Thermodynamics NEET JEE Chemistry Notes cover image with heat diagrams, molecules, and bold chapter title in a science-themed design
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Thermodynamics Class 11 Notes for NEET & JEE | Easy To Understand

We Need To Cover These Topics-

S. No.TopicSubtopics
1.Basic Concepts of Thermodynamics– System and Surroundings
– Types of Systems (Open, Closed, Isolated)
– State of a System
– Properties of a System (Extensive, Intensive)
– Thermodynamic Variables
2.Types of Processes– Isothermal
– Adiabatic
– Isobaric
– Isochoric
– Cyclic and Reversible/Irreversible Processes
3.Internal Energy (U)– Concept of Internal Energy
– Change in Internal Energy (ΔU)
– First Law of Thermodynamics (ΔU = q + w)
4.Work (w)– Work Done in Expansion/Compression of Gases (w = -PΔV)
– Work Done in Reversible Isothermal Expansion
5.Heat (q)– Sign Conventions for Heat and Work
– Heat Transfer in Various Processes
6.First Law of Thermodynamics– Mathematical Expression
– Applications
– Limitation
7.Enthalpy (H)– Definition: H = U + PV
– Change in Enthalpy (ΔH)
– Enthalpy of Reaction, Formation, Combustion, Fusion, etc.
8.Heat Capacity– Specific Heat Capacity
– Molar Heat Capacity
– Relation Between Cp and Cv (Cp – Cv = R)
9.Measurement of ΔU and ΔH– Calorimetry
– Bomb Calorimeter
– Relation Between ΔH and ΔU: ΔH = ΔU + ΔnRT
10.Hess’s Law of Constant Heat Summation– Statement and Application
– Solving Enthalpy Problems Using Hess’s Law
11.Enthalpies of Different Reactions– Enthalpy of Formation
– Enthalpy of Combustion
– Enthalpy of Atomization
– Enthalpy of Sublimation
– Enthalpy of Fusion & Vaporization
– Enthalpy of Neutralization
12.Spontaneity and Second Law of Thermodynamics– Spontaneous and Non-Spontaneous Processes
– Entropy (S): Concept and Units
– ΔS for System and Surroundings
13.Gibbs Free Energy (G)– G = H – TS
– ΔG = ΔH – TΔS
– Criteria for Spontaneity (ΔG < 0)
– Temperature Dependence of Spontaneity
14.Third Law of Thermodynamics– Statement
– Applications in Entropy Calculations
15.Important Thermodynamic Equations– ΔH = ΔU + ΔnRT
– q = m × C × ΔT
– ΔG = ΔH – TΔS
– Cp – Cv = R
– Relation between w, q, and ΔU

1. Introduction – Basic Concepts of Thermodynamics

🌟 Introduction to Thermodynamics

Thermodynamics is the branch of science that deals with the study of energy changes—especially heat and work—during physical and chemical processes.

It helps us answer:

  • How much heat is absorbed or evolved?
  • Will a chemical reaction occur on its own?
  • How much energy is usable or lost?

Thermodynamics does not depend on the microscopic behavior of molecules. It uses macroscopic observations like temperature, pressure, volume, etc.


🔹 1. System and Surroundings

🔸 System

A system is the specific part of the universe that we are studying or observing.
It can be a reaction vessel, gas in a cylinder, or water in a beaker.

Example:
If you’re studying the reaction between hydrogen and oxygen in a container, the container and its contents are the system.

🔸 Surroundings

Everything outside the system that can exchange energy (heat or work) with the system is called the surroundings.

Example:
Air around the beaker or lab environment.

🔸 Boundary

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

Boundaries can be:

  • Real (e.g., walls of a container)
  • Imaginary (e.g., for mathematical analysis)

🔹 2. Types of Systems

Systems are classified based on how they exchange energy and/or matter with the surroundings.

System TypeMatter ExchangeEnergy Exchange (Heat/Work)Example
Open SystemYesYesOpen cup of hot tea
Closed SystemNoYesGas in a piston (sealed)
Isolated SystemNoNoThermos flask (ideally)

📌 Summary:

  • Open System: Exchange of both energy & matter
  • Closed System: Exchange of energy only
  • Isolated System: No exchange of energy or matter

🔹 3. State of a System

A thermodynamic state is defined by the values of state variables like pressure (P), volume (V), temperature (T), and composition.

🔸 State Function:

These are properties that depend only on the initial and final states, not on the path taken.

🧪 Examples:

  • Internal energy (U)
  • Enthalpy (H)
  • Entropy (S)
  • Pressure (P)
  • Volume (V)
  • Temperature (T)

🔹 4. Properties of a System

Thermodynamic properties describe the system and can be of two types:

(A) Extensive Properties

  • Depend on the quantity of matter present.
  • They add up when you combine two systems.

🧪 Examples:

  • Mass
  • Volume
  • Internal Energy (U)
  • Enthalpy (H)

✅ If you double the mass, these also double.


(B) Intensive Properties

  • Do not depend on the quantity of matter.
  • They remain the same, even if the system size changes.

🧪 Examples:

  • Temperature
  • Pressure
  • Density
  • Refractive Index

✅ If you take half a glass of water, temperature remains the same.


🧠 Summary Table:

Property TypeDepends on Quantity?Additive?Examples
ExtensiveYesYesMass, Volume, Energy
IntensiveNoNoTemperature, Pressure, Density

🔹 5. Thermodynamic Variables (State Variables)

Thermodynamic variables define the state of a system. Common variables include:

VariableSymbolMeaningUnits
PressurePForce per unit areaatm, Pa, bar
VolumeVSpace occupied by the systemm³, L, cm³
TemperatureTDegree of hotnessK (Kelvin), °C
Internal EnergyUTotal energy inside systemJoules
EnthalpyHHeat at constant pressureJoules
EntropySMeasure of disorderJ/K

🔸 Explanation of Key Thermodynamic Variables

📌 Pressure (P)

  • Caused by the collision of gas molecules with the container walls.
  • Measured in atm, bar, pascal.

📌 Volume (V)

  • The amount of space a system occupies.
  • Measured in litres or cubic metres (m³).

📌 Temperature (T)

  • Average kinetic energy of molecules.
  • Measured in Kelvin for thermodynamic calculations.

📌 Internal Energy (U)

  • Total energy of all molecules: translational, rotational, vibrational.
  • Cannot be measured directly; only change (ΔU) is measurable.

📌 Enthalpy (H)

  • Useful for constant pressure processes:
    H=U+PV

📌 Entropy (S)

  • Degree of randomness or disorder.
  • More randomness = higher entropy.

🔹 Thermodynamic Equilibrium

A system is in thermodynamic equilibrium when:

  • No net flow of energy or matter occurs
  • The system is at mechanical, thermal, and chemical equilibrium

✅ Conditions:

  1. Thermal Equilibrium: Temperature is uniform
  2. Mechanical Equilibrium: No unbalanced forces
  3. Chemical Equilibrium: No net chemical change

🔹 Types of Processes in Thermodynamics

Though not part of this section directly, understanding types of processes is important.

ProcessConditionExample
IsothermalTemperature constant (T)Ideal gas expansion at constant T
AdiabaticNo heat exchange (q = 0)Rapid gas expansion in piston
IsobaricPressure constant (P)Heating gas in piston
IsochoricVolume constant (V)Heating in rigid container
ReversibleInfinitely slow processIdeal for max work output
IrreversibleReal, fast processesExplosion, sudden expansion

🧠 Important Thermodynamic Terms to Remember

TermDefinition
SystemPart of universe under study
SurroundingsEverything external to system
BoundaryInterface separating system and surroundings
State FunctionDepends only on initial and final state (e.g., ΔU, ΔH)
Path FunctionDepends on path taken (e.g., work, heat)
EquilibriumNo net change occurs in the system
Spontaneous ProcessOccurs naturally without external help
Non-Spontaneous ProcessRequires external energy to occur

🔍 Common Confusions Cleared

ConfusionClarification
Enthalpy vs Internal EnergyH=U+PVH = U + PVH=U+PV. Use H for constant pressure, U for general energy changes.
Open vs Closed vs IsolatedClosed = no matter exchange, Isolated = no exchange at all.
Extensive vs IntensiveUse “E” for extensive (mass, volume), “I” for intensive (pressure, temperature).
State vs Path FunctionsState: independent of process (ΔU), Path: depends on how you go (work, heat)

🧪 Application in Daily Life & Industry

  • Refrigerators use principles of thermodynamics to move heat out.
  • Engines convert chemical energy to mechanical work.
  • Biological systems like metabolism follow energy conservation laws.
  • Power plants use thermodynamic cycles to generate electricity.

📚 NEET & JEE Focused Practice Concepts

  • Identify system and surroundings in reactions.
  • Differentiate between open, closed, and isolated systems.
  • Classify properties as intensive or extensive.
  • Solve basic numerical on pressure, volume, temperature relations.
  • Understand state vs path functions through examples.

💡 Tips to Master This Topic

  1. Visualize real-life systems: Thermos = isolated, teacup = open.
  2. Remember SI units: Pressure (Pa), Temperature (K), Energy (J).
  3. Practice categorization: Intensive vs Extensive, System types.
  4. Relate to math: Use algebraic expressions for state variables.
  5. Revise definitions frequently to avoid confusion in exams.

✅ Final Revision Table

ConceptKey Points
System & SurroundingsSystem = what we study; surroundings = rest of the universe
Types of SystemsOpen (matter + energy), Closed (only energy), Isolated (none)
Properties of SystemIntensive = independent of mass; Extensive = dependent on mass
Thermodynamic VariablesP, V, T, U, H, S – define system state
State FunctionsDepend only on initial/final state (U, H, S)
Path FunctionsDepend on route/process taken (q, w)

🏁 Conclusion

Thermodynamics starts with understanding systems, their properties, and how they interact with surroundings. Mastering these basic concepts helps you lay the foundation for tougher topics like heat capacity, enthalpy changes, entropy, and spontaneity.

This topic is a guaranteed scorer in NEET & JEE, especially in conceptual and formula-based questions.
With clarity of system boundaries, variables, and property classification, you’ll be equipped to tackle even advanced problems with ease.


2. Thermodynamics – Types of Processes

📚 What is a Thermodynamic Process ?

A thermodynamic process refers to the change that takes place in a system when its state variables (pressure, volume, temperature, etc.) are altered.

In other words, when a system moves from one state to another, it undergoes a thermodynamic process.


🔹 1. Isothermal Process (Temperature Constant)

🔸 Definition:

A process in which the temperature (T) remains constant throughout the change is called an isothermal process.

🔁 T = constant,
ΔT = 0

Since temperature is constant, the internal energy (ΔU) of the system also remains constant for an ideal gas.

🔸 Equation used:

From the ideal gas law:
PV=nRT

Since T is constant:
P ∝ 1/V ​(Boyle’s Law)

🔸 Work done (w) in Isothermal process:

For reversible isothermal expansion/compression of an ideal gas: W = nRT ln(V2/V1​ ​​)

  • V1​ = initial volume
  • V1​ = final volume
  • R = universal gas constant
  • T = temperature

🔸 Graph (P-V curve):

  • A hyperbolic curve (because P ∝ 1/V)
  • Known as an isotherm

🔸 Real-life Examples:

  • Melting of ice at 0°C (temperature remains constant)
  • Evaporation of water at boiling point

🔹 2. Adiabatic Process (No Heat Exchange)

🔸 Definition:

A process in which no heat is exchanged between the system and surroundings is called an adiabatic process.

🔁 q = 0

All the energy changes occur in the form of work done.

🔸 Equation:

For an ideal gas undergoing adiabatic reversible process: PVγ=constantPV^\gamma = \text{constant}PVγ=constant

Where:

  • γ = CP/ CV (Ratio of heat capacities)

Also: TVγ−1 = constant and TγP1−γ = constant

🔸 Work done in adiabatic process:

W = P2​V2​− P1​V1​/ γ−1 ​or w= nR (T1​−T2​)​ / γ−1

🔸 Graph (P-V curve):

  • Steeper than isothermal curve

🔸 Real-life Examples:

  • Sudden compression/expansion of gas
  • Release of air from a tire

🔹 3. Isobaric Process (Pressure Constant)

🔸 Definition:

A process in which the pressure remains constant is called an isobaric process.

🔁 P = constant

In this case, the volume and temperature may change.

🔸 Equation:

From ideal gas law:
V ∝ T (at constant pressure) (Charles’s Law)

🔸 Work done:

W = P (V2−V1​)

Here,

  • V2​ = final volume
  • V1​ = initial volume
  • P = constant pressure

🔸 Enthalpy Change (ΔH):

Since P is constant: qP​=ΔH

🔸 Graph:

  • Straight horizontal line in P-V graph

🔸 Real-life Examples:

  • Boiling of water in open vessel
  • Cooking food in open pan

🔹 4. Isochoric Process (Volume Constant)

🔸 Definition:

A process in which volume remains constant is called an isochoric process.

🔁 V = constant

Since volume doesn’t change, no work is done by the system: w = 0

🔸 Enthalpy and Internal Energy:

  • Heat given is used to increase internal energy:

qV​=ΔU

🔸 Graph:

  • Vertical line in a P-V diagram (since volume is fixed)

🔸 Real-life Examples:

  • Heating a gas in a rigid container
  • Gasoline explosion in closed engine chamber

🔹 5. Cyclic Process

🔸 Definition:

A process in which the system returns to its original state after undergoing a series of changes is called a cyclic process.

🔁 Final state = Initial state

So, ΔU = 0, since internal energy is a state function.

🔸 Energy Relations:

  • Since ΔU = 0:

q=−w

That means:
Net heat supplied = Net work done by the system

🔸 Graph:

Forms a closed loop on a PV diagram.

🔸 Real-life Examples:

  • Working of heat engines (Carnot cycle, Otto cycle)
  • Refrigeration cycles

🔹 6. Reversible Process

🔸 Definition:

A process that occurs infinitely slowly, such that the system is always in equilibrium with its surroundings, is called a reversible process.

✅ Can be reversed by an infinitesimal change
Ideal case (not practically possible)

🔸 Features:

  • Infinitely slow
  • Maximum work is obtained
  • System remains close to equilibrium at all steps
  • Path is well defined

🔸 Graph:

  • Smooth, defined curve in PV diagram

🔸 Real-life Examples:

  • Theoretical Carnot cycle
  • Ideal gas expansion in controlled conditions

🔹 7. Irreversible Process

🔸 Definition:

A process that cannot be reversed exactly to regain the initial state is called an irreversible process.

❌ Happens quickly
❌ System not in equilibrium

🔸 Features:

  • Occurs spontaneously
  • Produces less work
  • Generates entropy
  • Sudden changes

🔸 Real-life Examples:

  • Combustion
  • Mixing of gases
  • Free expansion of gas into vacuum

📌 Comparison Table – Types of Thermodynamic Processes

ProcessConditionHeat (q)Work (w)Example
IsothermalT = constantq ≠ 0nRT ln(v2/V1​ ​​)Slow gas expansion at room temp
Adiabaticq = 0q = 0Depends on ΔTAir compression in cylinder
IsobaricP = constantq = ΔHP(V₂ – V₁)Water boiling in open vessel
IsochoricV = constantq = ΔUw = 0Gas heating in rigid container
CyclicFinal state = Initialq = -wNet area under PVCarnot or Otto engine
ReversibleInfinitely slowMax q/wMax work outputIdeal piston process (theoretical)
IrreversibleQuick, realq lostLess workExplosion, rapid gas flow

🧠 Key Formulas to Remember

  1. Isothermal Process (Ideal Gas): W = nRT ln(v2/V1​ ​​)
  2. Adiabatic Process (Ideal Gas): PVγ = constant
  3. Isobaric Process: W = P (V2−V1​), qP​ = ΔH
  4. Isochoric Process: w = 0, qV​=ΔU
  5. Cyclic Process: ΔU=0⇒q=−w

🏁 Final Summary – Fast Revision

ConceptSymbol/ConditionHeat Exchange (q)Work Done (w)Internal Energy Change (ΔU)
IsothermalT constantYesYes, nRT ln(v2/V1​ ​​)ΔU = 0
Adiabaticq = 0NoYesΔU = -w
IsobaricP constantYesYes, P (V2−V1​)ΔU = q – w
IsochoricV constantYesNo (w = 0)ΔU = q
CyclicInitial = Final StateYesYesΔU = 0
ReversibleIdealized slow processYesMaximum possibleDepends on type
IrreversibleSpontaneous, real processYesLess than reversibleDepends on type

💡 NEET/JEE Tips:

  • NEET: Focus on basic definitions, PV graphs, and conceptual clarity.
  • JEE: Practice numericals involving work done, heat exchanged, and internal energy change. Also, expect theory + numerical mix questions.

✅ Conclusion

Understanding different thermodynamic processes is the foundation for mastering energy changes in chemical reactions, engines, biological systems, and industrial setups. These concepts are not only scoring but also help in building logical thinking.

If you revise regularly, draw graphs, and practice numericals, you’ll never forget the core idea of how systems behave under different conditions of pressure, temperature, heat, and volume.


3. Thermodynamics – Internal Energy (U)

🔷 Introduction

Thermodynamics explains how energy is transferred or transformed in physical and chemical changes. A key concept in thermodynamics is Internal Energy (U) — the total energy of a system.

When energy is added or removed (as heat or work), the system’s internal energy changes. This change is governed by the First Law of Thermodynamics — a fundamental law of nature.


🔹 1. What is Internal Energy (U)?

🔸 Definition:

Internal Energy of a system is the total energy possessed by all the molecules within the system due to their:

  • Translational motion (movement)
  • Rotational motion
  • Vibrational motion
  • Intermolecular forces
  • Potential energy (due to position)
  • Electronic energy (in atoms or molecules)

🧠 Internal energy is a state function: It depends only on the state (P, V, T) of the system — not on how the system reached that state.


🔹 2. Key Characteristics of Internal Energy

FeatureDescription
SymbolU
NatureScalar quantity
UnitsJoule (J), calorie (1 cal = 4.184 J)
SI UnitJoule (J)
State Function?Yes – depends only on initial and final states
Extensive or Intensive?Extensive – depends on the amount of substance
Measurable?Absolute value cannot be measured, only change (ΔU) is measurable

🔹 3. Components of Internal Energy

Let’s break down what contributes to internal energy:

Type of EnergyExplanation
Translational EnergyDue to straight-line motion of molecules
Rotational EnergyDue to spinning/rotation of molecules
Vibrational EnergyDue to vibration of atoms within molecules
Potential EnergyDue to intermolecular forces
Electronic EnergyDue to electronic transitions or configurations
Chemical EnergyStored in bonds; released/absorbed during reactions

⚠️ In thermodynamics, we don’t split U into these parts during calculations. We treat U as a total quantity.


🔹 4. Change in Internal Energy (ΔU)

🔸 Definition:

When a system absorbs or loses energy, its internal energy changes. The change in internal energy is denoted by ΔU.

ΔU = Ufinal​ − Uinitial​

  • If ΔU > 0: Internal energy increases (system gains energy)
  • If ΔU < 0: Internal energy decreases (system loses energy)

🔸 Energy Transfer Modes:

A system’s internal energy can change by:

  1. Heat (q) – energy flow due to temperature difference
  2. Work (w) – mechanical energy done by or on the system

🔹 5. First Law of Thermodynamics

🔸 Statement:

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

In thermodynamics, this is expressed as: ΔU = q+w

Where:

  • ΔU = change in internal energy
  • q = heat added to the system
  • w = work done on the system

🔸 Sign Convention (Very Important for JEE/NEET)

QuantityPositive (+)Negative (−)
Heat (q)Heat added to the systemHeat released by the system
Work (w)Work done on the system (e.g., compression)Work done by the system (e.g., expansion)
ΔUIncrease in internal energyDecrease in internal energy

Use this convention for NEET/JEE numericals.


🔹 6. Heat, Work & Internal Energy: Conceptual View

🧪 Think of a balloon:

  • When you heat the balloon (q > 0), molecules move faster → ΔU increases.
  • If the balloon expands, it does work on surroundings (w < 0) → energy is lost as work.

So, ΔU = q + w includes both heat & mechanical energy effects.


🔹 7. Mathematical Derivation

For infinitesimal (small) changes: dU = dq+dw

At constant volume:

  • Volume is fixed → no work (w = 0)
  • So, dqV = dU ⇒ Heat at constant volume changes only internal energy

At constant pressure:

  • Some heat used to do expansion work
  • Remaining changes internal energy, qP​ = ΔH = ΔU+PΔV ⇒ ΔU = ΔH−PΔV

🔹 8. Applications of First Law

ApplicationExplanation
Heat EnginesConvert heat → work using internal energy changes
Refrigerators/ACsUse work input to extract heat (reverse of natural process)
Human MetabolismFood (chemical energy) → work + heat
Explosions or CombustionEnergy released rapidly, doing work and increasing T, P
Thermochemical EquationsUse ΔU and ΔH to track energy flow in chemical reactions

🔹 9. Internal Energy in Ideal Gas Systems

For 1 mole of ideal monoatomic gas: U = 3/2​ RT ⇒ ΔU = 3/2​R (T2​−T1​)

For n moles: ΔU = 3/2 ​nR (T2​−T1​)

📌 Valid only for monoatomic gases. For diatomic gases, use:

ΔU = 5/2 nRΔT (for diatomic gases at moderate temperatures)


🔹 10. Experimental Measurement of ΔU

🔸 Calorimetry:

Use a bomb calorimeter at constant volume:

  • Substance is burned inside a steel vessel (bomb)
  • System is isolated: q = ΔU

Used to find calorific value of fuels.


🔹 11. Real-Life Examples

ScenarioΔU Explanation
Heating waterq > 0 → ΔU increases
Burning petrol in engineFuel energy → work + heat → ΔU decreases
Air compression in pistonWork done on gas → molecules vibrate more → ΔU increases
Cooling of teaq < 0 (heat loss) → ΔU decreases

🔹 12. Graphical Representation

🔸 P-V Diagram for Constant Volume (w = 0):

Pressure
|
| ___________
| |
| |
|__________|___________________ Volume (constant)

Only heat changes internal energy.


🔹 13. Difference: Heat vs Work vs Internal Energy

ParameterHeat (q)Work (w)Internal Energy (U)
What it isEnergy transferred due to ΔTEnergy used to move boundaryTotal microscopic energy in system
Path/State?Path functionPath functionState function
SI UnitJouleJouleJoule
Sign Conventions+ if absorbed, − if released+ if on system, − if by system+ = gain, − = loss

🔹 14. NEET/JEE Exam Focus

TopicAsked InType of Question
ΔU = q + wNEET + JEEConceptual + numerical (calculation of ΔU, q or w)
CalorimetryNEETBased on experimental ΔU measurement
Heat/Work sign conventionsNEET + JEEAssertion-Reason, MCQ, theory
Monoatomic/Diatomic gas ΔUJEE (Adv)Numeric problems with temperature change

🧠 Tricks to Remember

  • “U = Total Hidden Energy” inside system
  • “ΔU = Money in Wallet” – increases by incoming cash (q), decreases by spending (w)
  • Internal energy depends only on T for ideal gases
  • Use correct signs for q and w — very important for numericals

🏁 Summary Table

ConceptKey Point
Internal Energy (U)Total energy inside the system (microscopic level)
Change in Internal EnergyΔU = Ufinal − Uinitial
First Law of ThermodynamicsΔU = q + w
Sign of q+ if heat absorbed, − if heat released
Sign of w+ if work done on system, − if work done by system
Ideal Gas RelationΔU = 3/2nRΔT (monoatomic)
State or Path FunctionInternal energy = State function
Work at Constant Volumew = 0, hence q = ΔU

✅ Conclusion

Internal Energy is a core concept that links heat, work, and energy conservation. Every chemical reaction or physical process involves changes in internal energy, making it central to thermodynamics.

The First Law of Thermodynamics empowers us to quantify and predict energy changes in any system — whether it’s a test tube or a car engine.

Mastering this topic gives you a strong base for solving complex problems and understanding real-world phenomena in chemistry, biology, and physics.


4. Thermodynamics – Work (w)

🔷 1. Introduction: What is Work in Thermodynamics?

In thermodynamics, work (w) is a form of energy transfer between a system and its surroundings due to volume change under an external pressure.

🧠 Think: When a gas expands, it pushes its surroundings (like a piston), and does work.

Work is not stored in a system; it occurs when the state changes.


🔷 2. Thermodynamic Definition of Work

📘 Work done by a gas at constant pressure:

w= −Pext​ΔV

Where:

  • w = Work done by the system (Joules)
  • Pext​ = External pressure (atm or Pa)
  • ΔV = V2−V1 = Change in volume

📌 Negative sign means:

  • If gas expands, system does positive work, but w<0
  • If gas compresses, surroundings do work on system, w>0

🔷 3. Sign Conventions in Thermodynamics

ConditionSign of Work (w)
Gas expands (ΔV>0)w<0: Work by system
Gas compresses (ΔV<0)w>0: Work on system

Also remember:

  • Heat added → q>0
  • Heat released → q<0

🔷 4. Units of Work

UnitSymbolConversion
JouleJSI Unit
L·atm1 L·atm = 101.3 J
Caloriecal1 cal = 4.184 J

🔷 5. Types of Work in Thermodynamics

  • Pressure-volume work (P-V work): Most common in gas reactions
  • Electrical work: In electrochemical cells
  • Non-mechanical work: Like stretching membranes, etc.

In NEET/JEE, focus is mostly on P-V work during expansion or compression of gases.


🔷 6. Work Done in Expansion/Compression of Gases

Let’s understand how a gas does work when its volume changes.

📘 Formula:

w= −Pext(V2−V1) = −Pext ΔV

Where:

  • V1​: Initial volume
  • V2​: Final volume

🔹 Conditions:

  • External pressure is constant
  • Process is irreversible
  • Usually quick change

🔷 7. Graphical View of Expansion/Compression

If volume is on x-axis and pressure is on y-axis, then area under curve = work.

  • Irreversible process: straight horizontal line (constant pressure)
  • Reversible process: curved line (pressure changes gradually)

🔷 8. Work Done in Reversible Isothermal Expansion

This is a very important derivation for JEE and NEET.

🔹 Conditions:

  • Reversible: The system is in equilibrium at every step
  • Isothermal: Constant temperature
  • Ideal Gas behavior

✅ Final Formula:

w= −nrt ln (V2/V1​​​)

Where:

  • n = number of moles
  • R = gas constant (8.314 J/mol·K)
  • T = absolute temperature in Kelvin
  • V1​, V2​ = initial and final volumes

🔷 9. Key Points about Reversible Isothermal Work

  • Most efficient process: gives maximum work
  • Work is path-dependent
  • Applicable only to ideal gases under reversible and isothermal conditions
  • If V2>V1​, then w<0 → gas does work (expands)
  • If V2<V1​, then w>0 → work done on gas (compressed)

🔷 10. Comparison: Reversible vs Irreversible Work

PropertyReversible ProcessIrreversible Process
PressureChanges infinitesimallyConstant external pressure
Work DoneMaximum (more negative)Less than reversible
Equation−nRT ln(V2​/V1​)−Pext ΔV
Real/IdealTheoretical/idealRealistic
EfficiencyHighLow

🔷 11. Relation with Heat (q) and Internal Energy (ΔU)

From First Law of Thermodynamics: ΔU = q+w

For Isothermal Process:

  • ΔU=0\Delta U = 0ΔU=0 (T constant)

q = −w

For Adiabatic Process:

  • q=0

ΔU = w

For Isochoric Process:

  • w=0

q = ΔU


🔷 12. Real-Life Examples of Thermodynamic Work

ExampleDescription
Piston in a car engineGas expansion does work
Inflating a balloonGas does work against atmosphere
Compressing a gas in cylinderWork done on the gas
BreathingLungs expand/contract: work done

🔷 13. Common Mistakes to Avoid

❌ Using wrong sign for work
✅ Remember: Expansion → w < 0, Compression → w > 0

❌ Using Celsius in gas law problems
✅ Always use Kelvin in calculations

❌ Confusing irreversible and reversible formulas
✅ Reversible: −nRTln(V2​/V1)
✅ Irreversible: −Pext​ΔV


🔷 14. Summary Table

Process TypeFormulaNotes
Isothermal (rev)−nRTln(V2​/V1)Max work, ΔU = 0
Irreversible (constant P)−Pext​ΔVFast process, less work
Adiabatic (rev)Complex, involves γNo heat exchange (q = 0)
Isochoricw=0Volume fixed

✅ Final Conclusion

Thermodynamic work (w) is central to understanding how systems transfer energy. Whether in engines, respiration, or laboratory experiments, mastering how work is done in gas expansion/compression is essential for NEET and JEE.

Practice numericals for perfection!

Use correct formulas:

Irreversible: w= −Pext​ΔV

Reversible Isothermal: w= −nRTln(V2​/V1​)

Pay attention to signs, units, and physical meaning


5. Thermodynamics – Heat (q)

🔷 1. Introduction to Heat in Thermodynamics

Heat (q) is a form of energy transfer that occurs due to a temperature difference between the system and its surroundings. It is not contained in the system — it’s energy in transit.

Heat always flows from hotter body to colder body.


🔷 2. Definition of Heat (q)

Heat is the energy that flows into or out of a system due to a temperature difference.

  • Represented by: q
  • It is not a state function
  • It is a path function (depends on how change happens)
  • Units: Joule (J) or calorie (1 cal = 4.184 J)

🔷 3. Heat vs Temperature

PropertyHeat (q)Temperature (T)
NatureEnergy in transitMeasure of average kinetic energy
SI UnitJoule (J)Kelvin (K)
Path/StatePath functionState function
Transfer ?Yes, from high to low TNo, it’s a property

🔷 4. Sign Conventions for Heat and Work

Understanding signs is very important in thermodynamics.

ProcessHeat (q)Work (w)
Heat added to systemq > 0
Heat released by systemq < 0
Work done by system (expansion)w < 0
Work done on system (compression)w > 0

📌 Always remember:

  • Positive q → system gains heat
  • Negative q → system loses heat
  • Positive w → work done on system
  • Negative w → work done by system

🔷 5. First Law of Thermodynamics (Recap)

ΔU = q+w

Where:

  • ΔU = change in internal energy
  • q = heat supplied
  • w = work done on system

This equation relates heat, work, and internal energy of a system.


🔷 6. Modes of Heat Transfer

There are 3 main modes of heat transfer in thermodynamics:

ModeDescriptionExample
ConductionHeat transfer through solids by particle vibrationMetal rod heated at one end
ConvectionTransfer in fluids by mass movement of particlesBoiling water
RadiationTransfer through electromagnetic wavesHeat from the Sun

In thermodynamics, we mainly focus on net heat energy transferred, not how it travels.


🔷 7. Heat Transfer in Different Thermodynamic Processes

Let’s understand how heat behaves in various processes:


🔹 (A) Isothermal Process (T = constant)

  • Since temperature is constant, internal energy (ΔU) = 0
  • Hence, from first law:

q = −w

  • All heat supplied is converted into work done by system
  • Example: Ideal gas expanding at constant temperature

🔹 (B) Adiabatic Process (q = 0)

  • No heat exchange with surroundings: q = 0
  • From first law:

ΔU = w

  • Any work done changes internal energy directly
  • Example: Sudden compression or expansion of gas

🔹 (C) Isochoric Process (ΔV = 0 → w = 0)

  • No change in volume ⇒ No work done
  • So:

q = ΔU

  • All heat supplied increases internal energy
  • Example: Heating a gas in a rigid container

🔹 (D) Isobaric Process (P = constant)

  • Work is done (w = PΔV)
  • Heat supplied goes into:
    • Changing internal energy
    • Doing work

q = ΔU + PΔV = ΔH

  • Hence, at constant pressure, heat = change in enthalpy (ΔH)

🔹 (E) Cyclic Process (System returns to original state)

  • ΔU=0
  • Hence:

q= −w

  • Heat added = Work done by system

🔷 8. Heat Capacity and Specific Heat

🔹 (A) Heat Capacity (C)

  • Amount of heat required to raise the temperature of a substance by 1°C or 1 K
  • Unit: J/K
  • Formula: q = CΔT

🔹 (B) Specific Heat Capacity (c)

  • Heat required to raise temperature of 1 gram of substance by 1°C

q = mcΔT

Where:

  • m = mass
  • c = specific heat
  • ΔT = change in temperature

Very useful in calorimetry-based NEET/JEE problems


🔷 9. Heat Transfer Examples and Formulas

Process TypeHeat Transfer EquationRemarks
Isothermalq= −wΔU = 0
Adiabaticq= 0ΔU = w
Isochoricq= ΔUw = 0
Isobaricq= ΔU+PΔV= ΔHWork + ΔU

🔷 10. Calorimetry – Measuring Heat

🔹 Principle of Calorimetry:

Heat lost by hot body = Heat gained by cold body

m1​c1​ (T1​−Tf​) = m2​c2​ (Tf​−T2​)

Where:

  • m = mass
  • c = specific heat
  • T = temperature

Used to calculate:

  • Specific heat
  • Final temperature
  • Heat exchanged

🔷 11. Constant Pressure vs Constant Volume Heat

PropertyConstant Volume (qv)Constant Pressure (qp)
Relationqv=ΔUqP=ΔH
UseBomb calorimeterCoffee cup calorimeter
Work done?No (ΔV = 0)Yes (w = PΔV)
Heat transferOnly internal energy changesEnthalpy and work both involved

🔷 12. Heat in Phase Change

  • During phase changes (melting, boiling), temperature remains constant
  • Heat absorbed or released is called latent heat

q=mL

Where:

  • m = mass
  • L = latent heat of transformation

🔷 13. Real-life Examples of Heat Transfer

SituationType of Heat Transfer
Boiling waterConvection
Iron rod in fireConduction
Sunlight warming earthRadiation
Cooking in a pressure cookerIsobaric + Conduction
Explosion in sealed containerIsochoric + Adiabatic

🔷 15. Graphical Understanding

🔹 Heat vs Temperature Graph (for phase change)

Temp ↑
|
| /\
| / \____ (plateau)
| / \
|_____/ \________
Heat added →
  • Plateaus: phase changes, T = constant, q ≠ 0
  • Slopes: q = mcΔT

🔷 16. Summary Tables

A. Heat in Common Processes

ProcessTVqwΔU
IsothermalConst.V↑/↓≠0≠00
AdiabaticT↑/↓V↑/↓0≠0≠0
IsochoricT↑/↓Const.= ΔU0≠0
IsobaricT↑/↓V↑/↓= ΔH≠0≠0

B. Sign Convention Recap

Heat/WorkSystem GainsSystem Loses
Heat (q)+
Work (w)– (expansion)+ (compression)

✅ Final Takeaways

Be thorough with NEET/JEE sign conventions, formulae & units

Heat (q) is energy in transit — always due to temperature difference

It is a path function and needs sign attention

Every thermodynamic process has unique q behavior

Use first law: ΔU = q + w to link heat, work & internal energy

Learn heat transfer through calorimetry, phase change, and specific heat


6. First Law of Thermodynamics

🔷 What is the First Law of Thermodynamics ?

The First Law of Thermodynamics is a law of energy conservation.

👉 It says:

“Energy can neither be created nor destroyed, only changed from one form to another.”

This means the total energy of a system always remains the same. If heat is given to a system, it either increases its internal energy or helps it do work.


🔶 Mathematical Formula

The First Law is written as: ΔU= q+w

Where:

  • ΔU = change in internal energy of the system
  • q = heat added to the system
  • w = work done on the system

🔹 Sign Conventions (Very Important)

QuantityPositive (+)Negative (–)
q (heat)Heat is given to the systemHeat is removed from the system
w (work)Work done on the system (compression)Work done by the system (expansion)

🔷 What is a System and Surroundings ?

  • System: The part we are studying (e.g., gas in a cylinder).
  • Surroundings: Everything outside the system.

Energy flows between the system and surroundings as heat or work.


🔷 Understanding Each Term

🔹 Internal Energy (U)

  • It is the total energy of all molecules inside the system (kinetic + potential).
  • It depends only on temperature.
  • It is a state function (depends only on start and end points, not the path).

🔹 Heat (q)

  • Heat is energy transferred due to temperature difference.
  • It is a path function (depends on how the change happens).
  • Heat flows from hot to cold.

🔹 Work (w)

  • Work is done when the system expands or gets compressed.
  • Expansion = system does work = negative
  • Compression = work is done on system = positive

🔷 Expansion/Compression Work

Let’s say gas is inside a piston.

  • If gas expands: w= −PΔV (Volume increases, system does work)
  • If gas is compressed: w= +PΔV (Volume decreases, work is done on system)

🔷 First Law in Different Processes

1️⃣ Isothermal Process (Constant Temperature)

  • ΔU = 0 → internal energy doesn’t change
  • So, q = –w

2️⃣ Adiabatic Process (No Heat)

  • q = 0
  • So, ΔU = w

3️⃣ Isochoric Process (Constant Volume)

  • No volume change → w = 0
  • So, ΔU = q

4️⃣ Isobaric Process (Constant Pressure)

  • Work = –PΔV
  • ΔU = q – PΔV

5️⃣ Cyclic Process (Complete Cycle)

  • System returns to original state
  • So, ΔU = 0
  • Therefore, q = –w

🔷 Units

  • Heat (q), Work (w), Internal Energy (ΔU) → unit = Joule (J)
  • All three have the same dimension: [ML²T⁻²]

🔷 Applications of the First Law of Thermodynamics

1️⃣ Heat Engines

  • Converts heat into mechanical work.
  • Example: car engine.
  • First Law helps calculate how much fuel is needed.

2️⃣ Refrigerators and Air Conditioners

  • Work is done to remove heat from inside to outside.
  • First Law is used to balance energy flow.

3️⃣ Human Body and Metabolism

  • Food gives energy → used for work and body heat.
  • Follows energy conservation.

4️⃣ Chemical Reactions

  • In exothermic or endothermic reactions, energy is released or absorbed.
  • First Law helps measure this change.

5️⃣ Power Plants

  • Electricity is generated using heat energy (steam turbines).
  • First Law helps manage fuel and heat efficiency.

🔷 Limitations of the First Law of Thermodynamics

While this law is very important, it also has some limitations:

❌ 1. Does Not Show Direction

  • It tells energy is conserved, but not whether heat flows from hot to cold or cold to hot.

❌ 2. Cannot Predict Spontaneity

  • It doesn’t tell whether a reaction or process will happen on its own.

❌ 3. No Information About Entropy

  • It doesn’t explain increase or decrease in disorder or randomness.

❌ 4. Allows 100% Conversion of Heat to Work

  • But in real life, not all heat can be converted to work.

❌ 5. Quality of Energy Not Mentioned

  • Energy might be present but not always useful (e.g., low-temperature heat).

🔷 First Law vs Second Law of Thermodynamics

FeatureFirst LawSecond Law
Main IdeaEnergy conservationDirection and feasibility
Spontaneous Process?Not explainedExplained
Entropy (Disorder)?Not explainedExplained
Practical UsefulnessBasic energy balanceReal-world behavior explained

🔷 Solved Examples for Practice

Example 1:

Heat given to system = 400 J,
Work done by system = 250 J
What is the change in internal energy?

Answer:
q = +400 J, w = –250 J
ΔU = q + w = 400 – 250 = 150 J


Example 2:

If ΔU = 0 and q = 300 J, then work?

Use:
ΔU = q + w
0 = 300 + w → w = –300 J
So, system did 300 J of work.


Example 3:

Adiabatic process, work = –200 J
Find ΔU.

q = 0 → ΔU = w = –200 J


🔷 First Law in Science Fields

SubjectExample
PhysicsEngines, machines
ChemistryReactions, enthalpy, bond energy
BiologyFood → energy in cells
EnvironmentHeat flow in Earth’s atmosphere

🔷 First Law in Thermodynamic Cycles

🔸 Carnot Cycle:

  • Ideal engine
  • Heat added, work done, heat removed
  • First Law applies to each step

🔸 Otto Cycle:

  • Used in petrol engines
  • Expansion + compression of gas

🔸 Refrigeration Cycle:

  • Work done to move heat from cold to hot area

In all cycles, total energy is conserved.


🔷 Quick Summary (Table)

FeatureDescription
Main FormulaΔU = q + w
Heat Addedq is positive
Work Done by Systemw is negative
Internal EnergyDepends only on temperature
Used InEngines, Refrigerators, Body, Chemistry
LimitationsNo direction, no entropy, 100% conversion

🔷 Visual Concept Map

First Law of Thermodynamics

├── Formula: ΔU = q + w
│ ├── q = heat (in or out)
│ └── w = work (on or by system)

├── Important Cases
│ ├── Isothermal → ΔU = 0
│ ├── Adiabatic → q = 0
│ ├── Isochoric → w = 0
│ ├── Cyclic → ΔU = 0

├── Applications
│ ├── Engines
│ ├── Refrigerators
│ └── Metabolism

└── Limitations
├── No direction
├── No entropy
└── No info about spontaneity

🔷 Final Tips for JEE/NEET Students

✅ Always remember sign conventions.
✅ Practice numerical questions from NCERT and previous year papers.
✅ Link the law with real-life systems like engines, fridges, human body.
✅ Understand limitations to learn about Second Law better.

7. 🔥 Enthalpy (H)

🔷 Introduction

In thermodynamics, energy plays a central role in chemical and physical changes.
We already know about internal energy (U). But when reactions happen at constant pressure (like most reactions around us), we use another quantity: Enthalpy (H).


🔷 What is Enthalpy (H) ?

Enthalpy is a thermodynamic quantity that represents the total heat content of a system.

🔹 Definitio n:

H= U+PV

Where:

  • H = Enthalpy
  • U = Internal energy
  • P = Pressure
  • V = Volume

This means:-

Enthalpy is the sum of internal energy (U) and the product of pressure and volume (PV).


🔹 Units of Enthalpy:

  • SI unit: Joule (J)
  • Also used: kilojoule (kJ) = 1000 J
  • Dimensions: [ML²T⁻²]

🔷 What is Change in Enthalpy (ΔH) ?

Since we cannot measure the actual value of enthalpy (H), we measure the change in enthalpy (ΔH). ΔH= Hfinal ​− Hinitial​

🔹 For a chemical reaction:

ΔH= Hproduct ​− Hreactant​


🔹 Relation between ΔH and Heat (q)

At constant pressure: ΔH= qp​

This means:

Change in enthalpy equals the heat absorbed or released at constant pressure.


🔹 Exothermic vs Endothermic

TypeΔH SignHeatExample
ExothermicNegative (−)ReleasedCombustion, Neutralization
EndothermicPositive (+)AbsorbedMelting, Photosynthesis

🔷 Why is Enthalpy Useful ?

  • Most chemical reactions happen at constant pressure.
  • It’s difficult to measure internal energy, but easier to measure ΔH.
  • Enthalpy tells us how much heat is involved in the process.

🔷 Derivation of H = U + PV

We know:

  • Work done by gas = PΔV
  • First law: ΔU = q−PΔV

At constant pressure: qp​=ΔU+PΔV

Now, define: ΔH = ΔU+PΔV ⇒ H = U+PV


🔷 Enthalpy Change in Physical Processes

1️⃣ Enthalpy of Fusion (ΔHfus)

  • Heat required to melt 1 mole of solid → liquid at constant pressure.

2️⃣ Enthalpy of Vaporization (ΔHvap)

  • Heat required to convert 1 mole of liquid → gas at constant pressure.

3️⃣ Enthalpy of Sublimation (ΔHsub)

  • Solid directly changes into gas.

🔷 Enthalpy Change in Chemical Processes

🔹 Enthalpy of Reaction (ΔHrxn)

  • Heat change when a chemical reaction happens.

ΔH= Hproduct ​− Hreactant


🔹 Enthalpy of Formation (ΔHf)

  • Heat change when 1 mole of a compound forms from its elements.

Standard Enthalpy of Formation = ΔHf

  • For all elements in standard state, ΔHf​=0

🔹 Enthalpy of Combustion (ΔHc)

  • Heat released when 1 mole of substance burns in oxygen.

Highly exothermic.


🔹 Enthalpy of Neutralization (ΔHneut)

  • Heat released when 1 mole of H⁺ reacts with 1 mole of OH⁻ to form water.

🔹 Enthalpy of Atomization (ΔHatom)

  • Heat needed to convert 1 mole of elements into atoms in gaseous form.

🔹 Enthalpy of Solution (ΔHsoln)

  • Heat change when a solute dissolves in solvent.
  • Can be exothermic or endothermic.

🔷 Standard Enthalpy Conditions

Standard state:

  • Temperature = 298 K (25°C)
  • Pressure = 1 atm
  • Pure substances

Symbol: ΔH∘\Delta H^\circΔH∘


🔷 Hess’s Law of Constant Heat Summation

It states:

If a reaction occurs in multiple steps, the total enthalpy change is the sum of enthalpy changes of all steps.

This helps in calculating enthalpy indirectly.


🔷 Bond Enthalpy

  • Energy needed to break 1 mole of bonds in gaseous molecules.

H2​(g) → 2H(g)  ;  D(H–H) = 435 kJ/mol

Can be used to estimate ΔH of reactions.


🔹 Formula:

ΔH = ∑Bond energies of reactants – ∑Bond energies of products


🔷 Enthalpy vs Internal Energy

PropertyEnthalpy (H)Internal Energy (U)
DefinitionH = U + PVTotal energy of molecules
Used When?Constant PressureAny condition
Includes Work?Yes (PV work)No
Measured By?Calorimetry at P = constantCalorimetry at V = constant

🔷 Enthalpy Diagrams

Use energy level diagrams to show:

  • Exothermic: Reactants higher than products (ΔH < 0)
  • Endothermic: Products higher than reactants (ΔH > 0)

🔷 Real-Life Examples of Enthalpy

  • Cooking: Heat used to convert raw → cooked food.
  • Sweating: Vaporization of sweat absorbs heat.
  • Fuels: Petrol/diesel releases enthalpy (energy).
  • Ice Melting: Enthalpy of fusion used to melt ice.

🔷 Practice Questions (JEE/NEET Style)

Q1. For a reaction, ΔH = +100 kJ. What type is it ?

Ans: Endothermic (heat absorbed)


Q2. What is ΔH for neutralization of HCl and NaOH ?

Ans: –57.1 kJ/mol (standard for strong acid-base)


Q3. Which enthalpy is used in evaporation of water ?

Ans: Enthalpy of vaporization (ΔH_vap)


Q4. If ΔH = –890 kJ for CH₄ combustion, what does it mean ?

Ans: 890 kJ heat is released → exothermic reaction.


🔷 Concept Map

Enthalpy (H)

├── Formula: H = U + PV

├── ΔH = q at constant pressure

├── Types:
│ ├── ΔH_rxn: Reaction
│ ├── ΔH_f: Formation
│ ├── ΔH_c: Combustion
│ ├── ΔH_fus: Fusion
│ ├── ΔH_vap: Vaporization
│ ├── ΔH_neut: Neutralization

├── Standard Conditions:
│ ├── Temp = 298 K
│ └── Pressure = 1 atm

└── Tools:
├── Hess's Law
└── Bond Enthalpy

🔷 Summary Table

TermMeaning
Enthalpy (H)Total heat content of system
FormulaH = U + PV
ΔH > 0Endothermic reaction
ΔH < 0Exothermic reaction
ΔH_rxnHeat of chemical reaction
ΔH_fHeat to form 1 mol of compound
ΔH_cHeat from burning 1 mol substance
ΔH_fusHeat to melt 1 mol solid
ΔH_vapHeat to vaporize 1 mol liquid
ΔH_neutHeat when acid reacts with base
ToolsHess’s Law, Bond Enthalpy

8. Heat Capacity

🔷 Introduction

Heat is a form of energy. But when we give heat to different substances, they don’t all heat up at the same rate. Why?

The answer lies in a property called Heat Capacity. This concept is very important in thermodynamics and helps us understand how different substances absorb or release heat.


🔷 1. What is Heat Capacity ?

🔹 Definition:

Heat capacity (C) is the amount of heat energy required to raise the temperature of a body (or system) by 1°C (or 1 K).

C = q/ΔT

Where:

  • C = heat capacity
  • q = heat supplied (in joules)
  • ΔT = temperature change (in °C or K)

🔹 Units and Dimensions:

  • SI Unit: J/K (Joule per Kelvin)
  • Dimensions: [ML²T⁻²K⁻¹]

🔹 Key Points:

  • Greater heat capacity = slower temperature rise
  • Lesser heat capacity = quicker temperature rise

Example:
Metal heats up faster than water because water has high heat capacity.


🔷 2. Specific Heat Capacity (c)

🔹 Definition:

Specific heat capacity is the amount of heat needed to raise the temperature of 1 gram of a substance by 1°C (or 1 K).

c = q/mΔT

Where:

  • c = specific heat capacity
  • q = heat added
  • m = mass (in grams)
  • ΔT = temperature change

🔹 Units:

  • SI Unit: J g⁻¹ K⁻¹
  • CGS Unit: cal g⁻¹ °C⁻¹

🔹 Example:

Water’s specific heat capacity = 4.18 J/g·K
→ This means: 4.18 J is needed to raise 1 gram of water by 1 K.


🔹 Importance in Real Life:

  • Water’s high specific heat helps regulate body temperature.
  • Used in climate control systems and engineering (coolants).

🔷 3. Molar Heat Capacity (Cₘ)

🔹 Definition:

Molar heat capacity is the amount of heat required to raise the temperature of 1 mole of a substance by 1°C (or 1 K).

cm = q/nΔT

Where:

  • Cₘ = molar heat capacity
  • q = heat supplied
  • n = number of moles
  • ΔT = change in temperature

🔹 Units:

  • SI Unit: J mol⁻¹ K⁻¹

🔹 Types of Molar Heat Capacity:

TypeSymbolCondition
Molar Heat Capacity at Constant VolumeCᵥVolume is fixed
Molar Heat Capacity at Constant PressureCₚPressure is fixed

🔷 4. Molar Heat Capacity at Constant Volume (Cᵥ)

🔹 Definition:

The amount of heat required to raise the temperature of 1 mole of a substance by 1 K at constant volume.

  • No work is done (ΔV = 0)
  • All heat increases internal energy

qv​ = nCv​ΔT


🔷 5. Molar Heat Capacity at Constant Pressure (Cₚ)

🔹 Definition:

The amount of heat required to raise the temperature of 1 mole of a substance by 1 K at constant pressure.

  • System does work due to expansion
  • More heat is needed than in constant volume

qp​= nCp​ΔT


🔹 So, always:

Cp>Cv


🔷 6. Relation Between Cp and Cv

🔹 For an ideal gas:

Cp​−Cv ​=R

Where:

  • R = universal gas constant = 8.314 J mol⁻¹ K⁻¹

🔹 Important for JEE/NEET:

This formula only works for ideal gases. For real gases, the relation is slightly different.


🔷 7. Ratio of Cp/Cv = γ (Gamma)

γ = Cp/Cv

This ratio is very important in adiabatic processes and in calculating speed of sound in gases.


🔹 γ Values for Gases:

Gas Typeγ Value (Cp/Cv)
Monoatomic (He, Ne)1.66
Diatomic (O₂, N₂)1.4
Polyatomic (CO₂, H₂O)1.33 or lower

🔷 8. Table: Quick Comparison of All Capacities

PropertyHeat Capacity (C)Specific Heat (c)Molar Heat (Cₘ)
Depends onEntire system1 gram1 mole
Formulaq/ΔTq/mΔTq/nΔT
Units (SI)J/KJ/g·KJ/mol·K
Use in equationsq = CΔTq = mcΔTq = nCₘΔT

🔷 9. Application of Specific and Molar Heat

1️⃣ In Physics

  • Calorimetry experiments
  • Predicting heat flow

2️⃣ In Chemistry

  • Reaction enthalpy calculations
  • Heat absorbed/released in chemical changes

3️⃣ In Biology

  • Thermoregulation in living beings

4️⃣ In Environment

  • Oceans absorb heat slowly due to water’s high specific heat → regulates Earth’s temperature.

🔷 10. Graphical Understanding

A plot of Temperature vs Heat for two different materials:

  • Steeper slope → lower specific heat
  • Flatter slope → higher specific heat

🔷 11. Experimental Measurement (Calorimetry)

Using a calorimeter, we can:

  • Measure heat change
  • Calculate specific or molar heat
  • Example: mixing hot and cold water and finding final temperature

🔷 12. Importance of Cp – Cv = R

  • Used in adiabatic processes
  • Helps find speed of sound
  • Used in engine cycles (Otto, Diesel)

🔷 3. Heat Capacity of Solids and Liquids

Solids:

  • Usually high heat capacity
  • Heat energy increases vibrations of atoms

Liquids:

  • Water has the highest specific heat → excellent coolant

🔷 14. Variation with Temperature

  • For gases: Cp and Cv slightly increase with temperature
  • For solids: Follow Dulong and Petit law (C ≈ 3R for metals)

🔷 15. Important Constants

ConstantValue
R (gas constant)8.314 J/mol·K
c (water)4.18 J/g·K
Cp (air)~29 J/mol·K
Cv (air)~20 J/mol·K

🔷 17. Common Misconceptions

❌ Cp and Cv are same — No! Cp > Cv always.
❌ All substances heat up at same rate — No! Depends on specific heat.
❌ Cp – Cv = R for all materials — No! Only for ideal gases.


🔷 18. Summary Table

ConceptKey FormulaSI Unit
Heat Capacity (C)q = CΔTJ/K
Specific Heat (c)q = mcΔTJ/g·K
Molar Heat Capacityq = nCΔTJ/mol·K
Cp – Cv RelationCp – Cv = RJ/mol·K
γ Ratioγ = Cp / CvDimensionless

🔷 19. Concept Map

Heat Capacity

├── Specific Heat (c)
│ └── Per gram basis

├── Molar Heat (Cₘ)
│ ├── Cp → Constant Pressure
│ └── Cv → Constant Volume

├── Cp – Cv = R (Ideal Gas Only)

└── Applications
├── Chemistry
├── Physics
├── Biology
└── Environment

🔷 Final Tips for JEE/NEET

✅ Learn and remember formulas
✅ Focus on units
✅ Understand Cp – Cv = R deeply
✅ Practice numerical problems regularly
✅ Revise γ values for different gases

9. Measurement of ΔU and ΔH

🔷 INTRODUCTION

In thermodynamics, energy changes occur in the form of:

  • ΔU = Change in internal energy
  • ΔH = Change in enthalpy

These are key to understanding how much energy is involved during a physical or chemical change.

This note explains:

  • How to measure ΔU and ΔH
  • How calorimetry is used
  • The working of a bomb calorimeter
  • Mathematical relation between ΔU and ΔH: ΔH = ΔU+ΔnRT

🔷 1. MEASUREMENT OF ΔU (Change in Internal Energy)

🔹 Internal Energy (U):

  • Total energy of all molecules in a system
  • Includes kinetic and potential energy

🔹 Change in Internal Energy:

ΔU = q+w

  • q = heat exchanged
  • w = work done
  • ΔU can be positive (increase in energy) or negative (decrease in energy)

🔹 At Constant Volume (V = constant):

  • No expansion work (w = 0)
  • So:

ΔU = qv​

This is why ΔU is measured at constant volume using a bomb calorimeter.


🔷 2. MEASUREMENT OF ΔH (Change in Enthalpy)

🔹 Enthalpy (H):

H= U + PV

  • H is the total heat content at constant pressure

🔹 Change in Enthalpy:

ΔH = ΔU+Δ(PV)

If pressure is constant, then: ΔH= ΔU+PΔV

At constant pressure: ΔH = qp

So, ΔH is measured by allowing the reaction to occur at constant pressure and calculating heat exchange.


🔷 3. CALORIMETRY

🔹 What is Calorimetry ?

Calorimetry is the science of measuring the heat of chemical reactions or physical changes.

It uses a device called a calorimeter to measure heat.


🔹 Principle of Calorimetry:

Based on the law of conservation of energy:

“Heat lost by hot body = Heat gained by cold body”

qlost  ​= qgained​


🔹 Calorimeter Setup:

  • A container made of insulating material
  • Contains water or solution
  • Thermometer to record temperature change

🔹 Formula Used:

q= mcΔT

Where:

  • q = heat (in joules)
  • m = mass (in grams)
  • c = specific heat capacity
  • ΔT = change in temperature

🔷 4. BOMB CALORIMETER

🔹 What is a Bomb Calorimeter ?

A bomb calorimeter is a special calorimeter used to measure the heat of combustion reactions at constant volume.

It helps measure ΔU directly.


🔹 Construction:

  • Strong steel container (the bomb)
  • Filled with oxygen gas
  • Sample is placed in the bomb
  • Bomb is placed in water bath
  • Thermometer and stirrer are attached

🔹 Working:

  1. Sample is ignited using electric spark
  2. It burns in oxygen → releases heat
  3. Heat is transferred to water
  4. Temperature rise is measured
  5. q = mcΔT is used to calculate heat

🔹 Important Point:

  • Bomb calorimeter measures ΔU (internal energy) because volume is constant.
  • Heat released in bomb calorimeter:

qv ​= ΔU = Ccalorimeter​ × ΔT

Where Ccalorimeter​ ​ = heat capacity of entire system


🔹 Example: Combustion of Glucose

C6​H12​O6​ + 6O2 ​→ 6CO2 ​+ 6H2​O

Energy released = measured using bomb calorimeter → ΔU


🔷 5. RELATION BETWEEN ΔH AND ΔU

At constant pressure: ΔH= ΔU+PΔV

From ideal gas law, PV= nRT⇒ Δ(PV)= Δ(nRT)

At constant T: ΔH= ΔU + ΔnR


🔹 Final Formula:

ΔH = ΔU + ΔnRT​

Where:

  • Δn = (moles of gaseous products – moles of gaseous reactants)
  • R = gas constant = 8.314 J/mol·K
  • T = temperature in K

🔹 When to Use This Formula ?

When reaction involves gases

Δn = change in number of moles of gaseous substances


🔷 6. DIFFERENCE BETWEEN ΔU AND ΔH

PropertyΔU (Internal Energy)ΔH (Enthalpy)
Measured atConstant volumeConstant pressure
Includes work?NoYes (PΔV work)
UnitskJ or JkJ or J
EquationΔU = qᵥΔH = qₚ
Used inBomb calorimeterSimple calorimeter

🔷 7. SUMMARY TABLE

TopicFormula/Key Point
ΔUΔU = q + w
ΔHΔH = ΔU + PΔV
Constant VolumeΔU = qv (measured in bomb calorimeter)
Constant PressureΔH = qp (measured in calorimeter)
Ideal Gas RelationΔH = ΔU + ΔnRT
Heat in calorimeterq = mcΔT
Bomb calorimeterMeasures heat of combustion = ΔU
Δn in formula(mol of gaseous products – mol of gaseous reactants)

🔷 8. GRAPHICAL UNDERSTANDING

Combustion Reaction
|--------------------------------|
| Bomb Calorimeter → measures ΔU|
| Constant Volume |
|--------------------------------|
| Simple Calorimeter → measures ΔH
| Constant Pressure |
|--------------------------------|

🔷 9. APPLICATIONS IN REAL LIFE

  • Food Industry: Measuring calories (energy in food)
  • Fuel Research: Heat values of petrol, diesel, etc.
  • Thermodynamics: In engines and power plants
  • Biology: Cellular respiration energy measurement

🔷 11. CONCEPT MAP

Heat Measurement

├── ΔU (Internal Energy)
│ └── Constant Volume → Bomb Calorimeter

├── ΔH (Enthalpy)
│ └── Constant Pressure → Simple Calorimeter

└── Relation
└── ΔH = ΔU + ΔnRT (for ideal gases)

🔷 12. FINAL TIPS FOR JEE/NEET

✅ Learn formula: ΔH = ΔU + ΔnRT
✅ Practice bomb calorimeter numericals
✅ Understand Δn = only for gaseous substances
✅ Practice heat calculation: q = mcΔT
✅ Revise First Law of Thermodynamics

10. Hess’s Law of Constant Heat Summation

🔷 INTRODUCTION

In chemistry, it is not always easy to measure the heat (enthalpy) change of a chemical reaction directly.
So, how can we calculate it indirectly?

The answer is — Hess’s Law. It is one of the most powerful tools in thermochemistry.


🔷 1. WHAT IS HESS’S LAW ?

🔹 Statement of Hess’s Law:

“If a chemical reaction can be expressed as a sum of two or more steps, then the total enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the individual steps.”

🔹 In Simple Words:

It does not matter how a reaction happens (in one step or many), the total heat change (ΔH) remains the same.


🔹 Mathematical Form:

ΔHtotal​ = ΔH1 ​+ ΔH2​ + ΔH3​ + …


🔷 2. WHY DOES HESS’S LAW WORK ?

Because enthalpy is a state function.

  • State function → depends only on initial and final state, not on the path.
  • Just like elevation change doesn’t depend on the path you take to climb a hill.

🔷 3. IMPORTANT POINTS

  • Hess’s Law works for enthalpy, but also valid for other state functions like entropy and Gibbs free energy.
  • It is extremely useful when:
    • Reaction happens in multiple steps
    • ΔH cannot be measured directly
    • Data for other steps is known

🔷 4. HESS’S LAW – EXAMPLES AND APPLICATIONS

Let’s see real cases where Hess’s law is used:

  • Enthalpy of formation of methane,
  • Formation of CO (carbon monoxide), etc.

🔷 5. STEPS TO SOLVE HESS’S LAW QUESTIONS

✅ Step 1: Write the given reactions

✅ Step 2: Identify the target equation

✅ Step 3: Modify equations

  • Multiply if required
  • Reverse if needed (change sign of ΔH)

✅ Step 4: Add equations

✅ Step 5: Add ΔH values


🔷 6. GRAPHICAL ILLUSTRATION

Imagine:

  • A → B directly (ΔH₁)
  • A → C → B in steps (ΔH₂ + ΔH₃)

Still: ΔH1=ΔH2+ΔH3

Graphically:

A -----→ B  (ΔH₁)
\
\
→ C ----→ B (ΔH₂ + ΔH₃)

🔷 7. APPLICATIONS OF HESS’S LAW

✅ 1. Calculating enthalpy of formation

  • When ΔH_f can’t be measured directly

✅ 2. Calculating enthalpy of transition

  • E.g., graphite to diamond

✅ 3. Enthalpy of hydration

  • From known enthalpies of solution and dissolution

✅ 4. Lattice energy calculations

  • Using Born–Haber cycle

✅ 5. Determining fuel efficiency

  • Comparing heat of combustion

🔷 8. SPECIAL CASES

🔹 HESS’S LAW FOR PHASE CHANGE:

Example: H2O(s)→H2O(g)

Can break into: H2​O(s)  →H2​O(l)(ΔH₁ = fusion)

H2​O(l)  →H2​O(g) (ΔH₂ = vaporization)

So: ΔH=ΔH1+ΔH2


🔷 9. BORN–HABER CYCLE (Application of Hess’s Law)

Used to calculate:

  • Lattice enthalpy of ionic compounds

Example: NaCl formation: Na(s)+1/2​Cl2(g)  →NaCl(s)

Break into:

  • Atomization
  • Ionization
  • Electron affinity
  • Lattice formation

Add all steps using Hess’s Law.


🔷 10. NUMERICAL PROBLEMS (JEE/NEET Style)


Q1. Given:

  • C + O₂ → CO₂; ΔH = –393.5 kJ
  • H₂ + ½O₂ → H₂O; ΔH = –285.8 kJ
  • CH₄ + 2O₂ → CO₂ + 2H₂O; ΔH = –890.3 kJ

Find ΔH for: C+2H2​→CH4​

Solution: Use Hess’s Law:
Answer: –74.8 kJ


Q2. Given:

  • CO + ½O₂ → CO₂; ΔH = –283.0 kJ
  • C + O₂ → CO₂; ΔH = –393.5 kJ

Find ΔH for: C + ½ O2→CO

Answer: –393.5+283.0=–110.5 kJ


Q3. Find ΔH for:

Na (s)+12Cl2→NaCl (s)

Given steps:

  • Na → Na⁺ + e⁻; ΔH = +496 kJ/mol
  • ½Cl₂ → Cl; ΔH = +122 kJ/mol
  • Cl + e⁻ → Cl⁻; ΔH = –349 kJ/mol
  • Lattice enthalpy = –786 kJ/mol

Answer: Add all = –517 kJ/mol


🔷 11. ADVANTAGES OF HESS’S LAW

✔ Saves experimental time
✔ Works when reaction is too dangerous or slow
✔ Used in calculating values not possible in labs
✔ Based on reliable mathematical logic


🔷 12. PRACTICE TIPS FOR JEE/NEET

✅ Master the skill of reversing and combining equations
✅ Always check physical states (s, l, g)
✅ Practice combustion and formation enthalpy questions
✅ Be quick in ΔnRT calculations if needed
✅ Use Hess’s law in bond enthalpy or lattice energy questions


🔷 13. CONCEPT MAP

Hess’s Law

├── Statement
│ └── Total ΔH is path-independent

├── Applications
│ ├── ΔH of formation
│ ├── ΔH of combustion
│ ├── Phase changes
│ └── Born–Haber cycle

├── Solving Method
│ ├── Reverse or multiply equations
│ └── Add reactions and ΔH

└── Basis
└── Enthalpy is a state function

🔷 14. SUMMARY TABLE

ConceptDescription/Formula
Hess’s LawΔH of net reaction = sum of stepwise ΔH
FormulaΔH_total = ΔH₁ + ΔH₂ + ΔH₃ + …
Target EquationThe final desired chemical equation
How to SolveReverse/multiply and add steps
Common Use CasesΔHf, ΔHcomb, lattice enthalpy, etc.
State FunctionEnthalpy depends only on start and end

🔷 15. FINAL REVISION

✅ Always remember:

  • State functions → path-independent
  • Hess’s Law is very useful in:
    • Multistep reactions
    • Reactions difficult to measure directly
  • Don’t forget to reverse ΔH when reversing equations

11. Enthalpies of Different Reactions

🔷 Introduction

In chemistry, enthalpy (ΔH) is used to measure the heat change during a reaction. But reactions happen in many different ways — forming compounds, burning fuel, melting ice, boiling water, or neutralizing acids.

Each of these reactions involves a different type of enthalpy.

In this note, you will learn:

  • Definitions of various types of enthalpies
  • Their equations and units
  • Simple examples
  • Applications in real life and competitive exams

🔷 What is Enthalpy ?

Enthalpy (H) is the total heat content of a system.

For any reaction: ΔH = Hproducts​ − Hreactants​

If:

  • ΔH < 0 → Exothermic (heat released)
  • ΔH > 0 → Endothermic (heat absorbed)

🔷 Types of Enthalpy Changes (ΔH)

There are six major types you must master:

  1. Enthalpy of Formation (ΔHf°)
  2. Enthalpy of Combustion (ΔHc°)
  3. Enthalpy of Atomization (ΔHat)
  4. Enthalpy of Sublimation (ΔHsub)
  5. Enthalpy of Fusion (ΔHfus) & Vaporization (ΔHvap)
  6. Enthalpy of Neutralization (ΔHneu)

Let’s explore each one in detail:


🔷 1. Enthalpy of Formation (ΔHf°)

🔹 Definition:

Heat change when 1 mole of a compound is formed from its elements in their standard states under standard conditions (298 K, 1 atm).

A + B → AB, ΔH=ΔHf


🔹 Standard State:

  • Most stable physical state at 25°C and 1 atm.
  • Example: O₂ (g), H₂ (g), C (graphite), N₂ (g)

🔹 Key Points:

  • For elements in standard state: ΔHf° = 0
  • ΔHf can be positive (endothermic) or negative (exothermic)

🔹 Applications:

  • Calculate ΔH for reactions using Hess’s Law
  • Thermodynamic stability of compounds

🔷 2. Enthalpy of Combustion (ΔHc°)

🔹 Definition:

Heat change when 1 mole of a substance completely burns in excess oxygen to form CO₂ and H₂O under standard conditions.


🔹 Formula:

Fuel + O2​ → CO2​ + H2​O, ΔH = ΔHc∘​


🔹 Key Points:

  • Always exothermic (ΔH < 0)
  • Units: kJ/mol
  • Measured using bomb calorimeter

🔹 Applications:

  • Compare fuel efficiency
  • Used in engine, explosives, and energy calculations

🔷 3. Enthalpy of Atomization (ΔHat)

🔹 Definition:

Heat required to convert 1 mole of an element in its standard state into individual gaseous atoms.


🔹 Formula:

X(s,l,g)​ →   X(g)  ​ΔH = ΔHat​


🔹 Key Points:

  • Always endothermic (energy is absorbed)
  • For diatomic molecules, divide bond dissociation enthalpy by 2
  • Used in bond enthalpy and Born-Haber cycles

🔷 4. Enthalpy of Sublimation (ΔHsub)

🔹 Definition:

Heat required to convert 1 mole of a solid directly into gaseous form without becoming liquid.


🔹 Formula:

X(s)​ → X(g),  ​ΔH = ΔHsub​


🔹 Key Points:

  • Always endothermic
  • Involves phase change (solid → gas)
  • Measured using calorimetry

🔹 Applications:

  • Sublimation-based purification
  • Calculations in phase diagrams

🔷 5. Enthalpy of Fusion (ΔHfus) & Vaporization (ΔHvap)

🔹 A. Enthalpy of Fusion (ΔHfus)

Heat required to melt 1 mole of solid into liquid at its melting point.

X(s)​ → X(l)​ ΔH= ΔHfus​


🔹 B. Enthalpy of Vaporization (ΔHvap)

Heat required to convert 1 mole of liquid to vapor at boiling point.

X(l)​  → X(g)  ​ΔH = ΔHvap


🔹 Key Points:

  • Both are endothermic
  • Fusion < Vaporization (it takes more energy to vaporize)
  • Useful in thermodynamic cycles, refrigeration, meteorology

🔷 6. Enthalpy of Neutralization (ΔHneu​)

🔹 Definition:

Heat change when 1 mole of water is formed during neutralization of an acid and base.


🔹 Formula:

Acid + Base → Salt + H2​O,  ΔH=ΔHneu​


🔹 Key Points:

  • Always exothermic (heat is released)
  • For strong acid + strong base, ΔHneu ≈ –57.1 kJ/mol
  • For weak acid/base, value is less because energy is used in ionization

🔷 Summary Table

Type of EnthalpySymbolDefinition (per mole)Sign
FormationΔHf°Elements → Compound+/–
CombustionΔHc°Compound + O₂ → CO₂ + H₂O
AtomizationΔHatElement (standard) → Gaseous atom+
SublimationΔHsubSolid → Gas+
FusionΔHfusSolid → Liquid+
VaporizationΔHvapLiquid → Gas+
NeutralizationΔHneuAcid + Base → Salt + H₂O

🔷 Graphical Representation

Solid ──ΔHfus──► Liquid ──ΔHvap──► Gas
↑ ↑
│ │
ΔHsub ΔHat

🔷 Real-Life Applications

Enthalpy TypeApplication
FormationThermodynamic data tables, reaction heat
CombustionFuel efficiency, calorific value
AtomizationBond energy, lattice energy
SublimationIodine purification, freeze-drying
Fusion/VaporizationWeather, cooling, melting, boiling
NeutralizationAcid-base titrations, buffer reactions

🔷 Solved Numerical Examples


Q1: Calculate heat released when 2 moles of CH₄ are burned (ΔHc° = –890.3 kJ/mol)

q=n×ΔH=2×–890.3=–1780.6 kJq = n × ΔH = 2 × –890.3 = –1780.6\ \text{kJ}q=n×ΔH=2×–890.3=–1780.6 kJ


Q2: Calculate ΔH for neutralization of 0.5 mol HCl with NaOH (ΔHneu = –57.1 kJ/mol)

q=0.5×(–57.1)=–28.55 kJq = 0.5 × (–57.1) = –28.55\ \text{kJ}q=0.5×(–57.1)=–28.55 kJ


Q3: Find total heat to convert 1 mol of ice at 0°C to steam at 100°C

Given:

  • ΔHfus = 6.01

🔷 Key Formulas to Remember

SituationFormula
Heat = mass × specific heat × ΔTq = mcΔT
Enthalpy change per moleΔH = q/n
Total heat in phase changeq = ΔH × number of moles
Neutralizationq = ΔHneu × moles of H₂O formed

🔷 Concept Map

Enthalpy Changes

├── Formation (ΔHf°)

├── Combustion (ΔHc°)

├── Atomization (ΔHat)

├── Sublimation (ΔHsub)

├── Fusion & Vaporization (ΔHfus, ΔHvap)

└── Neutralization (ΔHneu)

🔷 Final Tips for JEE/NEET

✅ Memorize definitions and units
✅ Practice numerical problems regularly
✅ Use ΔH signs correctly (exothermic = –ve, endothermic = +ve)
✅ Understand physical meanings, not just formulas
✅ Make flashcards for standard enthalpies

12. 🌠 Spontaneity and Second Law of Thermodynamics

🔷 1. INTRODUCTION

Not all processes in nature require external energy to happen. Ice melts, water flows downhill, iron rusts — these are all spontaneous processes.

But not all reactions happen on their own. Some need help (like electrolysis or pumping water uphill).

This natural tendency is studied under Spontaneity and the Second Law of Thermodynamics.


🔷 2. WHAT IS SPONTANEITY ?

🔹 Definition:

A process is called spontaneous if it occurs on its own, without any external help or energy.


🔹 Examples of Spontaneous Processes:

  • Ice melts at room temperature
  • Heat flows from hot to cold body
  • Salt dissolves in water
  • Iron rusts in moist air
  • Gas expands in a vacuum

🔹 Examples of Non-Spontaneous Processes:

  • Water flowing uphill
  • Gas compression without applying force
  • Electrolysis of water (needs electricity)

🔹 Important:

  • Spontaneous ≠ Fast (Rusting is slow but spontaneous)
  • Not all spontaneous reactions are exothermic
  • A reaction can be spontaneous even if ΔH > 0 (endothermic)

🔷 3. LIMITATION OF FIRST LAW OF THERMODYNAMICS

First law: ΔU=q+w

But it only talks about:

  • Energy conservation
  • Not about whether a process will happen or not

So, we need another principle — the Second Law of Thermodynamics


🔷 4. SECOND LAW OF THERMODYNAMICS

🔹 Statement:

“All spontaneous processes occur in a direction which increases the total entropy of the universe.”

Or,

“Heat cannot flow from a colder body to a hotter body on its own.”


🔹 In simple words:

  • Nature favors disorder (more randomness)
  • Systems tend to move toward maximum entropy

🔹 Other Forms of Second Law:

  • Clausius Statement:
    “Heat cannot spontaneously flow from cold to hot body.”
  • Kelvin-Planck Statement:
    “No process can convert heat completely into work.”

🔷 5. ENTROPY (S)

🔹 Definition:

Entropy is a measure of randomness or disorder of a system.


🔹 Symbol: S

🔹 Units:

  • SI: J·K⁻¹
  • CGS: cal·K⁻¹

🔹 Nature:

  • Entropy is a state function
  • Extensive property (depends on amount)
  • Higher entropy = more disorder

🔹 Order vs. Disorder:

StateEntropy Level
SolidLow entropy
LiquidMedium entropy
GasHigh entropy

🔹 Processes with Entropy Increase:

  • Melting (solid → liquid)
  • Boiling (liquid → gas)
  • Dissolution of salts in water
  • Mixing of gases
  • Expansion of gas

🔷 6. CHANGE IN ENTROPY (ΔS)

🔹 Formula:

ΔS= Sfinal ​− Sinitial​

  • ΔS > 0 → Entropy increases
  • ΔS < 0 → Entropy decreases

🔹 For Reversible Process:

ΔS= qrev​/T

Where:

  • qrev​ = heat exchanged reversibly
  • T = temperature in Kelvin

🔷 7. ENTROPY CHANGES IN SYSTEM AND SURROUNDINGS

The total entropy change is: ΔSuniverse​ = ΔSsystem​ + ΔSsurroundings​


🔹 For Spontaneous Process:

ΔSuniverse​ > 0

🔹 For Equilibrium:

ΔSuniverse​ = 0

🔹 For Non-Spontaneous Process:

ΔSuniverse​ < 0


🔹 Important Note:

Entropy of surroundings is often calculated using: ΔSsurroundings  ​= −qsystem​/T


🔷 8. ENTROPY CHANGE IN DIFFERENT PROCESSES

ProcessΔS
Solid → Liquid+ve
Liquid → Gas+ve
Gas → Liquid/Solid–ve
Increase in number of gas molecules+ve
Decrease in gas moles–ve

🔹 Example:

H2​(g) + ½ ​O2​(g) → H2​O(l)

ΔS = negative (because 1.5 mol gas → 0 gas mol)


🔷 9. ENTROPY AND REVERSIBILITY

  • Reversible process → ΔS = 0 (ideal)
  • Irreversible process → ΔS > 0

🔹 Real-life Examples:

SituationΔS
Boiling water+ve
Condensation of steam–ve
Mixing gases+ve
Freezing water–ve

🔷 10. ENTROPY IN CHEMICAL REACTIONS

Calculate ΔSreaction:- ΔSreaction ​= ∑Sproducts − ∑Sreactants​

Where:

  • S = standard molar entropy (from data tables)

🔷 11. ENTROPY & GIBBS FREE ENERGY (ΔG)

🔹 Combined Formula:

ΔG = ΔH − TΔS

Where:

  • ΔG = Gibbs Free Energy
  • ΔH = Enthalpy
  • T = Temperature (in K)
  • ΔS = Entropy

🔹 Gibbs Rule for Spontaneity:

ΔGProcess
< 0Spontaneous
= 0At equilibrium
> 0Non-spontaneous

🔷 12. SECOND LAW IN TERMS OF GIBBS FREE ENERGY

“At constant temperature and pressure, a process is spontaneous if ΔG < 0.”

This is the most useful form for chemical reactions.


🔷 13. APPLICATIONS IN DAILY LIFE

PhenomenonSpontaneous?Reason
Melting of iceYesEntropy increases
Cooking foodYesEnergy + entropy increase
Iron rustingYesExothermic + ΔS > 0
Water electrolysisNoRequires electricity
Diffusion of perfumeYesEntropy increases

Q. Predict spontaneity at high temperature:

ΔH = +ve, ΔS = +ve

Answer: Spontaneous at high temperature
Because: TΔS > ΔH


🔷 15. SUMMARY TABLE

ConceptFormula/Rule
Entropy (S)Measure of disorder
ΔS = qrev/TFor reversible processes
ΔSreactionΔS = ΣSproducts – ΣSreactants
Second LawΔSuniverse ≥ 0
Gibbs EnergyΔG = ΔH – TΔS
Spontaneity (Gibbs)ΔG < 0 → spontaneous
Surrounding EntropyΔSsurr = –qsys/T

🔷 16. CONCEPT MAP

Spontaneity

├── Spontaneous Process
│ ├── ΔG < 0
│ └── ΔS(universe) > 0

├── Entropy (S)
│ ├── Disorder Measure
│ ├── Units: J/K
│ └── ΔS = qrev/T

└── Second Law of Thermodynamics
├── Entropy of universe increases
├── ΔSuniverse = ΔSsystem + ΔSsurroundings
└── Basis for spontaneity

🔷 17. FINAL REVISION TIPS

✅ Understand entropy conceptually (not just as a formula)
✅ Memorize ΔG = ΔH – TΔS and Gibbs spontaneity rules
✅ Practice entropy-based numericals
✅ Watch for phase change questions — they often involve entropy
✅ Use entropy to explain mixing, diffusion, natural flow direction

13. ⚡ Gibbs Free Energy (G)

🔷 1. INTRODUCTION

In thermodynamics, we already know that energy is conserved (First Law) and that entropy increases in natural processes (Second Law). But scientists needed a more practical formula to predict if a chemical reaction will happen on its own.

That’s where Gibbs Free Energy (G) comes in.


🔷 2. WHAT IS GIBBS FREE ENERGY ?

🔹 Definition:

Gibbs Free Energy is the energy in a system available to do useful work at constant temperature and pressure.


🔹 Symbol: G

🔹 Unit: Joules (J) or kJ

🔹 Nature: State function (depends only on initial and final state)


🔹 Formula:

G= H−TS

Where:

  • G = Gibbs Free Energy
  • H = Enthalpy
  • T = Temperature in Kelvin
  • S = Entropy

🔷 3. PHYSICAL MEANING OF G = H – TS

This formula tells us that the energy available to do useful work is equal to the total energy (H) minus the energy lost to disorder (TS).

  • H (Enthalpy) = total energy content
  • TS (unusable energy) = energy lost due to entropy
  • So, G = Useful Energy

🔷 4. GIBBS FREE ENERGY CHANGE (ΔG)

In a chemical reaction, what matters is not just G, but the change in G.

🔹 Formula:

ΔG = ΔH−TΔS

Where:

  • ΔG = Change in Gibbs free energy
  • ΔH = Change in enthalpy
  • ΔS = Change in entropy
  • T = Absolute temperature (in K)

🔷 5. CRITERIA FOR SPONTANEITY (BASED ON ΔG)

Gibbs Free Energy tells us whether a reaction will happen on its own or not.

ΔG ValueType of ReactionMeaning
ΔG < 0SpontaneousReaction occurs naturally
ΔG = 0At EquilibriumNo net change
ΔG > 0Non-SpontaneousReaction won’t occur alone

🔹 Examples:

  • Ice melts at 0°C → ΔG < 0 → Spontaneous
  • Water electrolysis → ΔG > 0 → Needs electricity

🔷 6. GRAPHICAL UNDERSTANDING

Let’s understand Gibbs energy vs. reaction progress:

ΔG < 0 → Downhill slope → Reaction moves forward  
ΔG = 0 → Flat → Reaction in equilibrium
ΔG > 0 → Uphill → Reaction resists progress

🔷 7. ENTROPY AND GIBBS: LINK TO 2ND LAW

Second Law says: ΔSuniverse​ > 0 for spontaneous process

We can prove: ΔSuniverse  =−ΔG​/T​

So:

  • If ΔG < 0 → ΔSuniv > 0 → Spontaneous
  • If ΔG > 0 → ΔSuniv < 0 → Non-Spontaneous

This shows ΔG is directly linked to entropy of the universe.


🔷 8. TEMPERATURE DEPENDENCE OF SPONTANEITY

Let’s analyze: ΔG= ΔH−TΔS

Depending on the sign of ΔH and ΔS, temperature plays a big role.


🔹 Case 1: ΔH < 0, ΔS > 0

→ ΔG = –ve at all T
→ Reaction is always spontaneous

🧪 Example: Combustion of fuel


🔹 Case 2: ΔH > 0, ΔS > 0

→ ΔG = –ve at high T
→ Reaction is spontaneous at high temperature

🧪 Example: Melting of ice


🔹 Case 3: ΔH < 0, ΔS < 0

→ ΔG = –ve at low T
→ Reaction is spontaneous at low temperature

🧪 Example: Condensation of steam


🔹 Case 4: ΔH > 0, ΔS < 0

→ ΔG = +ve at all T
→ Reaction is never spontaneous

🧪 Example: Non-spontaneous decomposition


🔷 9. STANDARD GIBBS FREE ENERGY (ΔG°)

Standard conditions:

  • T = 298 K
  • P = 1 atm
  • Concentration = 1 M

🔹 Formula:

ΔG∘​ = ∑ΔG∘​products − ∑ΔG∘​reactants

🔹 Units: kJ/mol


🔷 10. GIBBS FREE ENERGY AND EQUILIBRIUM

🔹 Relation between ΔG and Equilibrium Constant (K):

ΔG∘ = −RT ln ⁡K

Where:

  • R = 8.314 J/mol·K
  • T = Temperature in Kelvin
  • K = Equilibrium constant

🔹 Interpretation:

ΔG°K ValueNature
ΔG° < 0K > 1Product-favored
ΔG° = 0K = 1Equilibrium
ΔG° > 0K < 1Reactant-favored

🔷 12. GIBBS ENERGY FOR PHASE TRANSITIONS

At phase change (like melting or boiling), system is in equilibrium.

So: ΔG= 0 = ΔH−TΔS → T = ΔH​/ΔS

This is used to calculate boiling or melting points.


🔷 13. COMPARISON WITH OTHER THERMODYNAMIC FUNCTIONS

FunctionSymbolMeaning
EnthalpyHTotal heat content
EntropySDegree of disorder
Internal EnergyUTotal energy in system
Gibbs EnergyGEnergy available to do work

🔷 14. REAL-LIFE APPLICATIONS OF GIBBS FREE ENERGY

ProcessΔGNature
Combustion of fuelNegativeSpontaneous
Rusting of ironNegativeSpontaneous
PhotosynthesisPositiveNon-spontaneous
Battery dischargeNegativeSpontaneous
Electrolysis of waterPositiveNeeds external energy

🔷 15. QUICK RECALL TABLE

ΔHΔSΔG (low T)ΔG (high T)Spontaneity
+Always spontaneous
+++Spontaneous at high T
+Spontaneous at low T
+++Never spontaneous

🔷 16. CONCEPT MAP

Gibbs Free Energy (G)

├── Formula: G = H - TS

├── ΔG = ΔH - TΔS
│ ├── < 0 → Spontaneous
│ ├── = 0 → Equilibrium
│ └── > 0 → Non-spontaneous

├── Temp. Dependence
│ ├── H > 0, S > 0 → High T: Spontaneous
│ ├── H < 0, S < 0 → Low T: Spontaneous

└── Linked to Equilibrium
└── ΔG° = –RT ln K

🔷 17. FINAL TIPS FOR JEE/NEET

✅ Always check sign of ΔH and ΔS first
✅ Use ΔG = ΔH – TΔS to predict spontaneity
✅ Practice units: J, kJ, K
✅ Link with entropy and equilibrium
✅ Memorize the temperature dependence table

14. 🔺 Third Law of Thermodynamics

✅ Statement of the Third Law:

“The entropy of a perfect crystalline substance is zero at absolute zero temperature (0 Kelvin).”

In simple words, when a substance is cooled to 0 K (–273.15 °C) and forms a perfect crystal, its atoms or molecules are completely ordered — no movement, no randomness. So, its entropy (S) becomes zero.


🧠 What is Entropy ?

Entropy (S) is a measure of disorder or randomness in a system.

  • More disorder → More entropy
  • Solids have low entropy, gases have high entropy

At 0 K in a perfect crystal, there is no disorder, so S = 0.


💎 What is a Perfect Crystal ?

A perfect crystal is a solid in which every particle is arranged in a perfect repeating pattern, with no defects or impurities.
At 0 K, these particles are completely still — they don’t vibrate or move.


📌 Why is the Third Law Important ?

The Third Law helps us:

  1. Set a starting point for entropy measurements
    → We can assign absolute values of entropy (not just relative changes)
  2. Use the equation: S=∫0T​(Cp/T) ​​dT to calculate the entropy of a substance at any temperature

🧪 Example:

NaCl (solid) at 0 K has perfectly arranged Na⁺ and Cl⁻ ions → S = 0
But as we heat it, atoms start to vibrate → entropy increases

So, entropy increases with temperature
That’s why all standard entropy values at 298 K are positive, like:

  • S°(H₂O) = 70 J/mol·K
  • S°(CO₂) = 213 J/mol·K

🔍 Application in Entropy Calculations

For a chemical reaction, entropy change is calculated using: ΔS∘​ =∑S∘​products ​− ∑S∘​reactants

These S° values are based on the Third Law, using S = 0 at 0 K as a reference point.


🌟 Residual Entropy

Some substances still have disorder at 0 K.
Example: Carbon monoxide (CO) molecules can orient in two ways → disorder remains → entropy is not zero. This leftover entropy is called Residual Entropy.


⚠️ Limitations of the Third Law

  • It applies only to perfect crystals
  • Not valid for amorphous solids or mixtures
  • Absolute zero (0 K) is not practically achievable

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