Class 11 Physics: Thermodynamics | Physics Q&A

CBSE Class 11 › Unit VIII: Thermodynamics

Thermodynamics

NCERT Chapter 11 • Laws of Thermodynamics, Processes, Heat Engines & Carnot Cycle

NCERT 2025–26 Unit VIII • ~10 Marks JEE Main • 2 Questions

Steampunk heat engine illustrating the conversion of heat into work. — Thermodynamics: The study of heat, work, and the laws that govern their conversion.

1. Thermal Equilibrium & Zeroth Law

Two systems are in thermal equilibrium if they have the same temperature and no net heat flows between them when connected by a conducting (diathermic) wall.

The Zeroth Law of Thermodynamics states:

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

This law justifies the concept of temperature as a physical quantity that determines the direction of heat flow.

Diagram demonstrating the Zeroth Law of Thermodynamics using three systems. — If A and B are in thermal equilibrium with C, they are in equilibrium with each other (*T_A = T_B = T_C*).

2. First Law of Thermodynamics

This is the principle of conservation of energy applied to thermodynamic systems.

ΔQ = ΔU + ΔW

ΔQ: Heat supplied to the system (+ve if absorbed).
ΔU: Change in internal energy (depends only on initial and final states).
ΔW: Work done by the system (+ve if system expands).

Work Done by a Gas

For a gas expanding from volume V₁ to V₂:

$W=\int^{V_2}_{V_1} PdV$

Geometrically, this is the area under the P–V curve.

Internal Energy (U): A state function. For an ideal gas, U depends only on temperature (T). In real gases, it also depends slightly on volume due to intermolecular forces.

Problem 1: A gas absorbs 200 J of heat and does 50 J of work. Find the change in internal energy.

Show Answer

Using 1st Law: ΔQ = ΔU + ΔW.
200 = ΔU + 50 ⇒ ΔU = 150 J.

3. Specific Heat Capacity

When heat ΔQ is supplied to a substance, its temperature changes by ΔT. We define:

Specific heat capacity (s): Heat required to raise the temperature of 1 kg of a substance by 1 K.
$s = \dfrac{1}{m} \dfrac{\Delta Q}{\Delta T}$ [Unit: J kg⁻¹ K⁻¹]
Molar specific heat capacity (C): Heat required to raise the temperature of 1 mole by 1 K.
$C = \dfrac{1}{\mu} \dfrac{\Delta Q}{\Delta T}$ [Unit: J mol⁻¹ K⁻¹]

Calorie Definition

Historically, 1 calorie was defined as the heat needed to raise 1 g of water from 14.5°C to 15.5°C.
Now, 1 cal = 4.186 J. In SI units, we use joules exclusively.

Specific Heat of Solids

Using the law of equipartition of energy, each atom in a solid has 6 degrees of freedom (3 kinetic + 3 potential). Thus, average energy per mole is:

U = 3RT ⇒ C = $\dfrac{\Delta U}{\Delta T}$ = 3R ≈ 25 J mol⁻¹ K⁻¹

This matches experimental values for most solids at room temperature (see table below).

Substance	Specific Heat (J kg⁻¹ K⁻¹)	Molar Heat (J mol⁻¹ K⁻¹)
Aluminium	900	24.4
Copper	385	24.5
Iron	470	26.0
Lead	130	27.0

Relation for Ideal Gases: $C_p - C_v = R$

For 1 mole of ideal gas:

At constant volume: $C_v = \left( \dfrac{\Delta U}{\Delta T} \right)_v$
At constant pressure: $C_p = \left( \dfrac{\Delta U}{\Delta T} \right)_p + P \left( \dfrac{\Delta V}{\Delta T} \right)_p = C_v + R$

Hence,

C_p − C_v = R

where γ = C_p/C_v is used in adiabatic processes.

Derivation: C_p − C_v = R for Ideal Gas 3 Marks • CBSE Important

🔍 Show Derivation

Step 1: Definition of C_v
At constant volume, no work is done (ΔV = 0), so from First Law:
ΔQ = ΔU + PΔV = ΔU + 0 = ΔU
Molar specific heat at constant volume:
$\bf{C_v = \left(\dfrac{ΔQ}{ΔT}\right)_v = \left(\dfrac{ΔU}{ΔT}\right)_v}$

Step 2: Definition of C_p

At constant pressure, work done by gas: $W = PΔV$

From First Law: $ΔQ = ΔU + PΔV$

Molar specific heat at constant pressure:

$C_p = \left(\dfrac{ΔQ}{ΔT}\right)_p = \left(\dfrac{ΔU}{ΔT}\right)_p + P\left(\dfrac{ΔV}{ΔT}\left)_p$

Step 3: For ideal gas, U depends only on T

Since internal energy of ideal gas depends only on temperature:

$\left(\dfrac{ΔU}{ΔT}\right)_p = \left(\dfrac{ΔU}{ΔT}\right)_v = C_v$

Step 4: Apply ideal gas equation

For 1 mole: $PV = RT$

At constant pressure: $P\left(\dfrac{ΔV}{ΔT}\right)_p = R$

Step 5: Combine results

$C_p = \left(\dfrac{ΔU}{ΔT}\right)_p + P\left(\dfrac{ΔV}{ΔT}\right)_p$

$C_p = C_v + R$

→ $\bf{C_p- C_v = R}$

Note:

This relation holds only for ideal gases. For real gases, $C_p-C_v > R$ due to intermolecular forces.

4. Thermodynamic State Variables & Equation of State

An equilibrium state of a system is fully described by **state variables**: pressure (P), volume (V), temperature (T), internal energy (U), and composition.

Extensive vs. Intensive Variables

Extensive: Depend on system size (e.g., mass, volume, internal energy). Halved if system is split.
Intensive: Independent of size (e.g., pressure, temperature, density). Unchanged if system is split.

In the First Law: $ΔQ = ΔU + PΔV$ , all terms are extensive. (P is intensive, ΔV is extensive ⇒ PΔV is extensive.)

Non-Equilibrium States

State variables are undefined for non-equilibrium states, such as:

Free expansion into vacuum (no uniform P or T)
Explosive chemical reactions (e.g., petrol-air mixture ignition)

Only after equilibrium is restored can we assign P, V, T.

5. Thermodynamic Processes

A process describes how a system changes from one equilibrium state to another.

P-V diagrams for Isobaric, Isochoric, Isothermal, and Adiabatic processes. — P–V diagrams visualize the path taken between thermodynamic states.

Process	Constraint	Relation	Work Done (W)
Isothermal	Constant Temperature (ΔT = 0)	PV = constant	nRT ln(V₂/V₁)
Adiabatic	No Heat Exchange (ΔQ = 0)	PV^γ = constant	(P₁V₁ − P₂V₂)/(γ − 1)
Isobaric	Constant Pressure (ΔP = 0)	V/T = constant	P(V₂ − V₁)
Isochoric	Constant Volume (ΔV = 0)	P/T = constant	0

Derivation of Work Done in Thermodynamic Processes 4 Marks • JEE Focus

🔍 Show Full Derivation

Foundation: Consider a gas in a cylinder with movable piston of area A. When gas expands by small distance dx, work done:
dW = Force × displacement = (P × A) × dx = P × (A dx) = P dV

Total work from $V_1 \to V_2$ : $W =\int^{V_2}_{V_1}PdV$

(Geometrically: area under P–V curve)

Isothermal Process (T = constant):

From ideal gas law: PV = nRT ⇒ P = nRT / V

→ $W =\int^{V_2}_{V_1}\dfrac{nRT}{V}dV = nRT \int^{V_2}_{V_1}\dfrac{dV}{v}$

→ $W = nRT \left[ln V\right]^{V_2}_{V_1}= nRT (\ln V_2 - \ln V_1)$

→ $W = nRT \ln\dfrac{V_2}{V_1}$

Adiabatic Process (Q = 0):

For adiabatic process: $PV^{\gamma} = K$ (constant, where $\gamma = \dfrac{C_p}{C_v}$

→ $W = \int^{V_2}_{V_1}K V^{-\gamma} dV = K \left[\dfrac{V^{1-\gamma}}{1-\gamma}\right]^{V_2}_{V_1}$

→ $W = \dfrac{K}{1-\gamma} (V^{1-\gamma}_2 - V^{1-\gamma}_1)$

Using $K = P_1V^{\gamma}_1 = P_2V^{\gamma}_2$ and simplifying:

→ $W = \dfrac{P_1V^{\gamma}_1V^{1-\gamma}_2 - P_2V^{\gamma}_2V^{1-\gamma}_1}{1-\gamma}$

→ $W = \dfrac{P_1V_1 - P_2V_2}{1-\gamma} = \dfrac{P_2V_2 - P_1V_1}{\gamma - 1}$

→ $\bf{W = \dfrac{P_1V_1 - P_2V_2}{1-\gamma}}$

Isobaric Process (P = constant):

$W = \int^{V_2}_{V_1}PdV = P \int^{V_2}_{V_1}dV$

→ $\bf{W = P (V_2-V_1)}$

Isochoric Process (V = constant):

Since $dV = 0$ throughout:

$W =\int^{V_2}_{V_1} P \cdot 0 = 0$

→ $\bf{W = 0}$

Key Insight:

Work depends on the path taken in P–V diagram (not just endpoints). That’s why W is not a state function, unlike internal energy (U).

Isothermal vs Adiabatic

Isothermal: Quasi-static process with thermal contact. ΔU = 0 (ideal gas). So, ΔQ = ΔW.
Adiabatic: Insulated system (no heat exchange). ΔQ = 0 ⇒ ΔW = −ΔU. Temperature changes during expansion/compression.

Problem 2: Calculate the work done when 1 mole of an ideal gas expands isothermally at 300 K from 10 L to 20 L. (R = 8.31 J/mol·K)

Show Answer

W = nRT ln(V₂/V₁) = 1 × 8.31 × 300 × ln(2) ≈ 8.31 × 300 × 0.693 ≈ 1728 J.

6. Reversible & Irreversible Processes

A process is reversible if both the system and surroundings can be restored to their original states with no other change in the universe.

Conditions for Reversibility

Must be quasi-static (infinitely slow, always in equilibrium)
Must have no dissipative effects (friction, viscosity, resistance)

Irreversible Processes (Most Natural Processes)

Irreversibility arises due to:

Non-equilibrium states: Free expansion, explosive reactions
Dissipative effects: Friction converts work to heat irreversibly

Examples: diffusion of gas, stirring liquid, cooling of hot object.

The Carnot engine is the only practical example of a reversible engine. All real engines are irreversible and less efficient.

7. Second Law & Heat Engines

The First Law allows energy conservation but doesn’t restrict the direction of processes. The Second Law introduces irreversibility and defines the natural direction of heat flow.

Statements of the Second Law

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.” → Efficiency η < 100%.
Clausius Statement: “No process is possible whose sole result is the transfer of heat from a colder object to a hotter object.” → Heat cannot spontaneously flow from cold to hot.

Heat Engine

A device that converts heat from a hot reservoir into useful work, rejecting some waste heat to a cold reservoir.

$\eta = 1-\dfrac{Q_2}{Q_1} = 1-\dfrac{T_2}{T_1}$ (for Carnot engine)

Where:

Q₁ = heat absorbed from source at temperature T₁ (in Kelvin)
Q₂ = heat rejected to sink at temperature T₂

8. Refrigerators & Heat Pumps

A refrigerator uses external work to extract heat from a cold interior and dump it into a warmer room.

Schematic comparison of energy flow in a Heat Engine and a Refrigerator. — A heat engine converts heat to work; a refrigerator uses work to extract heat.

Coefficient of Performance (COP or α)

$\alpha = \dfrac{Q_2}{W} = \dfrac{Q_2}{Q_1-Q_2} = \dfrac{T_2}{T_1-T_2}$

For a heat pump (used for heating), $COP = \dfrac{Q_1}{W} = \dfrac{T_1}{T_1-T_2}$ .

Problem 3: A refrigerator works between 0°C and 27°C. Calculate its coefficient of performance.

Show Answer

T₂ = 273 K (cold), T₁ = 300 K (hot).
α = T₂ / (T₁ − T₂) = 273 / (300 − 273) = 273 / 27 ≈ 10.1.

9. Carnot Engine & Cycle

The Carnot engine is an idealized reversible heat engine operating between two temperatures. It gives the maximum possible efficiency for any heat engine working between those temperatures.

P-V diagram of the Carnot Cycle showing the four reversible steps. — *The Carnot Cycle consists of two isothermal and two adiabatic processes.*

Four Reversible Stages

Isothermal Expansion (A→B): Absorbs heat Q₁ at high temperature T₁.
Adiabatic Expansion (B→C): Expands further; temperature drops from T₁ to T₂.
Isothermal Compression (C→D): Rejects heat Q₂ at low temperature T₂.
Adiabatic Compression (D→A): Compressed back; temperature rises from T₂ to T₁.

Derivation: Carnot Engine Efficiency η = 1 − T₂/T₁ 5 Marks • JEE Advanced

🔍 Show Full Derivation

Step 1: Isothermal Expansion (A→B) at T₁

Gas absorbs heat Q₁ from hot reservoir at temperature T₁.

Since isothermal: $ΔU = 0 ⇒ Q_1 = W_1}$

$Q_1 = nRT_1 \ln\dfrac{V_B}{V_A}$ …(1)

Step 2: Adiabatic Expansion (B→C)

No heat exchange ( $Q = 0$ ), temperature drops from $T_1$ to $T_2$ .

For adiabatic process: $T V^{\gamma - 1} = \text{constant}$

$T_1 V_B^{\gamma - 1} = T_2 V_C^{\gamma - 1}$

→ $\dfrac{T_1}{T_2} = \left( \dfrac{V_C}{V_B} \right)^{\gamma - 1}$ …(2)

Step 3: Isothermal Compression (C→D) at $T_2$

Gas rejects heat $Q_2$ to cold reservoir at temperature $T_2$ .

$Q_2 = n R T_2 \ln \left( \dfrac{V_C}{V_D} \right)$ …(3)

(Note: $Q_2$ is magnitude, actual heat rejected = $-Q_2$ )

Step 4: Adiabatic Compression (D→A)

Temperature rises from $T_2$ to $T_1$ .

$T_2 V_D^{\gamma - 1} = T_1 V_A^{\gamma - 1}$

→ $\dfrac{T_1}{T_2} = \left( \dfrac{V_D}{V_A} \right)^{\gamma - 1}$ …(4)

Step 5: Equate (2) and (4)

$\left( \dfrac{V_C}{V_B} \right)^{\gamma - 1} = \left( \dfrac{V_D}{V_A} \right)^{\gamma - 1}$

→ $\dfrac{V_C}{V_B} = \dfrac{V_D}{V_A}$

→ $\dfrac{V_B}{V_A} = \dfrac{V_C}{V_D}$ …(5)

Step 6: Efficiency $\eta =$ Work Output / Heat Input

$\eta = \dfrac{W}{Q_1} = \dfrac{Q_1 - Q_2}{Q_1} = 1 - \dfrac{Q_2}{Q_1}$

Step 7: Substitute $Q_1$ and $Q_2$ from (1) and (3)

$\eta = 1 - \dfrac{n R T_2 \ln \left( \dfrac{V_C}{V_D} \right)} {n R T_1 \ln \left( \dfrac{V_B}{V_A} \right)}$

$\eta = 1 - \dfrac{T_2}{T_1} \cdot \dfrac{\ln \left( \dfrac{V_C}{V_D} \right)} {\ln \left( \dfrac{V_B}{V_A} \right)}$

Step 8: Use relation (5)

Since $\dfrac{V_B}{V_A} = \dfrac{V_C}{V_D}$ , their logarithms are equal:

$\ln \left( \dfrac{V_B}{V_A} \right) = \ln \left( \dfrac{V_C}{V_D} \right)$

→ $\eta = 1 - \dfrac{T_2}{T_1} \cdot 1$

→ $\bf{ \eta = 1 - \dfrac{T_2}{T_1}}$

Key Points:

• Efficiency depends only on temperatures, not on working substance
• T₁ and T₂ must be in Kelvin
• Maximum possible efficiency for any heat engine between T₁ and T₂

Carnot’s Theorem

(a) No real engine operating between T₁ and T₂ can be more efficient than a Carnot engine.
(b) The Carnot efficiency is independent of the working substance.

Problem 4 (JEE): A Carnot engine absorbs 1000 J of heat at 127°C and rejects 600 J of heat. Find the temperature of the sink and efficiency.

Show Answer

T₁ = 127 + 273 = 400 K. Q₁ = 1000 J, Q₂ = 600 J.
Efficiency η = 1 − Q₂/Q₁ = 1 − 600/1000 = 0.4 = 40%.
Also, η = 1 − T₂/T₁ ⇒ 0.4 = 1 − T₂/400 ⇒ T₂ = 240 K = −33°C.

← Previous Chapter Next Chapter: Kinetic Theory →