 cm³ (liquid), 1671 cm³ (steam). At P = 1.013 × 10⁵ Pa:

ΔW = P(V_g − V_l) = 1.013 × 10⁵ × 1671 × 10⁻⁶ = 169.2 J

ΔU = ΔQ − ΔW = 2256 − 169.2 = 2086.8 J

Most of the heat (about 92.5 %) increases the internal energy of the water (overcoming molecular attraction when going from liquid to vapour); only about 7.5 % is spent in pushing back the atmosphere.

11.6 Specific Heat Capacity

If heat ΔQ raises the temperature of a substance from T to T + ΔT, its heat capacity is S = ΔQ/ΔT. Dividing by the mass m (kg) gives the specific heat capacity, which characterises the material and not the amount:

s = S/m = (1/m)(ΔQ/ΔT) [J kg⁻¹ K⁻¹]

If the amount is given in moles μ, the molar specific heat capacity is C = S/μ = (1/μ)(ΔQ/ΔT) [J mol⁻¹ K⁻¹]. Both depend on the nature of the substance, its temperature and, for gases, the conditions under which heat is supplied.

Solids: law of equipartition (Dulong–Petit)

An atom in a solid vibrates about its mean position. A one-dimensional oscillator has average energy 2 × ½ k_BT = k_BT, so in three dimensions the average energy per atom is 3k_BT. For one mole: U = 3k_BT N_A = 3RT. At constant pressure ΔV for a solid is negligible, so ΔQ ≈ ΔU and

C = ΔQ/ΔT = ΔU/ΔT = 3R ≈ 25 J mol⁻¹ K⁻¹

This agrees with experiment at ordinary temperatures (carbon, in the form of diamond, is an exception) and fails at low temperatures. Table: some measured values at room temperature.

Specific heat of water and the calorie

The old unit of heat, the calorie, was the heat needed to raise 1 g of water through 1 °C; since s of water varies slightly with temperature (Fig. 11.5), the calorie is precisely defined as the heat required to raise 1 g of water from 14.5 °C to 15.5 °C. In SI, s_water = 4186 J kg⁻¹ K⁻¹ = 4.186 J g⁻¹ K⁻¹, and 1 cal = 4.186 J. The “mechanical equivalent of heat” is just this conversion factor between two units of energy.

Cp − Cv = R for an ideal gas

For 1 mole of an ideal gas, ΔQ = ΔU + PΔV. At constant volume (ΔV = 0): C_v = (ΔU/ΔT). At constant pressure: C_p = (ΔU/ΔT) + P(ΔV/ΔT)_p. For an ideal gas U depends only on T, so the first term is the same in both, and PV = RT gives P(ΔV/ΔT)_p = R. Hence

C_p − C_v = R (11.8)

C_p exceeds C_v because at constant pressure part of the heat is used up in doing work of expansion, not in raising the temperature.

11.7 Thermodynamic State Variables and Equation of State

Every equilibrium state of a thermodynamic system is completely described by specific values of some macroscopic variables called state variables: for a gas, pressure, volume, temperature and mass (and composition for a mixture). A system is not always in equilibrium: a gas expanding freely into vacuum, or a petrol-air mixture exploding after a spark, has non-uniform pressure and temperature during the process, and cannot be described by state variables until it settles (Fig. 11.6).

Equation of state

The relation connecting the state variables is the equation of state. For an ideal gas it is PV = μRT. For a fixed amount of gas (given μ) only two variables are independent: say (P, V) or (T, V). The P–V curve at constant temperature is an isotherm. Real gases obey more complicated equations of state (e.g. van der Waals).

Extensive and intensive variables

Imagine dividing a system in equilibrium into two equal halves. Variables whose values stay the same in each half are intensive; variables whose values get halved are extensive.

Checking the consistency of an equation using this classification is good practice: in ΔQ = ΔU + PΔV every term is extensive (an intensive P multiplied by an extensive ΔV is extensive).

11.8 Thermodynamic Processes

11.8.1 Quasi-static process

A gas in thermal and mechanical equilibrium with its surroundings has pressure equal to the external pressure and temperature equal to that of the surroundings. If the external pressure is suddenly reduced, the piston accelerates, the gas is not in equilibrium during the change and cannot be described by (P, V, T). To keep the system always in equilibrium, the process must be carried out infinitely slowly: change the external pressure by an infinitesimal amount (for instance by removing sand grains one by one), and change the temperature by an infinitesimal amount by bringing the system in contact with reservoirs at progressively different temperatures (Fig. 11.7). Such a process is quasi-static: at every stage the system’s pressure and temperature differ only infinitesimally from those of the surroundings.

A quasi-static process is a hypothetical construct; in practice a process that is sufficiently slow, with no accelerated piston motion, large temperature gradients or turbulence, approximates it. Only for quasi-static processes can we draw a path on a P–V diagram.

The standard processes of an ideal gas (μ moles)

Isothermal process

T is constant, so P V = const. The gas is in thermal contact with a reservoir. For an ideal gas U depends only on T, so ΔU = 0 and the heat absorbed equals the work done. Work by μ moles expanding from V₁ to V₂:

W = ∫P dV = μRT ∫(dV/V) = μRT ln(V₂/V₁)

For expansion (V₂ > V₁) W and Q are positive; for compression both are negative (heat is released to the reservoir).

Adiabatic process

No heat enters or leaves the system (ΔQ = 0): the system is insulated, or the process is so fast that no heat is exchanged. By the first law, ΔU = −ΔW. For an ideal gas, with γ = C_p/C_v,

PV^γ = constant

Fig. 11.8 shows the P–V curves of an ideal gas for two isotherms and an adiabatic process connecting them. The adiabatic curve is steeper than the isotherm through the same point, because in an adiabatic expansion the temperature also falls and thus the pressure falls faster. Work done in going from (P₁, V₁, T₁) to (P₂, V₂, T₂):

W = ∫P dV = const ∫V^−γ dV = (P₁V₁ − P₂V₂)/(γ − 1) = μR(T₁ − T₂)/(γ − 1)

If the gas does work (W > 0) the temperature falls (T₂ < T₁). If work is done on the gas (adiabatic compression) the temperature rises. This is why a bicycle pump becomes hot, and why a gas cools when it expands rapidly (the principle of the cooling in aerosol sprays).

Isochoric and isobaric processes

Isochoric (V constant): ΔV = 0 so W = 0, and ΔQ = ΔU = μC_vΔT: all the heat goes into internal energy. Isobaric (P constant): W = P(V₂ − V₁) = μRΔT; the heat is ΔQ = μC_pΔT, of which μC_vΔT increases U and μRΔT does work.

Cyclic process and P–V diagrams

In a cyclic process the system returns to its initial state, so ΔU = 0 over a complete cycle and Q = W: the net heat absorbed is converted into net work, which equals the area enclosed by the loop on the P–V diagram (clockwise loop: positive work by the gas). The work done by the gas in any process is the area under the P–V curve and above the V-axis; it depends on the path, not only on the end points.

11.9 Second Law of Thermodynamics

The first law says energy is conserved, but it does not say which processes actually occur. A hot cup of tea cools by itself, but never gets hotter by absorbing heat from the cooler room; a falling stone heats the ground, but a heated ground never throws a stone up. The first law would allow these reverse processes; the second law forbids them.

• Kelvin–Planck statement. No process is possible whose sole result is the absorption of heat from a reservoir and the complete conversion of the heat into work.
• Clausius statement. No process is possible whose sole result is the transfer of heat from a colder object to a hotter object.

The two statements are equivalent. The word “sole” matters: in a heat engine, heat is absorbed from the source and converted to work, but part is also rejected to the sink; in a refrigerator heat is taken from the cold body, but work is supplied from outside.

Heat engine

A heat engine is a device in which a system undergoes a cyclic process, absorbing heat Q₁ from a hot reservoir (source), doing work W and rejecting heat Q₂ to a cold reservoir (sink). Over a cycle ΔU = 0, so W = Q₁ − Q₂. The efficiency is

η = W/Q₁ = 1 − Q₂/Q₁

η = 1 would need Q₂ = 0, which the second law forbids: no engine is 100 % efficient.

Refrigerator / heat pump

A refrigerator is a heat engine running backwards: it absorbs heat Q₂ from the cold reservoir, work W is done on it, and it releases Q₁ = Q₂ + W to the hot reservoir. Its coefficient of performance is α = Q₂/W. It cannot be infinite: the Clausius statement says W cannot be zero.

11.10 Reversible and Irreversible Processes

A thermodynamic process is reversible if it can be retraced in the opposite direction so that the system and the surroundings both return exactly to their original states, with no other change anywhere in the universe. A process that is not reversible is irreversible.

Why most processes are irreversible

• Spontaneous processes such as free expansion of a gas, a hot body cooling in cold surroundings, and mixing of two gases.
• Dissipative effects: friction, viscosity, electrical resistance, and other agencies that convert organised energy into heat. Work converted to heat by friction cannot be fully recovered as work.
• Non-equilibrium states during rapid changes: sudden compression or expansion, explosions, turbulence, large temperature differences.

A process is reversible only if it is quasi-static and has no dissipative effects. A quasi-static isothermal expansion of an ideal gas against a pressure that is infinitesimally smaller than the gas pressure, with a frictionless piston, is an example. All natural processes are irreversible to some degree; reversible processes are an idealisation that sets the upper limit on performance.

11.11 Carnot Engine

Given a hot reservoir at T₁ and a cold reservoir at T₂, what is the maximum efficiency possible for an engine operating between them? Sadi Carnot (1824), a French engineer, answered this with an ideal engine that is reversible. It must be reversible, so every heat exchange with a reservoir must occur at (almost) the same temperature as the reservoir (otherwise heat would flow across a finite temperature difference, which is irreversible), and any change of temperature from T₁ to T₂ (or back) must occur without heat exchange, i.e. adiabatically. Hence the Carnot cycle has four steps (Fig. 11.9), here with an ideal gas as the working substance:

1. Step 1 → 2: isothermal expansion at T₁. The gas absorbs heat Q₁ from the hot reservoir: Q₁ = μRT₁ ln(V₂/V₁) = W_1→2.
2. Step 2 → 3: adiabatic expansion from T₁ to T₂. No heat exchange; the gas does work, W_2→3 = μR(T₁ − T₂)/(γ − 1).
3. Step 3 → 4: isothermal compression at T₂. The gas releases heat Q₂ to the cold reservoir: Q₂ = μRT₂ ln(V₃/V₄).
4. Step 4 → 1: adiabatic compression from T₂ back to T₁; work is done on the gas, and the gas returns to its initial state.

Efficiency of the Carnot engine

Applying PV^γ = const to the two adiabats gives T₁V₂^γ−1 = T₂V₃^γ−1 and T₁V₁^γ−1 = T₂V₄^γ−1, hence V₂/V₁ = V₃/V₄. Therefore

Q₂/Q₁ = T₂/T₁

η = 1 − Q₂/Q₁ = 1 − T₂/T₁

The efficiency depends only on the temperatures of the two reservoirs, not on the working substance. It becomes 1 (100 %) only if T₂ = 0 K, which is unattainable, and η = 0 if T₁ = T₂ (no work can be extracted from a single reservoir). The temperatures must be in kelvin.

Carnot’s theorem

• (a) Working between two given temperatures T₁ and T₂ of the hot and cold reservoirs, no engine can have efficiency more than that of the Carnot engine.
• (b) The efficiency of the Carnot engine is independent of the nature of the working substance.

Proof of (a), the idea. Couple an irreversible engine I to a reversible engine R run backward as a refrigerator, both between the same reservoirs (Fig. 11.10). Let I absorb Q₁ and do work W′, and R take in Q₂′ and have work W done on it. If I were more efficient than R, then W′ > W, and the combined system would extract the heat W′ − W from the cold reservoir and convert it entirely into work, contradicting the Kelvin–Planck statement. So η_I ≤ η_R.

Refrigerator based on the Carnot cycle

Run the Carnot cycle backward: take Q₂ from the cold body at T₂, supply work W, release Q₁ = Q₂ + W at T₁. The coefficient of performance is α = Q₂/W = T₂/(T₁ − T₂). A refrigerator is most efficient when the temperature difference is small.

Chapter Summary and Formula Sheet

1. Zeroth law: two systems in thermal equilibrium with a third system separately are in thermal equilibrium with each other; it leads to the concept of temperature.
2. Internal energy is the sum of molecular kinetic and potential energies (excluding the overall motion of the system). Heat and work are two modes of energy transfer; neither is a state variable.
3. First law: ΔQ = ΔU + ΔW, the conservation of energy including heat and work.
4. Specific heat: s = (1/m)(ΔQ/ΔT); molar specific heat C = (1/μ)(ΔQ/ΔT); for solids C = 3R at ordinary temperatures. For an ideal gas C_p − C_v = R. 1 cal = 4.186 J.
5. State variables (P, V, T, m) describe equilibrium states; the equation of state (PV = μRT) relates them. Extensive: U, V, M. Intensive: P, T, ρ.
6. Quasi-static process: infinitely slow, system always in equilibrium with the surroundings.
7. Isothermal expansion of an ideal gas: Q = W = μRT ln(V₂/V₁). Adiabatic: PV^γ = const, W = μR(T₁ − T₂)/(γ − 1).
8. Second law: Kelvin–Planck and Clausius statements; heat cannot be completely converted to work in a cyclic process, and cannot flow by itself from cold to hot.
9. Carnot engine: reversible, two isotherms + two adiabats; η = 1 − T₂/T₁; no engine working between two given temperatures can be more efficient (Carnot’s theorem).

Formulae at a glance

Points to Ponder

1. Temperature is related to the mean internal energy of a system, not to the total energy of the system.
2. Equilibrium in thermodynamics: the macroscopic variables do not change in time. In a state of thermodynamic equilibrium, the microscopic constituents of a system are not in mechanical equilibrium.
3. Heat capacity, in general, depends on the process the system goes through when heat is supplied.
4. In isothermal quasi-static processes, heat is absorbed or given out by the system even though at every stage the gas has the same temperature as the surrounding reservoir. This is possible because of the infinitesimal difference in temperature between the system and the reservoir.
5. Heat is energy in transit; it is not stored in a body. A statement about “the heat in a body” is wrong; one speaks of the internal energy of the body.