Important Questions Class 11 Chemistry Chapter 5 Thermodynamics 2026–27

Thermodynamics explains energy changes in chemical and physical processes by studying macroscopic systems.
For CBSE Class 11 Chemistry, Important Questions Class 11 Chemistry Chapter 5 help revise formulas, derivations, numericals and spontaneity.

Thermodynamics is Chapter 5 in the updated Class 11 Chemistry NCERT textbook for 2026–27. It studies energy transformations in chemical and physical processes. The chapter deals with macroscopic systems and focuses on the initial and final states of a process, not on the speed or detailed path of change.

Use these Important Questions Class 11 Chemistry Chapter 5 to revise system and surroundings, open system, closed system, isolated system, internal energy, heat, work, first law of thermodynamics, enthalpy, calorimetry, Hess’s law, entropy, Gibbs free energy and spontaneity. These Thermodynamics Class 11 important questions with answers also include formula-based and numerical-style practice.

Key Takeaways

• Thermodynamics: It studies energy transformations in macroscopic systems.
• First law: It is expressed as ΔU = q + w.
• Enthalpy: It is defined as H = U + pV.
• Gibbs free energy: It predicts spontaneity using ΔG = ΔH − TΔS.

Confused between ΔU, ΔH, entropy and Gibbs free energy numericals?
Practise formula selection, sign conventions and NCERT-based Thermodynamics answers on the Extramarks Learning App. Sign Up Free

Formula Snapshot for Important Questions Class 11 Chemistry Chapter 5 Thermodynamics

Thermodynamics questions become easier when each formula is linked with its condition. Check whether the process is at constant volume, constant pressure, reversible, adiabatic or linked to spontaneity.

First law of thermodynamics:
ΔU = q + w

Enthalpy:
H = U + pV

Relation between enthalpy and internal energy at constant pressure:
ΔH = ΔU + pΔV

For gaseous reactions:
ΔH = ΔU + ΔnᵍRT

Heat capacity:
q = CΔT

Specific heat capacity:
q = mcΔT

Relation between molar heat capacities of an ideal gas:
Cₚ − Cᵥ = R

Entropy change for a reversible process:
ΔS = qᵣₑᵥ / T

Gibbs free energy:
ΔG = ΔH − TΔS

Very Short Answer Questions for Important Questions For Class 11 Chemistry Chapter 5 Thermodynamics

Thermodynamic terms build the base for formulas and numericals. These direct answers help revise system and surroundings, heat, work, internal energy and state functions quickly.

Q1. What is thermodynamics?

Thermodynamics is the study of energy transformations in physical and chemical processes.

It deals with macroscopic systems involving a large number of particles. It focuses on energy changes between initial and final states.

Q2. What is a thermodynamic system?

A thermodynamic system is the part of the universe selected for observation.

For example, a reaction mixture in a beaker can be treated as the system.

Q3. What are surroundings in thermodynamics?

Surroundings include everything outside the system that can interact with it.

The system and surroundings together form the universe.

Universe = System + Surroundings

Q4. What is a boundary?

A boundary is the real or imaginary wall that separates the system from its surroundings.

It helps track the movement of matter and energy into or out of the system.

Q5. What is an open system?

An open system can exchange both matter and energy with its surroundings.

Reactants kept in an open beaker form an open system.

Q6. What is a closed system?

A closed system can exchange energy but not matter with its surroundings.

Reactants kept in a closed conducting vessel form a closed system.

Q7. What is an isolated system?

An isolated system can exchange neither matter nor energy with its surroundings.

Reactants kept in a thermos flask or closed insulated vessel form an isolated system.

Objective Questions from Thermodynamics Class 11 Important Questions

Objective questions from Thermodynamics test exact definitions, formulas and sign conventions. These are useful for quick CBSE Class 11 Chemistry revision.

Q8. Choose the correct Coffee kept in an open cup is which type of system?

1. a) Closed system
   b) Isolated system
   c) Open system
   d) Adiabatic system
2. c) Open system.

Coffee in an open cup can exchange heat and water vapour with the surroundings.

Q9. Choose the correct A thermos flask is close to which type of system?

1. a) Open system
   b) Closed system
   c) Isolated system
   d) Homogeneous system
2. c) Isolated system.

A thermos flask is designed to minimise exchange of both matter and energy.

Q10. Fill in the blank: The first law of thermodynamics is written as ______.

ΔU = q + w.

This equation shows that change in internal energy depends on heat and work contributions.

Q11. Fill in the blank: Enthalpy is defined as H = ______.

U + pV.

Here, U is internal energy, p is pressure and V is volume.

Q12. True or False: Work is a state function.

False.

Work depends on the path followed during a process. Hence, it is not a state function.

Q13. Match the following: open system, closed system, isolated system and adiabatic process.

Open system exchanges matter and energy, closed system exchanges only energy, isolated system exchanges neither matter nor energy, and adiabatic process involves no heat transfer.

This matching checks the basic classification of thermodynamic systems and processes.

Short Answer Questions for Class 11 Chemistry Chapter 5 Important Questions

Short Answer Questions from Thermodynamics usually test definitions, sign conventions and formula conditions. Use exact terms such as internal energy, state function, enthalpy and heat capacity.

Q14. Differentiate between system and surroundings.

System is the part of the universe selected for thermodynamic study, while surroundings include everything else that can interact with the system.

For example, if a reaction mixture in a beaker is studied, the reaction mixture is the system.

The beaker and nearby environment form the surroundings. Together, system and surroundings constitute the universe.

Q15. Explain open, closed and isolated systems with examples.

An open system exchanges both matter and energy with surroundings.

Reactants in an open beaker form an open system.

A closed system exchanges energy but not matter. Reactants in a closed conducting vessel form a closed system.

An isolated system exchanges neither matter nor energy, as in a thermos flask.

Q16. Why is internal energy called a state function?

Internal energy is called a state function because its change depends only on the initial and final states of the system.

It does not depend on the path followed.

If a system changes from state A to state B, ΔU remains the same for all paths connecting those two states.

Q17. Why is work not a state function?

Work is not a state function because it depends on the path followed during a process.

Different paths between the same initial and final states can involve different amounts of work.

This is why work is called a path function.

Q18. Explain the sign convention for heat and work.

Heat is positive when it is transferred from surroundings to the system.

Heat is negative when it is transferred from the system to the surroundings.

Work is positive when work is done on the system. Work is negative when work is done by the system.

Q19. What is an adiabatic process?

An adiabatic process is a process in which no heat is exchanged between the system and surroundings.

For an adiabatic process:

q = 0

Hence, the first law becomes:

ΔU = wₐd

Q20. Why is no work done during free expansion of an ideal gas?

No work is done during free expansion because external pressure is zero.

Work done at constant external pressure is given by:

w = −pₑₓΔV

In free expansion:

pₑₓ = 0

Therefore:

w = 0

Q21. What is the difference between extensive and intensive properties?

Extensive properties depend on the amount of matter present in the system.

Examples include mass, volume, internal energy, enthalpy and heat capacity.

Intensive properties do not depend on the amount of matter. Examples include temperature, pressure and density.

Long Answer Questions on Thermodynamics, Enthalpy and Spontaneity

Long Answer Questions from Thermodynamics need formula statements, conditions and stepwise explanation. Strong answers show the relation between heat, work, internal energy, enthalpy, entropy and spontaneity.

Q22. State and explain the first law of thermodynamics.

The first law of thermodynamics states that energy can neither be created nor destroyed.

It can only be transformed from one form to another.

Mathematically, it is written as:

ΔU = q + w

Here, ΔU is change in internal energy, q is heat exchanged and w is work done. If heat is supplied to the system, q is positive. If work is done on the system, w is positive.

For an isolated system:

q = 0

w = 0

So:

ΔU = 0

This means the energy of an isolated system remains constant.

Q23. Derive the relation ΔH = ΔU + pΔV.

Enthalpy is defined as:

H = U + pV

For a finite change:

ΔH = ΔU + Δ(pV)

At constant pressure, p remains constant. Therefore:

Δ(pV) = pΔV

So:

ΔH = ΔU + pΔV

This relation shows that enthalpy change includes internal energy change and pressure-volume work.

Q24. Derive the relation ΔH = ΔU + ΔnᵍRT for gaseous reactions.

For gaseous reactions at constant temperature and pressure:

pV = nRT

For gaseous reactants:

pVᴬ = nᴬRT

For gaseous products:

pVᴮ = nᴮRT

Subtracting:

pVᴮ − pVᴬ = nᴮRT − nᴬRT

p(Vᴮ − Vᴬ) = (nᴮ − nᴬ)RT

pΔV = ΔnᵍRT

Substituting in:

ΔH = ΔU + pΔV

we get:

ΔH = ΔU + ΔnᵍRT

Here, Δnᵍ is the number of moles of gaseous products minus the number of moles of gaseous reactants.

Q25. Prove that Cₚ − Cᵥ = R for one mole of an ideal gas.

For one mole of an ideal gas:

H = U + pV

Since:

pV = RT

we get:

H = U + RT

For a temperature change:

ΔH = ΔU + RΔT

At constant pressure:

ΔH = CₚΔT

At constant volume:

ΔU = CᵥΔT

Substituting these values:

CₚΔT = CᵥΔT + RΔT

Dividing by ΔT:

Cₚ = Cᵥ + R

Therefore:

Cₚ − Cᵥ = R

Q26. State and explain Hess’s law of constant heat summation.

Hess’s law states that the enthalpy change of a reaction is the same whether the reaction occurs in one step or several steps.

This is because enthalpy is a state function.

If a reaction A → B occurs through several intermediate steps, then:

ΔᵣH = ΔᵣH₁ + ΔᵣH₂ + ΔᵣH₃

Hess’s law is useful when the enthalpy change of a reaction cannot be measured directly. It helps calculate reaction enthalpy using known thermochemical equations.

Q27. Explain entropy and its relation with spontaneity.

Entropy is a thermodynamic state function that measures randomness or disorder.

A process that increases the total entropy of the universe is favoured.

Entropy change is written as:

ΔS = qᵣₑᵥ / T

Gibbs free energy combines the effects of enthalpy and entropy:

ΔG = ΔH − TΔS

A process is spontaneous when:

ΔG < 0

Numerical Questions for Important Questions For Class 11 Chemistry Chapter 5 Thermodynamics - 2026-27

Numerical questions in Thermodynamics require formula selection before substitution. Check whether the condition is constant volume, constant pressure, isothermal or related to Gibbs free energy and spontaneity.

Q28. Calculate the work done when a gas expands against constant external pressure.

Work done against constant external pressure is calculated using:

w = −pₑₓΔV

Suppose a gas expands from 2 L to 10 L against 1 atm.

Vᵢ = 2 L
V𝒻 = 10 L

ΔV = V𝒻 − Vᵢ
ΔV = 10 − 2 = 8 L

w = −1 × 8
w = −8 L atm

The negative sign shows that work is done by the system.

Q29. Calculate ΔH from ΔU for a gaseous reaction.

For gaseous reactions:

ΔH = ΔU + ΔnᵍRT

Given:

ΔU = 37.904 kJ mol⁻¹
Δnᵍ = 1
R = 8.314 J K⁻¹ mol⁻¹
T = 373 K

First calculate:

ΔnᵍRT = 1 × 8.314 × 373

ΔnᵍRT = 3101 J mol⁻¹

Convert joules to kilojoules:

3101 J mol⁻¹ = 3.101 kJ mol⁻¹

Now:

ΔH = 37.904 + 3.101

ΔH = 41.005 kJ mol⁻¹

Q30. A reaction has ΔH = +20 kJ mol⁻¹ and ΔS = +100 J K⁻¹ mol⁻¹. Above what temperature will it become spontaneous?

The reaction becomes spontaneous when:

ΔG < 0

Use:

ΔG = ΔH − TΔS

At the point where spontaneity begins:

ΔG = 0

So:

0 = ΔH − TΔS

TΔS = ΔH

T = ΔH / ΔS

Convert ΔH into joules:

20 kJ mol⁻¹ = 20000 J mol⁻¹

Substitute:

T = 20000 J mol⁻¹ / 100 J K⁻¹ mol⁻¹

T = 200 K

The reaction becomes spontaneous above 200 K.

Class 11 Chemistry Chapter-Wise Important Questions