Chapter 11- Thermodynamics

1. Introduction: The Engine of the Universe | Class 11 Physics Chapter 11 Thermodynamics Notes

Good morning, class! Today, we are opening the door to one of the most powerful and influential branches of science: Thermodynamics. While our previous lessons in Mechanics dealt with visible objects like rolling balls, falling apples, and moving cars, Thermodynamics dives deeper. It deals with the invisible energy that drives those movements. It is the study of Heat, Work, and the Internal Energy of systems.

The name itself gives us a clue: Thermo means “Heat” and Dynamics means “Power” or “Motion”. Historically, this science wasn’t born in a quiet laboratory; it was born in the noisy, soot-filled factories of the Industrial Revolution. Engineers were desperate to answer one question: “How do we get more work out of a lump of coal?” They wanted to build better steam engines. From those practical questions, we discovered universal laws that govern everything from the engine of a Ferrari to the fusion reactions in the Sun, and even the digestion of the breakfast you ate this morning.

Teacher’s Insight: A common confusion arises between “Thermal Properties of Matter” (Chapter 11) and “Thermodynamics” (Chapter 12).

In Chapter 10, we looked at molecular details—why water expands when it freezes, or how molecules vibrate.

In Chapter 11 (Thermodynamics), we zoom out. We stop caring about individual molecules. We look at the “Big Picture” or Macroscopic Variables like Pressure, Volume, and Temperature. We treat the gas as a single “System” and study how it exchanges energy.

2. Fundamental Concepts: Speaking the Language

Before we can master the laws, we must agree on our vocabulary. Thermodynamics is very precise about definitions.

2.1 System and Surroundings

Types of thermodynamics

Imagine you are studying a balloon filled with helium.

  • The System: The specific part of the universe under study. In this case, the helium gas inside the balloon.
  • The Surroundings: Everything else in the universe. The rubber of the balloon, the air in the room, you, and the rest of the world.
  • The Boundary: The surface (real or imaginary) that separates the system from the surroundings. The nature of this boundary dictates what can happen.

2.2 Types of Walls (Boundaries)

The wall of a container determines if heat can enter or leave.

  • Adiabatic Wall: Think of this as a “perfect insulator.” It allows NO heat transfer. If a system is enclosed in adiabatic walls, its temperature can change only if we do work on it (like compressing it), not by heating it.

    Example: A high-quality thermos flask or Styrofoam cup.
  • Diathermic Wall: Think of this as a “perfect conductor.” It allows heat to flow freely. If two systems are separated by a diathermic wall, they will exchange heat until they reach the same temperature.

    Example: A thin copper sheet or an aluminum kettle.

2.3 Thermodynamic State Variables

To describe the condition (or “state”) of a system, we use measurable macroscopic quantities. These are divided into two categories:

Extensive Variables Intensive Variables
Depend on the size or mass of the system. Independent of the size or mass of the system.
Test: If you divide the system in half, these values get halved. Test: If you divide the system in half, these values remain unchanged.
Examples: Volume (V), Mass (m), Internal Energy (U). Examples: Pressure (P), Temperature (T), Density.

3. The Zeroth Law of Thermodynamics

Zeroth law of thermodynamics

The Zeroth Law of Thermodynamics

It sounds strange to start counting at zero, doesn’t it? This law was formulated by Ralph Fowler in 1931, long after the First and Second Laws were well-established. However, physicists realized that this concept was logically fundamental to the others, so they placed it at the beginning.

3.1 The Statement

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

3.2 Why is this significant?

This law gives us the scientific definition of Temperature.

Imagine System C is a Thermometer.

1. You place the thermometer (C) in contact with a cup of coffee (A). The reading stops changing at 70°C.

2. You place the same thermometer (C) in contact with a beaker of hot water (B). The reading stops changing at 70°C.

3. According to the Zeroth Law, we can now guarantee that if you mix A and B, there will be no net heat flow between them. They are in Thermal Equilibrium.

Therefore, Temperature is simply the physical property that determines whether a system is in equilibrium with another.


4. The First Law of Thermodynamics

This is the bedrock of physics. It is essentially the Law of Conservation of Energy applied to thermodynamic systems. It tells us that energy cannot be created or destroyed, only transformed.

4.1 Internal Energy (U)

Every substance possesses a hidden store of energy called Internal Energy. It is the sum of two things:

1. Kinetic Energy: The random motion of molecules (translation, rotation, vibration). This depends directly on Temperature.

2. Potential Energy: The energy due to intermolecular forces (attraction/repulsion). This depends on Volume (spacing between molecules).

Important for Ideal Gases: In an Ideal Gas, we assume there are no intermolecular forces (no Potential Energy). Therefore, the Internal Energy of an ideal gas depends ONLY on Temperature.

4.2 The Mathematical Form

ΔQ = ΔU + ΔW

This equation tells a simple accounting story. When you supply Heat energy (ΔQ) to a system, that energy can go to two places:

1. Increasing the system’s own Internal Energy (ΔU) (making it hotter).

2. Doing mechanical Work (ΔW) on the surroundings (like pushing a piston).

First law of  thermodynamics

Visual representation of the First Law. Heat = Change in Internal Energy + Work Done.

4.3 The Sign Convention (Crucial!)

If you get the signs wrong, your entire calculation will be wrong. In Physics (IUPAC convention for Heat, Mechanical convention for Work), we follow this:

  • Heat (ΔQ):
    • Heat given TO the system: Positive (+)
    • Heat taken FROM the system: Negative (-)
  • Work (ΔW):
    • Work done BY the system (Expansion): Positive (+). (The gas pushes the piston up).
    • Work done ON the system (Compression): Negative (-). (You push the piston down).
  • Internal Energy (ΔU):
    • Temperature Increases: Positive (+)
    • Temperature Decreases: Negative (-)

5. Specific Heat Capacities of Gases

In solids and liquids, heating is straightforward. But gases are compressible, which makes things complicated. The amount of heat required to raise the temperature of a gas depends heavily on how the gas is confined.

5.1 Cv (Molar Specific Heat at Constant Volume)

Imagine a gas trapped in a rigid steel box. It cannot expand.

Since Volume is constant, Work Done (ΔW = PΔV) is Zero.

From the First Law: ΔQ = ΔU.

This means 100% of the heat you supply goes directly into raising the temperature.

Formula: (ΔQ)v = n Cv ΔT

5.2 Cp (Molar Specific Heat at Constant Pressure)

Now, imagine the gas in a cylinder with a movable piston. As you heat it, the gas expands to keep the pressure constant.

The gas does Work (ΔW) by pushing the piston.

From the First Law: ΔQ = ΔU + ΔW.

Here, the heat you supply must do two jobs: raise the temperature (ΔU) AND pay for the work done (ΔW).

Formula: (ΔQ)p = n Cp ΔT

Why is Cp > Cv?

This is a favorite exam question! At constant pressure, you need “extra” heat to compensate for the energy lost in doing expansion work. At constant volume, no work is done, so you need less heat for the same temperature rise. Hence, Cp is always greater than Cv.

5.3 Mayer’s Relation

For one mole of an ideal gas, the relationship is precise:

Cp – Cv = R

Where R is the Universal Gas Constant (8.314 J mol-1 K-1).

We also define the Ratio of Specific Heats (γ):

γ = Cp / Cv

The value of γ depends on the atomicity of the gas:

Monatomic (He, Ar): γ = 1.67

Diatomic (N2, O2): γ = 1.4

Polyatomic (CO2): γ = 1.33


6. Thermodynamic Processes

A “Process” describes how a system changes from State 1 (P1, V1, T1) to State 2 (P2, V2, T2). We visualize these on a P-V Diagram (Indicator Diagram). The most important rule to remember is: The Area under the P-V Curve equals the Work Done.

PV diagram

Comparison of different thermodynamic processes on a P-V diagram.

6.1 Isothermal Process

Definition: A process occurring at Constant Temperature (ΔT = 0).

Condition: The process must happen very slowly, allowing heat exchange with the surroundings to maintain equilibrium. Also, the walls must be conducting (diathermic).

  • Equation of State: Boyle’s Law holds true: PV = Constant.
  • Internal Energy: Since T is constant, ΔU = 0.
  • First Law: ΔQ = 0 + ΔW  →  ΔQ = ΔW.

    Meaning: All heat supplied is converted completely into Work.
  • Work Formula: W = nRT ln(V2 / V1)

6.2 Adiabatic Process

Definition: A process where No Heat is Exchanged with the surroundings (ΔQ = 0).

Condition: The process happens very rapidly (sudden compression or expansion) or the system is perfectly insulated.

  • Equation of State: Poisson’s Law: PVγ = Constant.
  • First Law: 0 = ΔU + ΔW  →  ΔW = -ΔU.

    Meaning: If a gas expands adiabatically, it does work at the expense of its own internal energy. This causes the temperature to fall. (This is how clouds form!)
  • Slope: The Adiabatic curve is steeper than the Isothermal curve by a factor of γ.
  • Work Formula: W = (nR(T1 – T2)) / (γ – 1)

6.3 Isochoric Process

Definition: A process occurring at Constant Volume (ΔV = 0).

Example: Heating gas inside a sealed pressure cooker.

  • Work Done: Since the volume doesn’t change, the piston doesn’t move. W = PΔV = 0.
  • First Law: ΔQ = ΔU. All heat increases the temperature.

6.4 Isobaric Process

Definition: A process occurring at Constant Pressure (ΔP = 0).

Example: Boiling water in an open pot (Atmospheric pressure is constant).

  • Equation: Charles’s Law: V/T = Constant.
  • Work Done: W = P(V2 – V1) = nR(T2 – T1).

7. The Second Law of Thermodynamics

The First Law is perfect for balancing the energy books, but it misses something about the real world. The First Law says energy is conserved, but it doesn’t say anything about the direction of flow.

For example: A cup of hot coffee cools down in a room. The energy lost by the coffee is gained by the room. The First Law is satisfied.

But have you ever seen a cold cup of coffee suddenly suck heat from the room and get hot? The First Law allows this (energy would still be conserved), but nature forbids it. The Second Law explains why.

7.1 Kelvin-Planck Statement (Heat Engines)

“It is impossible to construct a heat engine that operates in a cycle and produces no other effect than the extraction of heat from a reservoir and the performance of an equivalent amount of work.”

Translation: You cannot build an engine that is 100% efficient. You can never convert Heat completely into Work. You must waste some heat to a colder sink. There is no “perfect” engine.

7.2 Clausius Statement (Refrigerators)

“It is impossible for heat to flow spontaneously from a colder body to a hotter body without any external work.”

Translation: Heat loves to flow from Hot to Cold. If you want to force it to go the other way (Cold to Hot, like inside a fridge), you must pay for it by doing Work (plugging in the fridge).

7.3 Reversible vs. Irreversible Processes

  • Reversible Process: An ideal process that can be retraced back to the initial state without leaving any trace on the universe. It must be quasi-static (extremely slow) and frictionless. (Does not exist in reality).
  • Irreversible Process: A real process where energy is dissipated (friction, turbulence). It cannot be exactly reversed. (All real-life processes are irreversible).

8. Heat Engines and the Carnot Cycle

Heat engine flow diagram

A Heat Engine is a device that continuously converts thermal energy into mechanical work. A standard engine has three parts:

1. Source (T1): High temperature reservoir (Provides heat Q1).

2. Working Substance: The gas or steam that expands.

3. Sink (T2): Low temperature reservoir (Absorbs waste heat Q2).

Efficiency (η) = Output / Input = W / Q1

Since W = Q1 – Q2, we can write:

η = 1 – (Q2 / Q1)

The Carnot Engine

Sadi Carnot, a French engineer, conceptualized the most efficient possible engine allowed by the laws of physics. It is an ideal, theoretical engine that operates on a reversible cycle known as the Carnot Cycle.

The cycle consists of four steps:

  1. Isothermal Expansion: Gas absorbs heat Q1 from Source at T1.
  2. Adiabatic Expansion: Gas expands further, cooling down from T1 to T2.
  3. Isothermal Compression: Gas rejects waste heat Q2 to Sink at T2.
  4. Adiabatic Compression: Gas is compressed, heating up from T2 back to T1.

Carnot’s Theorem: No engine operating between two temperatures can be more efficient than a Carnot engine operating between the same two temperatures. The efficiency depends only on the temperatures of the Source and Sink.

ηCarnot = 1 – (T2 / T1)

WARNING: In this formula, T1 and T2 MUST be in Kelvin.


9. Comprehensive Practice Problems

Physics is not a spectator sport. You must practice to understand. Here is a curated set of problems ranging from conceptual to calculation-heavy.

Part A: Conceptual Multiple Choice

  1. In an adiabatic process, the quantity which remains constant is:
    (a) Temperature
    (b) Pressure
    (c) Total heat of the system
    (d) Volume

    Answer: (c). In adiabatic, ΔQ = 0. The heat content doesn’t change due to transfer.
  2. The First Law of Thermodynamics is a special case of:
    (a) Newton’s Law
    (b) Law of Conservation of Energy
    (c) Charles’s Law
    (d) Law of Heat Exchange

    Answer: (b). It is strictly conservation of energy applied to heat and work.
  3. For an ideal gas, the internal energy depends only on:
    (a) Pressure
    (b) Volume
    (c) Temperature
    (d) Size of molecule

    Answer: (c). U is a function of T only for ideal gases.

Part B: Short Answer Questions

  1. Q: Why does air escaping from a burst tire feel cool?

    Answer: When a tire bursts, the air inside (which is at high pressure) expands suddenly. This is an Adiabatic Expansion because it happens too fast for heat to enter from outside. In adiabatic expansion, the gas does work (ΔW is positive) at the cost of its own internal energy (ΔU becomes negative). A drop in internal energy means a drop in temperature.

  2. Q: Can a room be cooled by leaving the door of an electric refrigerator open?

    Answer: No, the room will actually get hotter. A refrigerator pumps heat from its interior to the coils at the back. If the door is open, it takes heat from the room and releases it back into the room. Additionally, the electric motor generates its own heat. The net result is an increase in room temperature.

Part C: Numerical Problems (Detailed Solutions)

Problem 6: Calculation of Work

Question: An ideal gas expands from a volume of 5 Liters to 15 Liters at a constant pressure of 2 atmospheres. Calculate the work done by the gas. (Given: 1 atm = 1.01 × 105 Pa).

Isobaric expansion

Solution:

Step 1: Identify the process. Constant Pressure means it is an Isobaric Process.

Step 2: Convert everything to SI units.

Pressure P = 2 atm = 2 × 1.01 × 105 = 2.02 × 105 Pa.

Initial Volume V1 = 5 L = 5 × 10-3 m3.

Final Volume V2 = 15 L = 15 × 10-3 m3.

Step 3: Apply the Formula.

Work W = P(V2 – V1)

W = 2.02 × 105 × (15 × 10-3 – 5 × 10-3)

W = 2.02 × 105 × (10 × 10-3)

W = 2.02 × 102 × 10

W = 2020 Joules.

Problem 7: First Law Application

Question: A system absorbs 3000 J of heat. As a result, it expands doing 500 J of work on the surroundings. What is the change in the internal energy of the system?

Solution:

Step 1: Assign signs based on convention.

Heat absorbed: ΔQ = +3000 J.

Work done BY the system (expansion): ΔW = +500 J.

Step 2: Use the First Law equation.

ΔQ = ΔU + ΔW

3000 = ΔU + 500

Step 3: Solve for ΔU.

ΔU = 3000 – 500 = +2500 Joules.

(The positive sign indicates the temperature of the system increased).

Problem 8: Efficiency of a Heat Engine

Question: A Carnot engine absorbs 1000 J of heat energy from a reservoir at 127°C and rejects 600 J of heat during each cycle. Calculate (a) the efficiency of the engine, (b) the work done, and (c) the temperature of the sink.

Solution:

(a) Calculate Efficiency (η):

Given: Q1 (Input) = 1000 J, Q2 (Rejected) = 600 J.

η = 1 – (Q2 / Q1)

η = 1 – (600 / 1000) = 1 – 0.6 = 0.4 or 40%.

(b) Calculate Work Done (W):

W = Q1 – Q2

W = 1000 – 600 = 400 Joules.

(c) Calculate Sink Temperature (T2):

First, convert Source Temp T1 to Kelvin: 127°C + 273 = 400 K.

For a Carnot engine: Q2 / Q1 = T2 / T1

600 / 1000 = T2 / 400

0.6 = T2 / 400

T2 = 0.6 × 400 = 240 K.

Convert back to Celsius: 240 – 273 = -33°C.

Problem 9: Refrigerator Performance (COP)

Question: A refrigerator works between temperatures -3°C and 27°C. Calculate its Coefficient of Performance (COP).

Solution:

Step 1: Convert Temperatures to Kelvin. (Always do this first!)

Sink Temp T2 (Inside fridge) = -3 + 273 = 270 K.

Source Temp T1 (Room temp) = 27 + 273 = 300 K.

Step 2: Apply the COP Formula (β).

β = T2 / (T1 – T2)

Step 3: Calculate.

β = 270 / (300 – 270)

β = 270 / 30

β = 9.

(This means for every 1 Joule of electrical work, the fridge removes 9 Joules of heat. This is highly efficient!)

Read Also: 

Class-11 Chapter 10- Thermal properties of matter

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