Thermodynamics Class 11 Notes for NEET & JEE

We Need To Cover These Topics-

S. No.	Topic	Subtopics
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 Type	Matter Exchange	Energy Exchange (Heat/Work)	Example
Open System	Yes	Yes	Open cup of hot tea
Closed System	No	Yes	Gas in a piston (sealed)
Isolated System	No	No	Thermos 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 Type	Depends on Quantity?	Additive?	Examples
Extensive	Yes	Yes	Mass, Volume, Energy
Intensive	No	No	Temperature, Pressure, Density

🔹 5. Thermodynamic Variables (State Variables)

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

Variable	Symbol	Meaning	Units
Pressure	P	Force per unit area	atm, Pa, bar
Volume	V	Space occupied by the system	m³, L, cm³
Temperature	T	Degree of hotness	K (Kelvin), °C
Internal Energy	U	Total energy inside system	Joules
Enthalpy	H	Heat at constant pressure	Joules
Entropy	S	Measure of disorder	J/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:

Thermal Equilibrium: Temperature is uniform
Mechanical Equilibrium: No unbalanced forces
Chemical Equilibrium: No net chemical change

🔹 Types of Processes in Thermodynamics

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

Process	Condition	Example
Isothermal	Temperature constant (T)	Ideal gas expansion at constant T
Adiabatic	No heat exchange (q = 0)	Rapid gas expansion in piston
Isobaric	Pressure constant (P)	Heating gas in piston
Isochoric	Volume constant (V)	Heating in rigid container
Reversible	Infinitely slow process	Ideal for max work output
Irreversible	Real, fast processes	Explosion, sudden expansion

🧠 Important Thermodynamic Terms to Remember

Term	Definition
System	Part of universe under study
Surroundings	Everything external to system
Boundary	Interface separating system and surroundings
State Function	Depends only on initial and final state (e.g., ΔU, ΔH)
Path Function	Depends on path taken (e.g., work, heat)
Equilibrium	No net change occurs in the system
Spontaneous Process	Occurs naturally without external help
Non-Spontaneous Process	Requires external energy to occur

🔍 Common Confusions Cleared

Confusion	Clarification
Enthalpy vs Internal Energy	H=U+PVH = U + PVH=U+PV. Use H for constant pressure, U for general energy changes.
Open vs Closed vs Isolated	Closed = no matter exchange, Isolated = no exchange at all.
Extensive vs Intensive	Use “E” for extensive (mass, volume), “I” for intensive (pressure, temperature).
State vs Path Functions	State: 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

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

✅ Final Revision Table

Concept	Key Points
System & Surroundings	System = what we study; surroundings = rest of the universe
Types of Systems	Open (matter + energy), Closed (only energy), Isolated (none)
Properties of System	Intensive = independent of mass; Extensive = dependent on mass
Thermodynamic Variables	P, V, T, U, H, S – define system state
State Functions	Depend only on initial/final state (U, H, S)
Path Functions	Depend 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 = C_P/ C_V (Ratio of heat capacities)

Also: TV^γ−1= constant and T^γP¹−γ = constant

🔸 Work done in adiabatic process:

W = P₂V₂− P₁V₁/ 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: q_P=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:

q_V=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

Process	Condition	Heat (q)	Work (w)	Example
Isothermal	T = constant	q ≠ 0	nRT ln(v₂/V₁ )	Slow gas expansion at room temp
Adiabatic	q = 0	q = 0	Depends on ΔT	Air compression in cylinder
Isobaric	P = constant	q = ΔH	P(V₂ – V₁)	Water boiling in open vessel
Isochoric	V = constant	q = ΔU	w = 0	Gas heating in rigid container
Cyclic	Final state = Initial	q = -w	Net area under PV	Carnot or Otto engine
Reversible	Infinitely slow	Max q/w	Max work output	Ideal piston process (theoretical)
Irreversible	Quick, real	q lost	Less work	Explosion, rapid gas flow

🧠 Key Formulas to Remember

Isothermal Process (Ideal Gas): W = nRT ln(v₂/V₁ )
Adiabatic Process (Ideal Gas): PV^γ = constant
Isobaric Process: W = P (V₂−V₁), q_P = ΔH
Isochoric Process: w = 0, q_V=ΔU
Cyclic Process: ΔU=0⇒q=−w

🏁 Final Summary – Fast Revision

Concept	Symbol/Condition	Heat Exchange (q)	Work Done (w)	Internal Energy Change (ΔU)
Isothermal	T constant	Yes	Yes, nRT ln(v₂/V₁ )	ΔU = 0
Adiabatic	q = 0	No	Yes	ΔU = -w
Isobaric	P constant	Yes	Yes, P (V₂−V₁)	ΔU = q – w
Isochoric	V constant	Yes	No (w = 0)	ΔU = q
Cyclic	Initial = Final State	Yes	Yes	ΔU = 0
Reversible	Idealized slow process	Yes	Maximum possible	Depends on type
Irreversible	Spontaneous, real process	Yes	Less than reversible	Depends 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.

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🔹 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

Feature	Description
Symbol	U
Nature	Scalar quantity
Units	Joule (J), calorie (1 cal = 4.184 J)
SI Unit	Joule (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 Energy	Explanation
Translational Energy	Due to straight-line motion of molecules
Rotational Energy	Due to spinning/rotation of molecules
Vibrational Energy	Due to vibration of atoms within molecules
Potential Energy	Due to intermolecular forces
Electronic Energy	Due to electronic transitions or configurations
Chemical Energy	Stored 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 = U_final − U_initial

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:

Heat (q) – energy flow due to temperature difference
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)

Quantity	Positive (+)	Negative (−)
Heat (q)	Heat added to the system	Heat released by the system
Work (w)	Work done on the system (e.g., compression)	Work done by the system (e.g., expansion)
ΔU	Increase in internal energy	Decrease 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 = d_q+d_w

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, q_P = ΔH = ΔU+PΔV ⇒ ΔU = ΔH−PΔV

🔹 8. Applications of First Law

Application	Explanation
Heat Engines	Convert heat → work using internal energy changes
Refrigerators/ACs	Use work input to extract heat (reverse of natural process)
Human Metabolism	Food (chemical energy) → work + heat
Explosions or Combustion	Energy released rapidly, doing work and increasing T, P
Thermochemical Equations	Use 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 = 3/2R (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

Scenario	ΔU Explanation
Heating water	q > 0 → ΔU increases
Burning petrol in engine	Fuel energy → work + heat → ΔU decreases
Air compression in piston	Work done on gas → molecules vibrate more → ΔU increases
Cooling of tea	q < 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

Parameter	Heat (q)	Work (w)	Internal Energy (U)
What it is	Energy transferred due to ΔT	Energy used to move boundary	Total microscopic energy in system
Path/State?	Path function	Path function	State function
SI Unit	Joule	Joule	Joule
Sign Conventions	+ if absorbed, − if released	+ if on system, − if by system	+ = gain, − = loss

🔹 14. NEET/JEE Exam Focus

Topic	Asked In	Type of Question
ΔU = q + w	NEET + JEE	Conceptual + numerical (calculation of ΔU, q or w)
Calorimetry	NEET	Based on experimental ΔU measurement
Heat/Work sign conventions	NEET + JEE	Assertion-Reason, MCQ, theory
Monoatomic/Diatomic gas ΔU	JEE (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

Concept	Key Point
Internal Energy (U)	Total energy inside the system (microscopic level)
Change in Internal Energy	ΔU = U_final − U_initial
First Law of Thermodynamics	ΔU = 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 Relation	ΔU = 3/2nRΔT (monoatomic)
State or Path Function	Internal energy = State function
Work at Constant Volume	w = 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= −P_extΔV

Where:

w = Work done by the system (Joules)
P_ext = 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

Condition	Sign 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

Unit	Symbol	Conversion
Joule	J	SI Unit
L·atm		1 L·atm = 101.3 J
Calorie	cal	1 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= −P_ext(V₂−V₁) = −P_ext 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

Property	Reversible Process	Irreversible Process
Pressure	Changes infinitesimally	Constant external pressure
Work Done	Maximum (more negative)	Less than reversible
Equation	−nRT ln(V₂/V₁)	−P_ext ΔV
Real/Ideal	Theoretical/ideal	Realistic
Efficiency	High	Low

🔷 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:

ΔU = w

For Isochoric Process:

q = ΔU

🔷 12. Real-Life Examples of Thermodynamic Work

Example	Description
Piston in a car engine	Gas expansion does work
Inflating a balloon	Gas does work against atmosphere
Compressing a gas in cylinder	Work done on the gas
Breathing	Lungs 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: −P_extΔV

🔷 14. Summary Table

Process Type	Formula	Notes
Isothermal (rev)	−nRTln(V₂/V₁)	Max work, ΔU = 0
Irreversible (constant P)	−P_extΔV	Fast process, less work
Adiabatic (rev)	Complex, involves γ	No heat exchange (q = 0)
Isochoric	w=0	Volume 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= −P_extΔV

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

Property	Heat (q)	Temperature (T)
Nature	Energy in transit	Measure of average kinetic energy
SI Unit	Joule (J)	Kelvin (K)
Path/State	Path function	State function
Transfer ?	Yes, from high to low T	No, it’s a property

🔷 4. Sign Conventions for Heat and Work

Understanding signs is very important in thermodynamics.

Process	Heat (q)	Work (w)
Heat added to system	q > 0	—
Heat released by system	q < 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:

Mode	Description	Example
Conduction	Heat transfer through solids by particle vibration	Metal rod heated at one end
Convection	Transfer in fluids by mass movement of particles	Boiling water
Radiation	Transfer through electromagnetic waves	Heat 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 Type	Heat Transfer Equation	Remarks
Isothermal	q= −w	ΔU = 0
Adiabatic	q= 0	ΔU = w
Isochoric	q= ΔU	w = 0
Isobaric	q= ΔU+PΔV= ΔH	Work + ΔU

🔷 10. Calorimetry – Measuring Heat

🔹 Principle of Calorimetry:

Heat lost by hot body = Heat gained by cold body

m₁c₁ (T₁−T_f) = m₂c₂ (T_f−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

Property	Constant Volume (qv)	Constant Pressure (qp)
Relation	q_v=ΔU	q_P=ΔH
Use	Bomb calorimeter	Coffee cup calorimeter
Work done?	No (ΔV = 0)	Yes (w = PΔV)
Heat transfer	Only internal energy changes	Enthalpy 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

Situation	Type of Heat Transfer
Boiling water	Convection
Iron rod in fire	Conduction
Sunlight warming earth	Radiation
Cooking in a pressure cooker	Isobaric + Conduction
Explosion in sealed container	Isochoric + 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

Process	T	V	q	w	ΔU
Isothermal	Const.	V↑/↓	≠0	≠0	0
Adiabatic	T↑/↓	V↑/↓	0	≠0	≠0
Isochoric	T↑/↓	Const.	= ΔU	0	≠0
Isobaric	T↑/↓	V↑/↓	= ΔH	≠0	≠0

B. Sign Convention Recap

Heat/Work	System Gains	System 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)

Quantity	Positive (+)	Negative (–)
q (heat)	Heat is given to the system	Heat 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

Feature	First Law	Second Law
Main Idea	Energy conservation	Direction and feasibility
Spontaneous Process?	Not explained	Explained
Entropy (Disorder)?	Not explained	Explained
Practical Usefulness	Basic energy balance	Real-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

Subject	Example
Physics	Engines, machines
Chemistry	Reactions, enthalpy, bond energy
Biology	Food → energy in cells
Environment	Heat 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)

Feature	Description
Main Formula	ΔU = q + w
Heat Added	q is positive
Work Done by System	w is negative
Internal Energy	Depends only on temperature
Used In	Engines, Refrigerators, Body, Chemistry
Limitations	No 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= H_final − H_initial

🔹 For a chemical reaction:

ΔH= H_product − H_reactant

🔹 Relation between ΔH and Heat (q)

At constant pressure: ΔH= q_p

This means:

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

🔹 Exothermic vs Endothermic

Type	ΔH Sign	Heat	Example
Exothermic	Negative (−)	Released	Combustion, Neutralization
Endothermic	Positive (+)	Absorbed	Melting, 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: q_p=ΔU+PΔV

Now, define: ΔH = ΔU+PΔV ⇒ H = U+PV

🔷 Enthalpy Change in Physical Processes

1️⃣ Enthalpy of Fusion (ΔH_fus)

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

2️⃣ Enthalpy of Vaporization (ΔH_vap)

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

3️⃣ Enthalpy of Sublimation (ΔH_sub)

Solid directly changes into gas.

🔷 Enthalpy Change in Chemical Processes

🔹 Enthalpy of Reaction (ΔH_rxn)

Heat change when a chemical reaction happens.

ΔH= H_product − H_reactant

🔹 Enthalpy of Formation (ΔH_f)

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

Standard Enthalpy of Formation = ΔH^∘_f

For all elements in standard state, ΔH^∘_f=0

🔹 Enthalpy of Combustion (ΔH_c)

Heat released when 1 mole of substance burns in oxygen.

Highly exothermic.

🔹 Enthalpy of Neutralization (ΔH_neut)

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

🔹 Enthalpy of Atomization (ΔH_atom)

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

🔹 Enthalpy of Solution (ΔH_soln)

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.

H_2(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

Property	Enthalpy (H)	Internal Energy (U)
Definition	H = U + PV	Total energy of molecules
Used When?	Constant Pressure	Any condition
Includes Work?	Yes (PV work)	No
Measured By?	Calorimetry at P = constant	Calorimetry 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

Term	Meaning
Enthalpy (H)	Total heat content of system
Formula	H = U + PV
ΔH > 0	Endothermic reaction
ΔH < 0	Exothermic reaction
ΔH_rxn	Heat of chemical reaction
ΔH_f	Heat to form 1 mol of compound
ΔH_c	Heat from burning 1 mol substance
ΔH_fus	Heat to melt 1 mol solid
ΔH_vap	Heat to vaporize 1 mol liquid
ΔH_neut	Heat when acid reacts with base
Tools	Hess’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).

c_m = 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:

Type	Symbol	Condition
Molar Heat Capacity at Constant Volume	Cᵥ	Volume is fixed
Molar Heat Capacity at Constant Pressure	Cₚ	Pressure 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

q_v = nC_vΔT

🔷 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

q_p= nC_pΔT

🔹 So, always:

C_p>C_v

🔷 6. Relation Between C_p and C_v

🔹 For an ideal gas:

C_p−C_v=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 C_p/C_v= γ (Gamma)

γ = C_p/C_v

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

🔹 γ Values for Gases:

Gas Type	γ 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

Property	Heat Capacity (C)	Specific Heat (c)	Molar Heat (Cₘ)
Depends on	Entire system	1 gram	1 mole
Formula	q/ΔT	q/mΔT	q/nΔT
Units (SI)	J/K	J/g·K	J/mol·K
Use in equations	q = CΔT	q = mcΔT	q = 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 C_p – C_v = 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: C_p and C_v slightly increase with temperature
For solids: Follow Dulong and Petit law (C ≈ 3R for metals)

🔷 15. Important Constants

Constant	Value
R (gas constant)	8.314 J/mol·K
c (water)	4.18 J/g·K
C_p (air)	~29 J/mol·K
C_v (air)	~20 J/mol·K

🔷 17. Common Misconceptions

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

🔷 18. Summary Table

Concept	Key Formula	SI Unit
Heat Capacity (C)	q = CΔT	J/K
Specific Heat (c)	q = mcΔT	J/g·K
Molar Heat Capacity	q = nCΔT	J/mol·K
Cp – Cv Relation	C_p – C_v= R	J/mol·K
γ Ratio	γ = C_p / C_v	Dimensionless

🔷 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 C_p – C_v= 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 = q_v

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 = q_p

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”

q_lost = q_gained

🔹 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:

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

🔹 Important Point:

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

q_v = ΔU = C_calorimeter × ΔT

Where C_calorimeter = heat capacity of entire system

🔹 Example: Combustion of Glucose

C₆H₁₂O₆ + 6O₂→ 6CO₂+ 6H₂O

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

Property	ΔU (Internal Energy)	ΔH (Enthalpy)
Measured at	Constant volume	Constant pressure
Includes work?	No	Yes (PΔV work)
Units	kJ or J	kJ or J
Equation	ΔU = qᵥ	ΔH = qₚ
Used in	Bomb calorimeter	Simple calorimeter

🔷 7. SUMMARY TABLE

Topic	Formula/Key Point
ΔU	ΔU = q + w
ΔH	ΔH = ΔU + PΔV
Constant Volume	ΔU = q_v (measured in bomb calorimeter)
Constant Pressure	ΔH = q_p (measured in calorimeter)
Ideal Gas Relation	ΔH = ΔU + ΔnRT
Heat in calorimeter	q = mcΔT
Bomb calorimeter	Measures 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:

ΔH_total = 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: H₂O_(s) →H₂O_(l)(ΔH₁ = fusion)

H₂O_(l) →H₂O_(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)+1/2Cl₂_(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

Concept	Description/Formula
Hess’s Law	ΔH of net reaction = sum of stepwise ΔH
Formula	ΔH_total = ΔH₁ + ΔH₂ + ΔH₃ + …
Target Equation	The final desired chemical equation
How to Solve	Reverse/multiply and add steps
Common Use Cases	ΔH_f, ΔH_comb, lattice enthalpy, etc.
State Function	Enthalpy 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 = H_products − H_reactants

If:

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

🔷 Types of Enthalpy Changes (ΔH)

There are six major types you must master:

Enthalpy of Formation (ΔHf°)
Enthalpy of Combustion (ΔHc°)
Enthalpy of Atomization (ΔHat)
Enthalpy of Sublimation (ΔHsub)
Enthalpy of Fusion (ΔHfus) & Vaporization (ΔHvap)
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: ΔH_f° = 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 + H₂O, ΔH = ΔH_c^∘

🔹 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 = ΔH_at

🔹 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 (ΔH_sub)

🔹 Definition:

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

🔹 Formula:

X_(s) → X_(g), ΔH = ΔH_sub

🔹 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 (ΔH_fus)

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

X_(s) → X_(l) ΔH= ΔH_fus

🔹 B. Enthalpy of Vaporization (ΔH_vap)

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

X_(l) → X_(g) ΔH = ΔH_vap

🔹 Key Points:

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

🔷 6. Enthalpy of Neutralization (ΔH_neu)

🔹 Definition:

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

🔹 Formula:

Acid + Base → Salt + H₂O, ΔH=ΔH_neu

🔹 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 Enthalpy	Symbol	Definition (per mole)	Sign
Formation	ΔH_f^°	Elements → Compound	+/–
Combustion	ΔH_c^°	Compound + O₂ → CO₂ + H₂O	–
Atomization	ΔH_at	Element (standard) → Gaseous atom	+
Sublimation	ΔH_sub	Solid → Gas	+
Fusion	ΔH_fus	Solid → Liquid	+
Vaporization	ΔH_vap	Liquid → Gas	+
Neutralization	ΔH_neu	Acid + Base → Salt + H₂O	–

🔷 Graphical Representation

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

🔷 Real-Life Applications

Enthalpy Type	Application
Formation	Thermodynamic data tables, reaction heat
Combustion	Fuel efficiency, calorific value
Atomization	Bond energy, lattice energy
Sublimation	Iodine purification, freeze-drying
Fusion/Vaporization	Weather, cooling, melting, boiling
Neutralization	Acid-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

Situation	Formula
Heat = mass × specific heat × ΔT	q = mcΔT
Enthalpy change per mole	ΔH = q/n
Total heat in phase change	q = ΔH × number of moles
Neutralization	q = 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:

State	Entropy Level
Solid	Low entropy
Liquid	Medium entropy
Gas	High 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= S_final− S_initial

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

🔹 For Reversible Process:

ΔS= q_rev/T

Where:

q_rev = heat exchanged reversibly
T = temperature in Kelvin

🔷 7. ENTROPY CHANGES IN SYSTEM AND SURROUNDINGS

The total entropy change is: ΔS_universe = ΔS_system+ ΔS_surroundings

🔹 For Spontaneous Process:

ΔS_universe > 0

🔹 For Equilibrium:

ΔS_universe = 0

🔹 For Non-Spontaneous Process:

ΔS_universe < 0

🔹 Important Note:

Entropy of surroundings is often calculated using: ΔS_surroundings = −q_system/T

🔷 8. ENTROPY CHANGE IN DIFFERENT PROCESSES

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

🔹 Example:

H_2(g)+ ½ O_2(g) → H₂O_(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:

Situation	ΔS
Boiling water	+ve
Condensation of steam	–ve
Mixing gases	+ve
Freezing water	–ve

🔷 10. ENTROPY IN CHEMICAL REACTIONS

Calculate ΔSreaction:- ΔS_reaction = ∑S^∘_products − ∑S^∘_reactants

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:

ΔG	Process
< 0	Spontaneous
= 0	At equilibrium
> 0	Non-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

Phenomenon	Spontaneous?	Reason
Melting of ice	Yes	Entropy increases
Cooking food	Yes	Energy + entropy increase
Iron rusting	Yes	Exothermic + ΔS > 0
Water electrolysis	No	Requires electricity
Diffusion of perfume	Yes	Entropy increases

Q. Predict spontaneity at high temperature:

ΔH = +ve, ΔS = +ve

Answer: Spontaneous at high temperature
Because: TΔS > ΔH

🔷 15. SUMMARY TABLE

Concept	Formula/Rule
Entropy (S)	Measure of disorder
ΔS = qrev/T	For reversible processes
ΔSreaction	ΔS = ΣSproducts – ΣSreactants
Second Law	ΔSuniverse ≥ 0
Gibbs Energy	ΔG = ΔH – TΔS
Spontaneity (Gibbs)	ΔG < 0 → spontaneous
Surrounding Entropy	ΔSsurr = –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 Value	Type of Reaction	Meaning
ΔG < 0	Spontaneous	Reaction occurs naturally
ΔG = 0	At Equilibrium	No net change
ΔG > 0	Non-Spontaneous	Reaction 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: ΔS_universe > 0 for spontaneous process

We can prove: ΔS_universe =−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 Value	Nature
ΔG° < 0	K > 1	Product-favored
ΔG° = 0	K = 1	Equilibrium
ΔG° > 0	K < 1	Reactant-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

Function	Symbol	Meaning
Enthalpy	H	Total heat content
Entropy	S	Degree of disorder
Internal Energy	U	Total energy in system
Gibbs Energy	G	Energy available to do work

🔷 14. REAL-LIFE APPLICATIONS OF GIBBS FREE ENERGY

Process	ΔG	Nature
Combustion of fuel	Negative	Spontaneous
Rusting of iron	Negative	Spontaneous
Photosynthesis	Positive	Non-spontaneous
Battery discharge	Negative	Spontaneous
Electrolysis of water	Positive	Needs external energy

🔷 15. QUICK RECALL TABLE

ΔH	ΔS	ΔG (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:

Set a starting point for entropy measurements
→ We can assign absolute values of entropy (not just relative changes)
Use the equation: S=∫₀^T(C_p/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