Class 11 Physics Notes Chapter 4 (Chapter 4) – Physics Part-II Book
Alright class, let's get straight into the crucial concepts from Chapter 12, Thermodynamics, from your NCERT Class 11 Physics Part-II book. Now, I understand the chapter number might seem confusing as you mentioned 'Chapter 4', but Chapter 4 ('Motion in a Plane') is in Part I. Since you specified the Part-II book, we'll focus on the fourth chapter within Part II, which is Chapter 12: Thermodynamics. This is a very important chapter for various government exams, focusing on energy transformations involving heat and work. Pay close attention!
Chapter 12: Thermodynamics - Detailed Notes for Government Exam Preparation
1. Basic Concepts:
- Thermodynamics: The branch of physics dealing with heat and its relation to other forms of energy (like mechanical work) and energy transformations.
- System: A specific part of the universe under study (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 the surroundings.
- Thermodynamic State Variables: Parameters describing the state of a system.
- Intensive Variables: Independent of the system's size (e.g., Pressure (P), Temperature (T), Density).
- Extensive Variables: Dependent on the system's size (e.g., Volume (V), Mass (m), Internal Energy (U), Heat Capacity).
- Equation of State: A relation between state variables (e.g., PV = nRT for an ideal gas).
- Thermodynamic Equilibrium: A system is in equilibrium if its macroscopic variables (P, V, T, composition) do not change with time. Requires Thermal, Mechanical, and Chemical equilibrium.
2. 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. They are said to have the same Temperature.
- Zeroth Law: 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 provides the basis for temperature measurement.
3. Heat, Internal Energy, and Work:
- Internal Energy (U): The sum of kinetic and potential energies of all molecules within the system. It's a state function (depends only on the initial and final states, not the path taken). For an ideal gas, U depends only on Temperature (U ∝ T). ΔU = nCvΔT.
- Heat (Q): Energy transferred between 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.
- Sign Convention:
- Work (W): Energy transferred between system and surroundings by means other than temperature difference (e.g., expansion/compression of gas). It's a path function.
- Work done by the system (Expansion): W = ∫ P dV
- Sign Convention (Physics Convention):
- W > 0: Work done by the system (Expansion).
- W < 0: Work done on the system (Compression).
- (Note: Chemistry often uses the opposite sign convention for work. Be careful based on the context).
- Work done can be calculated as the area under the P-V diagram.
4. First Law of Thermodynamics:
- Statement: Energy conservation applied to thermodynamic processes. When heat (ΔQ) is supplied to a system, some of it increases the internal energy (ΔU), and the rest is used by the system to do work (ΔW).
- Equation: ΔQ = ΔU + ΔW
- ΔQ = Heat added to the system
- ΔU = Change in internal energy of the system
- ΔW = Work done by the system
- Alternative form (using work done on the system, W_on = -ΔW): ΔU = ΔQ + W_on
- Key Points:
- ΔU is path-independent (depends only on initial and final states).
- ΔQ and ΔW are path-dependent.
- The First Law establishes ΔU as a state function.
5. Specific Heat Capacity:
- Molar Specific Heat at Constant Volume (Cv): Heat required to raise the temperature of 1 mole of gas by 1K at constant volume.
- At constant volume, ΔW = 0. So, ΔQ = ΔU = nCvΔT.
- Cv = (1/n) (dU/dT)
- Molar Specific Heat at Constant Pressure (Cp): Heat required to raise the temperature of 1 mole of gas by 1K at constant pressure.
- At constant pressure, ΔQ = ΔU + PΔV = nCpΔT.
- Mayer's Relation (for ideal gas): Cp - Cv = R (where R is the universal gas constant).
- Adiabatic Exponent (γ): γ = Cp / Cv
- For monatomic gas (e.g., He, Ar): Cv = (3/2)R, Cp = (5/2)R, γ = 5/3 ≈ 1.67
- For diatomic gas (e.g., N2, O2, H2): Cv = (5/2)R, Cp = (7/2)R, γ = 7/5 = 1.4
- For polyatomic gas (e.g., CO2, NH3): Cv ≈ 3R, Cp ≈ 4R, γ ≈ 4/3 ≈ 1.33 (approximate, depends on molecular structure and temperature).
6. Thermodynamic Processes:
- Isothermal Process (T = constant):
- ΔT = 0 => ΔU = 0 (for ideal gas).
- First Law: ΔQ = ΔW. Heat supplied is used entirely to do work.
- Equation: PV = constant (Boyle's Law).
- Work Done: W = nRT ln(Vf / Vi) = nRT ln(Pi / Pf).
- Slow process, requires heat exchange with surroundings.
- Adiabatic Process (Q = constant, ΔQ = 0):
- No heat exchange with surroundings (perfectly insulated system or very fast process).
- First Law: 0 = ΔU + ΔW => ΔW = -ΔU. Work is done at the expense of internal energy.
- Equation: PV^γ = constant. Also, TV^(γ-1) = constant and P(1-γ)Tγ = constant.
- Work Done: W = (PiVi - PfVf) / (γ - 1) = nR(Ti - Tf) / (γ - 1).
- Adiabatic curve is steeper than isothermal curve on a P-V diagram (γ > 1).
- Isobaric Process (P = constant):
- Equation: V/T = constant (Charles's Law).
- Work Done: W = P(Vf - Vi) = nR(Tf - Ti).
- First Law: ΔQ = ΔU + PΔV = nCvΔT + nRΔT = n(Cv + R)ΔT = nCpΔT.
- Isochoric Process (V = constant):
- ΔV = 0 => Work Done W = 0.
- First Law: ΔQ = ΔU. Heat supplied increases internal energy.
- Equation: P/T = constant (Gay-Lussac's Law).
- ΔQ = ΔU = nCvΔT.
- Cyclic Process: System returns to its initial state after a series of changes.
- ΔU = 0 (Internal energy is a state function).
- First Law: ΔQ = ΔW. Net heat absorbed equals net work done by the system.
- Area enclosed by the cycle on a P-V diagram represents the net work done per cycle (Clockwise cycle: W > 0, Anticlockwise cycle: W < 0).
7. Heat Engines:
- Device converting heat energy into mechanical work continuously through a cyclic process.
- Components:
- Source (Hot Reservoir): At high temperature T1, supplies heat Q1.
- Working Substance: (e.g., gas) undergoes cyclic process.
- Sink (Cold Reservoir): At low temperature T2 (T2 < T1), rejects heat Q2.
- Process: Takes heat Q1 from source, converts part of it into work W, rejects remaining heat Q2 to sink.
- Efficiency (η): η = (Net Work Done / Heat Input) = W / Q1
- From First Law for a cycle: W = Q1 - Q2 (assuming Q1, Q2 are magnitudes).
- η = (Q1 - Q2) / Q1 = 1 - (Q2 / Q1)
- Efficiency is always less than 1 (or 100%).
8. Refrigerators and Heat Pumps:
- Devices working in reverse of a heat engine; transfer heat from a cold body to a hot body using external work.
- Refrigerator: Purpose is to cool the cold reservoir (inside of fridge).
- Takes heat Q2 from cold reservoir (T2), external work W is done on it, rejects heat Q1 to hot reservoir (surroundings, T1).
- Coefficient of Performance (β or COP_ref): β = (Heat Extracted / Work Done) = Q2 / W
- β = Q2 / (Q1 - Q2)
- Heat Pump: Purpose is to heat the hot reservoir (room in winter).
- Same mechanism, but focus is on heat delivered Q1.
- Coefficient of Performance (COP_HP): COP_HP = (Heat Delivered / Work Done) = Q1 / W
- COP_HP = Q1 / (Q1 - Q2)
- COP_HP = β + 1
- COP can be greater than 1.
9. Second Law of Thermodynamics:
- Specifies the direction of spontaneous thermodynamic processes. Puts limitations on heat-to-work conversion.
- Kelvin-Planck Statement: It is impossible to construct an engine, operating in a cycle, which will produce no effect other than extracting heat from a single reservoir and performing an equivalent amount of work. (Implies no 100% efficient heat engine, Q2 cannot be zero).
- Clausius Statement: It is impossible to construct a device that operates in a cycle and produces no effect other than the transfer of heat from a colder body to a hotter body without external work input. (Implies heat doesn't spontaneously flow from cold to hot; refrigerators need work input).
- Both statements are equivalent.
10. Reversible and Irreversible Processes:
- Reversible Process: A process that can be reversed such that both the system and the surroundings return to their original states without any other changes anywhere else in the universe.
- Idealized concept. Requires quasi-static (infinitely slow) execution and absence of dissipative forces (like friction, viscosity).
- Irreversible Process: Any process that is not reversible. All natural processes are irreversible (due to factors like friction, heat transfer across finite temperature difference, free expansion).
11. Carnot Engine and Carnot Cycle:
- An ideal, theoretical reversible heat engine operating between two temperatures T1 (source) and T2 (sink).
- Carnot Cycle (on P-V diagram):
- Isothermal Expansion (at T1, absorbs Q1)
- Adiabatic Expansion (Temp drops T1 -> T2)
- Isothermal Compression (at T2, rejects Q2)
- Adiabatic Compression (Temp rises T2 -> T1)
- Efficiency of Carnot Engine (η_c): Depends only on the absolute temperatures (in Kelvin) of the source and sink.
- η_c = 1 - (T2 / T1) (where T1 and T2 are in Kelvin).
- For a Carnot engine, Q2/Q1 = T2/T1.
- Carnot's Theorem:
- (a) No engine operating between two given temperatures can be more efficient than a reversible Carnot engine operating between the same two temperatures.
- (b) All reversible engines operating between the same two temperatures have the same efficiency, regardless of the working substance.
Key Formulas Summary:
- First Law: ΔQ = ΔU + ΔW
- Ideal Gas Internal Energy: ΔU = nCvΔT
- Mayer's Relation: Cp - Cv = R
- Adiabatic Exponent: γ = Cp / Cv
- Work (Isothermal): W = nRT ln(Vf / Vi)
- Work (Adiabatic): W = (PiVi - PfVf) / (γ - 1)
- Adiabatic Process Eq: PV^γ = constant
- Engine Efficiency: η = 1 - Q2/Q1
- Carnot Efficiency: η_c = 1 - T2/T1 (T in Kelvin)
- Refrigerator COP: β = Q2 / (Q1 - Q2) = T2 / (T1 - T2) (for Carnot Ref.)
- Heat Pump COP: COP_HP = Q1 / (Q1 - Q2) = T1 / (T1 - T2) (for Carnot HP)
Multiple Choice Questions (MCQs):
-
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 -
In an adiabatic process, which of the following remains constant?
a) Temperature
b) Pressure
c) Volume
d) Heat content (No heat exchange) -
For an ideal gas, the internal energy depends only on:
a) Pressure
b) Volume
c) Temperature
d) Number of moles -
The efficiency of a Carnot engine operating between temperatures T1 (source) and T2 (sink) is given by: (Temperatures in Kelvin)
a) 1 - T1/T2
b) 1 - T2/T1
c) T1/T2
d) T2/T1 -
During an isothermal expansion of an ideal gas:
a) Internal energy increases
b) Internal energy decreases
c) Internal energy remains constant
d) Work done is zero -
Mayer's relation for an ideal gas is:
a) Cp / Cv = R
b) Cp * Cv = R
c) Cp - Cv = R
d) Cp + Cv = R -
A refrigerator takes heat from a cold body and rejects it to a hot body. This process requires:
a) No external work
b) External work to be done on the system
c) Work to be done by the system
d) Heat input from a third body -
Which of the following thermodynamic processes is represented by the equation PV^γ = constant?
a) Isothermal
b) Isobaric
c) Isochoric
d) Adiabatic -
A cyclic process is one in which:
a) The initial and final temperatures are the same
b) The initial and final pressures are the same
c) The system returns to its initial thermodynamic state
d) No work is done -
The ratio of specific heats γ = Cp/Cv for a diatomic gas (like Oxygen or Nitrogen at moderate temperatures) is approximately:
a) 1.67
b) 1.40
c) 1.33
d) 1.00
Answers to MCQs:
- c) Conservation of energy
- d) Heat content (No heat exchange, ΔQ=0)
- c) Temperature
- b) 1 - T2/T1
- c) Internal energy remains constant (ΔT=0 for ideal gas => ΔU=0)
- c) Cp - Cv = R
- b) External work to be done on the system
- d) Adiabatic
- c) The system returns to its initial thermodynamic state
- b) 1.40 (or 7/5)
Study these notes thoroughly. Focus on understanding the definitions, laws, the characteristics of each thermodynamic process, and the formulas, especially the sign conventions and units. Good luck with your preparation!