Thermodynamics Class 11 Chemistry: Notes, Formulas & NEET Questions

 -Dr.Sanjaykumar Pawar 

THERMODYNAMICS – CLASS 11 CHEMISTRY 

Thermodynamics – Learn the laws of energy, heat, work, enthalpy, entropy and Gibbs free energy for NEET and JEE.

Introduction

Thermodynamics is the branch of science that deals with heat, work, energy and their transformations. In Chemistry, thermodynamics helps us understand why chemical reactions release or absorb energy, how energy changes occur during physical processes, and whether a reaction is thermodynamically favourable. Examples:

• Burning of fuels releases heat.
• Melting of ice absorbs heat.
• Expansion of gases involves work.
• Chemical reactions involve changes in internal energy and enthalpy.

1. Basic Terminology of Thermodynamics

1.1 System

A system is the part of the universe selected for thermodynamic study. Example: Gas enclosed in a cylinder.

1.2 Surroundings

Everything outside the system that can interact with it is called the surroundings.

1.3 Boundary

The real or imaginary surface separating the system from its surroundings is called the boundary.

Important:
Universe = System + Surroundings

2. Types of Thermodynamic Systems

2.1 Open System

An open system exchanges both matter and energy with its surroundings. Example: Water boiling in an open vessel.

2.2 Closed System

A closed system exchanges energy but not matter with its surroundings. Example: Gas enclosed in a sealed cylinder with a movable piston.

2.3 Isolated System

An isolated system exchanges neither matter nor energy with its surroundings. Example: An ideal thermally insulated and sealed container.

System	Matter Exchange	Energy Exchange
Open	Yes	Yes
Closed	No	Yes
Isolated	No	No

3. State of a System

The state of a thermodynamic system is described by measurable properties such as:

• Pressure (P)
• Volume (V)
• Temperature (T)
• Composition
• Amount of substance

These properties are called state variables.

State Functions

A state function depends only on the initial and final states, not on the path followed. Examples:

• Internal energy (U)
• Enthalpy (H)
• Entropy (S)
• Gibbs free energy (G)

Path Functions

Path functions depend on the process or path followed. Examples:

• Heat (q)
• Work (w)

Heat and work are not properties stored in a system. They describe energy transfer across the system boundary.

4. Types of Thermodynamic Processes

4.1 Isothermal Process

A process occurring at constant temperature.

ΔT = 0

For an ideal gas:

ΔU = 0

4.2 Adiabatic Process

A process in which no heat is exchanged with the surroundings.

q = 0

Therefore:

ΔU = w

4.3 Isochoric Process

A process occurring at constant volume.

ΔV = 0

w = 0

4.4 Isobaric Process

A process occurring at constant pressure.

ΔP = 0

4.5 Cyclic Process

A process in which the system returns to its initial state.

ΔU = 0

5. Internal Energy (U)

Internal energy is the total microscopic energy of a system, including molecular kinetic and potential energies. It is represented by U. The change in internal energy is:

ΔU = U₂ - U₁

Where:

• U₁ = Initial internal energy
• U₂ = Final internal energy

For an ideal gas, internal energy depends only on temperature. For n moles of a calorically perfect ideal gas:

ΔU = nCv,mΔT

Where Cv,m is molar heat capacity at constant volume.

6. First Law of Thermodynamics

The first law is based on the conservation of energy. Energy can neither be created nor destroyed. It can only be transferred or transformed.

ΔU = q + w

Where:

• ΔU = Change in internal energy
• q = Heat supplied to the system
• w = Work done on the system

Chemistry Sign Convention

Process	Sign
Heat absorbed by system	q = Positive
Heat released by system	q = Negative
Work done on system	w = Positive
Work done by system	w = Negative

Example:
A system absorbs 500 J of heat and performs 200 J of work on its surroundings. Given: q = +500 J w = -200 J ΔU = q + w ΔU = 500 - 200 Answer: ΔU = +300 J

7. Pressure-Volume Work

Work is done when a system expands or contracts against an external pressure. For constant external pressure:

w = -PextΔV

Or:

w = -Pext(V₂ - V₁)

Where:

• Pext = External pressure
• V₁ = Initial volume
• V₂ = Final volume
• ΔV = Change in volume

Expansion

When volume increases:

ΔV > 0 → w < 0

Compression

When volume decreases:

ΔV < 0 → w > 0

Example:
A gas expands from 2 L to 5 L against a constant external pressure of 2 atm. w = -Pext(V₂ - V₁) w = -2(5 - 2) w = -6 L atm Using 1 L atm = 101.325 J: w = -607.95 J

8. Reversible Isothermal Expansion of an Ideal Gas

For a reversible isothermal expansion of an ideal gas:

w = -nRT ln(V₂/V₁)

Alternatively:

w = -2.303 nRT log₁₀(V₂/V₁)

Where:

• n = Number of moles
• R = Gas constant
• T = Absolute temperature
• V₁ = Initial volume
• V₂ = Final volume

For an ideal gas in an isothermal process:

ΔU = 0

q = -w

9. Enthalpy (H)

Enthalpy is a thermodynamic state function defined as internal energy plus pressure multiplied by volume.

H = U + PV

Change in enthalpy:

ΔH = ΔU + Δ(PV)

At constant pressure, with only pressure-volume work:

qₚ = ΔH

For an ideal gas reaction:

ΔH = ΔU + Δn(g)RT

Where: Δn(g) = Moles of gaseous products - Moles of gaseous reactants.

Example:
For a reaction at 300 K, ΔU = 10 kJ and Δn(g) = 2. ΔH = ΔU + Δn(g)RT ΔH = 10 + (2 × 8.314 × 300)/1000 ΔH = 14.9884 kJ

10. Exothermic and Endothermic Reactions

Exothermic Reaction

A reaction that releases heat to its surroundings.

ΔH < 0

Example: Combustion of methane. CH₄ + 2O₂ → CO₂ + 2H₂O

Endothermic Reaction

A reaction that absorbs heat from its surroundings.

ΔH > 0

Example: Thermal decomposition of calcium carbonate. CaCO₃ → CaO + CO₂

Remember: Exothermic = Heat released. Endothermic = Heat absorbed.

11. Heat Capacity and Specific Heat

Heat capacity is the amount of heat required to raise the temperature of a system by one kelvin.

C = q/ΔT

Specific heat capacity:

c = q/(mΔT)

Therefore:

q = mcΔT

Where:

• q = Heat transferred (J)
• m = Mass (kg or g, consistent with c)
• c = Specific heat capacity
• ΔT = Temperature change (K)

SI unit of specific heat capacity: J kg⁻¹ K⁻¹

Molar Heat Capacity

Cₘ = q/(nΔT)

For an ideal gas:

Cp,m - Cv,m = R

Where:

• Cp,m = Molar heat capacity at constant pressure
• Cv,m = Molar heat capacity at constant volume

12. Calorimetry

Calorimetry is the measurement of heat exchanged during a physical or chemical process. For an isolated calorimetric setup:

Heat lost + Heat gained = 0

Or:

qsystem + qsurroundings = 0

For a temperature change:

q = mcΔT

13. Hess's Law of Constant Heat Summation

Hess's law states that the total enthalpy change of a reaction is independent of the route by which the reaction takes place.

ΔHtotal = ΔH₁ + ΔH₂ + ΔH₃

This is because enthalpy is a state function.

Example:
A reaction occurs in two steps: Step 1: ΔH₁ = -100 kJ Step 2: ΔH₂ = +40 kJ Total enthalpy change: ΔH = -100 + 40 ΔH = -60 kJ

14. Enthalpy of Formation

Standard enthalpy of formation is the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states under specified standard conditions.

ΔrH° = ΣνΔfH°(products) - ΣνΔfH°(reactants)

Where:

• ΔfH° = Standard enthalpy of formation
• ν = Stoichiometric coefficient

The standard enthalpy of formation of an element in its standard state is zero.

15. Bond Enthalpy

Bond enthalpy is the enthalpy required to break one mole of a specified bond in the gaseous state. For a reaction:

ΔH ≈ ΣBE(bonds broken) - ΣBE(bonds formed)

Breaking bonds requires energy. Forming bonds releases energy.

16. Entropy (S)

Entropy is a thermodynamic state function related to the dispersal of energy and the number of accessible microscopic arrangements of a system. Symbol: S For a reversible process:

dS = δqrev/T

For a finite reversible isothermal process:

ΔS = qrev/T

SI unit: J mol⁻¹ K⁻¹ for molar entropy. In general, entropy increases when:

• A solid melts into a liquid.
• A liquid vaporizes into a gas.
• A gas expands freely.

17. Second Law of Thermodynamics

The second law states that the total entropy of an isolated system never decreases during a spontaneous process.

ΔSuniverse ≥ 0

For a spontaneous irreversible process:

ΔSuniverse > 0

For a reversible process:

ΔSuniverse = 0

18. Gibbs Free Energy

Gibbs free energy is a thermodynamic state function used to determine spontaneity at constant temperature and pressure.

G = H - TS

Change in Gibbs free energy at constant temperature:

ΔG = ΔH - TΔS

Conditions for Spontaneity

ΔG	Nature of Process
Negative	Spontaneous
Positive	Non-spontaneous
Zero	Equilibrium

19. Gibbs Energy and Temperature

The relationship between enthalpy, entropy and temperature determines whether a process is spontaneous.

ΔH	ΔS	Spontaneity
Negative	Positive	At all temperatures
Positive	Negative	At no temperature
Negative	Negative	At low temperature
Positive	Positive	At high temperature

20. Gibbs Energy and Equilibrium

For a reaction:

ΔG = ΔG° + RT ln Q

At equilibrium:

ΔG = 0

ΔG° = -RT ln K

Where:

• ΔG° = Standard Gibbs energy change
• R = Universal gas constant
• T = Absolute temperature
• K = Equilibrium constant
• Q = Reaction quotient

IMPORTANT FORMULA REVISION TABLE

Topic	Formula
Internal Energy	ΔU = q + w
PV Work	w = -PextΔV
Reversible Isothermal Work	w = -nRT ln(V₂/V₁)
Enthalpy	H = U + PV
Enthalpy Change	ΔH = ΔU + Δn(g)RT
Heat Capacity	C = q/ΔT
Specific Heat	q = mcΔT
Hess's Law	ΔHtotal = ΣΔHsteps
Bond Enthalpy	Broken - Formed
Entropy	dS = δqrev/T
Gibbs Energy	ΔG = ΔH - TΔS
Equilibrium	ΔG° = -RT ln K

COMMON STUDENT MISTAKES

• Using Celsius instead of Kelvin in thermodynamic equations.
• Confusing heat with temperature.
• Using the wrong sign for expansion work.
• Assuming heat and work are state functions.
• Confusing ΔU with ΔH.
• Forgetting stoichiometric coefficients in Hess's law.
• Ignoring the units of entropy in Gibbs energy calculations.
• Assuming every exothermic reaction is spontaneous.

NEET PRACTICE QUESTIONS

Q1. Which of the following is a state function?
A. Heat
B. Work
C. Internal energy
D. Path length

Answer: C. Internal energy

Q2. The first law of thermodynamics is:
A. ΔU = q - w
B. ΔU = q + w
C. ΔU = q/w
D. ΔU = 0

Answer: B. ΔU = q + w

Q3. In an isochoric process:
A. Pressure is constant
B. Temperature is constant
C. Volume is constant
D. Heat is zero

Answer: C. Volume is constant

Q4. The work done in an expansion against constant external pressure is:
A. Positive
B. Negative
C. Zero
D. Infinite

Answer: B. Negative

Q5. For an ideal gas in an isothermal process:
A. ΔU = 0
B. ΔH is always positive
C. q = 0
D. w = 0

Answer: A. ΔU = 0

Q6. Which reaction is exothermic?
A. A reaction with positive ΔH
B. A reaction with negative ΔH
C. A reaction with zero ΔH
D. Every reaction

Answer: B. A reaction with negative ΔH

Q7. At constant pressure, with only PV work:
A. qp = ΔU
B. qp = ΔH
C. qp = 0
D. qp = w

Answer: B. qp = ΔH

Q8. Which condition indicates spontaneity at constant temperature and pressure?
A. ΔG > 0
B. ΔG < 0
C. ΔG = 100
D. ΔH = 0

Answer: B. ΔG < 0

Q9. The SI unit of molar entropy is:
A. J mol⁻¹ K⁻¹
B. J mol⁻¹
C. Pa
D. J K

Answer: A. J mol⁻¹ K⁻¹

Q10. At equilibrium, the Gibbs energy change of a reaction is:
A. Positive
B. Negative
C. Zero
D. Infinite

Answer: C. Zero

NUMERICAL PRACTICE

Q11. A system absorbs 800 J of heat and performs 300 J of work. Calculate ΔU.

Solution:

q = +800 J w = -300 J ΔU = 800 - 300 Answer = +500 J

Q12. A gas expands by 2 L against an external pressure of 3 atm. Calculate work.

Solution:

w = -PextΔV w = -3 × 2 w = -6 L atm Answer = -607.95 J

Q13. Calculate ΔG at 300 K if ΔH = 40 kJ/mol and ΔS = 100 J/mol K.

Solution:

ΔG = ΔH - TΔS Convert entropy contribution: TΔS = 300 × 100 = 30000 J/mol = 30 kJ/mol ΔG = 40 - 30 Answer = +10 kJ/mol The process is non-spontaneous in the forward direction under the specified conditions.

Q14. For an ideal gas reaction, ΔU = 5 kJ/mol and Δn(g) = 1 at 300 K. Calculate ΔH.

Solution:

ΔH = ΔU + Δn(g)RT ΔH = 5 + (1 × 8.314 × 300)/1000 Answer = 7.4942 kJ/mol

FINAL REVISION CHECKLIST

• Understand system and surroundings.
• Learn state and path functions.
• Master all thermodynamic processes.
• Understand the first law of thermodynamics.
• Practise pressure-volume work calculations.
• Learn enthalpy and heat capacity.
• Understand Hess's law.
• Study entropy and the second law.
• Master Gibbs free energy and spontaneity.
• Practise NEET numerical problems.

Conclusion

Thermodynamics provides the foundation for understanding energy changes in chemical reactions. Students should focus on the first law, enthalpy, Hess's law, entropy and Gibbs free energy. Understanding the sign conventions and practising numerical problems regularly will help students develop strong conceptual knowledge for CBSE and NEET examinations.

End of Lesson – Thermodynamics