Thermodynamics

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Section 11.2 & 11.3

Thermal Equilibrium & Zeroth Law

Thermal Equilibrium: A system is in thermodynamic equilibrium when its macroscopic variables (P, V, T, mass) do not change with time.

Adiabatic wall — insulating wall that does NOT allow heat flow. Diathermic wall — conducting wall that ALLOWS heat flow.

⚖️ Zeroth Law of Thermodynamics (R.H. Fowler, 1931)

"If two systems A and B are separately in thermal equilibrium with a third system C, then A and B are in thermal equilibrium with each other."

This law establishes the concept of temperature: it is the physical quantity that is equal for all systems in mutual thermal equilibrium.

🖼️ Insert Diagram: Systems A and B separated by adiabatic/diathermic wall with system C Recommended: Fig 11.1 and Fig 11.2 from NCERT — Zeroth Law illustration

🎯 Practice MCQs — Zeroth Law & Thermal Equilibrium

AAI ATC Style

1 The Zeroth Law of Thermodynamics leads to the concept of:

A Internal energy

B Entropy

C Temperature

D Work done

✅ Temperature. The Zeroth Law identifies temperature as the quantity that is equal for any two systems in thermal equilibrium. It gives us a logical basis to define and measure temperature.

2 A diathermic wall is one that:

A Does not allow heat flow between systems

B Allows heat flow between systems

C Allows only work to be done, not heat

D Maintains constant pressure

✅ Allows heat flow. Diathermic = conducting wall. Adiabatic = insulating wall. When a diathermic wall separates two systems, heat flows until they reach the same temperature (thermal equilibrium).

3 Which statement correctly describes thermodynamic equilibrium?

A Net force and torque on the system are zero

B Macroscopic variables of the system do not change with time

C All molecules are at rest

D Pressure and volume are equal

✅ Macroscopic variables (P, V, T, mass) do not change with time. This is thermodynamic equilibrium. Note: This is different from mechanical equilibrium (net force = 0). In thermodynamic equilibrium, molecules are still moving randomly.

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Section 11.4 & 11.5

Heat, Internal Energy & First Law

Internal Energy (U) = sum of kinetic + potential energies of all molecules. It is a state variable — depends only on the state, NOT on how that state was reached.

Heat (Q) and Work (W) are NOT state variables — they are energy in transit. "A gas has a certain amount of heat" is a meaningless statement; "a gas has a certain amount of internal energy" is correct.

🔋 First Law of Thermodynamics

ΔQ = ΔU + ΔW

ΔQ = heat supplied to the system | ΔU = change in internal energy | ΔW = work done BY the system

For a gas in a cylinder: ΔW = PΔV, so ΔQ = ΔU + PΔV

      🔑 Important Application: For 1g water converting to steam at 100°C: ΔQ = 2256 J (latent heat). ΔW = P(Vg−Vl) = 1.013×10⁵ × 1670×10⁻⁶ = 169.2 J. So ΔU = 2256 − 169.2 = 2086.8 J. Most heat goes into increasing internal energy, not doing work.
    

📌 Key Distinctions

Internal energy U — state variable (depends only on state, not path)
Heat Q and Work W — NOT state variables (path dependent)
ΔQ − ΔW = ΔU — always path independent (depends only on initial and final states)
For ideal gas isothermal process: ΔU = 0, so ΔQ = ΔW
For adiabatic process: ΔQ = 0, so ΔU = −ΔW (work done decreases internal energy)

🎯 Practice MCQs — First Law of Thermodynamics

AAI ATC Style

1 An electric heater supplies 100 W to a system. The system performs work at 75 J/s. The rate of increase of internal energy is:

A 175 W

B 75 W

C 25 W

D 100 W

✅ ΔU/t = ΔQ/t − ΔW/t = 100 − 75 = 25 W. From First Law: ΔQ = ΔU + ΔW. Rate form: dQ/dt = dU/dt + dW/dt → 100 = dU/dt + 75 → dU/dt = 25 W.

2 In an adiabatic process, 22.3 J of work is done ON the system. The change in internal energy is:

A −22.3 J

B 0 J

C +22.3 J

D Depends on temperature

✅ +22.3 J. In an adiabatic process, ΔQ = 0. Work done ON system = −ΔW = +22.3 J. From ΔQ = ΔU + ΔW: 0 = ΔU − 22.3 → ΔU = +22.3 J. Work done on the gas increases its internal energy (and temperature).

3 For a gas taken from state A to state B via two different paths, which quantity is the same for both paths?

A Heat absorbed (ΔQ)

B Work done (ΔW)

C Change in internal energy (ΔU)

D All three are path dependent

✅ ΔU is path independent because internal energy U is a state variable — its change depends only on the initial and final states. ΔQ and ΔW individually depend on the path, but their difference (ΔQ − ΔW = ΔU) is always path independent.

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Section 11.6

Specific Heat Capacity of Gases (Cp & Cv)

For gases, we define two molar specific heats depending on the process:

Cv = molar specific heat at constant volume. At constant volume, all heat goes into increasing internal energy: Cv = ΔU/ΔT

Cp = molar specific heat at constant pressure. At constant pressure, heat goes into both internal energy increase and work done.

Cp − Cv = R

Meyer's Relation | R = 8.31 J mol⁻¹ K⁻¹ (Universal gas constant) | Cp > Cv always for an ideal gas

Also, γ = Cp/Cv (ratio of specific heats). For monatomic gas: γ = 5/3. For diatomic gas: γ = 7/5 = 1.4.

For solids, molar heat capacity C = 3R ≈ 25 J/mol/K (law of Dulong-Petit, valid at ordinary temperatures).

🎯 Practice MCQs — Specific Heat of Gases

AAI ATC Style

1 The amount of heat supplied to 2×10⁻² kg of nitrogen (M = 28, R = 8.3 J/mol/K) at constant pressure to raise temperature by 45°C is: (Cp of N₂ = 29.1 J/mol/K)

A 374.1 J

B 748 J

C 933.2 J

D 1866 J

✅ μ = mass/M = 0.02/0.028 = 0.714 mol. ΔQ = μ × Cp × ΔT = 0.714 × 29.1 × 45 = 933.2 J. Always convert mass to moles first, then use ΔQ = μCpΔT for constant pressure process.

2 For an ideal gas, Cp − Cv equals:

A Zero

B 2R

C R

D R/2

✅ Cp − Cv = R = 8.31 J/mol/K. Meyer's relation. The extra R comes from the work done by the gas in expanding at constant pressure: P(ΔV/ΔT) = R for 1 mole of ideal gas. This is why Cp > Cv always.

3 The molar heat capacity predicted by Dulong-Petit law for solids at ordinary temperatures is:

A R

B 2R

C 3R ≈ 25 J/mol/K

D 5R/2

✅ C = 3R ≈ 24.93 ≈ 25 J/mol/K. From equipartition of energy: each atom has 3 vibrational modes, each contributing kBT to energy (kinetic + potential) → total U = 3RT per mole → C = dU/dT = 3R. This agrees with experiments for most metals at room temperature.

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Section 11.8

Thermodynamic Processes

A quasi-static process is an infinitely slow process where the system is always in equilibrium. Real processes approximate this when changes are slow and smooth.

Process	Fixed Quantity	Key Relation	Work Done
Isothermal	T = constant	PV = constant (Boyle's Law)	W = μRT ln(V₂/V₁)
Adiabatic	ΔQ = 0	PVᵞ = constant	W = μR(T₁−T₂)/(γ−1)
Isobaric	P = constant	V/T = constant (Charles' Law)	W = PΔV = μRΔT
Isochoric	V = constant	P/T = constant	W = 0
Cyclic	Returns to start	ΔU = 0	W = ΔQ (net)

      🔑 Key AAI ATC Insight — Adiabatic vs Isothermal on P-V Diagram: The adiabatic curve is steeper than the isothermal curve because γ > 1. In adiabatic expansion, temperature drops; in isothermal, temperature stays constant. For H₂ gas: γ = 7/5 = 1.4 (diatomic). This is used in jet engine compression cycles.
    

🖼️ Insert Diagram: P-V curves for isothermal and adiabatic processes — showing steeper adiabatic curve Recommended: Fig 11.8 from NCERT — P-V curves for isothermal and adiabatic processes

🎯 Practice MCQs — Thermodynamic Processes

AAI ATC Style

1 3 moles of hydrogen (γ = 7/5) in an insulated cylinder are compressed to half the original volume. The factor by which pressure increases is: (P₁V₁ᵞ = P₂V₂ᵞ)

A 2

B 2.5

C 2^1.4 ≈ 2.64

D 3.5

✅ P₁V₁ᵞ = P₂V₂ᵞ → P₂/P₁ = (V₁/V₂)ᵞ = 2^1.4 ≈ 2.64. Insulated walls → adiabatic process. γ for diatomic gas (H₂) = 7/5 = 1.4. For isothermal compression to half volume, P would only double (×2); for adiabatic, it's more (×2.64).

2 In an isothermal expansion of an ideal gas, which of the following is correct?

A ΔQ = 0 and ΔU = W

B ΔU = 0 and ΔQ = 0

C ΔU = 0 and ΔQ = W

D W = 0 and ΔQ = ΔU

✅ ΔU = 0 and ΔQ = W. Isothermal → T constant → for ideal gas, ΔU = 0 (internal energy depends only on T). From First Law: ΔQ = ΔU + W = 0 + W = W. All heat absorbed is converted to work in isothermal expansion.

3 In an isochoric (constant volume) process, the gas absorbs 500 J of heat. The work done BY the gas and change in internal energy are respectively:

A 500 J, 0 J

B 500 J, 500 J

C 0 J, 500 J

D 250 J, 250 J

✅ W = 0, ΔU = 500 J. Isochoric → V constant → ΔV = 0 → W = PΔV = 0. From First Law: ΔQ = ΔU + W → 500 = ΔU + 0 → ΔU = 500 J. All absorbed heat goes entirely into raising internal energy (and temperature).

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Section 11.9 & 11.10

Second Law of Thermodynamics

The Second Law identifies the direction of natural processes — it goes beyond energy conservation (First Law) to tell us which processes are actually possible.

📋 Kelvin-Planck Statement

"No process is possible whose sole result is the absorption of heat from a reservoir and complete conversion of the heat into work."

→ No heat engine can have 100% efficiency.

📋 Clausius Statement

"No process is possible whose sole result is the transfer of heat from a colder object to a hotter object."

→ A refrigerator cannot operate without external work input. Both statements are equivalent.

📌 Reversible vs Irreversible Processes

A reversible process is quasi-static AND non-dissipative (no friction, viscosity, etc.).
All natural/spontaneous processes are irreversible — free expansion, diffusion, heat flow, combustion.
Irreversibility arises from: (1) non-equilibrium states, (2) dissipative effects (friction, viscosity).
Reversibility is an idealization — real engines are always less efficient than Carnot engine.

🎯 Practice MCQs — Second Law

AAI ATC Style

1 Which of the following is NOT a consequence of the Second Law of Thermodynamics?

A Heat cannot spontaneously flow from cold to hot body

B Efficiency of a heat engine can never be 100%

C Energy is conserved in all processes

D All natural processes are irreversible

✅ Energy conservation is the First Law, not the Second Law. The Second Law adds additional restrictions on top of energy conservation — it tells us the direction and limitations of processes. The First Law doesn't prevent a book from jumping off a table; the Second Law does.

2 Free expansion of a gas in vacuum is an example of:

A Reversible process

B Quasi-static process

C Irreversible process

D Isothermal process

✅ Irreversible. Free expansion of a gas into vacuum takes the system through non-equilibrium states. It cannot spontaneously reverse. Also, no work is done (W = 0, since P_external = 0) and if insulated, ΔQ = 0, so ΔU = 0 and T remains constant.

3 Two cylinders A (gas at STP) and B (vacuum) of equal capacity are connected and opened. The temperature change of gas after reaching equilibrium is:

A Temperature increases

B Temperature decreases

C No change in temperature

D Temperature first increases then decreases

✅ No change in temperature. Free expansion into vacuum: ΔW = 0 (no opposing pressure), system is insulated (ΔQ = 0). First Law: ΔU = 0 → for ideal gas, ΔU = 0 means ΔT = 0. Intermediate states are non-equilibrium, so pressure and temperature are not defined during the process.

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Section 11.11

Carnot Engine — Maximum Efficiency

The Carnot engine is the most efficient possible heat engine operating between two temperatures T₁ (hot source) and T₂ (cold sink). Proposed by Sadi Carnot in 1824.

The Carnot cycle consists of four reversible steps:

Step 1→2: Isothermal expansion at T₁ — gas absorbs Q₁ from hot reservoir
Step 2→3: Adiabatic expansion — temperature drops from T₁ to T₂
Step 3→4: Isothermal compression at T₂ — gas releases Q₂ to cold reservoir
Step 4→1: Adiabatic compression — temperature rises from T₂ to T₁

η = 1 − T₂/T₁ = 1 − Q₂/Q₁

Carnot Efficiency | T₁ = Source temperature (K) | T₂ = Sink temperature (K) | Always < 1 (always less than 100%)

      🔑 Carnot's Theorem: (1) No engine working between two temperatures can have efficiency GREATER than the Carnot engine. (2) The efficiency of Carnot engine is INDEPENDENT of the working substance — it depends only on T₁ and T₂. This is why temperatures must be in Kelvin (absolute scale) in the formula.
    

🖼️ Insert Diagram: Carnot cycle P-V diagram — 4 steps: isothermal expansion, adiabatic expansion, isothermal compression, adiabatic compression Recommended: Fig 11.9 from NCERT — Carnot cycle for ideal gas

📌 Important Relationships — Carnot Engine

η = 1 − T₂/T₁ = W/Q₁ = (Q₁−Q₂)/Q₁
Q₁/Q₂ = T₁/T₂ — universal relation independent of working substance
For η = 1 (100% efficiency): T₂ = 0 K — impossible (absolute zero unattainable)
Reversed Carnot engine = Carnot refrigerator (COP = T₂/(T₁−T₂))
Real engines are always LESS efficient than Carnot engine due to irreversibilities

🎯 Practice MCQs — Carnot Engine

AAI ATC Style

1 A Carnot engine operates between source temperature 500 K and sink temperature 300 K. Its efficiency is:

A 60%

B 30%

C 40%

D 50%

✅ η = 1 − T₂/T₁ = 1 − 300/500 = 1 − 0.6 = 0.4 = 40%. Always use temperatures in Kelvin. The engine converts 40% of heat absorbed from the hot source into useful work. The remaining 60% is rejected to the cold sink.

2 A Carnot engine absorbs 1000 J from a hot reservoir at 600 K and rejects heat to a sink at 300 K. Work done per cycle and heat rejected are respectively:

A 250 J, 750 J

B 750 J, 250 J

C 500 J, 500 J

D 400 J, 600 J

✅ η = 1 − 300/600 = 0.5 = 50%. W = η × Q₁ = 0.5 × 1000 = 500 J. Q₂ = Q₁ − W = 1000 − 500 = 500 J. Also verifiable: Q₂/Q₁ = T₂/T₁ = 300/600 = 0.5 → Q₂ = 500 J. ✓

3 To increase the efficiency of a Carnot engine from 30% to 40%, while keeping the source temperature T₁ = 1000 K constant, the sink temperature T₂ should be changed from 700 K to:

A 600 K

B 600 K

C 500 K

D 400 K

✅ For η = 40%: 0.4 = 1 − T₂/1000 → T₂/1000 = 0.6 → T₂ = 600 K. To increase efficiency, we must lower the sink temperature (or raise source temperature). Lowering T₂ from 700 K to 600 K increases η from 30% to 40%.

Thermal Equilibrium & Zeroth Law

⚖️ Zeroth Law of Thermodynamics (R.H. Fowler, 1931)

🎯 Practice MCQs — Zeroth Law & Thermal Equilibrium

Heat, Internal Energy & First Law

🔋 First Law of Thermodynamics

📌 Key Distinctions

🎯 Practice MCQs — First Law of Thermodynamics

Specific Heat Capacity of Gases (Cp & Cv)

🎯 Practice MCQs — Specific Heat of Gases

Thermodynamic Processes

🎯 Practice MCQs — Thermodynamic Processes

Second Law of Thermodynamics

📋 Kelvin-Planck Statement

📋 Clausius Statement

📌 Reversible vs Irreversible Processes

🎯 Practice MCQs — Second Law

Carnot Engine — Maximum Efficiency

📌 Important Relationships — Carnot Engine

🎯 Practice MCQs — Carnot Engine

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