Thermodynamics

THERMAL EQUILIBRIUM

**Thermal equilibrium** is a state where the macroscopic variables (pressure, volume, temperature) of a system remain constant over time. A system is in thermal equilibrium when there is no net change in its properties.

**Key distinctions from mechanical equilibrium:**

Mechanical equilibrium: net external force and torque are zero

Thermal equilibrium: macroscopic variables do not change with time

When two systems A and B are separated by an **adiabatic wall** (insulating wall that prevents heat flow), any combination of their (P, V) values can exist in equilibrium with any other combination—they do not interact thermally.

When separated by a **diathermic wall** (conducting wall allowing heat flow), the systems spontaneously change their macroscopic variables until they reach a state where no further changes occur. At this point, they are in **thermal equilibrium with each other**.

**Example:** Two bodies at different temperatures placed in contact will exchange heat until their temperatures become equal and they reach thermal equilibrium.

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ZEROTH LAW OF THERMODYNAMICS

**Statement:** If system A is in thermal equilibrium with system C, and system B is separately in thermal equilibrium with system C, then A and B are in thermal equilibrium with each other.

**Mathematical expression:**

If TA = TC and TB = TC, then TA = TB

**Significance:**

Provides the foundation for the concept of **temperature** as a thermodynamic variable

Temperature is the physical quantity that is equal for all systems in mutual thermal equilibrium

Named "Zeroth" law by R.H. Fowler in 1931 because it was discovered after the First and Second Laws but was logically foundational

**Physical interpretation:**

The Zeroth Law establishes that temperature is a property that determines whether two bodies will be in thermal equilibrium. It allows us to compare temperatures of different bodies without necessarily bringing them into contact—we can use a third body (thermometer) to compare temperatures of other bodies.

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HEAT, INTERNAL ENERGY, AND WORK

Internal Energy (U)

**Definition:** Internal energy is the sum of kinetic and potential energies of all molecules in a system, calculated in the reference frame where the centre of mass is at rest.

**Key properties:**

Includes only disordered, random molecular motion (translational, rotational, vibrational)

Does NOT include the kinetic energy of the system as a whole

Is a **state variable**: depends only on the present state (P, V, T), not on the path taken to reach that state

For an ideal gas (neglecting intermolecular forces): U ∝ T

Changes in internal energy: ΔU depends only on initial and final states

**Real-life example:** When a bullet is fired from a gun, its kinetic energy as a projectile is not part of internal energy. When the bullet hits wood and stops, its kinetic energy converts to heat, increasing the internal energy (and temperature) of the bullet and wood.

Work (W)

**Definition:** Work is energy transfer to or from a system through mechanical means (moving pistons, compressing gas) that does not involve direct heat transfer due to temperature difference.

**Work done by a gas:** When a gas expands against constant external pressure P:

$$\Delta W = P \Delta V$$

where:

ΔW = work done by the system on surroundings (in joules)

P = external pressure (in Pa)

ΔV = change in volume (in m³)

**Sign convention:**

Positive work: system expands (ΔV > 0), gas does work on surroundings

Negative work: system compressed (ΔV < 0), surroundings do work on system

Heat (Q)

**Definition:** Heat is energy transfer from one system to another due to a temperature difference between them. It is energy in transit between systems.

**Critical distinction—Heat vs. Internal Energy:**

**Internal energy (U):** Property of a system in a given state; the system "has" internal energy

**Heat (Q):** Energy transfer process; heat is "supplied to" or "removed from" a system, never "contained in" a system

A statement like "gas in state A has heat Q" is **meaningless**

A statement like "heat Q is supplied to the gas" is **meaningful**

**Direction of heat flow:** Heat flows spontaneously from higher temperature to lower temperature until thermal equilibrium is reached.

**Two distinct modes of energy transfer:**

| Mode | Mechanism | Associated with |

|------|-----------|-----------------|

| **Heat** | Temperature difference | Conduction, convection, radiation |

| **Work** | Mechanical means (piston, paddle) | Pressure, volume change |

**Example:** Rubbing palms together generates heat through work (friction). A steam engine uses heat from steam to do work on pistons.

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FIRST LAW OF THERMODYNAMICS

**Statement:** Energy supplied to a system equals the increase in its internal energy plus the work done by the system on the surroundings.

**Mathematical expression:**

$$\Delta Q = \Delta U + \Delta W$$

**Alternative form:**

$$\Delta Q - \Delta W = \Delta U$$

**Variable definitions:**

ΔQ = heat supplied to the system (J)

ΔU = change in internal energy (J)

ΔW = work done by the system on surroundings (J)

**Derivation/Logical basis:**

The First Law is a direct application of conservation of energy to thermodynamic systems:

Energy entering system = increase in internal energy + energy leaving as work

No energy is created or destroyed, only transformed between forms

**Key insight:** While ΔQ and ΔW individually depend on the path taken from initial to final state, their combination (ΔQ − ΔW) is path-independent because ΔU is path-independent.

First Law for Constant Pressure Processes

When heat is supplied to a system at constant external pressure P:

$$\Delta Q = \Delta U + P \Delta V$$

**Derivation:**

Work done by system: ΔW = PΔV (for constant pressure)

From First Law: ΔQ = ΔU + ΔW

Substituting: ΔQ = ΔU + PΔV

Worked Example: Phase Change of Water

**Problem:** Calculate the change in internal energy when 1 g of water converts from liquid to vapour at atmospheric pressure.

**Given:**

Latent heat of vaporization: ΔQ = 2256 J/g = 2256 J

Liquid volume: Vl = 1 cm³ = 1 × 10⁻⁶ m³

Vapour volume: Vg = 1671 cm³ = 1.671 × 10⁻³ m³

Atmospheric pressure: P = 1.013 × 10⁵ Pa

**Solution:**

Work done by the system:

$$\Delta W = P(V_g - V_l)$$

$$\Delta W = 1.013 \times 10^5 \times (1.671 \times 10^{-3} - 1 \times 10^{-6})$$

$$\Delta W = 1.013 \times 10^5 \times 1.671 \times 10^{-3}$$

$$\Delta W = 169.2 \text{ J}$$

Change in internal energy:

$$\Delta U = \Delta Q - \Delta W$$

$$\Delta U = 2256 - 169.2$$

$$\Delta U = 2086.8 \text{ J}$$

**Interpretation:** About 92% of the heat supplied goes into increasing internal energy (breaking molecular bonds), while only 8% does work against atmospheric pressure.

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SPECIFIC HEAT CAPACITY

Heat Capacity of a Substance

**Definition:** The heat capacity S of a substance is the amount of heat required to raise its temperature by 1 K (or 1°C).

**Formula:**

$$S = \frac{\Delta Q}{\Delta T}$$

where:

S = heat capacity (J K⁻¹)

ΔQ = heat supplied (J)

ΔT = temperature change (K)

**Property:** S is proportional to mass of substance; larger objects have larger heat capacities.

Specific Heat Capacity

**Definition:** Specific heat capacity (s) is the heat capacity per unit mass—the heat required to raise the temperature of 1 kg of a substance by 1 K.

**Formula:**

$$s = \frac{S}{m} = \frac{1}{m} \frac{\Delta Q}{\Delta T}$$

where:

s = specific heat capacity (J kg⁻¹ K⁻¹)

m = mass (kg)

ΔQ = heat supplied (J)

ΔT = temperature change (K)

**Rearranged form:** $$\Delta Q = m \cdot s \cdot \Delta T$$

**Properties:**

Depends on the nature/material of the substance

Depends on temperature (though often treated as constant for small ΔT)

Independent of the amount of substance

Intensive property (does not change with quantity)

Molar Specific Heat Capacity

**Definition:** Molar specific heat capacity (C) is the heat capacity per mole of substance.

**Formula:**

$$C = \frac{1}{\mu} \frac{\Delta Q}{\Delta T}$$

where:

C = molar specific heat capacity (J mol⁻¹ K⁻¹)

μ = number of moles (mol)

ΔQ = heat supplied (J)

ΔT = temperature change (K)

**Rearranged form:** $$\Delta Q = \mu \cdot C \cdot \Delta T$$

**Relationship between s and C:**

$$C = s \times M$$

where M = molar mass (kg/mol)

**Properties:**

Depends on the substance and temperature

**Additional condition required:** Depends on how heat is supplied (constant pressure Cp or constant volume Cv)

For gases: Cp > Cv (at constant pressure, gas expands, doing work; more heat needed)

Typical Specific Heat Capacities at Room Temperature

| Substance | Specific Heat Capacity (J kg⁻¹ K⁻¹) |

|-----------|------|

| Water | 4200 |

| Ice | 2100 |

| Aluminum | 900 |

| Iron | 450 |

| Mercury | 140 |

| Copper | 385 |

**Observation:** Water has unusually high specific heat capacity, making it useful for heat storage and temperature regulation.

Worked Example: Specific Heat Capacity

**Problem:** How much heat is required to raise the temperature of 2 kg of water from 20°C to 80°C? (s for water = 4200 J kg⁻¹ K⁻¹)

**Solution:**

$$\Delta Q = m \cdot s \cdot \Delta T$$

$$\Delta T = 80 - 20 = 60 \text{ K}$$

$$\Delta Q = 2 \times 4200 \times 60$$

$$\Delta Q = 504,000 \text{ J} = 504 \text{ kJ}$$

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THERMODYNAMIC STATE VARIABLES AND EQUATION OF STATE

Thermodynamic State Variables

**Definition:** State variables are macroscopic properties whose values depend only on the current state of the system, not on the history or path taken to reach that state.

**Primary state variables for a gas:**

**Pressure (P):** Force per unit area exerted by gas molecules on container walls (Pa)

**Volume (V):** Space occupied by the gas (m³)

**Temperature (T):** Measure of average kinetic energy of molecules (K)

**Internal energy (U):** Sum of kinetic and potential energies of molecules (J)

**Number of moles (n) or mass (m):** Amount of substance

**Path-dependent quantities (NOT state variables):**

Heat (Q): depends on process followed

Work (W): depends on process followed

Entropy (S): state variable for entropy specifically

**Mathematical significance:** For any state variable X:

ΔX is path-independent

If initial state = final state (closed cycle), ΔX = 0

Only differences have physical meaning

Equation of State

**Definition:** An equation of state is a mathematical relationship connecting state variables (P, V, T, n) for a substance.

**For an ideal gas:**

$$PV = nRT$$

where:

P = pressure (Pa)

V = volume (m³)

n = number of moles (mol)

R = universal gas constant = 8.314 J mol⁻¹ K⁻¹

T = absolute temperature (K)

**Alternative forms:**

$$PV = NkT$$ (where N = number of molecules, k = Boltzmann constant = 1.38 × 10⁻²³ J K⁻¹)

$$P = \rho \frac{RT}{M}$$ (where ρ = density, M = molar mass)

**Validity:** Ideal gas equation is valid for:

Gases at low pressure (molecules far apart)

High temperatures (kinetic energy >> intermolecular forces)

Real gases approach ideal behavior under these conditions

**Real gases:** Use van der Waals equation (not in CBSE Class 11 scope):

$$\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT$$

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THERMODYNAMIC PROCESSES

Types of Thermodynamic Processes

A thermodynamic process describes how a system changes from one state to another. Different constraints during the process define different types.

Isobaric Process (Constant Pressure)

**Definition:** A process occurring at constant pressure (P = constant).

**Characteristics:**

Pressure of the system remains constant throughout

Volume and temperature change

Equation: PV/T = constant

From ideal gas law: V/T = constant (for fixed n)

**Work done by system:**

$$\Delta W = P \Delta V$$

**First Law application:**

$$\Delta Q = \Delta U + P \Delta V$$

**Example:** Heating water at atmospheric pressure until it boils; expansion of gas in a cylinder with a freely movable piston.

**P-V diagram:** Horizontal line

Isochoric Process (Constant Volume)

**Definition:** A process in which volume remains constant (V = constant).

**Characteristics:**

Volume of system does not change

Pressure and temperature change

No work is done by or on the system: ΔW = 0

From ideal gas law: P/T = constant (for fixed n)

**Work done:**

$$\Delta W = P \Delta V = 0$$

**First Law application:**

$$\Delta Q = \Delta U$$

All heat supplied goes entirely into changing internal energy.

**Example:** Heating gas in a rigid container; heating water in a closed, rigid vessel.

**P-V diagram:** Vertical line

Isothermal Process (Constant Temperature)

**Definition:** A process where temperature remains constant (T = constant).

**Characteristics:**

Temperature of the system does not change

Internal energy does not change: ΔU = 0 (for ideal gas)

Pressure and volume change such that PV = constant

From ideal gas law: PV = nRT = constant

**Work done by system:**

$$\Delta W = \int P \, dV = nRT \ln\left(\frac{V_f}{V_i}\right) = P_i V_i \ln\left(\frac{V_f}{V_i}\right)$$

**First Law application:**

$$\Delta Q = \Delta W$$

(Heat supplied equals work done by system)

**Example:** Slow expansion or compression of gas in contact with a heat reservoir; isothermal expansion of ideal gas.

**P-V diagram:** Hyperbola (rectangular hyperbola PV = constant)

**Worked example:** An ideal gas at pressure P₁ = 10⁵ Pa and volume V₁ = 1 m³ undergoes isothermal expansion to volume V₂ = 2 m³. Calculate work done by the gas.

Solution:

$$\Delta W = P_1 V_1 \ln\left(\frac{V_2}{V_1}\right) = 10^5 \times 1 \times \ln(2)$$

$$\Delta W = 10^5 \times 0.693 = 69,300 \text{ J} = 69.3 \text{ kJ}$$

Adiabatic Process (No Heat Exchange)

**Definition:** A process where no heat is exchanged with surroundings (ΔQ = 0).

**Characteristics:**

Thermally insulated system

All work done by system comes from internal energy change

Temperature, pressure, and volume all change

Relation: TVᵞ⁻¹ = constant or PVᵞ = constant

where γ = Cp/Cv (heat capacity ratio, typically 1.4 for diatomic gases)

**First Law application:**

$$\Delta Q = 0$$

$$0 = \Delta U + \Delta W$$

$$\Delta U = -\Delta W$$

Internal energy change equals negative of work done (work reduces internal energy).

**Example:** Rapid compression or expansion of gas; sound wave propagation; adiabatic cooling in expanding gas.

**P-V diagram:** Steeper curve than isothermal (because T changes).

Cyclic Process

**Definition:** A process where the system returns to its initial state.

**Characteristics:**

Initial state = final state

For any state variable: ΔX = 0 for complete cycle

ΔU = 0 (returns to same internal energy)

From First Law: ΔQ = ΔW

Heat supplied in cycle = net work done in cycle

**Example:** Carnot cycle, Otto cycle (engine cycles).

**P-V diagram:** Closed loop; area enclosed = net work done in cycle.

**Worked example:** A cyclic process has: stage 1 (isobaric expansion): ΔQ₁ = 100 J; stage 2 (isothermal compression): ΔQ₂ = −30 J. Calculate work done in the cycle.

Solution:

$$\Delta Q_{total} = \Delta Q_1 + \Delta Q_2 = 100 - 30 = 70 \text{ J}$$

$$\Delta U_{total} = 0$$ (cyclic process)

$$\Delta W_{total} = \Delta Q_{total} = 70 \text{ J}$$

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SECOND LAW OF THERMODYNAMICS

Statement and Significance

**Statement (Clausius form):** Heat cannot spontaneously flow from a colder body to a hotter body without external work being done on the system.

**Statement (Kelvin-Planck form):** It is impossible to construct a heat engine that operates in a cycle and converts heat entirely into work without rejecting some heat to a colder reservoir.

**Physical meaning:**

Not all energy is available to do useful work

Some energy must be dissipated as heat to the environment

Processes have a preferred direction (thermodynamic arrow of time)

Disorder (entropy) of isolated systems tends to increase

Entropy

**Definition:** Entropy (S) is a measure of disorder or randomness in a system. It quantifies the number of microscopic arrangements corresponding to a macroscopic state.

**Thermodynamic definition (for reversible processes):**

$$dS = \frac{\delta Q_{rev}}{T}$$

$$\Delta S = \int \frac{dQ_{rev}}{T}$$

**Second Law in terms of entropy:**

For any spontaneous process in an isolated system:

$$\Delta S > 0$$ (entropy increases)

For reversible process:

$$\Delta S = 0$$

**Key insights:**

Entropy is a state variable (depends only on current state)

In isolated systems, entropy never decreases

Entropy explains irreversibility: time's arrow points toward increasing entropy

Heat naturally flows from hot to cold because this increases total entropy of universe

Examples of Entropy Increase

**Example 1—Diffusion:** Gas molecules spreading throughout container increases entropy (more disorder).

**Example 2—Heat transfer:** When hot water at temperature Th releases heat to cold water at Tc:

Hot water loses entropy: ΔS_hot < 0

Cold water gains more entropy: ΔS_cold > 0 (because T_c is smaller)

Total entropy increases: ΔS_total = ΔS_hot + ΔS_cold > 0

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REVERSIBLE AND IRREVERSIBLE PROCESSES

Reversible Processes

**Definition:** A reversible process is one that can be reversed such that both the system and surroundings return to their initial states, with no net change in the universe.

**Characteristics:**

Proceeds infinitely slowly through equilibrium states

No dissipative forces (friction, viscosity, thermal conductivity effects are negligible)

Entropy of universe remains constant: ΔS_universe = 0

System always infinitesimally close to equilibrium

Work output is maximum possible

**Conditions for reversibility:**

Quasi-static process (infinitely slow)

No irreversible phenomena (friction, turbulence, etc.)

System always in thermal and mechanical equilibrium

**Examples:**

Isothermal expansion against constant external pressure equal to gas pressure

Adiabatic expansion with no friction

Infinitely slow movement of piston

**Practical reality:** Completely reversible processes do not exist; they are idealizations and limits.

Irreversible Processes

**Definition:** An irreversible process is one that cannot return the system and surroundings to initial states without permanent changes occurring.

**Characteristics:**

Occurs at finite rates (not infinitely slow)

Involves dissipative forces and irreversible phenomena

Entropy of universe increases: ΔS_universe > 0

System undergoes finite jumps away from equilibrium

Work output is less than maximum possible

Direction of process is determined by entropy increase

**Examples:**

Expansion of gas into vacuum (free expansion)

Friction converting kinetic energy to heat

Heat transfer across finite temperature difference

Inelastic collision

Stirring of a liquid

Mixing of gases

**Example—Free expansion:**

When a gas expands into a vacuum:

ΔU = 0 (no work done, isolated system, ΔT constant for ideal gas)

ΔS > 0 (volume increased, disorder increases)

Cannot reverse spontaneously

Irreversible process

Comparison

| Property | Reversible | Irreversible |

|----------|-----------|--------------|

| Process rate | Infinitely slow | Finite rate |

| Equilibrium | Always maintained | Maintained only at start/end |

| Friction/dissipation | None | Present |

| ΔS_universe | = 0 | > 0 |

| Work output | Maximum | Less than maximum |

| Spontaneity | No (artificial) | Natural, spontaneous |

| Return to initial state | Possible (with external work) | Not possible without external work |

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CARNOT ENGINE

Definition and Concept

**Carnot engine:** An idealized heat engine operating between two heat reservoirs (one hot at temperature Th, one cold at temperature Tc) that operates through a reversible cyclic process (Carnot cycle).

**Historical significance:** Conceived by Sadi Carnot in 1824 to determine the maximum possible efficiency of any heat engine.

Carnot Cycle

The Carnot cycle consists of four reversible processes:

**Stage 1—Isothermal expansion (A → B):**

Temperature: T_h (constant)

Gas expands against external pressure

Heat absorbed from hot reservoir: Q_h

Work done by gas: W₁ = Q_h

Internal energy change: ΔU = 0

**Stage 2—Adiabatic expansion (B → C):**

No heat exchange: Q = 0

Temperature decreases from T_h to T_c

Work done by gas: W₂

Internal energy decreases: ΔU < 0

**Stage 3—Isothermal compression (C → D):**

Temperature: T_c (constant)

External work compresses gas

Heat rejected to cold reservoir: Q_c

Work done on gas: W₃ = Q_c

Internal energy change: ΔU = 0

**Stage 4—Adiabatic compression (D → A):**

No heat exchange: Q = 0

Temperature increases from T_c to T_h

Work done on gas: W₄

Internal energy increases: ΔU > 0

**P-V diagram:** Shows the characteristic shape with two isothermal curves and two adiabatic curves forming a closed loop.

Efficiency of Carnot Engine

**Definition:** Efficiency (η) is the ratio of useful work output to heat input.

**Formula:**

$$\eta = \frac{W_{net}}{Q_h} = \frac{Q_h - Q_c}{Q_h} = 1 - \frac{Q_c}{Q_h}$$

where:

W_net = net work done by engine in one cycle

Q_h = heat absorbed from hot reservoir

Q_c = heat rejected to cold reservoir

**For a Carnot engine (reversible process):**

$$\eta_{Carnot} = 1 - \frac{T_c}{T_h}$$

where T_c and T_h are absolute temperatures (Kelvin).

**Key observations:**

Depends only on the temperatures of the two reservoirs

Independent of the working substance

Maximum possible efficiency at given temperatures

Maximum efficiency approaches 100% only as T_c → 0 (absolute zero, unattainable)

Increases with larger temperature difference (T_h − T_c)

**Critical result:** No heat engine operating between two temperatures can have efficiency greater than a Carnot engine operating between the same temperatures:

$$\eta \leq \eta_{Carnot} = 1 - \frac{T_c}{T_h}$$

Coefficient of Performance of Carnot Heat Pump/Refrigerator

For a Carnot refrigerator (reverse cycle):

**Coefficient of Performance (COP):**

$$COP = \frac{Q_c}{W} = \frac{T_c}{T_h - T_c}$$

Higher COP means more heat removed per unit work input.

Worked Example: Carnot Engine Efficiency

**Problem:** A Carnot engine operates between a hot reservoir at 500 K and a cold reservoir at 300 K. Calculate:

(a) Maximum efficiency

(b) If 1000 J of heat is absorbed from hot reservoir, calculate work done and heat rejected.

**Solution:**

(a) Maximum efficiency:

$$\eta = 1 - \frac{T_c}{T_h} = 1 - \frac{300}{500} = 1 - 0.6 = 0.4 = 40\%$$

(b) With Q_h = 1000 J:

Work done:

$$W = \eta \times Q_h = 0.4 \times 1000 = 400 \text{ J}$$

Heat rejected:

$$Q_c = Q_h - W = 1000 - 400 = 600 \text{ J}$$

Verification: $$\frac{Q_c}{Q_h} = \frac{600}{1000} = 0.6 = \frac{T_c}{T_h} = \frac{300}{500}$$ ✓

Carnot Engine vs. Real Engines

| Property | Carnot Engine | Real Engines |

|----------|---------------|-------------|

| Reversibility | Completely reversible | Irreversible |

| Efficiency | Maximum possible | Less than Carnot |

| Friction losses | None | Significant |

| Achievability | Theoretical ideal | Practical but not Carnot |

| Cycle duration | Infinitely long (very slow) | Finite, rapid cycles |

**Real engines (Otto, Diesel, Rankine):** Operate irreversibly with lower efficiency than corresponding Carnot engines at same temperatures.

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KEY FORMULAS SUMMARY

| Concept | Formula | SI Units |

|---------|---------|----------|

| First Law | ΔQ = ΔU + ΔW | J |

| Work at constant P | ΔW = PΔV | J (Pa·m³) |

| Specific heat | ΔQ = msΔT | J |

| Molar heat capacity | ΔQ = μCΔT | J |

| Ideal gas law | PV = nRT | Pa·m³ |

| Carnot efficiency | η = 1 − Tc/Th | dimensionless |

| Entropy change | ΔS = Q_rev/T | J/K |

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EXAM-IMPORTANT POINTS

**Conceptual understanding (not just formulas):**

Heat and work are modes of energy transfer; internal energy is a state variable

Second Law determines direction of spontaneous processes

Reversible processes set maximum efficiency limits

Entropy measures disorder and explains irreversibility

**Common mistakes to avoid:**

Confusing heat with internal energy

Using Carnot efficiency for real engines

Incorrect sign conventions for work

Forgetting to use absolute temperature (Kelvin)

**Numerical problem types:**

Phase changes using ΔQ = ΔU + PΔV

Thermodynamic processes with P-V diagrams

Carnot engine efficiency calculations

Multi-step processes (isobaric, isochoric, isothermal)