Important Questions Class 11 Physics Chapter 11: Thermodynamics

Thermodynamics studies heat, temperature, work and energy transfer in bulk systems.
Heat and work are modes of energy transfer, while internal energy is a state variable.

Thermodynamics explains how heat converts into work and how work changes the internal energy of a system. Important Questions Class 11 Physics Chapter 11 help students practise thermal equilibrium, zeroth law, heat, internal energy, first law, specific heat, state variables, isothermal process, adiabatic process, second law, reversible processes and Carnot engine. The CBSE 2026 chapter builds the base for heat engines, refrigerators, ideal gas processes and energy conservation in thermal systems.

Key Takeaways

• Zeroth Law: Two systems in thermal equilibrium with a third system are in equilibrium with each other.
• First Law: Heat supplied to a system follows ΔQ = ΔU + ΔW.
• Ideal Gas Process: Isothermal expansion gives W = μRT ln(V2/V1).
• Carnot Engine: Maximum efficiency between two reservoirs is η = 1 − T2/T1.

Important Questions Class 11 Physics Chapter 11 Structure 2026

Important Questions Class 11 Physics Chapter 11 with Answers

Thermodynamics uses macroscopic variables such as pressure, volume, temperature and internal energy.
Students should separate state variables from path-dependent quantities before solving numericals.
These thermodynamics class 11 important questions follow the NCERT 2026 sequence.

1. What does Important Questions Class 11 Physics Chapter 11 mainly test?

Important Questions Class 11 Physics Chapter 11 mainly test thermal equilibrium, first law, thermodynamic processes, second law and Carnot engine. The chapter connects heat, work and internal energy.

1. Equilibrium Skill: Use temperature as the thermal equilibrium marker.
2. Energy Skill: Apply ΔQ = ΔU + ΔW.
3. Process Skill: Identify isothermal, adiabatic, isobaric and isochoric processes.
4. Engine Skill: Use η = 1 − T2/T1 for Carnot engine.
5. Final Result: The chapter tests heat-work conversion and thermal laws.

2. What is thermodynamics in Class 11 Physics?

Thermodynamics is the branch of physics that studies heat, temperature and energy conversion. It deals with bulk systems using macroscopic variables.

1. Main Quantities: Heat, work, temperature and internal energy.
2. System Scale: Bulk matter.
3. Description: Uses pressure, volume, temperature and mass.
4. Final Result: Thermodynamics is a macroscopic science.

3. Why is thermodynamics called a macroscopic science?

Thermodynamics is called macroscopic because it does not track individual molecules. It describes systems through measurable bulk quantities.

1. Gas Example: Uses P, V and T.
2. Microscopic Detail: Molecular velocities are not specified.
3. State Description: Few variables define equilibrium.
4. Final Result: Thermodynamics avoids molecular-level tracking.

Thermal Equilibrium Class 11 Questions

Thermal equilibrium occurs when systems in contact stop exchanging heat.
It is identified by equality of temperature.
These thermal equilibrium class 11 questions explain adiabatic walls, diathermic walls and equilibrium states.

4. What is thermal equilibrium?

Thermal equilibrium is the state in which two systems in thermal contact have the same temperature. No net heat flows between them.

1. Thermal Contact: Heat exchange is possible.
2. Condition: Temperatures become equal.
3. Result: Heat flow stops.
4. Final Result: Thermal equilibrium means equal temperature.

5. What is thermodynamic equilibrium?

Thermodynamic equilibrium is a state where macroscopic variables do not change with time. Pressure, volume and temperature remain fixed.

1. Pressure: Constant with time.
2. Volume: Constant with time.
3. Temperature: Constant with time.
4. Final Result: Equilibrium state has time-independent macroscopic variables.

6. What is an adiabatic wall?

An adiabatic wall is an insulating wall that does not allow heat flow. Systems separated by it cannot exchange heat.

1. Heat Transfer: Not allowed.
2. Material Nature: Insulating.
3. Thermal Contact: Absent.
4. Final Result: Adiabatic wall blocks heat transfer.

7. What is a diathermic wall?

A diathermic wall is a conducting wall that allows heat flow. Systems separated by it can reach thermal equilibrium.

1. Heat Transfer: Allowed.
2. Material Nature: Conducting.
3. Effect: Temperatures can equalise.
4. Final Result: Diathermic wall permits heat transfer.

8. How is thermal equilibrium different from mechanical equilibrium?

Thermal equilibrium requires equal temperature, while mechanical equilibrium requires balanced forces and torques. They describe different conditions.

1. Thermal Equilibrium: No heat flow.
2. Mechanical Equilibrium: Net force and torque are zero.
3. Common Feature: Both describe no macroscopic change.
4. Final Result: Thermal equilibrium depends on temperature equality.

Zeroth Law of Thermodynamics Class 11 Questions

The zeroth law gives the formal basis of temperature measurement.
It explains why a thermometer can compare temperatures.
These zeroth law of thermodynamics class 11 questions cover thermal contact and temperature scale.

9. State the zeroth law of thermodynamics.

The zeroth law states that two systems in thermal equilibrium with a third system are in thermal equilibrium with each other. It defines temperature.

1. System A: In equilibrium with C.
2. System B: In equilibrium with C.
3. Conclusion: A and B are in equilibrium.
4. Final Result: Zeroth law defines equality of temperature.

10. Why is the zeroth law important?

The zeroth law is important because it introduces temperature as a measurable thermodynamic variable. It makes thermometry possible.

1. Thermometer Role: Acts as the third system.
2. Temperature Equality: Indicates thermal equilibrium.
3. Measurement: Allows comparison of hotness.
4. Final Result: Zeroth law forms the basis of thermometers.

11. How does the zeroth law define temperature?

The zeroth law defines temperature as the quantity that becomes equal for systems in thermal equilibrium. Equal temperature means no net heat flow.

1. Equilibrium Condition: No heat exchange.
2. Common Quantity: Temperature.
3. Thermal Contact: Systems settle to the same value.
4. Final Result: Temperature marks thermal equilibrium.

12. Why was it named zeroth law?

It was named zeroth law because it was formulated after the first and second laws but is more fundamental. R.H. Fowler named it in 1931.

1. Historical Order: Stated later.
2. Logical Order: Comes before first law.
3. Concept Introduced: Temperature.
4. Final Result: Zeroth law logically precedes the first law.

Heat Internal Energy and Work Class 11 Questions

Heat and work are ways of transferring energy to or from a system.
Internal energy belongs to the state of the system.
These heat internal energy and work class 11 questions clarify the most common conceptual trap.

13. What is internal energy?

Internal energy is the sum of molecular kinetic and potential energies inside a system. It excludes motion of the whole system.

1. Molecular Kinetic Energy: Due to random molecular motion.
2. Molecular Potential Energy: Due to intermolecular interactions.
3. Whole-Body Motion: Not included.
4. Final Result: Internal energy is a state variable.

14. Why is heat not a state variable?

Heat is not a state variable because it describes energy transfer due to temperature difference. It does not belong to a stored state.

1. Heat Meaning: Energy in transit.
2. Path Role: Depends on process.
3. System State: Does not contain heat.
4. Final Result: Heat is a path-dependent transfer quantity.

15. Why is work not a state variable?

Work is not a state variable because it depends on the process used. It is energy transferred by mechanical means.

1. Work Meaning: Energy transfer by force or volume change.
2. Path Dependence: Different paths give different work.
3. State Role: Not stored in a system.
4. Final Result: Work is path dependent.

16. What is the difference between heat and internal energy?

Heat is energy transferred due to temperature difference, while internal energy is energy stored in molecular motion and interaction. Only internal energy is a state variable.

1. Heat: Transfer quantity.
2. Internal Energy: State quantity.
3. Example: A gas has internal energy, not heat.
4. Final Result: Heat changes internal energy.

17. What is work done by gas at constant pressure?

Work done by gas at constant pressure is W = PΔV. It is positive during expansion.

1. Pressure: P.
2. Volume Change: ΔV.
3. Formula Used: W = PΔV.
4. Final Result: Constant pressure work equals PΔV.

18. What is the sign of work in thermodynamics?

Work is positive when done by the system and negative when done on the system. This chapter uses ΔQ = ΔU + ΔW.

1. Expansion: Work done by gas is positive.
2. Compression: Work done on gas is negative for system work.
3. Sign Convention: W means work by system.
4. Final Result: Positive W means work by the system.

First Law of Thermodynamics Class 11 Questions

The first law applies conservation of energy to heat and work.
It tells how heat supplied divides into internal energy change and work.
These first law of thermodynamics class 11 questions cover signs, equations and applications.

19. State the first law of thermodynamics.

The first law states that heat supplied to a system equals change in internal energy plus work done by the system. Its formula is ΔQ = ΔU + ΔW.

1. Heat Supplied: ΔQ.
2. Internal Energy Change: ΔU.
3. Work Done by System: ΔW.
4. Final Result: First law is energy conservation for thermal systems.

20. What is the alternative form of the first law?

The alternative form is ΔU = ΔQ − ΔW. It gives the change in internal energy directly.

1. Heat Input: Raises internal energy.
2. Work Output: Reduces energy left inside.
3. Formula: ΔU = ΔQ − ΔW.
4. Final Result: Internal energy change equals heat minus work.

21. What happens when ΔU = 0?

When ΔU = 0, heat supplied equals work done by the system. This happens in isothermal expansion of an ideal gas.

1. Condition: No internal energy change.
2. First Law: ΔQ = ΔW.
3. Example: Isothermal ideal gas process.
4. Final Result: All supplied heat becomes work when ΔU = 0.

22. What happens when volume remains constant?

When volume remains constant, work done is zero. Heat supplied changes only internal energy.

1. Isochoric Process: ΔV = 0.
2. Work: W = PΔV = 0.
3. First Law: ΔQ = ΔU.
4. Final Result: At constant volume, heat changes internal energy.

23. A gas absorbs 500 J heat and does 200 J work. Find ΔU.

The change in internal energy is 300 J. Use ΔU = ΔQ − ΔW.

1. Given Data:
   ΔQ = 500 J
   ΔW = 200 J
2. Formula Used: ΔU = ΔQ − ΔW.
3. Calculation:
   ΔU = 500 − 200
   ΔU = 300 J
4. Final Result: ΔU = 300 J.

24. A gas has 150 J work done on it and loses 50 J heat. Find ΔU.

The change in internal energy is 100 J. Work done on the gas increases internal energy.

1. Given Data:
   Heat lost: ΔQ = −50 J
   Work by gas: ΔW = −150 J
2. Formula Used: ΔU = ΔQ − ΔW.
3. Calculation:
   ΔU = −50 − (−150)
   ΔU = 100 J
4. Final Result: ΔU = 100 J.

Specific Heat Capacity Class 11 Questions

Specific heat capacity tells how much heat changes temperature.
For gases, the value depends on whether pressure or volume remains constant.
These specific heat capacity class 11 questions cover heat capacity, molar heat capacity and Cp − Cv.

25. What is heat capacity?

Heat capacity is heat required to raise the temperature of a substance by one kelvin. Its symbol is S.

1. Heat Supplied: ΔQ.
2. Temperature Change: ΔT.
3. Formula Used: S = ΔQ/ΔT.
4. Final Result: Heat capacity measures heat needed per kelvin.

26. What is specific heat capacity?

Specific heat capacity is heat required to raise temperature of unit mass by one kelvin. Its unit is J kg^-1 K^-1.

1. Mass: m.
2. Heat Supplied: ΔQ.
3. Formula Used: s = ΔQ/(mΔT).
4. Final Result: Specific heat capacity is heat per kg per kelvin.

27. What is molar specific heat capacity?

Molar specific heat capacity is heat required to raise one mole of a substance by one kelvin. Its unit is J mol^-1 K^-1.

1. Moles: μ.
2. Heat Supplied: ΔQ.
3. Formula Used: C = ΔQ/(μΔT).
4. Final Result: Molar heat capacity is heat per mole per kelvin.

28. What is specific heat capacity of water in SI units?

Specific heat capacity of water is 4186 J kg^-1 K^-1. It is also 4.186 J g^-1 K^-1.

1. SI Value: 4186 J kg^-1 K^-1.
2. Gram Form: 4.186 J g^-1 K^-1.
3. Old Unit Link: 1 cal = 4.186 J.
4. Final Result: Water has high specific heat capacity.

29. Why should coolant have high specific heat capacity?

Coolant should have high specific heat capacity because it can absorb large heat with small temperature rise. This keeps machinery safer.

1. Heat Absorption: High for same temperature rise.
2. Temperature Control: Prevents overheating.
3. Example: Water works as a good coolant.
4. Final Result: High specific heat improves cooling.

30. What is the relation between Cp and Cv for an ideal gas?

For an ideal gas, Cp − Cv = R. Here R is the universal gas constant.

1. Constant Volume: No expansion work.
2. Constant Pressure: Gas does expansion work.
3. Extra Heat: Equals R per mole per kelvin.
4. Final Result: Cp exceeds Cv by R.

Thermodynamic State Variables Class 11 Questions

State variables describe equilibrium states of a thermodynamic system.
Their values do not depend on the path used to reach the state.
These thermodynamic state variables class 11 questions cover state functions and equation of state.

31. What are thermodynamic state variables?

Thermodynamic state variables are macroscopic quantities that describe the equilibrium state of a system. Examples include P, V, T and U.

1. Pressure: P.
2. Volume: V.
3. Temperature: T.
4. Internal Energy: U.
5. Final Result: State variables describe equilibrium states.

32. Why is internal energy a state variable?

Internal energy is a state variable because it depends only on the current state. It does not depend on the path taken.

1. Initial State: Has a defined U.
2. Final State: Has a defined U.
3. Change: ΔU depends only on initial and final states.
4. Final Result: Internal energy is path independent.

33. What is an equation of state?

An equation of state relates thermodynamic state variables. For an ideal gas, it is PV = μRT.

1. State Variables: P, V and T.
2. Gas Amount: μ.
3. Ideal Gas Equation: PV = μRT.
4. Final Result: Equation of state connects state variables.

34. What is an extensive variable?

An extensive variable depends on the size or amount of the system. Volume, mass and internal energy are examples.

1. Volume: Depends on system size.
2. Mass: Depends on amount of matter.
3. Internal Energy: Doubles when system doubles.
4. Final Result: Extensive variables scale with system size.

35. What is an intensive variable?

An intensive variable does not depend on the size of the system. Pressure, temperature and density are examples.

1. Pressure: Same after equal division.
2. Temperature: Same after equal division.
3. Density: Same for uniform material.
4. Final Result: Intensive variables do not scale with size.

36. Is heat a state variable?

No, heat is not a state variable. It depends on the path of energy transfer.

1. Heat: Energy in transit.
2. Process Dependence: Different paths give different heat.
3. State Description: Heat is not stored in a system.
4. Final Result: Heat is path dependent.

Thermodynamic Processes Class 11 Questions

A thermodynamic process changes a system from one state to another.
Special processes depend on which quantity remains constant.
These thermodynamic processes class 11 questions cover quasi-static, isothermal, adiabatic, isobaric and isochoric cases.

37. What is a quasi-static process?

A quasi-static process is an infinitely slow process in which the system remains nearly in equilibrium throughout. Pressure and temperature differences stay infinitesimal.

1. Speed: Extremely slow.
2. Equilibrium: Maintained at every stage.
3. Use: Allows well-defined P, V and T.
4. Final Result: Quasi-static process passes through equilibrium states.

38. What is an isothermal process?

An isothermal process is a process in which temperature remains constant. For an ideal gas, internal energy remains unchanged.

1. Constant Quantity: Temperature.
2. Ideal Gas: ΔU = 0.
3. Gas Law: PV = constant.
4. Final Result: Isothermal process has constant temperature.

39. What is an adiabatic process?

An adiabatic process is a process in which no heat flows between system and surroundings. It satisfies Q = 0.

1. Heat Transfer: Zero.
2. Insulation: System is thermally isolated.
3. Ideal Gas Relation: PV^γ = constant.
4. Final Result: Adiabatic process has no heat exchange.

40. What is an isobaric process?

An isobaric process occurs at constant pressure. Work done is W = P(V2 − V1).

1. Constant Quantity: Pressure.
2. Work Formula: W = PΔV.
3. Heat Use: Changes internal energy and does work.
4. Final Result: Isobaric process has constant pressure.

41. What is an isochoric process?

An isochoric process occurs at constant volume. Work done is zero because ΔV = 0.

1. Constant Quantity: Volume.
2. Work Formula: W = PΔV.
3. Result: W = 0.
4. Final Result: Isochoric process has constant volume.

42. What is a cyclic process?

A cyclic process returns a system to its initial state. Net change in internal energy is zero.

1. Initial State: Same as final state.
2. State Variable: U returns to original value.
3. Result: ΔU = 0.
4. Final Result: Cyclic process has zero net internal energy change.

Isothermal Process Class 11 Questions

In an isothermal process, temperature stays constant throughout the change.
For an ideal gas, heat supplied becomes work done by the gas.
These isothermal process class 11 questions cover Boyle’s law, work and internal energy.

43. What is the work done in isothermal expansion of an ideal gas?

Work done in isothermal expansion is W = μRT ln(V2/V1). It is positive when V2 > V1.

1. Temperature: Constant.
2. Ideal Gas Law: P = μRT/V.
3. Work Integral: W = ∫P dV.
4. Final Result: W = μRT ln(V2/V1).

44. Why is ΔU zero in an isothermal process for an ideal gas?

ΔU is zero because internal energy of an ideal gas depends only on temperature. Temperature remains constant.

1. Ideal Gas: U depends only on T.
2. Isothermal Process: ΔT = 0.
3. Result: ΔU = 0.
4. Final Result: Isothermal ideal gas has no internal energy change.

45. What happens to heat in isothermal expansion?

Heat supplied in isothermal expansion becomes work done by the gas. The internal energy remains unchanged.

1. First Law: ΔQ = ΔU + ΔW.
2. Isothermal Ideal Gas: ΔU = 0.
3. Result: ΔQ = ΔW.
4. Final Result: Heat absorbed equals work done.

46. One mole of ideal gas expands isothermally at 300 K from 2 L to 4 L. Find work done.

The work done is 1728 J approximately. Use W = μRT ln(V2/V1).

1. Given Data:
   μ = 1 mol
   R = 8.31 J mol^-1 K^-1
   T = 300 K
   V2/V1 = 4/2 = 2
2. Formula Used: W = μRT ln(V2/V1).
3. Calculation:
   W = 1 × 8.31 × 300 × ln 2
   W = 1728 J
4. Final Result: W ≈ 1728 J.

Adiabatic Process Class 11 Questions

In an adiabatic process, heat exchange with surroundings stays zero.
Work changes internal energy and temperature.
These adiabatic process class 11 questions cover PV^γ, compression and expansion.

47. What is the adiabatic relation for an ideal gas?

The adiabatic relation is PV^γ = constant. Here γ = Cp/Cv.

1. Heat Transfer: Q = 0.
2. Specific Heat Ratio: γ = Cp/Cv.
3. Equation: PV^γ = constant.
4. Final Result: Adiabatic ideal gas follows PV^γ = constant.

48. What happens during adiabatic expansion?

During adiabatic expansion, gas does work and its internal energy decreases. Its temperature falls.

1. Heat Exchange: Zero.
2. Work: Done by gas.
3. Internal Energy: Decreases.
4. Final Result: Adiabatic expansion cools the gas.

49. What happens during adiabatic compression?

During adiabatic compression, work is done on the gas. Its internal energy and temperature increase.

1. Heat Exchange: Zero.
2. Work: Done on gas.
3. Internal Energy: Increases.
4. Final Result: Adiabatic compression heats the gas.

50. How does pressure change when adiabatic volume is halved?

Pressure increases by a factor 2^γ. Use P1V1^γ = P2V2^γ.

1. Given: V2 = V1/2.
2. Relation: P1V1^γ = P2(V1/2)^γ.
3. Result: P2/P1 = 2^γ.
4. Final Result: Pressure becomes 2^γ times.

51. Why is adiabatic curve steeper than isothermal curve?

Adiabatic curve is steeper because pressure rises faster during compression. Temperature also increases during adiabatic compression.

1. Isothermal Relation: PV = constant.
2. Adiabatic Relation: PV^γ = constant.
3. Since: γ > 1.
4. Final Result: Adiabatic pressure changes faster with volume.

Second Law of Thermodynamics Class 11 Questions

The second law limits which energy processes can occur naturally.
It rules out perfect heat engines and perfect refrigerators.
These second law of thermodynamics class 11 questions cover Kelvin-Planck and Clausius statements.

52. Why is the second law needed?

The second law is needed because the first law only conserves energy. It does not decide the natural direction of processes.

1. First Law: Allows energy balance.
2. Observed Nature: Some energy-balanced processes never occur.
3. Second Law: Gives direction and limits.
4. Final Result: Second law restricts possible processes.

53. State Kelvin-Planck statement of second law.

Kelvin-Planck statement says no process can absorb heat from one reservoir and convert it completely into work. A perfect heat engine is impossible.

1. Heat Source: Single reservoir.
2. Desired Output: Complete work conversion.
3. Restriction: Not possible.
4. Final Result: No heat engine has 100% efficiency.

54. State Clausius statement of second law.

Clausius statement says no process can transfer heat from a colder body to a hotter body as its sole result. External work is needed.

1. Cold Body: Lower temperature.
2. Hot Body: Higher temperature.
3. Natural Flow: Heat flows hot to cold.
4. Final Result: Heat cannot flow cold to hot by itself.

55. Why can a heat engine not have 100% efficiency?

A heat engine cannot have 100% efficiency because some heat must be rejected to a colder reservoir. This follows from the second law.

1. Heat Input: Taken from source.
2. Work Output: Less than heat input.
3. Heat Rejection: Needed for cyclic operation.
4. Final Result: Perfect heat engine is impossible.

56. Why can a refrigerator not work without external work?

A refrigerator cannot work without external work because it transfers heat from cold body to hot body. This is opposite to natural heat flow.

1. Cold Region: Heat is removed.
2. Hot Region: Heat is delivered.
3. Work Input: Required by second law.
4. Final Result: Refrigerator needs work input.

Reversible and Irreversible Processes Class 11 Questions

Most natural processes are irreversible because of friction, viscosity or finite gradients.
Reversible processes are ideal and must be quasi-static without dissipation.
These questions help distinguish real and ideal thermodynamic processes.

57. What is a reversible process?

A reversible process can be reversed so that system and surroundings return to original states. It leaves no other change anywhere.

1. Path: Can be exactly retraced.
2. System: Returns to initial state.
3. Surroundings: Also return to initial state.
4. Final Result: Reversible process is an ideal process.

58. What is an irreversible process?

An irreversible process cannot be reversed without leaving changes in system or surroundings. Most natural processes are irreversible.

1. Examples: Free expansion and diffusion.
2. Dissipation: Friction and viscosity cause irreversibility.
3. Direction: Happens spontaneously one way.
4. Final Result: Irreversible processes dominate nature.

59. What conditions are needed for a reversible process?

A reversible process must be quasi-static and free from dissipative effects. Friction, viscosity and finite gradients must be absent.

1. Condition 1: Infinitely slow change.
2. Condition 2: No friction.
3. Condition 3: No finite temperature difference.
4. Final Result: Reversible process needs equilibrium at every stage.

60. Why is free expansion irreversible?

Free expansion is irreversible because the gas passes through non-equilibrium states. It cannot return spontaneously to its original volume.

1. Process: Gas expands into vacuum.
2. Intermediate States: Pressure not well-defined.
3. Return: Needs external work.
4. Final Result: Free expansion is not reversible.

Carnot Engine Class 11 Questions

Carnot engine gives the maximum efficiency possible between two reservoirs.
It uses two isothermal and two adiabatic processes.
These Carnot engine class 11 questions cover cycle steps, efficiency and limitations.

61. What is a Carnot engine?

A Carnot engine is a reversible heat engine operating between two temperatures. It works through the Carnot cycle.

1. Hot Reservoir: Temperature T1.
2. Cold Reservoir: Temperature T2.
3. Cycle: Two isothermal and two adiabatic processes.
4. Final Result: Carnot engine is an ideal reversible engine.

62. What are the four steps of Carnot cycle?

The four steps are isothermal expansion, adiabatic expansion, isothermal compression and adiabatic compression. They form a closed cycle.

1. Step 1: Isothermal expansion at T1.
2. Step 2: Adiabatic expansion from T1 to T2.
3. Step 3: Isothermal compression at T2.
4. Step 4: Adiabatic compression from T2 to T1.
5. Final Result: Carnot cycle has two isotherms and two adiabatics.

63. What is efficiency of Carnot engine?

Carnot engine efficiency is η = 1 − T2/T1. Temperatures must be in kelvin.

1. Source Temperature: T1.
2. Sink Temperature: T2.
3. Formula Used: η = 1 − T2/T1.
4. Final Result: Carnot efficiency depends only on reservoir temperatures.

64. Why is Carnot efficiency independent of working substance?

Carnot efficiency is independent of working substance because all reversible engines between same reservoirs have the same maximum efficiency. Only T1 and T2 matter.

1. Engine Type: Reversible.
2. Reservoirs: Same temperatures.
3. Efficiency Formula: η = 1 − T2/T1.
4. Final Result: Working substance does not affect Carnot efficiency.

65. A Carnot engine works between 600 K and 300 K. Find efficiency.

The efficiency is 50%. Use η = 1 − T2/T1.

1. Given Data:
   T1 = 600 K
   T2 = 300 K
2. Formula Used: η = 1 − T2/T1.
3. Calculation:
   η = 1 − 300/600
   η = 0.5
4. Final Result: Efficiency = 50%.

66. Can Carnot engine have 100% efficiency?

No, Carnot engine cannot have 100% efficiency unless T2 = 0 K. Absolute zero cannot be reached in practice.

1. Formula: η = 1 − T2/T1.
2. For η = 1: T2 must be 0 K.
3. Reality: T2 remains above 0 K.
4. Final Result: 100% heat engine efficiency is impossible.

NCERT Class 11 Physics Chapter 11 Questions

NCERT questions test heat capacity, first law, adiabatic compression, free expansion and work from graphs.
Students should use sign convention carefully in every numerical.
These NCERT Class 11 Physics Chapter 11 questions follow the 2026 exercise pattern.

67. A geyser heats water at 3 L/min from 27°C to 77°C. Find heat supplied per minute.

The heat supplied per minute is 6.28 × 10^5 J. Use Q = mcΔT.

1. Given Data:
   Volume = 3 L
   Mass = 3 kg
   c = 4186 J kg^-1 K^-1
   ΔT = 50 K
2. Formula Used: Q = mcΔT.
3. Calculation:
   Q = 3 × 4186 × 50
   Q = 6.28 × 10^5 J
4. Final Result: Heat supplied per minute = 6.28 × 10^5 J.

68. What is fuel consumption rate if heat of combustion is 4.0 × 10^4 J/g?

The fuel consumption rate is 15.7 g/min. Divide heat per minute by heat of combustion.

1. Heat Needed: 6.28 × 10^5 J/min.
2. Heat of Combustion: 4.0 × 10^4 J/g.
3. Calculation:
   Rate = 6.28 × 10^5/(4.0 × 10^4)
   Rate = 15.7 g/min
4. Final Result: Fuel consumption rate = 15.7 g/min.

69. Nitrogen mass is 2.0 × 10^-2 kg. Find moles if molar mass is 28 g mol^-1.

The amount of nitrogen is 0.714 mol. Use μ = mass/molar mass.

1. Given Data:
   Mass = 2.0 × 10^-2 kg = 20 g
   Molar mass = 28 g mol^-1
2. Formula Used: μ = m/M.
3. Calculation:
   μ = 20/28
   μ = 0.714 mol
4. Final Result: μ = 0.714 mol.

70. Heat nitrogen by 45°C at constant pressure. Take Cp = 7R/2. Find heat supplied.

The heat supplied is 934 J approximately. Use Q = μCpΔT.

1. Given Data:
   μ = 0.714 mol
   Cp = 7R/2
   R = 8.3 J mol^-1 K^-1
   ΔT = 45 K
2. Formula Used: Q = μCpΔT.
3. Calculation:
   Q = 0.714 × 3.5 × 8.3 × 45
   Q ≈ 934 J
4. Final Result: Heat supplied ≈ 934 J.

71. In an adiabatic process, 22.3 J work is done on gas. Find ΔU.

The change in internal energy is 22.3 J. In an adiabatic process, ΔQ = 0.

1. Given Data:
   Work done on gas = 22.3 J.
2. Work by Gas: ΔW = −22.3 J.
3. Formula Used: ΔU = ΔQ − ΔW.
4. Calculation:
   ΔU = 0 − (−22.3)
   ΔU = 22.3 J
5. Final Result: ΔU = 22.3 J.

72. If same state change absorbs 9.35 cal heat, find work done by system.

The work done by system is 16.9 J. Internal energy change remains 22.3 J.

1. Given Data:
   ΔQ = 9.35 cal = 9.35 × 4.19 J
   ΔQ = 39.18 J
   ΔU = 22.3 J
2. Formula Used: ΔW = ΔQ − ΔU.
3. Calculation:
   ΔW = 39.18 − 22.3
   ΔW = 16.88 J
4. Final Result: Work done by system ≈ 16.9 J.

73. A 100 W heater supplies heat while system does work at 75 J/s. Find rate of increase of internal energy.

The internal energy increases at 25 J/s. Use first law rate form.

1. Heat Rate: 100 J/s.
2. Work Rate: 75 J/s.
3. Formula Used: dU/dt = dQ/dt − dW/dt.
4. Calculation:
   dU/dt = 100 − 75
   dU/dt = 25 J/s
5. Final Result: Internal energy increases at 25 J/s.

74. A gas expands freely into vacuum in an insulated container. What is work done?

The work done is zero. In free expansion, external pressure is zero.

1. External Pressure: Zero.
2. Formula: W = PextΔV.
3. Heat Exchange: Zero in insulated setup.
4. Final Result: Free expansion does zero work.

75. Does free expansion lie on a P-V-T surface during intermediate states?

No, intermediate states of free expansion do not lie on a P-V-T surface. The gas is not in equilibrium during expansion.

1. Process: Sudden expansion into vacuum.
2. Pressure: Not uniform during expansion.
3. State Variables: Not well-defined.
4. Final Result: Intermediate free expansion states are non-equilibrium states.

CBSE Class 11 Physics Chapter-Wise Important Questions