class 11 notes

Class 11 Physics Notes Chapter 12 (Chapter 12) – Examplar Problems (English) Book

Philoid

Philoid

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Examplar Problems (English)
Alright class, let's dive into Chapter 12: Thermodynamics from your NCERT Exemplar. This is a crucial chapter, not just for your Class 11 understanding, but also forms the bedrock for many concepts tested in various government examinations. Pay close attention to the definitions, laws, and processes involved.

Chapter 12: Thermodynamics - Detailed Notes

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).
  • Thermodynamic System: A specific quantity of matter or a region in space chosen for study, separated from its surroundings by a boundary.
  • Surroundings: Everything external to the system that can interact with it.
  • Boundary: The real or imaginary surface separating the system from its surroundings.
  • Types of Systems:
    • Open System: Can exchange both energy (heat and work) and matter with the surroundings (e.g., boiling water in an open beaker).
    • Closed System: Can exchange energy but not matter with the surroundings (e.g., gas enclosed in a cylinder with a piston).
    • Isolated System: Can exchange 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 equilibrium state of a thermodynamic system.
  • Examples: Pressure (P), Volume (V), Temperature (T), Internal Energy (U), Entropy (S - introduced later).
  • State Functions: Properties whose value depends only on the current state of the system, not on the path taken to reach that state (e.g., P, V, T, U).
  • Path Functions: Properties whose value depends on the path followed during a process (e.g., Heat (Q), Work (W)).

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 they are brought into thermal contact. This implies they are at the same temperature.
  • Zeroth Law: 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), Heat (Q), and Work (W)

  • Internal Energy (U): The sum of kinetic energies (translational, rotational, vibrational) and potential energies (intermolecular forces) of the molecules within the system. It's a state function and depends only on the state (e.g., temperature for an ideal gas). ΔU = U_final - U_initial. For an ideal gas, U depends only on T.
  • Heat (Q): Energy transferred between the system and surroundings due to a temperature difference. It's a path function.
    • Sign Convention:
      • Q > 0: Heat supplied to the system.
      • Q < 0: Heat rejected by the system.
  • Work (W): Energy transferred between the system and surroundings by means other than temperature difference (e.g., expansion or compression of gas). It's a path function.
    • Sign Convention (Physics):
      • W > 0: Work done by the system (e.g., expansion).
      • W < 0: Work done on the system (e.g., compression).
    • Work Done by Gas: For a gas expanding/compressing from V1 to V2, W = ∫(P dV) from V1 to V2. Graphically, it's the area under the P-V diagram.

5. First Law of Thermodynamics

  • Statement: The heat (ΔQ) supplied to a system is partly used to increase its internal energy (ΔU) and partly used in doing work (ΔW) against the surroundings.
  • Equation: ΔQ = ΔU + ΔW
  • Basis: It's essentially the law of conservation of energy applied to thermodynamic systems.
  • Note: Always use consistent sign conventions for Q, U, and W. ΔU depends only on initial and final states, while ΔQ and ΔW depend on the process.

6. 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 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 Kelvin). Unit: J mol⁻¹ K⁻¹.
  • For Gases: Gases have two principal specific heat capacities:
    • Molar Specific Heat at Constant Volume (Cv): Heat required per mole per degree rise in temperature when volume is constant. ΔQ = nCvΔT. Since ΔW=0 at constant volume, ΔU = nCvΔT.
    • Molar Specific Heat at Constant Pressure (Cp): Heat required per mole per degree rise in temperature when pressure is constant. ΔQ = nCpΔT. At constant pressure, ΔW = PΔV = nRΔT. So, nCpΔT = nCvΔT + nRΔT.
  • Mayer's Relation: For an ideal gas, Cp - Cv = R (where R is the universal gas constant).
  • Adiabatic Exponent (Gamma, γ): Ratio of specific heats, γ = Cp / Cv.
    • γ depends on the atomicity of the gas (Monoatomic: γ=5/3; Diatomic: γ=7/5; Polyatomic: γ varies).

7. Thermodynamic Processes

  • A process occurs when a system changes from one state to another.

    • a) Isochoric Process (Volume Constant):

      • V = constant, ΔV = 0.
      • Work Done: ΔW = PΔV = 0.
      • First Law: ΔQ = ΔU.
      • Heat supplied = Change in internal energy = nCvΔT.
      • P/T = constant (Gay-Lussac's Law).
      • P-V Diagram: Vertical line.

    • b) Isobaric Process (Pressure Constant):

      • P = constant, ΔP = 0.
      • Work Done: ΔW = P(Vf - Vi) = nR(Tf - Ti).
      • First Law: ΔQ = ΔU + ΔW.
      • Heat supplied: ΔQ = nCpΔT.
      • Change in Internal Energy: ΔU = nCvΔT.
      • V/T = constant (Charles's Law).
      • P-V Diagram: Horizontal line.

    • c) Isothermal Process (Temperature Constant):

      • T = constant, ΔT = 0.
      • For an ideal gas, Internal Energy U depends only on T, so ΔU = 0.
      • First Law: ΔQ = ΔW. (Heat supplied is used entirely to do work).
      • Work Done: ΔW = nRT ln(Vf / Vi) = nRT ln(Pi / Pf).
      • PV = constant (Boyle's Law).
      • P-V Diagram: Hyperbola.

    • d) Adiabatic Process (No Heat Exchange):

      • ΔQ = 0. System is thermally insulated.
      • First Law: 0 = ΔU + ΔW => ΔW = -ΔU. (Work is done at the cost of internal energy).
      • Equations governing the process:
        • PV^γ = constant
        • TV^(γ-1) = constant
        • P^(1-γ) T^γ = constant
      • Work Done: ΔW = (PiVi - PfVf) / (γ-1) = nR(Ti - Tf) / (γ-1).
      • P-V Diagram: Steeper curve than isothermal (Slope_adiabatic = γ * Slope_isothermal).

    • e) Cyclic Process:

      • System returns to its initial state after a series of changes.
      • Initial state = Final state.
      • Change in Internal Energy: ΔU = 0 (since U is a state function).
      • First Law: ΔQ = ΔW.
      • Net work done during the cycle equals the net heat absorbed.
      • P-V Diagram: Closed loop. Area enclosed by the loop represents the net work done in the cycle (Clockwise cycle: W > 0, Net work done by system; Anticlockwise cycle: W < 0, Net work done on system).

8. Heat Engines

  • A device that converts heat energy into mechanical work continuously through a cyclic process.
  • Components:
    • Source (Hot Reservoir): Supplies heat (QH) at a high temperature (TH).
    • Working Substance: Undergoes a cyclic process (e.g., gas in a cylinder).
    • Sink (Cold Reservoir): Rejects unused heat (QL) at a lower temperature (TL).
  • Process: Absorbs QH from source -> Performs work W -> Rejects QL to sink.
  • Work Done (per cycle): W = QH - QL (from First Law for a cycle, ΔU=0).
  • Thermal Efficiency (η): Ratio of net work done by the engine in one cycle to the heat absorbed from the source in that cycle.
    • η = W / QH = (QH - QL) / QH = 1 - (QL / QH).
  • Efficiency is always less than 1 (or 100%) because QL cannot be zero (Second Law).

9. Refrigerators and Heat Pumps

  • Devices that work in reverse to heat engines; they transfer heat from a colder body to a hotter body using external work.
  • Refrigerator: Purpose is to cool the cold reservoir (e.g., inside of the fridge).
    • Absorbs heat QL from cold reservoir (TL).
    • External work W is done on the system.
    • Rejects heat QH = QL + W to the hot reservoir (TH, surroundings).
    • Coefficient of Performance (β or COP_ref): Ratio of heat extracted from the cold reservoir to the work done.
      • β = QL / W = QL / (QH - QL).
  • Heat Pump: Purpose is to heat the hot reservoir (e.g., room in winter).
    • Absorbs heat QL from cold reservoir (TL, outside air).
    • External work W is done on the system.
    • Rejects heat QH = QL + W to the hot reservoir (TH, room).
    • Coefficient of Performance (COP_HP): Ratio of heat delivered to the hot reservoir to the work done.
      • COP_HP = QH / W = QH / (QH - QL).
  • Relation: COP_HP = QH / W = (QL + W) / W = (QL / W) + 1 = β + 1.

10. Second Law of Thermodynamics

  • Deals with the direction of natural processes and the limitations on converting heat into work.
  • Kelvin-Planck Statement: It is impossible to construct a device operating in a cycle that will produce no effect other than extraction of heat from a single reservoir and perform an equivalent amount of work. (Implication: No heat engine can have 100% efficiency, QL cannot be zero).
  • Clausius Statement: It is impossible to construct a device operating in a cycle that will produce no effect other than the transfer of heat from a colder body to a hotter body without external work. (Implication: Heat cannot spontaneously flow from cold to hot; refrigerators need external work).
  • Both statements are equivalent.

11. 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 initial states, with no other change anywhere else in the universe.
    • Conditions: Must be quasi-static (extremely slow), no dissipative forces (like friction, viscosity).
    • Idealization, not perfectly achievable in reality. Examples: Slow isothermal/adiabatic expansion/compression without friction.
  • Irreversible Process: Any process that is not reversible.
    • All natural processes are irreversible (e.g., heat flow due to finite temperature difference, free expansion, dissolving salt, friction).

12. Carnot Engine and Carnot Cycle

  • An idealized reversible heat engine operating between two temperatures TH and TL.
  • Carnot Cycle (for ideal gas): Consists of four reversible processes:
    1. Isothermal Expansion (at TH, absorbs QH)
    2. Adiabatic Expansion (Temp drops from TH to TL)
    3. Isothermal Compression (at TL, rejects QL)
    4. Adiabatic Compression (Temp rises from TL to TH)
  • Efficiency of Carnot Engine (η_Carnot): Depends only on the temperatures of the source (TH) and sink (TL) (in Kelvin).
    • For a Carnot engine, QL / QH = TL / TH.
    • η_Carnot = 1 - (QL / QH) = 1 - (TL / TH).
  • Carnot's Theorem:
    1. No engine operating between two given heat reservoirs can be more efficient than a Carnot engine operating between the same two reservoirs.
    2. The efficiency of all reversible engines operating between the same two temperatures is the same, irrespective of the working substance.
  • Provides the theoretical maximum possible efficiency for any heat engine.

Multiple Choice Questions (MCQs)

  1. Which of the following statements is correct regarding the First Law of Thermodynamics?
    (a) It is applicable to reversible processes only.
    (b) It introduces the concept of internal energy.
    (c) It introduces the concept of entropy.
    (d) It is a restatement of the principle of conservation of energy.

  2. In an adiabatic expansion of an ideal gas:
    (a) Temperature increases.
    (b) Work is done on the gas.
    (c) Internal energy of the gas decreases.
    (d) Heat is supplied to the gas.

  3. The area enclosed by a cyclic process on a P-V diagram represents:
    (a) The change in internal energy.
    (b) The net heat supplied to the system.
    (c) The net work done during the cycle.
    (d) The total heat rejected by the system.

  4. For an ideal gas, the relation Cp - Cv = R is known as:
    (a) Carnot's theorem
    (b) Mayer's relation
    (c) Clausius statement
    (d) Kelvin-Planck statement

  5. During an isothermal process for an ideal gas:
    (a) Heat exchange is zero (ΔQ = 0).
    (b) Work done is zero (ΔW = 0).
    (c) Change in internal energy is zero (ΔU = 0).
    (d) Both pressure and volume remain constant.

  6. A Carnot engine operates between temperatures 600 K (source) and 300 K (sink). Its efficiency is:
    (a) 100%
    (b) 75%
    (c) 50%
    (d) 25%

  7. Which of the following processes is irreversible?
    (a) Slow isothermal compression of an ideal gas without friction.
    (b) Slow adiabatic expansion of an ideal gas without friction.
    (c) Heat transfer through a finite temperature difference.
    (d) Electrolysis (if reversed, regenerates original substances).

  8. The coefficient of performance (β) of a refrigerator is given by (QL=heat extracted from cold, QH=heat rejected to hot, W=work done):
    (a) QH / W
    (b) QL / W
    (c) W / QL
    (d) W / QH

  9. On a P-V diagram, the curve for an adiabatic process is steeper than the curve for an isothermal process because:
    (a) γ = 1
    (b) γ = 0
    (c) γ > 1
    (d) γ < 1

  10. According to the Zeroth Law of Thermodynamics, if system A is in thermal equilibrium with system C, and system B is also in thermal equilibrium with system C, then:
    (a) System A must be identical to system B.
    (b) Systems A and B are in thermal equilibrium with each other.
    (c) Systems A and B cannot be in thermal equilibrium with each other.
    (d) System C must have the highest temperature.

Answers to MCQs:

  1. (d)
  2. (c)
  3. (c)
  4. (b)
  5. (c)
  6. (c) [η = 1 - TL/TH = 1 - 300/600 = 1 - 0.5 = 0.5 or 50%]
  7. (c)
  8. (b)
  9. (c) [Slope_adiabatic = -γ(P/V), Slope_isothermal = -(P/V). Since γ > 1, adiabatic slope is steeper.]
  10. (b)

Make sure you understand the underlying concepts behind each point and formula. Practice numerical problems based on these processes and laws, especially those involving P-V diagrams and efficiency calculations. Good luck with your preparation!

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