What is the Ideal Gas Equation?

The Ideal Gas Equation is PV = nRT, where P is pressure, V is volume, n is number of moles, R is universal gas constant (8.314 J/mol·K or 0.0821 L·atm/mol·K), and T is absolute temperature in Kelvin. It combines all gas laws and describes behavior of ideal gases.

What is Van der Waals equation?

Van der Waals equation is (P + an²/V²)(V - nb) = nRT, which accounts for real gas behavior. The term 'a' corrects for intermolecular attractions and 'b' corrects for finite molecular volume. At high temperature and low pressure, real gases behave ideally.

What is Graham's Law of Diffusion?

Graham's Law states that the rate of diffusion or effusion of a gas is inversely proportional to the square root of its molar mass: r₁/r₂ = √(M₂/M₁). Lighter gases diffuse faster than heavier gases.

What are critical constants?

Critical constants are the temperature (Tc), pressure (Pc), and volume (Vc) at which the liquid and gas phases become indistinguishable. Above Tc, a gas cannot be liquefied by pressure alone. For Van der Waals gas: Tc = 8a/27Rb, Pc = a/27b², Vc = 3nb.

What is Dalton's Law of Partial Pressures?

Dalton's Law states that the total pressure of a mixture of non-reacting gases equals the sum of partial pressures of individual gases: P_total = P₁ + P₂ + P₃ + ... Also, partial pressure P_i = x_i × P_total, where x_i is mole fraction.

Eudiometry is the analysis of gaseous mixtures involving measurement of volume changes during chemical reactions. It uses Gay-Lussac's law that gases combine in simple whole number ratios by volume. Common applications include combustion analysis and determining molecular formulas.

States of Matter & Eudiometry for JEE 2025-26 | Complete Notes with Formulas & Solved Examples

Introduction to Gaseous State

The gaseous state is the simplest state of matter to study because gas molecules have negligible intermolecular forces and occupy the entire container. Understanding gases is fundamental to chemistry and physics.

1.1 Characteristics of Gases

Key Properties of Gases

No definite shape or volume
Highly compressible
Exert pressure uniformly
Low density compared to solids/liquids

Mix completely (homogeneous mixtures)
Random molecular motion
Large intermolecular distances
Weak intermolecular forces

1.2 Measurable Properties of Gases

Property	Symbol	SI Unit	Other Units
Pressure	P	Pascal (Pa) = N/m²	atm, bar, mmHg, torr
Volume	V	m³	L, mL, cm³
Temperature	T	Kelvin (K)	°C, °F
Amount	n	mole (mol)	-

1.3 Pressure Conversions (Must Memorize!)

Standard Pressure Conversions

1 atm = 101325 Pa

1 atm = 101.325 kPa

1 atm = 1.01325 bar

1 atm = 760 mmHg

1 atm = 760 torr

1 bar = 10⁵ Pa

💡 STP vs NTP

STP (Standard T & P)

T = 273.15 K (0°C)

P = 1 bar = 10⁵ Pa

Molar volume = 22.7 L/mol

NTP (Normal T & P)

T = 293 K (20°C)

P = 1 atm = 101325 Pa

Molar volume = 24.0 L/mol

Old STP: 0°C and 1 atm → Molar volume = 22.4 L/mol (still commonly used in JEE)

Gas Laws

Gas laws describe the relationships between pressure, volume, temperature, and amount of gas. These are the most frequently tested concepts in JEE - expect at least 2-3 direct questions!

2.1 Boyle's Law (1662)

At Constant T & n

Pressure-Volume Relationship

"At constant temperature, the volume of a fixed amount of gas is inversely proportional to pressure."

P \propto \frac{1}{V} \quad \text{or} \quad PV = \text{constant}

\[P_1V_1 = P_2V_2\]

Graph: P vs V → Rectangular hyperbola (isotherm)

Graph: P vs 1/V → Straight line through origin

Graph: PV vs P → Horizontal line

2.2 Charles' Law (1787)

At Constant P & n

Volume-Temperature Relationship

"At constant pressure, the volume of a fixed amount of gas is directly proportional to absolute temperature."

V \propto T \quad \text{or} \quad \frac{V}{T} = \text{constant}

\frac{V_1}{T_1} = \frac{V_2}{T_2}

Graph: V vs T(K) → Straight line through origin (isobar)

Graph: V vs T(°C) → Straight line intersecting at -273.15°C

2.3 Gay-Lussac's Law (Pressure Law)

At Constant V & n

Pressure-Temperature Relationship

"At constant volume, the pressure of a fixed amount of gas is directly proportional to absolute temperature."

P \propto T \quad \text{or} \quad \frac{P}{T} = \text{constant}

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

Graph: P vs T(K) → Straight line through origin (isochore)

2.4 Avogadro's Law (1811)

At Constant P & T

Volume-Amount Relationship

"Equal volumes of all gases at same T and P contain equal number of molecules."

V \propto n \quad \text{or} \quad \frac{V}{n} = \text{constant}

\frac{V_1}{n_1} = \frac{V_2}{n_2}

At STP (0°C, 1 atm): Molar volume = 22.4 L/mol for all ideal gases

📝 Solved Example 1 (JEE Main Pattern)

Question: A gas occupies 500 mL at 27°C and 700 mmHg. What will be its volume at STP?

Solution:

Given: V₁ = 500 mL, T₁ = 27°C = 300 K, P₁ = 700 mmHg

STP: T₂ = 273 K, P₂ = 760 mmHg

Using Combined Gas Law:

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

V_2 = \frac{P_1V_1T_2}{T_1P_2} = \frac{700 \times 500 \times 273}{300 \times 760}

V_2 = 419.1 \text{ mL}

2.5 Dalton's Law of Partial Pressures

For a Mixture of Non-Reacting Gases

"Total pressure of a mixture of non-reacting gases equals sum of their partial pressures."

P_{total} = P_1 + P_2 + P_3 + ...

Partial Pressure Formula

P_i = x_i \times P_{total}

where x_i = mole fraction of gas i

Mole Fraction

x_i = \frac{n_i}{n_{total}} = \frac{P_i}{P_{total}}

⚠️ Aqueous Tension

When gas is collected over water, total pressure includes water vapor pressure (aqueous tension):

P_{dry gas} = P_{total} - P_{water vapor}

At 25°C, P(H₂O) ≈ 23.8 mmHg

Ideal Gas Equation

The Ideal Gas Equation combines all gas laws into one powerful equation. This is the MOST IMPORTANT equation in this chapter - used in nearly every gas problem!

3.1 The Ideal Gas Equation

The Master Equation

\[PV = nRT\]

Pressure

Volume

Moles

Gas Constant

Temperature (K)

3.2 Values of Gas Constant R

Value of R	Units	When to Use
8.314	J mol⁻¹ K⁻¹	P in Pa, V in m³ (SI units)
0.0821	L atm mol⁻¹ K⁻¹	P in atm, V in L (most common in JEE)
0.0831	L bar mol⁻¹ K⁻¹	P in bar, V in L
82.1	mL atm mol⁻¹ K⁻¹	P in atm, V in mL
1.987 ≈ 2	cal mol⁻¹ K⁻¹	For energy calculations

3.3 Different Forms of Ideal Gas Equation

In terms of Mass

PV = \frac{m}{M}RT

m = mass, M = molar mass

In terms of Density

\[PM = dRT\]

d = density = m/V

Combined Gas Law

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

For fixed amount of gas

In terms of Molecules

\[PV = NkT\]

N = number of molecules, k = R/N_A = Boltzmann constant

💡 Calculating Molar Mass from Density

M = \frac{dRT}{P}

Very useful for finding molar mass of unknown gas from its density!

📝 Solved Example 2 (JEE Main Pattern)

Question: Calculate the density of CO₂ at 27°C and 1 atm pressure.

Solution:

Given: P = 1 atm, T = 27°C = 300 K, M(CO₂) = 44 g/mol

R = 0.0821 L atm mol⁻¹ K⁻¹

Using PM = dRT:

d = \frac{PM}{RT} = \frac{1 \times 44}{0.0821 \times 300}

d = 1.786 \text{ g/L}

Kinetic Theory of Gases

The Kinetic Theory explains gas behavior at the molecular level. It provides the theoretical basis for gas laws and helps calculate molecular speeds. High-scoring topic for numerical problems!

4.1 Postulates of Kinetic Theory

Key Assumptions

Gas consists of large number of tiny particles (molecules) in random motion
Volume of molecules is negligible compared to container volume
No intermolecular forces of attraction or repulsion
Collisions are perfectly elastic (no energy loss)
Average kinetic energy is proportional to absolute temperature
Molecules exert pressure by colliding with container walls

4.2 Kinetic Energy Equations

Important Kinetic Energy Formulas

Average KE per Molecule

KE_{avg} = \frac{3}{2}kT

k = Boltzmann constant = 1.38 × 10⁻²³ J/K

Average KE per Mole

KE_{mole} = \frac{3}{2}RT

R = 8.314 J mol⁻¹ K⁻¹

Total KE of n moles

KE_{total} = \frac{3}{2}nRT

From Gas Equation

PV = \frac{1}{3}mNc^2 = \frac{2}{3}(KE_{total})

⚠️ Key Point

At same temperature, all gases have same average kinetic energy regardless of their molar mass!

KE depends only on T, not on nature of gas.

4.3 Molecular Speeds

Three Types of Molecular Speeds

RMS Speed (u_rms)

u_{rms} = \sqrt{\frac{3RT}{M}}

Root Mean Square

Average Speed (u_avg)

u_{avg} = \sqrt{\frac{8RT}{\pi M}}

Arithmetic Mean

Most Probable (u_mp)

u_{mp} = \sqrt{\frac{2RT}{M}}

Maximum on curve

Relationship (Must Memorize!)

u_{mp} : u_{avg} : u_{rms} = 1 : 1.128 : 1.224

u_{mp} : u_{avg} : u_{rms} = \sqrt{2} : \sqrt{\frac{8}{\pi}} : \sqrt{3}

u_mp < u_avg < u_rms

💡 Memory Trick for Speed Ratio

"MAR" = Most probable, Average, RMS (in increasing order)
Ratio: 1 : 1.13 : 1.22 (approximately)

📝 Solved Example 3

Question: Calculate the RMS speed of O₂ molecules at 27°C.

Solution:

Given: T = 27°C = 300 K, M(O₂) = 32 g/mol = 0.032 kg/mol

R = 8.314 J mol⁻¹ K⁻¹

u_{rms} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3 \times 8.314 \times 300}{0.032}}

u_{rms} = \sqrt{233,343.75} = 483.1 \text{ m/s}

u_{rms} \approx 483 \text{ m/s}

Diffusion and Effusion

Graham's Law relates the rate of diffusion/effusion to molar mass. This is a favorite topic for JEE with many conceptual and numerical questions.

5.1 Definitions

Diffusion

Spontaneous mixing of gases due to random molecular motion

Example: Perfume spreading in a room

Effusion

Escape of gas through a tiny hole (smaller than mean free path)

Example: Air leaking from a punctured tire

5.2 Graham's Law

Graham's Law of Diffusion/Effusion

"Rate of diffusion/effusion is inversely proportional to square root of molar mass (or density)"

r \propto \frac{1}{\sqrt{M}} \propto \frac{1}{\sqrt{d}}

\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} = \sqrt{\frac{d_2}{d_1}}

Different Forms of Graham's Law

In terms of Volume

\frac{V_1/t_1}{V_2/t_2} = \sqrt{\frac{M_2}{M_1}}

In terms of Time

\frac{t_1}{t_2} = \sqrt{\frac{M_1}{M_2}}

For same volume

In terms of Distance

\frac{d_1}{d_2} = \sqrt{\frac{M_2}{M_1}}

Traveled in same time

In terms of Moles

\frac{n_1/t}{n_2/t} = \sqrt{\frac{M_2}{M_1}}

📝 Solved Example 4 (JEE Main Pattern)

Question: 50 mL of gas A diffuses through a membrane in the same time as 40 mL of gas B. If molar mass of B is 64 g/mol, find molar mass of A.

Solution:

Given: V_A = 50 mL, V_B = 40 mL (same time)

M_B = 64 g/mol, M_A = ?

\frac{r_A}{r_B} = \frac{V_A}{V_B} = \sqrt{\frac{M_B}{M_A}}

\frac{50}{40} = \sqrt{\frac{64}{M_A}}

\left(\frac{5}{4}\right)^2 = \frac{64}{M_A}

M_A = \frac{64 \times 16}{25} = 40.96 \text{ g/mol}

M_A \approx 41 \text{ g/mol}

💡 Quick Application

Separating Isotopes: Graham's law is used to separate U-235 from U-238 in UF₆ gas. Since M(²³⁵UF₆) < M(²³⁸UF₆), ²³⁵UF₆ diffuses slightly faster.

Real Gases & Van der Waals Equation

Real gases deviate from ideal behavior, especially at high pressure and low temperature. Van der Waals equation accounts for these deviations. This is an important topic for JEE Advanced!

6.1 Causes of Deviation

Ideal Gas Assumptions

Molecules have zero volume
No intermolecular forces

These assumptions fail at high P and low T

Real Gas Reality

Molecules have finite volume (b correction)
Attractive forces exist (a correction)

Van der Waals corrected both

6.2 Van der Waals Equation

The Real Gas Equation

\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT

For 1 mole:

\left(P + \frac{a}{V^2}\right)(V - b) = RT

'a' - Pressure Correction

Accounts for intermolecular attractions

Higher 'a' → stronger attraction → easier to liquefy

Units: atm L² mol⁻² or Pa m⁶ mol⁻²

'b' - Volume Correction

Accounts for molecular volume (excluded volume)

b = 4 × actual molecular volume

Units: L mol⁻¹ or m³ mol⁻¹

6.3 Compressibility Factor (Z)

Definition

Z = \frac{PV}{nRT} = \frac{PV_m}{RT}

where V_m = molar volume

Z = 1

Ideal gas behavior

Z < 1

Attractive forces dominate

Gas more compressible than ideal

Z > 1

Repulsive forces dominate

Gas less compressible than ideal

6.4 Boyle Temperature

Temperature at which Real Gas Behaves Ideally

T_B = \frac{a}{Rb}

At Boyle temperature, Z = 1 over a wide range of pressures (at moderate pressures)

💡 When do Real Gases Behave Ideally?

High temperature: KE >> Intermolecular attractions
Low pressure: Molecules far apart, volume negligible
At Boyle temperature: Deviations due to 'a' and 'b' cancel out

Critical Constants & Liquefaction

Critical constants define the conditions above which a gas cannot be liquefied. Understanding these concepts is crucial for JEE Advanced numerical problems.

7.1 Critical Constants

Definitions

Critical Temperature (Tc)

Temperature above which gas cannot be liquefied by pressure alone

Critical Pressure (Pc)

Pressure required to liquefy gas at Tc

Critical Volume (Vc)

Volume of 1 mole at Tc and Pc

7.2 Critical Constants from Van der Waals Constants

Important Formulas (JEE Advanced)

Critical Volume

\[V_c = 3b\]

Critical Pressure

P_c = \frac{a}{27b^2}

Critical Temperature

T_c = \frac{8a}{27Rb}

Critical Compressibility Factor

Z_c = \frac{P_cV_c}{RT_c} = \frac{3}{8} = 0.375

This is same for all Van der Waals gases (theoretical value)

Reverse Formulas: Van der Waals from Critical Constants

a = 3P_cV_c^2 = \frac{27R^2T_c^2}{64P_c}

b = \frac{V_c}{3} = \frac{RT_c}{8P_c}

7.3 Liquefaction of Gases

Methods of Liquefaction

1. Increase Pressure

Only works below critical temperature

2. Decrease Temperature

Remove kinetic energy from molecules

3. Joule-Thomson Effect

Cooling due to adiabatic expansion (below inversion temperature)

4. Adiabatic Demagnetization

Used for very low temperatures

⚠️ Inversion Temperature

T_i = \frac{2a}{Rb} = 2T_B = \frac{27T_c}{4}

• Below T_i: Joule-Thomson expansion causes cooling
• Above T_i: Joule-Thomson expansion causes heating
• H₂ and He have very low T_i, so they are pre-cooled before liquefaction

Liquid State

Liquids have properties intermediate between gases and solids. Understanding surface tension, viscosity, and vapor pressure is important for JEE.

8.1 Properties of Liquids

Property	Definition	Effect of Temperature
Vapor Pressure	Pressure exerted by vapor in equilibrium with liquid	Increases with T (exponentially)
Surface Tension (γ)	Force per unit length on surface (N/m)	Decreases with T
Viscosity (η)	Resistance to flow (Pa·s or poise)	Decreases with T

8.2 Vapor Pressure

Clausius-Clapeyron Equation

\ln\frac{P_2}{P_1} = \frac{\Delta H_{vap}}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)

or in differential form:

\frac{d(\ln P)}{dT} = \frac{\Delta H_{vap}}{RT^2}

8.3 Surface Tension

Key Formulas

Surface Tension

\gamma = \frac{F}{l} = \frac{\text{Work}}{\text{Area}}

Units: N/m or J/m²

Capillary Rise

h = \frac{2\gamma \cos\theta}{r\rho g}

θ = contact angle, r = radius

8.4 Viscosity

Coefficient of Viscosity (η)

F = \eta A \frac{dv}{dx}

where F = viscous force, A = area, dv/dx = velocity gradient

SI Unit: Pa·s (Pascal-second) or kg m⁻¹ s⁻¹

CGS Unit: poise (P), 1 P = 0.1 Pa·s

Eudiometry (Gas Analysis)

Eudiometry involves analysis of gaseous mixtures by measuring volume changes during chemical reactions. This is a HIGH-SCORING topic for JEE with predictable problem patterns!

9.1 Basic Principle

Gay-Lussac's Law of Combining Volumes

"Gases combine in simple whole number ratios by volume (at same T and P)"

Key Principle:

At constant T and P, volume ratio = mole ratio

\frac{V_A}{V_B} = \frac{n_A}{n_B}

9.2 Common Combustion Reactions

Reaction	Balanced Equation	Volume Ratio
Hydrogen combustion	2H₂ + O₂ → 2H₂O	2:1:2
Methane combustion	CH₄ + 2O₂ → CO₂ + 2H₂O	1:2:1:2
Carbon monoxide	2CO + O₂ → 2CO₂	2:1:2
General hydrocarbon	C_xH_y + (x + y/4)O₂ → xCO₂ + (y/2)H₂O	1:(x+y/4):x:(y/2)

9.3 Absorption of Gases

Common Absorbing Agents

Gas	Absorbed by
CO₂	KOH, NaOH, Ca(OH)₂
O₂	Alkaline pyrogallol
CO	Ammoniacal Cu₂Cl₂ solution
SO₂, Cl₂	KOH, NaOH
H₂O vapor	Anhydrous CaCl₂, P₂O₅, conc. H₂SO₄

9.4 Steps to Solve Eudiometry Problems

🎯 Problem-Solving Strategy

Write the balanced equation with volume coefficients
Assume volume of unknown gas = V
Calculate O₂ required for complete combustion
Calculate volume of CO₂ and H₂O produced
After cooling, H₂O condenses (volume negligible)
Use volume contractions to find V

📝 Solved Example 5 (JEE Main/Advanced Pattern)

Question: 20 mL of a gaseous hydrocarbon requires 60 mL of O₂ for complete combustion. After combustion and cooling, 40 mL of CO₂ is produced. Find the molecular formula of the hydrocarbon.

Solution:

Step 1: Let hydrocarbon be C_xH_y

C_xH_y + \left(x + \frac{y}{4}\right)O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O

Step 2: Using volume ratios

Volume of hydrocarbon : Volume of CO₂ = 1 : x

\frac{20}{40} = \frac{1}{x} \Rightarrow x = 2

Step 3: Using O₂ volume

Volume of hydrocarbon : Volume of O₂ = 1 : (x + y/4)

\frac{20}{60} = \frac{1}{x + y/4}

x + \frac{y}{4} = 3

2 + \frac{y}{4} = 3 \Rightarrow y = 4

\text{Molecular Formula: } C_2H_4 \text{ (Ethene)}

📝 Solved Example 6 (Volume Contraction)

Question: 100 mL of a mixture of CO and CO₂ is passed through excess KOH. Volume reduced to 60 mL. Find composition of mixture.

Solution:

KOH absorbs CO₂ but not CO

Volume of CO₂ absorbed = 100 - 60 = 40 mL

Volume of CO remaining = 60 mL

Composition:

CO = 60 mL (60%)

CO₂ = 40 mL (40%)

📝 Previous Year Questions Analysis

JEE Main (Last 5 Years)

✓ Ideal Gas Equation: 30%
✓ Gas Laws & Dalton's Law: 25%
✓ Graham's Law: 15%
✓ Kinetic Theory (Speeds): 15%
✓ Real Gases & Critical Constants: 10%
✓ Eudiometry: 5%

JEE Advanced (Last 5 Years)

✓ Van der Waals & Critical Constants: 30%
✓ Kinetic Theory (Derivations): 25%
✓ Eudiometry Problems: 20%
✓ Compressibility Factor: 15%
✓ Mixed Concept Problems: 10%

Weightage Analysis

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

JEE Advanced: 10-15 marks (3-4 questions)

Difficulty Level: Medium

Time Required: 3.5-4 hours practice

Download Complete PYQ PDF (2015-2024)

🎯 Practice Problem Set

Level 1: Basic (JEE Main Standard)

A gas occupies 2 L at 27°C and 1 atm. Find volume at STP.
Calculate density of N₂ at 25°C and 2 atm pressure.
Find RMS speed of H₂ molecules at 300 K.
50 mL of H₂ diffuses through a membrane in 20 min. How long will 40 mL of O₂ take?
Calculate total pressure if 2 L of N₂ at 1 atm and 3 L of O₂ at 2 atm are mixed in 5 L vessel.
Find molar mass of gas with density 2.5 g/L at 27°C and 1 atm.
Calculate average kinetic energy of 1 mole of gas at 27°C.
What is partial pressure of O₂ if mole fraction is 0.21 and total pressure is 1 atm?

Level 2: Intermediate (JEE Main/Advanced)

At what temperature will RMS speed of O₂ equal that of H₂ at 300 K?
For a gas, a = 3.6 atm L² mol⁻² and b = 0.04 L mol⁻¹. Calculate Boyle temperature.
If Z = 1.2 for a gas at certain conditions, is volume more or less than ideal?
Calculate critical temperature if a = 1.5 atm L² mol⁻² and b = 0.05 L mol⁻¹.
A mixture of H₂ and O₂ in 2:1 volume ratio effuses through an orifice. What is composition of effused gas?
20 mL hydrocarbon + 70 mL O₂ → 40 mL CO₂ + 40 mL H₂O. Find formula.
At T = Tc, what is compressibility factor for Van der Waals gas?
Ratio of u_rms of O₂ to u_avg of O₂ at same temperature is?

Level 3: Advanced (JEE Advanced/Olympiad)

Derive relation between critical constants and Van der Waals constants.
100 mL mixture of CO, CH₄ and N₂ is exploded with 100 mL O₂. After explosion and cooling, 80 mL remains. On passing through KOH, 50 mL remains. Find composition.
At what temperature will u_avg of N₂ equal u_mp of O₂ at 300 K?
For a gas obeying P(V-b) = RT, find expression for isothermal compressibility.
Two vessels of 1 L each contain O₂ at 1 atm and N₂ at 2 atm at same T. If connected, find partial pressures after equilibrium.
Calculate inversion temperature for N₂ if a = 1.39 atm L² mol⁻² and b = 0.039 L mol⁻¹.
Derive expression for Z in terms of Van der Waals constants at high pressure limit.
A mixture of gases A and B effuses through a small hole. If initially x_A = 0.5, find x_A in effused mixture (M_A = 4M_B).

Get Solutions PDF with Step-by-Step Working

📋 Quick Formula Sheet - States of Matter

Gas Laws

Boyle: PV = constant (T, n const)

Charles: V/T = constant (P, n const)

Gay-Lussac: P/T = constant (V, n const)

Avogadro: V/n = constant (P, T const)

Combined: P₁V₁/T₁ = P₂V₂/T₂

Ideal Gas Equation

PV = nRT

PM = dRT (d = density)

R = 8.314 J/mol·K = 0.0821 L·atm/mol·K

At STP: V_m = 22.4 L/mol

Kinetic Theory

KE per mol = (3/2)RT

u_rms = √(3RT/M)

u_avg = √(8RT/πM)

u_mp = √(2RT/M)

u_mp : u_avg : u_rms = 1 : 1.128 : 1.224

Dalton & Graham's Laws

P_total = P₁ + P₂ + P₃ + ...

P_i = x_i × P_total

r₁/r₂ = √(M₂/M₁) = √(d₂/d₁)

t₁/t₂ = √(M₁/M₂)

Van der Waals

(P + an²/V²)(V - nb) = nRT

Z = PV/nRT

T_B = a/Rb (Boyle temp)

T_i = 2a/Rb (Inversion temp)

Critical Constants

V_c = 3b

P_c = a/27b²

T_c = 8a/27Rb

Z_c = P_cV_c/RT_c = 3/8

📥 Download Printable Formula Sheet PDF

States of Matter & Eudiometry - Complete Guide for JEE 2025-26

Why States of Matter is Crucial for JEE?

States of Matter is a high-scoring chapter in JEE Chemistry with 6-10% weightage. It combines concepts from both chemistry and physics, making it crucial for:

Ideal Gas Equation - Foundation for all gas calculations
Kinetic Theory - Understanding molecular behavior
Van der Waals - Real gas behavior (JEE Advanced favorite)
Eudiometry - Direct application-based problems

In JEE Advanced, this chapter often has multi-concept questions combining gas laws, kinetic theory, and thermodynamics. The numerical problems are straightforward if formulas are memorized.

Key Formulas to Memorize

1. Gas Laws & Ideal Gas

• PV = nRT (Ideal gas equation)
• PM = dRT (Density form)
• P₁V₁/T₁ = P₂V₂/T₂ (Combined gas law)
• P_total = P₁ + P₂ + ... (Dalton's law)

2. Kinetic Theory

• u_rms = √(3RT/M)
• u_avg = √(8RT/πM)
• u_mp = √(2RT/M)
• KE = (3/2)RT per mole

3. Real Gases

• (P + an²/V²)(V - nb) = nRT
• T_c = 8a/27Rb, P_c = a/27b², V_c = 3b
• Z = PV/nRT (Compressibility factor)

📚 How to Study States of Matter Effectively?

For JEE Main Students:

Time Required: 3-4 days (3 hours/day)
Master Ideal Gas Equation - 70% problems use it
Learn all gas laws with their conditions
Practice Graham's law problems thoroughly
Memorize molecular speed relationships
Solve at least 50 numerical problems

For JEE Advanced Students:

Time Required: 5-6 days (4 hours/day)
Deep dive into Van der Waals equation
Master critical constants derivations
Practice compressibility factor graphs
Solve complex eudiometry problems
Study Joule-Thomson effect and inversion temperature

⚠️ Common Mistakes to Avoid in JEE Exam

❌
Using wrong value of R: Match R with your units! Use 0.0821 for P(atm), V(L) or 8.314 for P(Pa), V(m³)
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Temperature in Celsius: ALWAYS convert to Kelvin before using gas equations (T_K = T_C + 273)
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Confusing speed types: u_mp < u_avg < u_rms (remember "MAR" in increasing order)
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Graham's law ratio inverted: Rate ∝ 1/√M, so lighter gas diffuses FASTER
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Ignoring water vapor in eudiometry: After cooling, H₂O condenses - don't include its volume in final gas volume

📊 JEE Previous Year Question Analysis (2015-2024)

Year	JEE Main	JEE Advanced	Topic Focus
2024	2 Questions (8 marks)	3 Questions (11 marks)	Ideal gas, Van der Waals, Eudiometry
2023	3 Questions (12 marks)	2 Questions (8 marks)	Graham's law, Kinetic theory, Critical constants
2022	2 Questions (8 marks)	4 Questions (14 marks)	Dalton's law, Compressibility factor, RMS speed
2021	2 Questions (8 marks)	3 Questions (10 marks)	Gas laws, Molecular speeds comparison
2020	3 Questions (12 marks)	2 Questions (7 marks)	Real gases, Boyle temperature, Eudiometry

Trend: JEE Main focuses on direct formula applications while JEE Advanced asks conceptual questions on deviations from ideal behavior. Eudiometry problems appear regularly in both papers. Questions on comparing properties of different gases are common.

📌 Important Constants for Quick Reference

Gas Constant R

8.314 J/mol·K

0.0821 L·atm/mol·K

Boltzmann Constant

k = 1.38 × 10⁻²³ J/K

Avogadro's Number

Nₐ = 6.022 × 10²³

Molar Volume (STP)

22.4 L/mol

1 atm

101325 Pa

760 mmHg

1 bar

10⁵ Pa

STP Conditions

273 K, 1 atm

Critical Z

Z_c = 3/8 = 0.375

📑 Quick Navigation - States of Matter & Eudiometry

States of Matter & Eudiometry JEE Main & Advanced 2025-26

Introduction to Gaseous State

1.1 Characteristics of Gases

Key Properties of Gases

1.2 Measurable Properties of Gases

1.3 Pressure Conversions (Must Memorize!)

Standard Pressure Conversions

💡 STP vs NTP

STP (Standard T & P)

NTP (Normal T & P)

Gas Laws

2.1 Boyle's Law (1662)

Pressure-Volume Relationship

2.2 Charles' Law (1787)

Volume-Temperature Relationship

2.3 Gay-Lussac's Law (Pressure Law)

Pressure-Temperature Relationship

2.4 Avogadro's Law (1811)

Volume-Amount Relationship

📝 Solved Example 1 (JEE Main Pattern)

2.5 Dalton's Law of Partial Pressures

For a Mixture of Non-Reacting Gases

Partial Pressure Formula

Mole Fraction

⚠️ Aqueous Tension

Ideal Gas Equation

3.1 The Ideal Gas Equation

The Master Equation

3.2 Values of Gas Constant R

3.3 Different Forms of Ideal Gas Equation

In terms of Mass

In terms of Density

Combined Gas Law

In terms of Molecules

💡 Calculating Molar Mass from Density

📝 Solved Example 2 (JEE Main Pattern)

Kinetic Theory of Gases

4.1 Postulates of Kinetic Theory

Key Assumptions

4.2 Kinetic Energy Equations

Important Kinetic Energy Formulas

Average KE per Molecule

Average KE per Mole

Total KE of n moles

From Gas Equation

⚠️ Key Point

4.3 Molecular Speeds

Three Types of Molecular Speeds

RMS Speed (u_rms)

Average Speed (u_avg)

Most Probable (u_mp)

Relationship (Must Memorize!)

💡 Memory Trick for Speed Ratio

📝 Solved Example 3

Diffusion and Effusion

5.1 Definitions

Diffusion

Effusion

5.2 Graham's Law

Graham's Law of Diffusion/Effusion

Different Forms of Graham's Law

In terms of Volume

In terms of Time

In terms of Distance

In terms of Moles

📝 Solved Example 4 (JEE Main Pattern)

💡 Quick Application

Real Gases & Van der Waals Equation

6.1 Causes of Deviation

Ideal Gas Assumptions

Real Gas Reality

6.2 Van der Waals Equation

The Real Gas Equation

'a' - Pressure Correction

'b' - Volume Correction

6.3 Compressibility Factor (Z)

Definition

Z = 1

Z < 1