class 11 notes

Class 11 Physics Notes Chapter 6 (Chapter 6) – Physics Part-II Book

Philoid

Philoid

— 8 min read

Physics Part-II
Alright class, let's dive into Chapter 6: Thermodynamics from your NCERT Class 11 Physics Part-II book. This is a crucial chapter, not just for your Class 11 exams but also for various government competitive exams like NEET, JEE, NDA, and others where Physics is a component. Pay close attention to the concepts, laws, and formulas.

Chapter 6: Thermodynamics - Detailed Notes for Government Exam Preparation

1. Introduction

  • Thermodynamics: The branch of physics that deals with the concepts of heat, temperature, and the inter-conversion of heat and other forms of energy (like mechanical work).
  • System: A specific part of the universe under investigation (e.g., gas in a cylinder).
  • Surroundings: Everything outside the system that can interact with it.
  • Boundary: The real or imaginary surface separating the system from its surroundings.
  • Types of Systems:
    • Open System: Exchanges both energy (heat/work) and matter with the surroundings (e.g., boiling water in an open beaker).
    • Closed System: Exchanges only energy, not matter, with the surroundings (e.g., gas enclosed in a cylinder with a piston).
    • Isolated System: Exchanges neither energy nor matter with the surroundings (e.g., tea in a perfect thermos flask - an idealization).

2. Thermodynamic State Variables

  • Variables that describe the state of a thermodynamic system (e.g., Pressure (P), Volume (V), Temperature (T), Internal Energy (U), Entropy (S)).
  • State Variables: Depend only on the current state of the system, not on the path taken to reach that state (P, V, T, U, S).
  • Path Variables: Depend on the path taken during a process (e.g., Work (W), Heat (Q)).
  • Equation of State: A mathematical relation connecting state variables (e.g., Ideal Gas Law: PV = nRT).

3. Thermal Equilibrium and Zeroth Law of Thermodynamics

  • Thermal Equilibrium: Two systems are in thermal equilibrium if there is no net flow of heat between them when brought into thermal contact. This implies they are at the same temperature.
  • Zeroth Law of Thermodynamics: If two systems (say 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 provides the basis for the concept of temperature.

4. Internal Energy (U)

  • The sum of molecular kinetic and potential energies in the frame of reference relative to which the centre of mass of the system is at rest.
  • For an ideal gas, internal energy depends only on its temperature (U ∝ T). It's independent of pressure or volume.
  • Internal energy is a state function. Change in internal energy (ΔU) depends only on the initial and final states (ΔU = U_final - U_initial).

5. Heat (Q) and Work (W)

  • Heat (Q): Energy transferred between a system and its surroundings due to a temperature difference.
    • Sign Convention:
      • Q > 0: Heat supplied to the system.
      • Q < 0: Heat rejected by the system.
  • Work (W): Energy transferred between a system and its surroundings due to mechanisms other than temperature difference (e.g., expansion or compression of gas).
    • Work done by the gas during expansion (volume increases): ΔW = ∫ P dV
    • Sign Convention (Physics convention, often used in competitive exams):
      • W > 0: Work done by the system (e.g., expansion).
      • W < 0: Work done on the system (e.g., compression).
      • (Note: Chemistry often uses the opposite sign convention for work. Be careful about the context).
  • Both Heat and Work are path functions.

6. First Law of Thermodynamics

  • Based on the principle of conservation of energy.
  • Statement: When heat (ΔQ) is supplied to a system, some of it increases the internal energy (ΔU) of the system, and the rest is used by the system to do work (ΔW) on the surroundings.
  • Mathematical Form: ΔQ = ΔU + ΔW
    • ΔQ = Heat supplied to the system
    • ΔU = Change in internal energy of the system
    • ΔW = Work done by the system
  • This law applies to all processes in nature.

7. Specific Heat Capacity

  • Specific Heat Capacity (s or c): Amount of heat required to raise the temperature of unit mass of a substance by one degree Celsius (or one Kelvin). Unit: J kg⁻¹ K⁻¹.
  • Molar Specific Heat Capacity (C): Amount of heat required to raise the temperature of one mole of a substance by one degree Celsius (or one Kelvin). Unit: J mol⁻¹ K⁻¹.
  • For Gases: Specific heat can have many values depending on the process. Two important ones are:
    • Molar Specific Heat at Constant Volume (Cv): Heat supplied per mole per degree rise in temperature at constant volume.
      • At constant volume, ΔW = 0. So, from First Law: (ΔQ)_v = ΔU = nCvΔT.
      • For monoatomic ideal gas: Cv = (3/2)R
      • For diatomic ideal gas (moderate temp.): Cv = (5/2)R
    • Molar Specific Heat at Constant Pressure (Cp): Heat supplied per mole per degree rise in temperature at constant pressure.
      • At constant pressure: (ΔQ)_p = ΔU + PΔV = nCpΔT.
      • For monoatomic ideal gas: Cp = (5/2)R
      • For diatomic ideal gas (moderate temp.): Cp = (7/2)R
  • Mayer's Formula: Relation between Cp and Cv for an ideal gas:
    • Cp - Cv = R (where R is the Universal Gas Constant ≈ 8.314 J mol⁻¹ K⁻¹)
  • Adiabatic Exponent (Gamma, γ): Ratio of specific heats.
    • γ = Cp / Cv
    • For monoatomic gas: γ = 5/3 ≈ 1.67
    • For diatomic gas: γ = 7/5 = 1.4
    • For polyatomic gas: γ = (4+f)/(3+f) where f is vibrational degrees of freedom (or use Cp/Cv values directly).

8. Thermodynamic Processes

  • Isochoric Process (Constant Volume):
    • V = Constant (ΔV = 0)
    • Work Done (ΔW) = ∫ P dV = 0
    • First Law: ΔQ = ΔU = nCvΔT
    • P/T = Constant (Gay-Lussac's Law)
  • Isobaric Process (Constant Pressure):
    • P = Constant (ΔP = 0)
    • Work Done (ΔW) = P(V₂ - V₁) = PΔV = nRΔT
    • First Law: ΔQ = ΔU + PΔV = nCpΔT
    • V/T = Constant (Charles's Law)
  • Isothermal Process (Constant Temperature):
    • T = Constant (ΔT = 0)
    • Internal Energy change for ideal gas (ΔU) = nCvΔT = 0
    • First Law: ΔQ = ΔW
    • Work Done (ΔW) = nRT ln(V₂/V₁) = nRT ln(P₁/P₂) = 2.303 nRT log₁₀(V₂/V₁) = 2.303 nRT log₁₀(P₁/P₂)
    • PV = Constant (Boyle's Law)
  • Adiabatic Process (No Heat Exchange):
    • ΔQ = 0
    • First Law: 0 = ΔU + ΔW => ΔW = -ΔU = -nCvΔT
    • Work Done (ΔW) = (P₁V₁ - P₂V₂) / (γ - 1) = nR(T₁ - T₂) / (γ - 1)
    • Equation: PV^γ = Constant; TV^(γ-1) = Constant; P(1-γ)Tγ = Constant
    • Adiabatic process is steeper than isothermal process on a P-V diagram (Slope_adiabatic = -γ(P/V), Slope_isothermal = -(P/V)).

9. Heat Engines

  • A device that converts heat energy into mechanical work cyclically.
  • Components: Source (hot reservoir at T₁), Working Substance, Sink (cold reservoir at T₂).
  • Process: Takes heat Q₁ from source, converts part of it into work W, rejects remaining heat Q₂ to sink.
  • Efficiency (η): Ratio of work output to heat input.
    • W = Q₁ - Q₂ (from energy conservation in a cycle, ΔU=0)
    • η = W / Q₁ = (Q₁ - Q₂) / Q₁ = 1 - (Q₂ / Q₁)
    • Efficiency is always less than 1 (or 100%).

10. Refrigerators and Heat Pumps

  • Work in reverse to heat engines. Use external work (W) to transfer heat from a cold reservoir to a hot reservoir.
  • Refrigerator: Purpose is to cool the cold reservoir (extract heat Q₂).
    • Coefficient of Performance (β or COP_ref): β = Q₂ / W = Q₂ / (Q₁ - Q₂)
  • Heat Pump: Purpose is to heat the hot reservoir (deliver heat Q₁).
    • Coefficient of Performance (COP_hp): COP_hp = Q₁ / W = Q₁ / (Q₁ - Q₂)
    • Relation: COP_hp = 1 + COP_ref

11. Second Law of Thermodynamics

  • Deals with the direction of natural processes and the impossibility of certain processes, even if they conserve energy (allowed by First Law).
  • Kelvin-Planck Statement: It is impossible to construct a device operating in a cycle that will produce no effect other than extracting heat from a single reservoir and performing an equivalent amount of work. (Essentially, no heat engine can have 100% efficiency).
  • Clausius Statement: It is impossible to construct a device operating in a cycle that will produce no effect other than transferring heat from a colder body to a hotter body without external work. (Heat doesn't flow spontaneously from cold to hot).
  • Both statements are equivalent.

12. Reversible and Irreversible Processes

  • Reversible Process: A process that can be reversed in such a way that both the system and the surroundings return to their original states, with no other change anywhere else in the universe.
    • Idealized concept. Requires process to be quasi-static (infinitely slow) and non-dissipative (no friction, viscosity etc.).
  • Irreversible Process: Any process that is not reversible.
    • All natural processes are irreversible (due to friction, heat transfer across finite temperature difference, free expansion etc.).

13. Carnot Engine and Carnot Cycle

  • An idealized reversible heat engine operating between two temperatures T₁ (source) and T₂ (sink).
  • Carnot Cycle: Consists of four reversible processes:
    1. Isothermal Expansion (at T₁)
    2. Adiabatic Expansion (T₁ → T₂)
    3. Isothermal Compression (at T₂)
    4. Adiabatic Compression (T₂ → T₁)
  • Carnot's Theorem:
    • No engine operating between two given temperatures can be more efficient than a Carnot engine operating between the same temperatures.
    • The efficiency of all reversible engines operating between the same two temperatures is the same, irrespective of the working substance.
  • Efficiency of Carnot Engine (η_c):
    • η_c = 1 - (Q₂ / Q₁) = 1 - (T₂ / T₁)
    • Temperatures T₁ and T₂ must be in Kelvin (absolute scale).
    • This gives the maximum possible efficiency for any engine operating between T₁ and T₂.

Important Points for Exams:

  • Memorize the sign conventions for Q and W.
  • Know the First Law equation (ΔQ = ΔU + ΔW) and how it applies to different processes.
  • Understand the characteristics (conditions, work done formula, governing equation) of Isochoric, Isobaric, Isothermal, and Adiabatic processes. Be able to identify them on P-V diagrams.
  • Remember Mayer's relation (Cp - Cv = R) and the values of Cv, Cp, and γ for monoatomic and diatomic ideal gases.
  • Know the formulas for efficiency of a heat engine (η = 1 - Q₂/Q₁ = 1 - T₂/T₁ for Carnot) and COP of refrigerators/heat pumps.
  • Understand the statements of the Second Law.
  • Remember that T in efficiency/COP formulas must be in Kelvin.

Multiple Choice Questions (MCQs)

  1. In an adiabatic process, which of the following quantities remains constant?
    a) Temperature
    b) Pressure
    c) Volume
    d) Heat content (No heat exchange)

  2. The First Law of Thermodynamics is a statement of:
    a) Conservation of heat
    b) Conservation of work
    c) Conservation of energy
    d) Conservation of momentum

  3. For an ideal gas, the change in internal energy during an isothermal process is:
    a) Positive
    b) Negative
    c) Zero
    d) Dependent on the pressure change

  4. Mayer's formula for an ideal gas is:
    a) Cp / Cv = R
    b) Cp + Cv = R
    c) Cp - Cv = R
    d) Cv - Cp = R

  5. During which thermodynamic process does the volume remain constant?
    a) Isothermal
    b) Isobaric
    c) Adiabatic
    d) Isochoric

  6. The efficiency of a Carnot engine operating between a source temperature T₁ and sink temperature T₂ is given by:
    a) 1 - T₁/T₂
    b) 1 - T₂/T₁
    c) T₁/T₂
    d) T₂/T₁

  7. A system is given 400 Joules of heat and it does 100 Joules of work. What is the change in the internal energy of the system?
    a) 500 J
    b) 300 J
    c) -300 J
    d) -500 J

  8. Which of the following is NOT a state function?
    a) Internal Energy (U)
    b) Temperature (T)
    c) Work (W)
    d) Pressure (P)

  9. The value of the adiabatic exponent (γ = Cp/Cv) for a diatomic gas like Oxygen (O₂) at moderate temperatures is approximately:
    a) 1.67
    b) 1.4
    c) 1.33
    d) 1.5

  10. According to the Kelvin-Planck statement of the second law of thermodynamics:
    a) Heat cannot flow spontaneously from a colder to a hotter body.
    b) It is impossible to construct a 100% efficient heat engine.
    c) The entropy of the universe always increases.
    d) Internal energy depends only on temperature for an ideal gas.

Answers to MCQs:

  1. d) Heat content (No heat exchange, ΔQ = 0)
  2. c) Conservation of energy
  3. c) Zero (For an ideal gas, U depends only on T. If T is constant, ΔU = 0)
  4. c) Cp - Cv = R
  5. d) Isochoric
  6. b) 1 - T₂/T₁ (Temperatures must be in Kelvin)
  7. b) 300 J (ΔQ = ΔU + ΔW => 400 J = ΔU + 100 J => ΔU = 300 J)
  8. c) Work (W) (Work and Heat are path functions)
  9. b) 1.4 (For diatomic gas, Cp = (7/2)R, Cv = (5/2)R, γ = Cp/Cv = 7/5 = 1.4)
  10. b) It is impossible to construct a 100% efficient heat engine.

Study these notes thoroughly. Focus on understanding the underlying principles rather than just memorizing formulas. Good luck with your preparation!

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