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Thermodynamics

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Chapter Notes

Overview
This chapter introduces the fundamental laws of thermodynamics, which govern the conversion of heat into work and vice versa. It builds on the thermal properties of matter from the previous chapter and establishes the macroscopic principles of energy transfer. The one big idea a student must walk away with is that energy is conserved (First Law), but natural processes have a direction and a limit on efficiency (Second Law), and that heat and work are two distinct modes of energy transfer that change a system's internal energy.
Key Concepts (34)
Thermodynamics
The branch of physics that deals with the concepts of heat and temperature and the inter-conversion of heat and other forms of energy.
e.g. When we rub our palms together, work done produces heat; in a steam engine, heat of steam does work to move pistons.
Thermal equilibrium
A state in which the macroscopic variables that characterise the system do not change in time, and when two systems are in contact via a diathermic wall, no net flow of heat occurs between them.
e.g. A gas inside a closed rigid container, completely insulated from its surroundings, with fixed pressure, volume, and temperature that do not change with time.
Zeroth Law of Thermodynamics
Two systems in thermal equilibrium with a third system separately are in thermal equilibrium with each other.
e.g. If a cup of tea (A) and a glass of milk (B) are each in thermal equilibrium with a thermometer (C), then A and B are also in thermal equilibrium with each other.
Temperature
The thermodynamic variable whose value is equal for two systems in thermal equilibrium; it is a marker of the 'hotness' of a body and determines the direction of flow of heat.
e.g. A hot iron rod has a higher temperature than a cold metal block; heat flows from the rod to the block until temperatures equalise.
Internal energy (U)
The sum of the kinetic energies and potential energies of the molecules of a system, in the frame of reference relative to which the centre of mass of the system is at rest.
e.g. For an ideal gas, internal energy is the sum of kinetic energies due to translational, rotational, and vibrational motion of its molecules.
Heat (Q)
Energy in transit due to a temperature difference between the system and the surroundings.
e.g. When a hot metal block is placed in cold water, heat flows from the block to the water until thermal equilibrium is reached.
Work (W)
Energy transfer brought about by means that do not involve a temperature difference, such as moving a piston by raising or lowering a weight.
e.g. Pushing the piston of a gas cylinder down does work on the gas, increasing its internal energy.
First Law of Thermodynamics
The general principle of conservation of energy applied to a system: ∆Q = ∆U + ∆W, i.e., the heat supplied to the system goes partly to increase its internal energy and partly to do work on the surroundings.
e.g. For 1 g of water turning to steam, ∆Q = 2256 J, ∆W = 169.2 J, so ∆U = 2086.8 J.
Specific heat capacity (s)
The heat capacity per unit mass of a substance: s = (1/m)(∆Q/∆T), with unit J kg⁻¹ K⁻¹.
e.g. The specific heat capacity of water is 4186 J kg⁻¹ K⁻¹.
Molar specific heat capacity (C)
The heat capacity per mole of a substance: C = (1/µ)(∆Q/∆T), with unit J mol⁻¹ K⁻¹.
e.g. For a monatomic ideal gas, Cv = (3/2)R and Cp = (5/2)R.
State variables
Macroscopic variables that describe an equilibrium state of a system, such as pressure, volume, temperature, internal energy, and entropy.
e.g. Pressure (P), volume (V), and temperature (T) are state variables for a gas.
Equation of state
The connection between the state variables of a system; for an ideal gas, it is PV = µRT.
e.g. For a fixed amount of gas, PV = constant at constant temperature (Boyle's Law).
Extensive variables
Variables that indicate the 'size' of the system and are additive for subsystems; e.g., internal energy U, volume V, total mass M.
e.g. If a gas is divided into two equal parts, the volume of each part is half the original volume.
Intensive variables
Variables that do not depend on the size of the system and remain unchanged when the system is divided; e.g., pressure P, temperature T, density ρ.
e.g. If a gas is divided into two equal parts, the pressure and temperature remain the same in each part.
Quasi-static process
An idealised process in which at every stage the system is in an equilibrium state, occurring infinitely slowly so that the system remains in thermal and mechanical equilibrium with its surroundings throughout.
e.g. A gas expanding very slowly in a cylinder with a frictionless piston, where the external pressure is reduced infinitesimally at each step.
Isothermal process
A thermodynamic process in which the temperature of the system is kept fixed throughout.
e.g. The expansion of a gas in a metallic cylinder placed in a large reservoir of fixed temperature.
Adiabatic process
A thermodynamic process in which the system is insulated from the surroundings and no heat flows between the system and the surroundings (∆Q = 0).
e.g. The rapid compression of air in a diesel engine cylinder, causing a temperature rise without heat exchange.
Isobaric process
A thermodynamic process in which the pressure of the system remains constant.
e.g. Heating a gas in a cylinder with a movable piston that maintains constant external pressure.
Isochoric process
A thermodynamic process in which the volume of the system remains constant.
e.g. Heating a gas in a rigid, sealed container.
Cyclic process
A thermodynamic process in which the system returns to its initial state after a series of changes; ∆U = 0 for a cyclic process.
e.g. The Carnot cycle, where a gas undergoes isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression to return to its initial state.
Second Law of Thermodynamics
A principle that gives a fundamental limitation to the efficiency of a heat engine and the coefficient of performance of a refrigerator; it disallows many phenomena consistent with the First Law.
e.g. A book lying on a table cannot spontaneously jump to a height by cooling the table, even though energy conservation would allow it.
Kelvin-Planck statement
No process is possible whose sole result is the absorption of heat from a reservoir and the complete conversion of the heat into work.
e.g. A heat engine cannot have 100% efficiency; some heat must always be rejected to a cold reservoir.
Clausius statement
No process is possible whose sole result is the transfer of heat from a colder object to a hotter object.
e.g. A refrigerator requires work to transfer heat from a cold interior to a warmer room; it cannot do so spontaneously.
Reversible process
A thermodynamic process that can be turned back such that both the system and the surroundings return to their original states, with no other change anywhere else in the universe; it must be quasi-static and non-dissipative.
e.g. A quasi-static isothermal expansion of an ideal gas in a cylinder fitted with a frictionless movable piston.
Irreversible process
A spontaneous process in nature that cannot be reversed to bring both the system and surroundings to their initial states without other changes; arises from non-equilibrium states or dissipative effects like friction.
e.g. Free expansion of a gas into vacuum, or a moving body coming to a stop due to friction.
Carnot engine
A reversible heat engine operating between two temperatures (hot reservoir at T₁ and cold reservoir at T₂) that achieves the highest possible efficiency, given by η = 1 − T₂/T₁.
e.g. An ideal gas undergoing the Carnot cycle: isothermal expansion at T₁, adiabatic expansion to T₂, isothermal compression at T₂, and adiabatic compression back to T₁.
Carnot cycle
The sequence of four reversible steps (two isothermal and two adiabatic) that constitute the operation of a Carnot engine.
e.g. Step 1→2: Isothermal expansion at T₁; Step 2→3: Adiabatic expansion to T₂; Step 3→4: Isothermal compression at T₂; Step 4→1: Adiabatic compression back to T₁.
Carnot's theorem
Working between two given temperatures T₁ and T₂, no engine can have efficiency more than that of the Carnot engine, and the efficiency of the Carnot engine is independent of the working substance.
Caloric
An obsolete concept in which heat was regarded as a fine invisible fluid that flowed from a hot body to a cold body; discarded in favour of the modern concept of heat as a form of energy.
e.g. Count Rumford's cannon-boring experiment showed that heat produced depended on work done, not on sharpness of the drill, disproving the caloric theory.
Adiabatic wall
An insulating wall that does not allow flow of energy (heat) from one system to another.
e.g. The partition between two gases in Fig. 11.1(a) that prevents heat flow.
Diathermic wall
A conducting wall that allows energy flow (heat) from one system to another.
e.g. The partition between two gases in Fig. 11.1(b) that allows heat flow until thermal equilibrium is reached.
Heat capacity (S)
The amount of heat required to raise the temperature of a substance by 1 K: S = ∆Q/∆T.
e.g. For 1 g of water, S = 4.186 J/K.
Mechanical equivalent of heat
The amount of work needed to produce 1 cal of heat; it is a conversion factor between calorie and joule (1 cal = 4.186 J).
e.g. In SI units, the term is superfluous because heat and work are both measured in joules.
Isotherm
A pressure-volume curve for a fixed temperature in a P-V diagram.
e.g. For an ideal gas, an isotherm is a rectangular hyperbola given by PV = constant.
Formulas & Equations (14)
∆Q = ∆U + ∆W
Reads as: Delta Q equals delta U plus delta W
Variables: ∆Q = heat supplied to the system (J), ∆U = change in internal energy (J), ∆W = work done by the system (J)
Use when: Applying the First Law of Thermodynamics to any thermodynamic process
⚠ Students often forget the sign convention: ∆W is work done BY the system, so for work done ON the system, ∆W is negative.
∆Q = ∆U + P∆V
Reads as: Delta Q equals delta U plus P delta V
Variables: P = constant pressure (Pa), ∆V = change in volume (m³)
Use when: Calculating heat transfer for a gas in a cylinder with a movable piston at constant pressure
⚠ Using this formula when pressure is not constant; it applies only for isobaric processes.
C = (1/µ)(∆Q/∆T)
Reads as: C equals one over mu times delta Q over delta T
Variables: C = molar specific heat capacity (J mol⁻¹ K⁻¹), µ = number of moles, ∆Q = heat supplied (J), ∆T = temperature change (K)
Use when: Defining molar specific heat capacity for any substance
⚠ Confusing C with specific heat capacity s; C is per mole, s is per kilogram.
s = (1/m)(∆Q/∆T)
Reads as: s equals one over m times delta Q over delta T
Variables: s = specific heat capacity (J kg⁻¹ K⁻¹), m = mass (kg), ∆Q = heat supplied (J), ∆T = temperature change (K)
Use when: Calculating specific heat capacity of a substance
⚠ Forgetting to use mass in kg; using grams gives wrong numerical value.
C_p − C_v = R
Reads as: C p minus C v equals R
Variables: C_p = molar specific heat at constant pressure (J mol⁻¹ K⁻¹), C_v = molar specific heat at constant volume (J mol⁻¹ K⁻¹), R = universal gas constant (8.314 J mol⁻¹ K⁻¹)
Use when: Relating molar specific heats of an ideal gas
⚠ Forgetting that this relation holds only for ideal gases, not for real gases or solids.
PV = µRT
Reads as: P V equals mu R T
Variables: P = pressure (Pa), V = volume (m³), µ = number of moles, R = universal gas constant (8.314 J mol⁻¹ K⁻¹), T = absolute temperature (K)
Use when: Ideal gas equation of state for any equilibrium state
⚠ Using Celsius instead of Kelvin for temperature; T must be in Kelvin.
W = µRT ln(V₂/V₁) (isothermal process)
Reads as: W equals mu R T natural log of V two over V one
Variables: W = work done by gas (J), µ = number of moles, R = universal gas constant, T = constant temperature (K), V₂ = final volume (m³), V₁ = initial volume (m³)
Use when: Calculating work done in an isothermal expansion or compression of an ideal gas
⚠ Using log base 10 instead of natural log (ln); the formula uses natural logarithm.
P V^γ = constant (adiabatic process)
Reads as: P V to the power gamma equals constant
Variables: P = pressure (Pa), V = volume (m³), γ = C_p/C_v (ratio of specific heats, dimensionless)
Use when: Relating pressure and volume in an adiabatic process of an ideal gas
⚠ Forgetting that γ is different for monatomic (γ = 5/3), diatomic (γ = 7/5), and polyatomic gases.
P₁ V₁^γ = P₂ V₂^γ
Reads as: P one V one to the gamma equals P two V two to the gamma
Variables: P₁, V₁ = initial pressure and volume; P₂, V₂ = final pressure and volume; γ = C_p/C_v
Use when: Solving problems involving adiabatic changes of state for an ideal gas
⚠ Applying this relation to isothermal processes; it is valid only for adiabatic processes.
W = (P₁V₁ − P₂V₂)/(γ − 1) = µR(T₁ − T₂)/(γ − 1) (adiabatic process)
Reads as: W equals P one V one minus P two V two over gamma minus one, equals mu R times T one minus T two over gamma minus one
Variables: W = work done by gas (J), P₁, V₁, T₁ = initial state variables, P₂, V₂, T₂ = final state variables, γ = C_p/C_v, µ = number of moles, R = universal gas constant
Use when: Calculating work done in an adiabatic process of an ideal gas
⚠ Sign error: if work is done by the gas (expansion), T₂ < T₁, so W is positive; if work is done on the gas (compression), T₂ > T₁, so W is negative.
W = P(V₂ − V₁) = µR(T₂ − T₁) (isobaric process)
Reads as: W equals P times V two minus V one, equals mu R times T two minus T one
Variables: W = work done by gas (J), P = constant pressure (Pa), V₂, V₁ = final and initial volumes (m³), µ = number of moles, R = universal gas constant, T₂, T₁ = final and initial temperatures (K)
Use when: Calculating work done in an isobaric process
⚠ Forgetting that the formula uses change in volume, not absolute volume.
η = 1 − Q₂/Q₁ (heat engine efficiency)
Reads as: Eta equals one minus Q two over Q one
Variables: η = efficiency (dimensionless, often expressed as %), Q₁ = heat absorbed from hot reservoir (J), Q₂ = heat rejected to cold reservoir (J)
Use when: Calculating efficiency of any heat engine
⚠ Confusing Q₁ and Q₂: Q₁ is heat input (from hot reservoir), Q₂ is heat output (to cold reservoir).
η = 1 − T₂/T₁ (Carnot engine efficiency)
Reads as: Eta equals one minus T two over T one
Variables: η = Carnot efficiency (dimensionless), T₁ = temperature of hot reservoir (K), T₂ = temperature of cold reservoir (K)
Use when: Calculating the maximum possible efficiency of a heat engine operating between two reservoirs
⚠ Using Celsius temperatures; T₁ and T₂ must be in Kelvin.
U = 3RT (for a mole of a solid, Dulong-Petit law)
Reads as: U equals three R T
Variables: U = internal energy per mole (J mol⁻¹), R = universal gas constant (8.314 J mol⁻¹ K⁻¹), T = absolute temperature (K)
Use when: Estimating molar specific heat of solids at room temperature (C ≈ 3R ≈ 25 J mol⁻¹ K⁻¹)
⚠ Forgetting that this is an approximation that breaks down at low temperatures.
How It Works
Establishing Thermal Equilibrium via Zeroth Law
  1. Step 1: Systems A and B are separated by an adiabatic wall, but each is in contact with a third system C via a conducting wall.
  2. Step 2: The states of A and B change until both come to thermal equilibrium with C.
  3. Step 3: The adiabatic wall between A and B is replaced by a conducting wall, and C is insulated from A and B by an adiabatic wall.
  4. Step 4: It is observed that the states of A and B change no further; they are in thermal equilibrium with each other.
💡 The Zeroth Law is the basis for the concept of temperature; it allows us to use a thermometer (system C) to compare temperatures of other bodies.
First Law of Thermodynamics (Energy Conservation)
  1. Step 1: Identify the system and its surroundings.
  2. Step 2: Determine the heat supplied to the system (∆Q) and the work done by the system (∆W).
  3. Step 3: Apply ∆Q = ∆U + ∆W, where ∆U is the change in internal energy.
  4. Step 4: For a gas in a cylinder at constant pressure, use ∆Q = ∆U + P∆V.
💡 ∆U depends only on initial and final states (path-independent), but ∆Q and ∆W depend on the path. The combination ∆Q − ∆W is path-independent.
Isothermal Process for an Ideal Gas
  1. Step 1: Keep the system in contact with a reservoir at constant temperature T.
  2. Step 2: Allow the gas to expand or compress slowly (quasi-statically) so that temperature remains constant.
  3. Step 3: For an ideal gas, internal energy U depends only on temperature, so ∆U = 0.
  4. Step 4: From the First Law, ∆Q = ∆W = µRT ln(V₂/V₁).
💡 In an isothermal expansion, the gas absorbs heat and does work; in an isothermal compression, work is done on the gas and heat is released.
Adiabatic Process for an Ideal Gas
  1. Step 1: Insulate the system from the surroundings so that no heat exchange occurs (∆Q = 0).
  2. Step 2: Allow the gas to expand or compress; the temperature changes because work is done.
  3. Step 3: For an ideal gas, the relation P V^γ = constant holds, where γ = C_p/C_v.
  4. Step 4: Work done is given by W = (P₁V₁ − P₂V₂)/(γ − 1) = µR(T₁ − T₂)/(γ − 1).
💡 In adiabatic expansion, temperature decreases (T₂ < T₁); in adiabatic compression, temperature increases (T₂ > T₁). The adiabatic curve is steeper than the isothermal curve on a P-V diagram.
Carnot Cycle (Reversible Heat Engine)
  1. Step 1 (Isothermal expansion at T₁): The gas absorbs heat Q₁ from the hot reservoir and does work, expanding from V₁ to V₂.
  2. Step 2 (Adiabatic expansion): The gas expands further from V₂ to V₃, doing work, and its temperature falls from T₁ to T₂.
  3. Step 3 (Isothermal compression at T₂): The gas releases heat Q₂ to the cold reservoir, and work is done on the gas, compressing it from V₃ to V₄.
  4. Step 4 (Adiabatic compression): The gas is compressed from V₄ to V₁, work is done on the gas, and its temperature rises from T₂ back to T₁.
💡 The Carnot cycle is the most efficient cycle possible between two temperatures. Its efficiency η = 1 − T₂/T₁ is independent of the working substance.
Determining Extensive vs Intensive Variables
  1. Step 1: Consider a system in equilibrium and imagine dividing it into two equal parts.
  2. Step 2: If a variable's value gets halved in each part, it is extensive (e.g., volume, internal energy, mass).
  3. Step 3: If a variable's value remains unchanged in each part, it is intensive (e.g., pressure, temperature, density).
💡 In thermodynamic equations, both sides must be extensive. For example, in ∆Q = ∆U + P∆V, the product of intensive P and extensive ∆V is extensive.
Exam Traps — Don't Make These Mistakes
⚠ Confusing Heat with Internal Energy
✗ 'A gas in a given state has a certain amount of heat.'
→ Heat is energy in transit, not a state variable. A system in a given state has internal energy, not heat. Saying 'a gas has heat' is as meaningless as saying 'a gas has work'.
✓ 'A gas in a given state has a certain amount of internal energy.' Heat is supplied to or released by the system during a process.
⚠ Misapplying the First Law Sign Convention
✗ '∆Q = ∆U + ∆W, where ∆W is work done on the system.'
→ In the standard NCERT convention, ∆W is work done BY the system. If work is done ON the system, ∆W is negative. Using the wrong sign leads to incorrect calculations.
✓ '∆Q = ∆U + ∆W, where ∆W is work done BY the system. For work done on the system, ∆W is negative.'
⚠ Using Celsius Instead of Kelvin in Gas Laws
✗ 'PV = µRT with T = 27°C'
→ The ideal gas equation and all thermodynamic formulas require absolute temperature in Kelvin. Using Celsius gives a completely wrong numerical value.
✓ 'Convert Celsius to Kelvin: T(K) = T(°C) + 273.15. For example, 27°C = 300 K.'
⚠ Confusing Isothermal and Adiabatic Process Equations
✗ 'For an adiabatic process, PV = constant.'
→ PV = constant is Boyle's Law for an isothermal process. For an adiabatic process, the correct relation is PV^γ = constant, where γ = C_p/C_v.
✓ 'For an isothermal process: PV = constant. For an adiabatic process: PV^γ = constant.'
⚠ Assuming All Processes are Reversible
✗ 'The free expansion of a gas is reversible.'
→ Free expansion is irreversible because the gas passes through non-equilibrium states and cannot be returned to its initial state without external work. Reversible processes must be quasi-static and non-dissipative.
✓ 'Free expansion is an irreversible process. A reversible process must be quasi-static and have no dissipative effects like friction.'
⚠ Misinterpreting Carnot Efficiency as Actual Efficiency
✗ 'The efficiency of any heat engine is 1 − T₂/T₁.'
→ The formula η = 1 − T₂/T₁ gives the maximum possible efficiency of a Carnot engine (reversible). Real engines have lower efficiency due to irreversibilities.
✓ 'The Carnot efficiency η = 1 − T₂/T₁ is the maximum possible efficiency for an engine operating between temperatures T₁ and T₂. Actual engines have efficiency less than this.'
Compare & Contrast
Heat exchange with surroundings
Isothermal process
Adiabatic process
In an isothermal process, the system is in thermal contact with a reservoir and heat flows to keep temperature constant (∆T = 0). In an adiabatic process, the system is insulated and no heat flows (∆Q = 0), so temperature changes.
Ability to return to initial state without other changes
Reversible process
Irreversible process
A reversible process can be reversed to bring both the system and surroundings back to their original states with no other change; it must be quasi-static and non-dissipative. An irreversible process cannot be reversed without leaving changes elsewhere; most natural processes (e.g., free expansion, friction) are irreversible.
Mode of energy transfer
Heat (Q)
Work (W)
Heat is energy transfer due to a temperature difference between the system and surroundings. Work is energy transfer brought about by means that do not involve a temperature difference (e.g., moving a piston). Both are path-dependent and are not state variables.
Dependence on system size
Extensive variable
Intensive variable
Extensive variables (e.g., volume, internal energy, mass) depend on the size of the system and are additive. Intensive variables (e.g., pressure, temperature, density) do not depend on system size and remain unchanged when the system is divided.
Exam Focus
Marks pattern: Typically 1-2 MCQs (1 mark each), 1 short answer question (2-3 marks) on definitions or derivations, and 1 long answer question (5 marks) on Carnot cycle or application of First Law to thermodynamic processes.
Most asked: First Law of Thermodynamics and its application to various processes · Carnot engine and its efficiency · Isothermal and adiabatic processes for an ideal gas · Zeroth Law and concept of temperature · Specific heat capacities of gases (C_p − C_v = R)
Don't skip: Derive the expression for work done in an isothermal process for an ideal gas.; Derive the relation C_p − C_v = R for an ideal gas.; Explain the Carnot cycle and derive its efficiency.; State and explain the First Law of Thermodynamics with its applications to isothermal and adiabatic processes.
One-Line Revision (read aloud before the exam)
  1. Thermodynamics is the branch of physics dealing with heat, temperature, and the inter-conversion of heat and other forms of energy.
  2. Thermal equilibrium is a state where macroscopic variables do not change with time and no net heat flows between systems in contact.
  3. Zeroth Law: If A and B are in thermal equilibrium with C separately, then A and B are in thermal equilibrium with each other.
  4. Temperature is the thermodynamic variable that is equal for systems in thermal equilibrium.
  5. Internal energy U is the sum of molecular kinetic and potential energies in the centre-of-mass frame; it is a state variable.
  6. Heat is energy in transit due to temperature difference; work is energy transfer without temperature difference; both are path-dependent.
  7. First Law of Thermodynamics: ∆Q = ∆U + ∆W (energy conservation).
  8. For an ideal gas, C_p − C_v = R.
  9. In an isothermal process for an ideal gas, ∆U = 0 and Q = W = µRT ln(V₂/V₁).
  10. In an adiabatic process for an ideal gas, ∆Q = 0 and PV^γ = constant.
  11. A quasi-static process is infinitely slow and the system remains in equilibrium at every stage.
  12. Second Law (Kelvin-Planck): No process can completely convert heat into work without other effects.
  13. Second Law (Clausius): No process can transfer heat from a colder to a hotter body without other effects.
  14. A reversible process must be quasi-static and non-dissipative; most natural processes are irreversible.
  15. Carnot engine is a reversible engine operating between two temperatures with efficiency η = 1 − T₂/T₁.
  16. Carnot's theorem: No engine operating between two temperatures can be more efficient than a Carnot engine.
  17. Extensive variables (e.g., V, U, M) depend on system size; intensive variables (e.g., P, T, ρ) do not.
  18. The Carnot cycle consists of two isothermal and two adiabatic processes.
  19. For a solid, molar specific heat at room temperature is approximately 3R (Dulong-Petit law).
  20. The mechanical equivalent of heat (1 cal = 4.186 J) is a conversion factor between calorie and joule.

MCQs — 10 Questions with Answers

Q1. Two systems A and B are separated by an adiabatic wall. Which statement is true?

  • A. Systems A and B must be at the same temperature
  • B. Any state of A can be in equilibrium with any state of B ✓
  • C. Heat will flow from A to B until temperatures equalise
  • D. The internal energy of A must equal the internal energy of B

Answer: B — An adiabatic wall prevents heat flow, so A and B can have any combination of states and remain in equilibrium without exchanging energy.

Q2. According to the Zeroth Law of Thermodynamics, if object X has the same temperature as object Z, and object Y has the same temperature as object Z, then:

  • A. X and Y will exchange heat when brought into contact
  • B. X and Y have the same temperature and will be in thermal equilibrium ✓
  • C. X and Y must be made of the same material
  • D. X and Y cannot be brought into thermal contact without changing Z

Answer: B — The Zeroth Law directly states that thermal equilibrium is transitive: if A = C and B = C, then A = B in terms of temperature.

Q3. In Rumford's cannon-boring experiment, the heat produced was observed to depend on:

  • A. The sharpness of the drill
  • B. The amount of work done by the horses turning the drill ✓
  • C. The material composition of the cannon
  • D. The initial temperature of the cannon

Answer: B — Rumford found that heat depended on work input, not drill sharpness, proving heat is energy transfer rather than a pre-existing fluid.

Q4. A gas is compressed inside a rigid, thermally insulated container. Which statement is correct?

  • A. Heat is added to the gas, so Q > 0
  • B. Work is done on the gas (W < 0), so internal energy increases even though Q = 0 ✓
  • C. Internal energy decreases because the gas volume decreases
  • D. Temperature remains constant because the container is rigid

Answer: B — In an adiabatic compression (Q = 0), work is done on the gas (W < 0 by sign convention), so by ΔU = Q − W, the internal energy increases and temperature rises.

Q5. A bullet moving at 100 m/s strikes a wooden block and comes to rest. From a thermodynamic perspective, which energy conversion occurs?

  • A. Potential energy of the bullet converts to heat
  • B. Kinetic energy of bulk motion converts to internal energy (heat) of bullet and wood ✓
  • C. Thermal energy of the bullet converts to mechanical motion of the wood
  • D. The bullet's internal energy decreases because it stops moving

Answer: B — The ordered kinetic energy of the bullet's macroscopic motion is converted into disordered internal energy (heat) in the bullet and surrounding wood, increasing their temperatures.

Q6. Two systems are separated by a diathermic wall. After a long time, the pressure and volume of both systems change until thermal equilibrium is reached. Why do the macroscopic variables change?

  • A. The systems exchange mass through the diathermic wall
  • B. Heat flows between systems due to temperature difference until temperatures equalise ✓
  • C. Friction between the walls causes the gas molecules to move faster
  • D. The diathermic wall expands or contracts, doing work on the systems

Answer: B — A diathermic wall conducts heat; temperature difference drives spontaneous heat flow until both systems reach the same temperature and thermal equilibrium is established.

Q7. Which of the following is NOT a macroscopic variable used to describe the thermodynamic state of a gas?

  • A. Pressure (P)
  • B. Average kinetic energy of a single molecule ✓
  • C. Volume (V)
  • D. Temperature (T)

Answer: B — Macroscopic variables (P, V, T, mass, composition) are directly measurable and do not require knowledge of individual molecules; average molecular kinetic energy is microscopic.

Q8. Assertion (A): Temperature is a measure of the internal disordered motion of particles in a system. Reason (R): The temperature of a moving bullet does not change simply because the bullet is in motion relative to the ground.

  • A. Both A and R are true, and R correctly explains A ✓
  • B. Both A and R are true, but R does not explain A
  • C. A is true, but R is false
  • D. A is false, but R is true

Answer: A — Both statements are true: temperature reflects disordered particle motion, not bulk motion; the bullet's bulk kinetic energy does not affect its temperature—only internal disorder does.

Q9. A gas undergoes an expansion against constant external pressure. If 500 J of heat is added to the gas and the gas does 300 J of work on its surroundings, what is the change in internal energy?

  • A. 800 J
  • B. 200 J ✓
  • C. −200 J
  • D. −800 J

Answer: B — Using the First Law: ΔU = Q − W = 500 J − 300 J = 200 J; positive change means internal energy increases.

Q10. Which of the following best explains why the caloric theory of heat was abandoned?

  • A. Caloric was discovered to be too heavy to flow between objects
  • B. Temperature measurements became more precise and showed heat could increase without adding caloric fluid
  • C. Rumford's observation that heat generation depended on work input, not on material removed, contradicted the fluid concept ✓
  • D. Caloric could not explain why hot objects eventually cool down

Answer: C — Rumford's cannon-boring experiment showed heat production was proportional to work, not to drill sharpness or amount of material; this proved heat is energy, not a substance.

Flashcards

What is thermal equilibrium?

A state where two systems in contact through a diathermic wall have equal temperatures and no net heat flow occurs.

State the Zeroth Law of Thermodynamics.

If system A is in thermal equilibrium with C, and B is in thermal equilibrium with C, then A is in thermal equilibrium with B.

What distinguishes an adiabatic wall from a diathermic wall?

An adiabatic wall prevents heat flow between systems; a diathermic wall allows heat to flow freely.

What was Rumford's key observation in his cannon-boring experiment?

Heat produced depended on work done by the drill, not on the sharpness of the drill, proving heat is energy not a fluid.

What is the main difference between thermodynamics and mechanics?

Mechanics studies motion of bodies under forces; thermodynamics studies internal macroscopic state and temperature of systems.

Define internal energy in thermodynamics.

Internal energy is the total energy of disordered (random) motion of particles within a system, related to its temperature.

What are the two independent variables used to describe a gas in thermodynamics?

Pressure (P) and volume (V) are typically chosen as independent variables for a given mass of gas.

Why is the modern concept of heat superior to the caloric theory?

Heat as energy explains why work and heat are interconvertible, unlike the obsolete idea of heat as an indestructible fluid.

In a bullet piercing wood and stopping, what changes and what does not?

The bullet's mechanical kinetic energy changes (becomes zero); its internal energy and temperature increase due to work-to-heat conversion.

What macroscopic variables define the state of a gas in thermodynamics?

Pressure, volume, temperature, mass, and composition are the primary measurable macroscopic variables.

Important Board Questions

Define thermal equilibrium and explain with one example how a diathermic wall differs from an adiabatic wall in achieving thermal equilibrium. [2 marks]

State that thermal equilibrium means no change in macroscopic variables over time. Example: two gases at different temperatures separated by a diathermic wall → heat flows until T₁ = T₂; by an adiabatic wall → no heat flow, systems remain at different T.

State the Zeroth Law of Thermodynamics and explain why it provides an operational definition of temperature. Use the example of three systems A, B, and C. [5 marks]

State the law: if A in equilibrium with C and B in equilibrium with C, then A in equilibrium with B. Explanation: the law defines a property (temperature) that is equal when systems are in thermal equilibrium; all three systems have the same temperature when connected via diathermic walls; temperature becomes the measurable quantity that characterises thermal equilibrium without requiring knowledge of molecular motion.

Describe Rumford's cannon-boring experiment and explain how it disproved the caloric theory of heat. Derive or explain how this experiment leads to the modern understanding that heat is a form of energy. [6 marks]

Describe: boring a cannon produced heat enough to boil water; heat produced depended on work (horses turning drill), not on drill sharpness. Disproof of caloric: a sharper drill would extract more 'caloric fluid' (caloric theory prediction), but this was not observed. Modern conclusion: work and heat are interconvertible forms of energy; mechanical work of drilling is converted into internal energy (heat); this supports the First Law ΔU = Q − W, where both Q and W are energy transfers and U is state function depending only on temperature.

Assertion-Reason Questions for Thermodynamics

Q1. Assertion (A):

Heat is a form of energy that can be converted from mechanical work, as demonstrated by Rumford's cannon-boring experiment.

Reason (R):

The experiment showed that the amount of heat produced depended on the work done by the drilling mechanism, not on the properties of the drill itself.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true, and the reason correctly explains why heat must be a form of energy: if it were the caloric fluid (as previously believed), the drill's sharpness would matter, but the experiment proved only work matters. This directly supports the energy conversion principle in the assertion.

Q2. Assertion (A):

Two systems separated by an adiabatic wall cannot reach thermal equilibrium with each other.

Reason (R):

An adiabatic wall is insulating and does not allow heat flow between systems.

Answer: C — A is true but R is false — The assertion is true (adiabatic walls prevent thermal equilibrium), but the reasoning is incomplete. The reason states why there's no energy transfer, but doesn't explain that thermal equilibrium is defined by equal temperatures—which cannot occur without thermal contact. The wall prevents not just heat flow but the very condition of equilibrium.

Q3. Assertion (A):

The Zeroth Law of Thermodynamics establishes that temperature is a fundamental property that characterizes the thermal state of a system.

Reason (R):

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 must be in thermal equilibrium with each other.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — Both assertion and reason are true—the reason correctly describes the Zeroth Law's transitivity property. However, the reason does not directly explain how temperature is defined; it only implies that a common property (temperature) exists. The assertion and reason are true but the logical connection is indirect rather than explanatory.

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Case-Based Questions (CBQ) for Thermodynamics

Case Study:

A laboratory technician in Delhi's National Physical Laboratory conducts an experiment with two copper blocks: Block A (500 g, initial temperature 80°C) and Block B (500 g, initial temperature 20°C). Both blocks are placed in contact with each other inside a perfectly insulated box made of thermally non-conducting material. The blocks are separated internally by a thin conducting copper plate that allows heat transfer between them. After 30 minutes, both blocks reach the same temperature. The technician records temperature readings at regular intervals and observes no heat loss to the surroundings.

Q1. What thermodynamic principle explains why both blocks eventually reach the same temperature?

  • A. First Law of Thermodynamics
  • B. Zeroth Law of Thermodynamics ✓
  • C. Second Law of Thermodynamics
  • D. Conservation of mechanical energy

Answer: B

Q2. If the insulating box were replaced with a diathermic container open to the lab environment at 25°C, what would be the final equilibrium state?

  • A. Both blocks would remain at their initial temperatures
  • B. Both blocks would reach 25°C (lab temperature) after sufficient time ✓
  • C. Block A would cool to 25°C while Block B would remain at 20°C
  • D. The blocks would continue to exchange heat indefinitely

Answer: B

Q3. Why is it essential that the insulating box is made of adiabatic material in this experiment?

  • A. To prevent heat loss to the surroundings and ensure accurate measurement of internal thermal equilibrium ✓
  • B. To allow heat to flow between the blocks
  • C. To increase the heat capacity of the system
  • D. To ensure the blocks reach different final temperatures

Answer: A

Q4. Approximately what is the final equilibrium temperature of both blocks, assuming no heat loss to surroundings and equal heat capacities?

  • A. 50°C ✓
  • B. 45°C
  • C. 60°C
  • D. 35°C

Answer: A

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