What is the difference between enthalpy and entropy?

Enthalpy (H) is the heat content of a system at constant pressure, while Entropy (S) is a measure of disorder or randomness in a system. Enthalpy change (ΔH) determines whether a reaction is exothermic or endothermic, while entropy change (ΔS) determines the spontaneity along with temperature.

What is Gibbs free energy and its significance?

Gibbs free energy (G) is a thermodynamic potential that predicts the spontaneity of a process at constant temperature and pressure. The equation is ΔG = ΔH - TΔS. If ΔG 0, non-spontaneous; if ΔG = 0, the system is at equilibrium.

What is Hess's Law of constant heat summation?

Hess's Law states that the total enthalpy change for a reaction is the same whether it occurs in one step or multiple steps. This is because enthalpy is a state function and depends only on initial and final states, not on the path taken.

What are state functions and path functions?

State functions depend only on the initial and final states of the system, not on the path taken (e.g., U, H, S, G, P, V, T). Path functions depend on the path taken during the process (e.g., q - heat, w - work). The differential of state functions is exact, while for path functions it is inexact.

How many marks does thermodynamics carry in JEE?

Thermodynamics carries 8-12% weightage in JEE Main (2-3 questions, 8-12 marks) and 10-15% in JEE Advanced (3-4 questions, 12-16 marks). It's one of the highest-scoring chapters in Physical Chemistry with good numerical problems.

Thermodynamics for JEE 2025-26 | Complete Notes with Formulas, Laws & Solved Examples

Q: What is the first law of thermodynamics?

The first law of thermodynamics states that energy can neither be created nor destroyed, only converted from one form to another. Mathematically: ΔU = q + w, where ΔU is change in internal energy, q is heat absorbed, and w is work done on the system.

Basic Concepts and Terminology

Thermodynamics is the branch of physical chemistry that deals with energy changes accompanying physical and chemical transformations. Understanding the basic terminology is crucial for mastering this chapter.

1.1 System and Surroundings

System

The part of the universe under thermodynamic study or observation.

Example: A gas in a cylinder, a chemical reaction in a beaker

Surroundings

Everything else in the universe except the system.

Note: System + Surroundings = Universe

1.2 Types of Systems

Type	Matter Exchange	Energy Exchange	Example
Open System	✓ Yes	✓ Yes	Boiling water in open container
Closed System	✗ No	✓ Yes	Gas in a sealed cylinder with movable piston
Isolated System	✗ No	✗ No	Hot coffee in a perfect thermos flask

1.3 State Functions vs Path Functions

State Functions

Properties that depend only on the state of the system, not on how it reached that state.

P - Pressure
V - Volume
T - Temperature
U - Internal Energy
H - Enthalpy
S - Entropy
G - Gibbs Free Energy

For state functions: dX is an exact differential

Path Functions

Properties that depend on the path taken during the process.

q - Heat
w - Work

For path functions: đq and đw are inexact differentials

Important Note:

Though q and w are path functions, their sum (q + w) equals ΔU, which is a state function!

1.4 Extensive and Intensive Properties

Property Type	Definition	Examples	Behavior
Extensive	Depend on amount of matter	Mass, Volume, U, H, S, G, Heat capacity	Additive (double the amount → double the property)
Intensive	Independent of amount of matter	Temperature, Pressure, Density, Refractive index, Viscosity	Non-additive (remains same on doubling amount)

💡 Quick Memory Trick

"MATHS VU" - Extensive properties
Mass, Amount, Total energy, Heat capacity, Surface area, Volume, U (Internal energy)

"TEMPER" - Intensive properties
Temperature, Electromotive force, Molarity, Pressure, Emissivity, Refractive index

1.5 Thermodynamic Processes

Isothermal Process

T = constant

\Delta U = 0\] \[q = -w\] \[PV = \text{constant}

Example: Phase changes (melting, boiling) at constant T

Adiabatic Process

q = 0 (No heat exchange)

\Delta U = w\] \[PV^{\gamma} = \text{constant}\] \[TV^{\gamma-1} = \text{constant}

Example: Sudden compression/expansion

Isobaric Process

P = constant

q_p = \Delta H\] \[w = -P\Delta V\] \[\frac{V}{T} = \text{constant}

Example: Heating of gas in open container

Isochoric Process

V = constant

w = 0\] \[q_v = \Delta U\] \[\frac{P}{T} = \text{constant}

Example: Heating in sealed rigid container

⚠️ JEE Important Note

Sign Convention:
• Heat absorbed by system: q > 0 (positive)
• Heat released by system: q < 0 (negative)
• Work done ON system: w > 0 (positive)
• Work done BY system: w < 0 (negative)

Zeroth Law of Thermodynamics

The Zeroth Law establishes the concept of temperature and thermal equilibrium. Though it seems simple, it's the foundation for temperature measurement.

Statement of Zeroth Law

"If two systems (A and B) are separately in thermal equilibrium with a third system (C), then A and B are also in thermal equilibrium with each other."

Mathematical Form:

\text{If } T_A = T_C \text{ and } T_B = T_C \text{, then } T_A = T_B

2.1 Concept of Temperature

Temperature

A measure of the average kinetic energy of particles in a substance.

Intensive property
State function
Determines direction of heat flow
Same for all systems in thermal equilibrium

Thermal Equilibrium

State when two systems have the same temperature and no net heat flow occurs between them.

Condition:

No change in any macroscopic property over time

💡 Why It's Called "Zeroth" Law?

This law was formulated after the First and Second Laws were already established. Since it's more fundamental than both, it was called the "Zeroth" Law to indicate its foundational nature.

2.2 Application: Thermometers

📝 Real-World Application

How does a thermometer work?

1. The thermometer (system C) is placed in contact with the body (system A) whose temperature needs to be measured.

2. Heat exchange occurs until thermal equilibrium is reached: T_A = T_C

3. The thermometer is calibrated against a standard (system B at known temperature)

4. By Zeroth Law, if T_C = T_B, then T_A = T_B

Thus, we can measure temperature indirectly using thermal equilibrium!

First Law of Thermodynamics

The First Law is essentially the law of conservation of energy applied to thermodynamic systems. This is one of the most important and frequently tested topics in JEE.

Statement of First Law

"Energy can neither be created nor destroyed; it can only be converted from one form to another."

Mathematical Form:

\Delta U = q + w

ΔU = Change in internal energy (state function)

q = Heat absorbed by the system (path function)

w = Work done on the system (path function)

⚠️ Sign Convention (Very Important for JEE)

Quantity	Positive (+)	Negative (-)
q (Heat)	Absorbed by system (endothermic)	Released by system (exothermic)
w (Work)	Done ON the system (compression)	Done BY the system (expansion)
ΔU	Internal energy increases	Internal energy decreases

3.1 Internal Energy (U)

What is Internal Energy?

The total energy stored within a system, including:

Kinetic Energy: Energy due to motion of molecules
Potential Energy: Energy due to intermolecular forces
Bond Energy: Energy stored in chemical bonds
Nuclear Energy: Energy in atomic nuclei

Important Points:

U is a state function and extensive property
Absolute value of U cannot be determined, only ΔU
For ideal gas: U depends only on temperature
For ideal gas: ΔU = nC_vΔT

3.2 Work in Thermodynamics

Pressure-Volume Work

For expansion/compression against external pressure:

w = -\int P_{\text{ext}} dV

Different Cases:

1. Constant External Pressure (Irreversible):

w = -P_{\text{ext}}(V_2 - V_1) = -P_{\text{ext}}\Delta V

2. Isothermal Reversible (Ideal Gas):

w = -nRT\ln\frac{V_2}{V_1} = -2.303nRT\log\frac{V_2}{V_1}

3. Free Expansion (P_ext = 0):

\[w = 0\]

4. Isochoric Process (ΔV = 0):

\[w = 0\]

📝 Solved Example 1 (JEE Main 2023 Type)

Question: Calculate the work done when 1 mole of an ideal gas expands isothermally and reversibly from 10 L to 20 L at 300 K.

Solution:

Given: n = 1 mol, V₁ = 10 L, V₂ = 20 L, T = 300 K

R = 8.314 J mol⁻¹ K⁻¹

For isothermal reversible expansion:

w = -2.303 \times nRT \times \log\frac{V_2}{V_1}

w = -2.303 \times 1 \times 8.314 \times 300 \times \log\frac{20}{10}

w = -2.303 \times 8.314 \times 300 \times \log 2

w = -2.303 \times 8.314 \times 300 \times 0.3010

w = -1729 \text{ J} = -1.729 \text{ kJ}

Note: Negative sign indicates work is done BY the system (expansion).

3.3 Heat Capacity

Heat Capacity at Constant Volume (C_v)

q_v = nC_v\Delta T = \Delta U

At constant volume: w = 0, so all heat goes to internal energy

For ideal gas:

Monoatomic: C_v = (3/2)R

Diatomic: C_v = (5/2)R

Polyatomic: C_v = 3R

Heat Capacity at Constant Pressure (C_p)

q_p = nC_p\Delta T = \Delta H

At constant pressure: heat goes to both internal energy and work

For ideal gas:

Monoatomic: C_p = (5/2)R

Diatomic: C_p = (7/2)R

Polyatomic: C_p = 4R

Mayer's Equation (Very Important for JEE)

\[C_p - C_v = R\]

Heat Capacity Ratio (γ):

\gamma = \frac{C_p}{C_v}

Monoatomic gas: γ = 5/3 = 1.67

Diatomic gas: γ = 7/5 = 1.4

Polyatomic gas: γ = 4/3 = 1.33

📝 Solved Example 2 (JEE Advanced Pattern)

Question: 2 moles of an ideal gas at 300 K is heated at constant volume until pressure doubles. Calculate q, w, and ΔU. (C_v = (3/2)R)

Solution:

Step 1: Find final temperature

At constant volume: P ∝ T

\frac{P_1}{T_1} = \frac{P_2}{T_2}

\frac{P}{300} = \frac{2P}{T_2}

T_2 = 600 \text{ K}

Step 2: Calculate work

At constant volume: ΔV = 0

\[w = 0\]

Step 3: Calculate ΔU

\Delta U = nC_v\Delta T

\Delta U = 2 \times \frac{3}{2}R \times (600-300)

\Delta U = 3 \times 8.314 \times 300

\Delta U = 7483 \text{ J} = 7.483 \text{ kJ}

Step 4: Calculate q using First Law

\Delta U = q + w

\[7483 = q + 0\]

q = 7483 \text{ J} = 7.483 \text{ kJ}

Answer: q = 7.483 kJ, w = 0, ΔU = 7.483 kJ

Enthalpy and Thermochemistry

Enthalpy is one of the most important concepts in thermodynamics, especially for chemical reactions occurring at constant pressure. This topic carries significant weightage in both JEE Main and Advanced.

4.1 Definition of Enthalpy

What is Enthalpy (H)?

Enthalpy is the heat content of a system at constant pressure. It's a state function and extensive property.

\[H = U + PV\]

For a process at constant pressure:

\Delta H = \Delta U + P\Delta V

\Delta H = q_p

Heat absorbed/released at constant pressure equals enthalpy change

4.2 Relationship between ΔH and ΔU

For Chemical Reactions

\Delta H = \Delta U + \Delta n_g RT

Δn_g = (moles of gaseous products) - (moles of gaseous reactants)

R = 8.314 J mol⁻¹ K⁻¹ or 2 cal mol⁻¹ K⁻¹

T = Temperature in Kelvin

Special Cases:

If Δn_g = 0: ΔH = ΔU
If Δn_g > 0: ΔH > ΔU
If Δn_g < 0: ΔH < ΔU
For reactions involving only solids/liquids: ΔH ≈ ΔU

📝 Solved Example 3

Question: For the reaction N₂(g) + 3H₂(g) → 2NH₃(g), ΔH = -92.4 kJ at 298 K. Calculate ΔU.

Solution:

Step 1: Calculate Δn_g

\Delta n_g = n_{\text{products}} - n_{\text{reactants}}

\Delta n_g = 2 - (1 + 3) = 2 - 4 = -2

Step 2: Use the relationship

\Delta H = \Delta U + \Delta n_g RT

\Delta U = \Delta H - \Delta n_g RT

\Delta U = -92.4 - (-2) \times 8.314 \times 10^{-3} \times 298

\Delta U = -92.4 + 4.96

\Delta U = -87.44 \text{ kJ}

4.3 Types of Enthalpy Changes

Type	Symbol	Definition	Example
Formation	ΔH_f°	Heat change when 1 mole of compound is formed from elements in standard state	C + O₂ → CO₂
Combustion	ΔH_c°	Heat released when 1 mole burns completely in oxygen	CH₄ + 2O₂ → CO₂ + 2H₂O
Neutralization	ΔH_neut°	Heat released when 1 mole H⁺ neutralizes 1 mole OH⁻	HCl + NaOH → NaCl + H₂O (-57.1 kJ)
Atomization	ΔH_a°	Heat required to break 1 mole into isolated atoms	CH₄(g) → C(g) + 4H(g)
Bond Dissociation	ΔH_bond	Energy required to break 1 mole of specific bonds	H-H → 2H (436 kJ)
Sublimation	ΔH_sub°	Heat required to convert 1 mole solid → gas	I₂(s) → I₂(g)
Fusion	ΔH_fus°	Heat required to melt 1 mole solid → liquid	H₂O(s) → H₂O(l) (6.01 kJ)
Vaporization	ΔH_vap°	Heat required to convert 1 mole liquid → gas	H₂O(l) → H₂O(g) (40.66 kJ)
Solution	ΔH_sol°	Heat change when 1 mole dissolves in solvent	NaCl(s) → Na⁺(aq) + Cl⁻(aq)
Hydration	ΔH_hyd°	Heat released when 1 mole ion hydrates	Na⁺(g) + aq → Na⁺(aq)

💡 Standard Enthalpy (ΔH°)

The superscript "°" indicates standard conditions:

Pressure: 1 bar (≈ 1 atm)
Temperature: Usually 298 K (25°C)
Concentration: 1 M for solutions
Elements in their most stable state (allotrope)

ΔH_f° of elements in standard state = 0

4.4 Exothermic vs Endothermic Reactions

Exothermic (ΔH < 0)

Heat is released to surroundings

Characteristics:

ΔH is negative
Products have lower energy than reactants
Temperature of surroundings increases
Generally spontaneous at low temperature

Examples:

Combustion reactions
Neutralization
Condensation, Freezing
Respiration

Endothermic (ΔH > 0)

Heat is absorbed from surroundings

Characteristics:

ΔH is positive
Products have higher energy than reactants
Temperature of surroundings decreases
Generally non-spontaneous at low temperature

Examples:

Decomposition reactions
Evaporation, Melting, Sublimation
Photosynthesis
Dissolution of NH₄Cl

Second Law of Thermodynamics

The Second Law introduces the concept of entropy and explains the direction of spontaneous processes. It's crucial for understanding why certain reactions occur naturally while others don't.

Statements of Second Law

Clausius Statement:

Heat cannot spontaneously flow from a colder body to a hotter body without external work.

Kelvin-Planck Statement:

It is impossible to convert heat completely into work in a cyclic process without some heat being rejected to surroundings.

Entropy Statement:

The entropy of an isolated system always increases for spontaneous processes.

\Delta S_{\text{universe}} > 0 \text{ (spontaneous)}

⚠️ Key Implication

No heat engine can be 100% efficient. Some energy must always be "wasted" as heat to surroundings. This is why perpetual motion machines are impossible!

Entropy (S)

Entropy is a measure of disorder or randomness in a system. It's one of the most conceptually challenging yet important topics in thermodynamics for JEE.

6.1 Definition and Calculation

Entropy Change (ΔS)

\Delta S = \frac{q_{\text{rev}}}{T}

For reversible process at constant temperature

For Isothermal Process (Ideal Gas):

\Delta S = nR\ln\frac{V_2}{V_1} = nR\ln\frac{P_1}{P_2}

For Phase Transition:

\Delta S = \frac{\Delta H_{\text{phase}}}{T}

For Heating at Constant Pressure:

\Delta S = nC_p\ln\frac{T_2}{T_1}

6.2 Entropy and Spontaneity

Total Entropy Change

\Delta S_{\text{total}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}}

Criteria for Spontaneity:

✓ If ΔS_total > 0: Process is spontaneous
✓ If ΔS_total < 0: Process is non-spontaneous
✓ If ΔS_total = 0: System is at equilibrium

💡 Entropy Trends (JEE Important)

State: S_gas >> S_liquid > S_solid
Temperature: Entropy increases with temperature
Complexity: More complex molecules have higher entropy
Volume: Entropy increases with volume (for gases)
Mixing: Entropy of mixture > sum of pure components

Gibbs Free Energy (G)

Gibbs Free Energy is the most practical criterion for predicting spontaneity of processes at constant temperature and pressure. This is a high-scoring topic in JEE with direct numerical problems.

7.1 Definition and Gibbs Equation

Gibbs Free Energy

\[G = H - TS\]

For a process:

\Delta G = \Delta H - T\Delta S

This is the Gibbs-Helmholtz equation

7.2 Spontaneity Criteria

ΔG Value	Process Type	Energy	Example
ΔG < 0 (Negative)	Spontaneous (Exergonic)	Free energy released	Water flowing downhill
ΔG > 0 (Positive)	Non-spontaneous (Endergonic)	Free energy required	Water pumped uphill
ΔG = 0 (Zero)	Equilibrium	No net change	Ice-water at 0°C

7.3 Effect of Temperature on Spontaneity

ΔH	ΔS	ΔG = ΔH - TΔS	Spontaneity
-	+	Always negative	Spontaneous at all temperatures
+	-	Always positive	Non-spontaneous at all temperatures
-	-	Negative at low T, positive at high T	Spontaneous at low temperature only
+	+	Positive at low T, negative at high T	Spontaneous at high temperature only

📝 Solved Example 4 (JEE Main 2024 Type)

Question: For a reaction at 300 K: ΔH = 50 kJ/mol, ΔS = 100 J/K·mol. Predict whether the reaction is spontaneous at this temperature. At what temperature will it become spontaneous?

Solution:

Part 1: Check spontaneity at 300 K

\Delta G = \Delta H - T\Delta S

\Delta G = 50000 - 300 \times 100

\Delta G = 50000 - 30000 = 20000 \text{ J} = 20 \text{ kJ}

Since ΔG > 0, reaction is NON-SPONTANEOUS at 300 K

Part 2: Find temperature for spontaneity

For spontaneity: ΔG < 0, or at equilibrium ΔG = 0

\Delta H - T\Delta S = 0

T = \frac{\Delta H}{\Delta S}

T = \frac{50000}{100} = 500 \text{ K}

Answer: Reaction becomes spontaneous at T > 500 K

7.4 Standard Gibbs Free Energy

Calculating ΔG° for Reactions

\Delta G° = \sum \Delta G°_f\text{(products)} - \sum \Delta G°_f\text{(reactants)}

Similar to Hess's Law for enthalpy

Relationship with Equilibrium Constant:

\Delta G° = -RT\ln K = -2.303RT\log K

If K > 1: ΔG° < 0 (products favored)
If K < 1: ΔG° > 0 (reactants favored)
If K = 1: ΔG° = 0 (equal amounts)

Hess's Law of Constant Heat Summation

Hess's Law is one of the most important applications of the First Law. It's extensively used in JEE for calculating enthalpy changes of reactions that cannot be measured directly.

Statement of Hess's Law

"The total enthalpy change in a chemical reaction is the same whether it occurs in one step or several steps."

Why?

Because enthalpy (H) is a state function - it depends only on initial and final states, not on the path taken.

8.1 Applications of Hess's Law

1. Calculate Enthalpy of Formation

\Delta H°_{\text{reaction}} = \sum \Delta H°_f\text{(products)} - \sum \Delta H°_f\text{(reactants)}

2. Calculate Enthalpy from Bond Energies

\Delta H°_{\text{reaction}} = \sum \text{B.E.(bonds broken)} - \sum \text{B.E.(bonds formed)}

3. Born-Haber Cycle

Used to calculate lattice energy of ionic compounds using Hess's Law

📝 Solved Example 5 (JEE Advanced 2023 Type)

Question: Calculate the enthalpy of formation of methane (CH₄) given:

C(s) + O₂(g) → CO₂(g); ΔH₁ = -393.5 kJ

H₂(g) + ½O₂(g) → H₂O(l); ΔH₂ = -285.8 kJ

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l); ΔH₃ = -890.3 kJ

Solution:

Step 1: Write target equation

C(s) + 2H₂(g) → CH₄(g); ΔH_f = ?

Step 2: Manipulate given equations

Equation 1 (keep as is):

C(s) + O₂(g) → CO₂(g); ΔH₁ = -393.5 kJ

Equation 2 (multiply by 2):

2H₂(g) + O₂(g) → 2H₂O(l); 2ΔH₂ = -571.6 kJ

Equation 3 (reverse):

CO₂(g) + 2H₂O(l) → CH₄(g) + 2O₂(g); -ΔH₃ = +890.3 kJ

Step 3: Add all three equations

C(s) + O₂(g) → CO₂(g)

+ 2H₂(g) + O₂(g) → 2H₂O(l)

+ CO₂(g) + 2H₂O(l) → CH₄(g) + 2O₂(g)

C(s) + 2H₂(g) → CH₄(g)

Step 4: Calculate ΔH_f

\Delta H_f = \Delta H_1 + 2\Delta H_2 - \Delta H_3

\Delta H_f = -393.5 + (-571.6) - (-890.3)

\Delta H_f = -393.5 - 571.6 + 890.3

\Delta H_f = -74.8 \text{ kJ/mol}

💡 Tips for Hess's Law Problems

Reverse equation: Change sign of ΔH
Multiply equation by n: Multiply ΔH by n
Cancel common species: They should appear on opposite sides
Check final equation: Must match target equation exactly
Units: Always check if per mole or total

📝 Previous Year Questions Analysis

JEE Main (Last 5 Years)

✓ Enthalpy Calculations: 35%
✓ Hess's Law: 25%
✓ Gibbs Energy & Spontaneity: 20%
✓ Entropy: 15%
✓ Concepts & Definitions: 5%

JEE Advanced (Last 5 Years)

✓ Multi-concept problems: 40%
✓ Born-Haber Cycle: 25%
✓ ΔH-ΔU relationships: 20%
✓ Entropy calculations: 10%
✓ Thermodynamic cycles: 5%

Weightage Analysis

JEE Main: 8-12 marks (2-3 questions)

JEE Advanced: 12-16 marks (3-4 questions)

Difficulty Level: Medium to Hard

Time Required: 3-4 hours practice

Download Complete PYQ PDF (2015-2024)

🎯 Practice Problem Set

Level 1: Basic (JEE Main Standard)

Define: (a) Intensive property (b) State function (c) Isothermal process (d) Entropy
Calculate work done when 2 moles of ideal gas expands from 5L to 10L at 2 atm constant pressure
For reaction: N₂ + O₂ → 2NO, if ΔH = 180 kJ, find ΔU at 298K
Identify exothermic processes: (a) Melting (b) Combustion (c) Condensation (d) Dissolution of NaOH
Calculate ΔH when 1 mol ice at 0°C converts to water at 0°C. (ΔH_fus = 6.01 kJ/mol)
State the Second Law of Thermodynamics
Which has higher entropy: (a) H₂O(l) or H₂O(g)? (b) Diamond or graphite?
Calculate Cp if Cv = 20.8 J/mol·K for an ideal gas

Level 2: Intermediate (JEE Main/Advanced)

Calculate work done in isothermal reversible expansion of 1 mole ideal gas from 10L to 30L at 300K
For a reaction: ΔH = -100 kJ, ΔS = -200 J/K. Find temperature at which ΔG = 0
Given: C(s) + ½O₂ → CO (ΔH₁ = -110 kJ), CO + ½O₂ → CO₂ (ΔH₂ = -283 kJ). Find ΔH_f of CO₂
2 moles of gas at 300K heated to 400K at constant volume. If Cv = (3/2)R, find q, w, and ΔU
Calculate entropy change when 1 mole ice at 0°C melts. (ΔH_fus = 6.01 kJ/mol)
For PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), Kp = 2 at 500K. Calculate ΔG° for this reaction
Calculate bond energy of C-H if ΔH_f(CH₄) = -75 kJ, sublimation of C = 715 kJ, B.E.(H-H) = 436 kJ
A gas expands from 2L to 5L against constant pressure of 3 atm. Calculate work in Joules

Level 3: Advanced (JEE Advanced/Olympiad)

Derive the relationship between Cp and Cv for an ideal gas starting from first principles
Calculate lattice energy of NaCl using Born-Haber cycle given appropriate data
For adiabatic expansion of ideal gas, prove that TVᵞ⁻¹ = constant
Calculate maximum work obtainable when 1 mole ideal gas expands isothermally at 300K from 1 atm to 0.1 atm
Given resonance energy of benzene, calculate its heat of combustion
Two moles of gas undergo: (a) isothermal expansion (b) adiabatic expansion to same final volume. Compare work done
Calculate ΔG° at 373K if ΔH° = 40.66 kJ and ΔS° = 109 J/K for H₂O(l) → H₂O(g). Explain result
For Carnot engine between 500K and 300K, calculate efficiency and maximum work from 1000J heat

Get Solutions PDF with Step-by-Step Working

Year	JEE Main	JEE Advanced	Topic Focus
2024	2 Questions (8 marks)	3 Questions (12 marks)	ΔG-K relationship, Hess's Law
2023	3 Questions (12 marks)	4 Questions (14 marks)	Entropy, Born-Haber cycle
2022	2 Questions (8 marks)	3 Questions (11 marks)	ΔH-ΔU, Spontaneity
2021	3 Questions (12 marks)	3 Questions (10 marks)	Work calculations, Enthalpy

📑 Quick Navigation - Thermodynamics

Thermodynamics JEE Main & Advanced 2025-26

Basic Concepts and Terminology

1.1 System and Surroundings

System

Surroundings

1.2 Types of Systems

1.3 State Functions vs Path Functions

State Functions

Path Functions

1.4 Extensive and Intensive Properties

💡 Quick Memory Trick

1.5 Thermodynamic Processes

Isothermal Process

Adiabatic Process

Isobaric Process

Isochoric Process

⚠️ JEE Important Note

Zeroth Law of Thermodynamics

Statement of Zeroth Law

2.1 Concept of Temperature

Temperature

Thermal Equilibrium

💡 Why It's Called "Zeroth" Law?

2.2 Application: Thermometers

📝 Real-World Application

First Law of Thermodynamics

Statement of First Law

⚠️ Sign Convention (Very Important for JEE)

3.1 Internal Energy (U)

What is Internal Energy?

3.2 Work in Thermodynamics

Pressure-Volume Work

📝 Solved Example 1 (JEE Main 2023 Type)

3.3 Heat Capacity

Heat Capacity at Constant Volume (Cv)

Heat Capacity at Constant Pressure (Cp)

Mayer's Equation (Very Important for JEE)

📝 Solved Example 2 (JEE Advanced Pattern)

Enthalpy and Thermochemistry

4.1 Definition of Enthalpy

What is Enthalpy (H)?

4.2 Relationship between ΔH and ΔU

For Chemical Reactions

📝 Solved Example 3

4.3 Types of Enthalpy Changes

💡 Standard Enthalpy (ΔH°)

4.4 Exothermic vs Endothermic Reactions

Exothermic (ΔH < 0)

Endothermic (ΔH > 0)

Second Law of Thermodynamics

Statements of Second Law

⚠️ Key Implication

Entropy (S)

6.1 Definition and Calculation

Entropy Change (ΔS)

6.2 Entropy and Spontaneity

Total Entropy Change

💡 Entropy Trends (JEE Important)

Gibbs Free Energy (G)

7.1 Definition and Gibbs Equation

Gibbs Free Energy

7.2 Spontaneity Criteria

7.3 Effect of Temperature on Spontaneity

📝 Solved Example 4 (JEE Main 2024 Type)

7.4 Standard Gibbs Free Energy

Calculating ΔG° for Reactions

Hess's Law of Constant Heat Summation

Statement of Hess's Law

8.1 Applications of Hess's Law

1. Calculate Enthalpy of Formation

2. Calculate Enthalpy from Bond Energies

3. Born-Haber Cycle

📝 Solved Example 5 (JEE Advanced 2023 Type)

💡 Tips for Hess's Law Problems

📝 Previous Year Questions Analysis

JEE Main (Last 5 Years)

JEE Advanced (Last 5 Years)

Top 12 Most Repeated Question Types

Weightage Analysis

Heat Capacity at Constant Volume (C_v)

Heat Capacity at Constant Pressure (C_p)