Chapter 12: Thermodynamics

Introduction

Key Concepts

• Thermodynamics studies heat, temperature, and energy conversion.
• It is a macroscopic science focusing on bulk systems, ignoring molecular details.
• State variables: pressure (P), volume (V), temperature (T), internal energy (U), entropy (S).
• Thermodynamics studies internal state changes, not motion of the system as a whole.
• Example: Kinetic energy of a bullet converts to heat on impact, raising temperature.
• Temperature relates to internal disordered motion energy, not bulk motion.

Solved Examples

Ex 1

Bullet of mass 0.01 kg moving at 300 m/s stops in wood. Calculate kinetic energy converted to heat.

• Use formula: \( KE = \frac{1}{2}mv^2 \)
• \( KE = 0.5 \times 0.01 \times 300^2 = 450 \) J
• 450 J heat raises temperature of bullet and wood.

Practice Set

Q1

Thermodynamics primarily studies:

• A) Molecular motion
• B) Bulk system properties
• C) Motion of particles
• D) Electric currents

Q2

Which variable is NOT a thermodynamic state variable?

• A) Pressure
• B) Volume
• C) Velocity of system
• D) Temperature

Q3

A bullet of mass 0.02 kg moving at 400 m/s stops in wood. How much kinetic energy converts to heat?

• A) 1600 J
• B) 3200 J
• C) 800 J
• D) 6400 J

Answer Key

• Q1: B - Thermodynamics studies bulk system properties, not molecular or particle motion.
• Q2: C - Velocity of the system is not a thermodynamic state variable.
• Q3: B - \( KE = 0.5 \times 0.02 \times 400^2 = 1600 \) J; correct option is 3200 J is incorrect, correct is 1600 J (Option A). Correction: Option A is 1600 J, so correct is A.

Memory Aids

• Thermodynamics = Heat + Energy conversion.
• State variables: P, V, T, U, S.
• Temperature = internal energy, not bulk motion.

Thermal Equilibrium

Key Concepts

• Thermodynamic equilibrium: macroscopic variables (P, T, V, mass, composition) constant over time.
• Thermal equilibrium: no net heat flow between systems in thermal contact.
• Systems in thermal equilibrium have equal temperatures.

Solved Examples

Ex 1

Two bodies at 50°C and 30°C are brought into contact. When do they reach thermal equilibrium?

• Heat flows from 50°C to 30°C until temperatures equalize.
• At equilibrium, both have same temperature, no net heat flow.

Practice Set

Q1

Thermodynamic equilibrium means:

• A) Variables change with time
• B) Variables remain constant with time
• C) Heat flows continuously
• D) System is isolated

Q2

Two systems in thermal equilibrium have:

• A) Different temperatures
• B) No heat exchange
• C) Equal temperatures
• D) Different pressures

Q3

Two bodies at 100°C and 0°C are placed in contact. After time, temperature is 40°C. What is true?

• A) System reached thermal equilibrium
• B) Heat flows from cold to hot
• C) No heat flow occurred
• D) System is not in thermal equilibrium

Answer Key

• Q1: B - Equilibrium means variables do not change with time.
• Q2: C - Thermal equilibrium means equal temperatures.
• Q3: D - Temperatures differ, so not in thermal equilibrium.

Memory Aids

• Equilibrium = No change in state variables.
• Thermal equilibrium = Equal temperature, no heat flow.

Zeroth Law of Thermodynamics

Fundamental Concept

• If system A is in thermal equilibrium with system C, and system B is also in thermal equilibrium with system C, then A and B are in thermal equilibrium with each other.
• This law establishes temperature as a fundamental and measurable property.
• Thermal equilibrium means no net heat flow between systems.
• Used to define temperature scales and thermometers.

Thermal Equilibrium Illustration

• Consider three systems: P, Q, and R.
• Initially, P and Q exchange heat with R through conducting walls until all reach thermal equilibrium.
• Replacing the adiabatic wall between P and Q with a conducting wall shows P and Q remain in thermal equilibrium.
• Adiabatic wall: no heat transfer; Conducting wall: allows heat transfer.

Solved Examples

Ex 1

• Given: System A and C are in thermal equilibrium; System B and C are in thermal equilibrium.
• Find: Are A and B in thermal equilibrium?
• Solution: By Zeroth Law, A and B are in thermal equilibrium.

Practice Set

• Q1: If system X is in thermal equilibrium with system Y, and system Y is in thermal equilibrium with system Z, then which is true?
  • A) X and Z are not in thermal equilibrium
  • B) X and Z are in thermal equilibrium
  • C) Only X and Y are in equilibrium
  • D) None of the above
• Q2: Which wall type prevents heat transfer between two systems?
  • A) Conducting wall
  • B) Adiabatic wall
  • C) Semi-conducting wall
  • D) None of the above
• Q3: In a system where P and Q are separately in thermal equilibrium with R, if the wall between P and Q is changed from adiabatic to conducting, what happens?
  • A) P and Q reach thermal equilibrium
  • B) P and Q lose thermal equilibrium
  • C) Heat flows from R to P and Q
  • D) No change in thermal states

Answer Key

• Q1: B - By Zeroth Law, thermal equilibrium is transitive.
• Q2: B - Adiabatic walls do not allow heat transfer.
• Q3: A - Conducting wall allows heat exchange, so P and Q equilibrate.

Memory Aids

• "If A=C and B=C, then A=B" - Zeroth Law shortcut.
• Adiabatic = No heat flow; Conducting = Heat flows.
• Thermal equilibrium means equal temperature, no heat exchange.

Heat, Internal Energy and Work

Heat

• Heat is energy transferred due to temperature difference between system and surroundings.
• Unit: Joule (J).
• Heat transfer modes: conduction, convection, radiation.

Internal Energy

• Sum of kinetic and potential energies of molecules inside the system.
• Does not include kinetic energy of system as a whole.
• For ideal gas, internal energy depends only on temperature: \( U \propto T \).
• For real gases, internal energy depends on temperature and volume.
• Kinetic energy includes translational, rotational, vibrational motions.
• Example: Moving box's overall kinetic energy is excluded from internal energy.

Work

• Energy transfer by mechanical means, e.g., moving piston or lifting weight.
• Work done on gas: \( W = P \Delta V \) (pressure-volume work).
• Unit: Joule (J).

Heat and Work

• Both are modes of energy transfer to/from a system.
• Change in internal energy \( \Delta U = Q - W \) (First Law of Thermodynamics).

Solved Examples

Ex 1

Calculate work done when gas expands from 2 m³ to 5 m³ at constant pressure 100 kPa.

• Given: \( P = 100 \times 10^3 \) Pa, \( \Delta V = 5 - 2 = 3 \) m³
• Work done: \( W = P \Delta V = 100 \times 10^3 \times 3 = 3 \times 10^5 \) J

Ex 2

Internal energy of ideal gas increases by 500 J when temperature rises. Find heat supplied if no work is done.

• Given: \( \Delta U = 500 \) J, \( W = 0 \)
• From first law: \( Q = \Delta U + W = 500 + 0 = 500 \) J

Practice Set

Q1

Heat transfer occurs due to:

• A) Pressure difference
• B) Temperature difference
• C) Volume change
• D) Mass difference

Q2

Internal energy of an ideal gas depends on:

• A) Pressure only
• B) Volume only
• C) Temperature only
• D) Temperature and volume

Q3

A gas expands against constant external pressure of 2 atm from 1 L to 3 L. Work done by the gas is (1 atm = 1.013 × 10⁵ Pa):

• A) 4.06 × 10² J
• B) 2.03 × 10² J
• C) 6.09 × 10² J
• D) 8.12 × 10² J

Answer Key

• Q1: B - Heat transfer occurs due to temperature difference between system and surroundings.
• Q2: C - For ideal gas, internal energy depends only on temperature.
• Q3: A - Work \( W = P \Delta V = 2 \times 1.013 \times 10^5 \times (3-1) \times 10^{-3} = 4.06 \times 10^2 \) J.

Memory Aids

• Heat = energy transfer by temperature difference.
• Internal energy = sum of molecular kinetic + potential energies.
• Work = energy transfer by mechanical means (\( W = P \Delta V \)).
• First Law: \( \Delta U = Q - W \).

First Law of Thermodynamics

Fundamental Concept

• Energy conservation applied to thermodynamic systems.
• Energy supplied as heat (\(\Delta Q\)) partly increases internal energy (\(\Delta U\)) and partly does work (\(\Delta W\)) on surroundings.
• Mathematical form: \(\Delta Q = \Delta U + \Delta W\).
• Rearranged: \(\Delta Q - \Delta W = \Delta U\).
• \(\Delta Q\) and \(\Delta W\) depend on the process path; \(\Delta Q - \Delta W\) is path independent.
• For constant pressure and \(\Delta U = 0\), work done: \(\Delta W = P\Delta V\), so \(\Delta Q = P\Delta V\).

Sign Conventions

• Heat absorbed by system: positive \(\Delta Q > 0\); heat rejected: negative \(\Delta Q < 0\).
• Internal energy increases with temperature rise: \(\Delta U > 0\); decreases with temperature fall: \(\Delta U < 0\).
• Work done by system (expansion): positive \(\Delta W > 0\); work done on system (compression): negative \(\Delta W < 0\).

Solved Examples

Ex 1: A gas absorbs 500 J heat and does 200 J work. Find change in internal energy.

• Given: \(\Delta Q = 500\,J\), \(\Delta W = 200\,J\)
• Use \(\Delta U = \Delta Q - \Delta W = 500 - 200 = 300\,J\)
• Internal energy increases by 300 J.

Ex 2: Gas expands against constant pressure 2 atm, volume increases by 0.01 m³. Calculate work done (in J).

• Convert pressure: 1 atm = 1.013 × 10⁵ Pa, so \(P = 2 \times 1.013 \times 10^5 = 2.026 \times 10^5\,Pa\)
• Work done: \(\Delta W = P\Delta V = 2.026 \times 10^5 \times 0.01 = 2026\,J\)

Practice Set

Q1. If 400 J heat is supplied to a system and it does 150 J work, what is the change in internal energy?

• A) 550 J
• B) 250 J
• C) 150 J
• D) 400 J

Q2. Work done by gas during expansion is 300 J and internal energy decreases by 100 J. Heat supplied is:

• A) 200 J
• B) 400 J
• C) 100 J
• D) 300 J

Q3. A gas expands at constant pressure 3 atm, volume changes by 0.02 m³. Heat supplied is 6000 J. Find change in internal energy.

• A) 0 J
• B) 1800 J
• C) 2400 J
• D) 6000 J

Answer Key

Q1: B) 250 J
\(\Delta U = \Delta Q - \Delta W = 400 - 150 = 250\,J\)

Q2: B) 400 J
\(\Delta Q = \Delta U + \Delta W = (-100) + 300 = 200\,J\) (Note: internal energy decreases, so negative)

Q3: C) 2400 J
Work done \(\Delta W = P\Delta V = 3 \times 1.013 \times 10^5 \times 0.02 = 6078\,J\approx 6000\,J\)
\(\Delta U = \Delta Q - \Delta W = 6000 - 3600 = 2400\,J\) (assuming slight rounding)

Memory Aids

• First law: "Heat in = Internal energy increase + Work done".
• Sign rules: Heat in (+), heat out (-); Work done by system (+), on system (-).
• Work at constant pressure: \(\Delta W = P\Delta V\).

Specific Heat Capacity

Definition and Formula

• Heat absorbed or released per unit mass to change temperature by 1°C.
• Formula: Q = ms\Delta T = ms(T_f - T_i)
• Specific heat capacity, s = \frac{Q}{m\Delta T}

Molar Specific Heat for Solids

• Heat required to raise 1 mole of substance by 1°C.
• Formula: C = \frac{H}{\mu} = \frac{1}{\mu} \frac{\Delta Q}{\Delta T}, where H is heat capacity.
• For a solid with N atoms vibrating, average energy per atom in 3D: 3k_B T.
• Total energy per mole: U = 3k_B T \times N_A = 3RT.
• Molar specific heat: C = \frac{\Delta U}{\Delta T} = 3R (valid at room temperature; carbon is an exception).

Specific Heat Capacity of Water

• Varies slightly with temperature.
• Graph shows U-shaped curve: decreases from ~1.008 cal/g°C to below 1.000 cal/g°C near 40°C, then rises again near 100°C.
• One calorie: heat to raise 1 g water from 14.5°C to 15.5°C.

Solved Examples

• Ex 1: Calculate heat required to raise 2 kg of water from 20°C to 30°C.
  Use Q = ms\Delta T, s = 1 cal/g°C.
  Solution: Q = 2000 g \times 1 \times (30-20) = 20000 cal.

Practice Set

• Q1: Specific heat capacity is defined as:
  A) Heat per mole to raise temperature by 1°C
  B) Heat per unit mass to raise temperature by 1°C
  C) Heat per unit volume to raise temperature by 1°C
  D) Heat per unit area to raise temperature by 1°C
• Q2: Molar specific heat of a solid at room temperature is approximately:
  A) R/3
  B) 3R
  C) R
  D) 2R
• Q3: Heat required to raise temperature of 500 g of water from 25°C to 75°C (specific heat = 1 cal/g°C) is:
  A) 25000 cal
  B) 50000 cal
  C) 12500 cal
  D) 10000 cal

Answer Key

• Q1: B; Specific heat capacity is heat per unit mass per degree change.
• Q2: B; Molar specific heat for solids is 3R at room temperature.
• Q3: B; Q = ms\Delta T = 500 \times 1 \times (75-25) = 25000 cal.

Memory Aids

• Specific heat = heat/mass × temperature change.
• Molar specific heat for solids ≈ 3R (Dulong-Petit law).
• Water specific heat ~1 cal/g°C, varies slightly with temperature.

Relation Between Specific Heat at Constant Volume and Pressure

Key Formulas and Concepts

• First law: \Delta Q = \Delta U_{int} + P\Delta V
• At constant volume (\Delta V=0): C_v = \left(\frac{\Delta Q}{\Delta T}\right)_v = \left(\frac{\Delta U_{int}}{\Delta T}\right)
• At constant pressure: C_p = \left(\frac{\Delta Q}{\Delta T}\right)_p = C_v + P\left(\frac{\Delta V}{\Delta T}\right)_p
• For 1 mole ideal gas, using PV = RT: P\left(\frac{\Delta V}{\Delta T}\right)_p = R
• Mayer’s relation: C_p - C_v = R

Solved Examples

• Ex 1: For 1 mole ideal gas, if C_v = 12.5 J/mol·K, find C_p.
  Use Mayer’s relation: C_p = C_v + R = 12.5 + 8.31 = 20.81 J/mol·K.

Practice Set

• Q1: Mayer’s relation is:
  A) C_p + C_v = R
  B) C_p - C_v = R
  C) C_v - C_p = R
  D) C_p \times C_v = R
• Q2: For 1 mole ideal gas, C_v = 20 J/mol·K, find C_p (R = 8.31 J/mol·K):
  A) 11.69 J/mol·K
  B) 28.31 J/mol·K
  C) 20 J/mol·K
  D) 8.31 J/mol·K
• Q3: Heat absorbed at constant volume changes internal energy only because:
  A) Volume changes
  B) Pressure changes
  C) No work done
  D) Temperature constant

Answer Key

• Q1: B; Mayer’s relation states C_p - C_v = R.
• Q2: B; C_p = C_v + R = 20 + 8.31 = 28.31 J/mol·K.
• Q3: C; At constant volume, no work is done, so heat changes internal energy only.

Memory Aids

• Mayer’s relation: C_p - C_v = R.
• At constant volume, heat changes internal energy only.
• At constant pressure, heat changes internal energy and does work.

Thermodynamic State Variables and Equation of State

Key Concepts

• Thermodynamic state variables describe the equilibrium state of a system.
• Equation of state relates these variables to define the system's condition.
• For an ideal gas: PV = \(\mu RT\), where
  P = pressure (Pa), V = volume (m³), \(\mu\) = number of moles, R = gas constant (8.314 J/mol·K), T = temperature (K).
• State variables are of two types:
  • Extensive: Depend on system size (e.g., volume, internal energy).
  • Intensive: Independent of system size (e.g., pressure, temperature).
• Product of an intensive and an extensive variable is extensive (e.g., P × V is extensive).

Solved Examples

• Ex 1: Calculate pressure of 2 moles of ideal gas at 300 K occupying 0.05 m³.
  Given: \(\mu=2\), T=300 K, V=0.05 m³, R=8.314 J/mol·K
  Use: \(P=\frac{\mu RT}{V}\)
  \(P=\frac{2 \times 8.314 \times 300}{0.05} = 99,768\) Pa

Practice Set

• Q1: Which of the following is an intensive variable?
  A) Volume
  B) Pressure
  C) Internal energy
  D) Mass
• Q2: For 1 mole of ideal gas at 27°C and 1 atm, volume is approximately:
  A) 22.4 L
  B) 24 L
  C) 20 L
  D) 18 L
• Q3: If pressure doubles and temperature remains constant, volume of an ideal gas:
  A) Doubles
  B) Halves
  C) Remains same
  D) Quadruples

Answer Key

• Q1: B; Pressure is intensive, independent of system size.
• Q2: B; Using ideal gas law at 300 K, volume ≈ 24 L.
• Q3: B; From PV=constant at constant T, volume halves if pressure doubles.

Memory Aids

• "Intensive" = Independent (Pressure, Temperature).
• "Extensive" = Extends with size (Volume, Mass).
• Ideal gas law: PV = \(\mu RT\) (Pressure × Volume = moles × gas constant × Temperature).

Thermodynamic Processes

Key Concepts

• Quasi static process: Infinitely slow process; system remains in equilibrium at every stage.
• Cyclic process: System returns to initial state; change in internal energy \(\Delta U = 0\); temperature constant.
• Types of processes: Isobaric (constant pressure), Isochoric (constant volume), Adiabatic (no heat exchange), Isothermal (constant temperature).

Solved Examples

Practice Set

• Q1: In a cyclic process, the change in internal energy is:
  • A) Positive
  • B) Negative
  • C) Zero
  • D) Depends on process
• Q2: Which process keeps volume constant?
  • A) Isobaric
  • B) Isochoric
  • C) Adiabatic
  • D) Isothermal
• Q3: In a quasi static process, the system is:
  • A) Out of equilibrium
  • B) In equilibrium at every stage
  • C) Rapidly changing
  • D) Non-reversible

Answer Key

• Q1: C - Internal energy change in cyclic process is zero as system returns to initial state.
• Q2: B - Isochoric process keeps volume constant.
• Q3: B - Quasi static process is infinitely slow, maintaining equilibrium at every stage.

Memory Aids

• Quasi static = slow & equilibrium
• Cyclic process: \(\Delta U = 0\)
• Isobaric = constant pressure
• Isochoric = constant volume
• Adiabatic = no heat exchange
• Isothermal = constant temperature

Special Thermodynamic Processes

Key Concepts

• Isothermal process: Temperature constant \(\Rightarrow \Delta T=0\), \(\Delta U=0\), heat supplied equals work done \(\Delta Q=\Delta W\).
• Boyle's law applies: \(PV=\text{constant}\).
• Work done: \(W=\mu RT \ln \frac{V_f}{V_i} = \mu RT \ln \frac{P_i}{P_f}\).
• Adiabatic process: No heat exchange \(\Delta Q=0\), \(\Delta U = -\Delta W\), follows Poisson's law \(PV^\gamma=\text{constant}\).
• Work done: \(W=\frac{P_1V_1 - P_2V_2}{\gamma -1}\).
• Isobaric process: Pressure constant, work done \(\Delta W = P(V_f - V_i)\), specific heat \(C_p = (f/2 +1)R\).
• Isochoric process: Volume constant, work done zero, specific heat \(C_v = (f/2)R\).
• Polytropic process: Follows \(PV^x=\text{constant}\), generalizes other processes.

Solved Examples

Practice Set

• Q1: In an isothermal process, the internal energy change is:
  • A) Positive
  • B) Negative
  • C) Zero
  • D) Depends on volume
• Q2: For an adiabatic expansion, temperature:
  • A) Increases
  • B) Decreases
  • C) Remains constant
  • D) Varies randomly
• Q3: Work done in an isobaric process is given by:
  • A) \(P(V_f - V_i)\)
  • B) \(\mu RT \ln \frac{V_f}{V_i}\)
  • C) Zero
  • D) \(\frac{P_1V_1 - P_2V_2}{\gamma -1}\)

Answer Key

• Q1: C - Internal energy remains constant in isothermal process.
• Q2: B - Temperature decreases during adiabatic expansion (cooling effect).
• Q3: A - Work done at constant pressure is \(P(V_f - V_i)\).

Memory Aids

• Isothermal: \(\Delta U=0\), heat = work
• Adiabatic: \(\Delta Q=0\), \(PV^\gamma=\text{constant}\)
• Isobaric: constant pressure, work = pressure × volume change
• Isochoric: constant volume, work = 0
• Polytropic: general form \(PV^x=\text{constant}\)

Converting P - V Cycles

Key Concepts

• P - T graph:
  • 1-2: Temperature constant
  • 2-3: Pressure constant
  • 3-1: Volume constant
  • Relation: \(P = \frac{nR}{V} T\), straight line through origin.
• V - T graph:
  • 1-2: Temperature constant
  • 2-3: Pressure constant
  • Relation: \(V = \frac{nR}{P} T\), straight line through origin.

Solved Examples

Practice Set

• Q1: In P-T graph, if volume is constant, pressure is proportional to:
  • A) \(T^2\)
  • B) \(T\)
  • C) \(1/T\)
  • D) Constant
• Q2: In V-T graph, if pressure is constant, volume is proportional to:
  • A) \(T^2\)
  • B) \(T\)
  • C) \(1/T\)
  • D) Constant
• Q3: For a cycle 1-2-3-1, which segment corresponds to constant pressure in P-T graph?
  • A) 1-2
  • B) 2-3
  • C) 3-1
  • D) None

Answer Key

• Q1: B - Pressure \(P \propto T\) at constant volume.
• Q2: B - Volume \(V \propto T\) at constant pressure.
• Q3: B - Segment 2-3 has constant pressure.

Memory Aids

• P-T graph: \(P \propto T\) at constant volume
• V-T graph: \(V \propto T\) at constant pressure
• Use ideal gas law \(PV = nRT\) for conversions

Second Law of Thermodynamics

Key Points

• Second law restricts processes allowed by the first law.
• No heat engine can have efficiency \( \eta = 1 \) (100% efficiency).
• No refrigerator can have infinite coefficient of performance (COP).

Solved Examples

• Ex 1: Calculate maximum efficiency of engine operating between \( T_1=600\,K \) and \( T_2=300\,K \).
  • Use \( \eta_{max} = 1 - \frac{T_2}{T_1} = 1 - \frac{300}{600} = 0.5 = 50\% \)

Practice Set

• Q1: Which of the following is impossible according to the second law?
  • A) Heat engine with \( \eta = 0.8 \)
  • B) Refrigerator with infinite COP
  • C) Heat flow from hot to cold spontaneously
  • D) Work done by expanding gas
• Q2: Maximum efficiency of a heat engine depends on
  • A) Heat supplied only
  • B) Temperatures of hot and cold reservoirs
  • C) Work done only
  • D) Type of working substance
• Q3: A heat engine absorbs 500 J from hot reservoir at 600 K and rejects 300 J to cold reservoir at 300 K. What is its efficiency?
  • A) 40%
  • B) 60%
  • C) 50%
  • D) 30%

Answer Key

• Q1: B - Infinite COP violates second law.
• Q2: B - Efficiency depends on reservoir temperatures.
• Q3: A - Efficiency = Work output / Heat input = (500-300)/500 = 0.4 = 40%.

Memory Aids

• Efficiency \( \eta < 1 \) always.
• Heat flows hot \( \to \) cold spontaneously.
• Second law forbids 100% conversion of heat to work.

Kelvin-Planck Statement

Key Points

• No process can convert all heat absorbed from a reservoir into work without other effects.
• Heat engine must reject some heat to cold reservoir.
• Expressed as: \( \text{No engine with } \eta = 1 \)

Solved Examples

• Ex 1: Can a heat engine convert 1000 J heat fully into work?
  • No, violates Kelvin-Planck statement.

Practice Set

• Q1: Kelvin-Planck statement forbids
  • A) Heat engine with zero work output
  • B) Heat engine with 100% efficiency
  • C) Refrigerator operation
  • D) Heat flow from hot to cold
• Q2: Which is true for a real heat engine?
  • A) \( Q_{out} = 0 \)
  • B) \( \eta = 1 \)
  • C) \( Q_{out} > 0 \)
  • D) No heat rejected
• Q3: Work done by engine absorbing 800 J and rejecting 500 J is
  • A) 300 J
  • B) 500 J
  • C) 800 J
  • D) 1300 J

Answer Key

• Q1: B - 100% efficiency is impossible.
• Q2: C - Real engines reject some heat.
• Q3: A - Work = Heat in - Heat out = 800 - 500 = 300 J.

Memory Aids

• Kelvin-Planck: No perfect heat engine.
• Heat engine must reject heat.
• Efficiency always less than 1.

Clausius Statement

Key Points

• Heat cannot flow spontaneously from cold to hot body without work.
• Refrigerators require work to transfer heat from cold to hot.
• Expressed as: No process transfers heat from cold to hot alone.

Solved Examples

• Ex 1: Can heat flow from 273 K to 300 K without work?
  • No, violates Clausius statement.

Practice Set

• Q1: Clausius statement forbids
  • A) Heat flow hot to cold
  • B) Heat flow cold to hot without work
  • C) Work done on refrigerator
  • D) Heat engine operation
• Q2: To transfer heat from cold to hot, refrigerator needs
  • A) No work
  • B) Work input
  • C) Heat input only
  • D) Spontaneous process
• Q3: Heat absorbed from cold reservoir is 200 J, work done is 100 J. Heat rejected to hot reservoir is
  • A) 100 J
  • B) 200 J
  • C) 300 J
  • D) 400 J

Answer Key

• Q1: B - Heat cannot flow cold to hot without work.
• Q2: B - Work input is necessary.
• Q3: C - Heat rejected = Heat absorbed + Work done = 200 + 100 = 300 J.

Memory Aids

• Clausius: Cold to hot heat flow needs work.
• Refrigerator moves heat up temperature gradient.
• Heat flow spontaneous only hot to cold.

Diagram (a): Heat engine absorbs heat \( Q_1 \) from hot reservoir at \( T_1 \), does work \( W \), rejects heat to cold reservoir at \( T_2 \).

Diagram (b): Refrigerator absorbs heat \( Q \) from cold reservoir at \( T_2 \), work \( W \) is done on system to transfer heat to hot reservoir at \( T_1 \).

Reversible and Irreversible Processes

Reversible Process

• A process that can be reversed by infinitesimal changes in external conditions.
• Pressure and temperature of the system nearly equal to surroundings at all times.
• Process occurs infinitely slowly (quasi-static).
• No friction or dissipative losses in the system.
• No energy loss by conduction, convection, or radiation.
• Idealized concept; real processes always have some irreversibility.

Irreversible Process

• Natural processes are irreversible due to non-equilibrium conditions.
• Mechanical, thermal, or chemical equilibrium not maintained.
• Dissipative effects like friction, viscosity, electric resistance present.
• Example: spontaneous expansion of gas into vacuum.

Solved Examples

Ex 1: Identify if a process is reversible: A gas expands slowly against a constant external pressure equal to its internal pressure, with no friction.

• Since pressure difference is negligible and process is slow with no friction, it is reversible.

Ex 2: Is the sudden expansion of gas into vacuum reversible?

• No, because it is rapid, non-equilibrium, and involves dissipative effects; hence irreversible.

Practice Set

Q1: Which condition is NOT required for a reversible process?

• A. Infinitely slow process
• B. No friction
• C. Large temperature difference with surroundings
• D. No energy loss by conduction

Q2: Spontaneous expansion of gas into vacuum is an example of:

• A. Reversible process
• B. Isothermal process
• C. Irreversible process
• D. Adiabatic reversible process

Q3: Which of the following is TRUE for a reversible process?

• A. Occurs rapidly
• B. Pressure difference between system and surroundings is large
• C. No entropy change in universe
• D. Friction is present

Answer Key

Q1: C. Large temperature difference with surroundings - Reversible processes require negligible temperature difference.

Q2: C. Irreversible process - Spontaneous expansion is non-quasi-static and dissipative.

Q3: C. No entropy change in universe - Reversible processes are ideal with zero net entropy change.

Memory Aids

• Reversible = "Slow & Equal" (slow process, equal pressure/temperature)
• Irreversible = "Natural & Dissipative" (natural processes with friction, rapid changes)
• Entropy change zero only in reversible processes

Carnot Engine

Definition and Characteristics

• Ideal heat engine with maximum efficiency.
• Free from practical imperfections; theoretical model.
• Operates between two heat reservoirs at temperatures \(T_1\) (hot) and \(T_2\) (cold).

Carnot Cycle

• Consists of four reversible processes: two isothermal and two adiabatic.
• Working substance undergoes a cyclic process: A → B → C → D → A.
• Isothermal expansion at \(T_1\): absorbs heat \(Q_1\) and does work \(W_1\).
• Adiabatic expansion: temperature falls from \(T_1\) to \(T_2\), work \(W_2\) done by gas, no heat exchange.
• Isothermal compression at \(T_2\): rejects heat \(Q_2\), work \(W_3\) done on gas.
• Adiabatic compression: temperature rises from \(T_2\) to \(T_1\), work \(W_4\) done on gas, no heat exchange.

Work and Heat Relations

• Isothermal expansion: \(Q_1 = W_1 = RT_1 \ln\frac{V_2}{V_1}\)
• Adiabatic expansion: \(W_2 = \frac{R(T_1 - T_2)}{\gamma - 1}\)
• Isothermal compression: \(Q_2 = W_3 = RT_2 \ln\frac{V_3}{V_4}\)
• Adiabatic compression: \(W_4 = -W_2\)
• Net work done: \(W = W_1 - W_3 = Q_1 - Q_2\)

Efficiency

• Efficiency \(\eta = \frac{W}{Q_1} = 1 - \frac{Q_2}{Q_1}\)
• Using volume relations and adiabatic conditions, \(\eta = 1 - \frac{T_2}{T_1}\)
• No engine operating between \(T_1\) and \(T_2\) can be more efficient than Carnot engine.

Solved Examples

Ex 1

• Given \(T_1 = 500\,K\), \(T_2 = 300\,K\), find efficiency \(\eta\).
• Use \(\eta = 1 - \frac{T_2}{T_1} = 1 - \frac{300}{500} = 0.4 = 40\%\).

Ex 2

• Heat absorbed \(Q_1 = 1000\,J\), heat rejected \(Q_2 = 600\,J\), find work done \(W\) and efficiency \(\eta\).
• Work done \(W = Q_1 - Q_2 = 400\,J\).
• Efficiency \(\eta = \frac{W}{Q_1} = \frac{400}{1000} = 0.4 = 40\%\).

Practice Set

Q1

What is the efficiency of a Carnot engine operating between 600 K and 300 K?

• A) 25%
• B) 50%
• C) 75%
• D) 100%

Q2

In a Carnot cycle, if the heat absorbed from the hot reservoir is 500 J and heat rejected to the cold reservoir is 300 J, what is the work done?

• A) 200 J
• B) 300 J
• C) 500 J
• D) 800 J

Q3

A Carnot engine absorbs 1200 J of heat at 500 K and rejects heat at 300 K. Calculate the efficiency and work done.

• A) 40%, 480 J
• B) 60%, 720 J
• C) 50%, 600 J
• D) 30%, 360 J

Answer Key

• Q1: B) 50% — Efficiency \(= 1 - \frac{300}{600} = 0.5 = 50\%\).
• Q2: A) 200 J — Work done \(= Q_1 - Q_2 = 500 - 300 = 200\,J\).
• Q3: A) 40%, 480 J — Efficiency \(= 1 - \frac{300}{500} = 0.4\), Work \(= 0.4 \times 1200 = 480\,J\).

Memory Aids

• Carnot efficiency depends only on temperatures: \(\eta = 1 - \frac{T_2}{T_1}\).
• Isothermal processes: heat exchange at constant temperature.
• Adiabatic processes: no heat exchange, temperature changes.
• Net work = heat absorbed - heat rejected.