Gaseous State Class 11 Chemistry: Notes, Formulas & Numericals

 Dr.Sanjaykumar Pawar 

GASEOUS STATE – COMPLETE BEGINNER NOTES 

Gaseous State – Understand gas laws, molecular motion, ideal gas behaviour and kinetic theory with easy Class 11 Chemistry notes.

Introduction to Gaseous State

The gaseous state is one of the three fundamental states of matter. Gases have no fixed shape or volume. They occupy the entire container in which they are placed. Gas molecules are continuously moving in random directions. They collide with one another and with the walls of the container. The behaviour of gases can be explained using gas laws and the kinetic molecular theory.

Learning Objectives

After completing this lesson, students will be able to:

• Understand the fundamental gas laws.
• Apply the ideal gas equation.
• Calculate molecular speed and kinetic energy.
• Understand real gases and Van der Waals equation.
• Explain critical temperature and pressure.
• Calculate mean free path and collision frequency.
• Solve basic numerical problems.

1. Boyle's Law

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

PV = Constant

P₁V₁ = P₂V₂

Where:

• P = Pressure of gas
• V = Volume of gas
• P₁, V₁ = Initial pressure and volume
• P₂, V₂ = Final pressure and volume

Conditions: Temperature and amount of gas must remain constant.

Simple Explanation: When pressure increases, volume decreases. When pressure decreases, volume increases.

Example:
A gas occupies 4 L at 2 atm pressure. What will be its volume at 4 atm? P₁V₁ = P₂V₂ 2 × 4 = 4 × V₂ V₂ = 2 L

2. Charles's Law

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

V/T = Constant

V₁/T₁ = V₂/T₂

Where:

• V = Volume of gas
• T = Absolute temperature in Kelvin

Conditions: Pressure and amount of gas remain constant.

Always convert temperature into Kelvin before applying gas laws. K = °C + 273.15

Example:
A gas occupies 2 L at 300 K. Find its volume at 600 K at constant pressure. V₂ = V₁ × T₂/T₁ V₂ = 2 × 600/300 V₂ = 4 L

3. Gay-Lussac's Law (Pressure Law)

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

P/T = Constant

P₁/T₁ = P₂/T₂

Where:

• P = Pressure
• T = Absolute temperature

Conditions: Volume and amount of gas remain constant.

Example:
A gas has a pressure of 2 atm at 300 K. Find its pressure at 450 K. P₂ = P₁ × T₂/T₁ P₂ = 2 × 450/300 P₂ = 3 atm

4. Avogadro's Law

At constant temperature and pressure, the volume of a gas is directly proportional to the number of moles.

V ∝ n

V/n = Constant

V₁/n₁ = V₂/n₂

Where:

• V = Volume
• n = Number of moles

Important Concept: Equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules. One mole contains approximately 6.022 × 10²³ particles.

If 2 moles of a gas occupy 10 L, then 4 moles at the same temperature and pressure occupy: V₂ = 10 × 4/2 V₂ = 20 L

5. Ideal Gas Equation

The ideal gas equation combines Boyle's law, Charles's law and Avogadro's law.

PV = nRT

Other useful forms:

PV = (w/M)RT

P = (d/M)RT

Symbols and Meanings:

Symbol	Meaning
P	Pressure
V	Volume
n	Number of moles
R	Universal gas constant
T	Absolute temperature
w	Mass of gas
M	Molar mass
d	Density of gas

Important Value: R = 8.314 J mol⁻¹ K⁻¹

Example:
Calculate the volume of 1 mole of an ideal gas at 273.15 K and 1 atm pressure. PV = nRT V = nRT/P V = (1 × 0.082057 × 273.15)/1 V ≈ 22.4 L

6. Graham's Law of Diffusion

Diffusion is the process by which gas molecules spread and mix with one another. According to Graham's law, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass, under the same conditions.

r₁/r₂ = √(M₂/M₁)

Where:

• r₁, r₂ = Rates of diffusion
• M₁, M₂ = Molar masses of gases

Example:
Compare the diffusion rates of hydrogen and oxygen. Molar mass of H₂ = 2 g/mol Molar mass of O₂ = 32 g/mol r(H₂)/r(O₂) = √(32/2) = √16 = 4 Hydrogen diffuses four times faster than oxygen.

Lighter gas molecules generally diffuse faster than heavier gas molecules.

7. Dalton's Law of Partial Pressures

According to Dalton's law, the total pressure of a mixture of non-reacting gases is equal to the sum of their individual partial pressures.

P(total) = P₁ + P₂ + P₃ + ...

Partial pressure of a gas:

Pᵢ = Xᵢ × P(total)

Where: Xᵢ = Mole fraction of the gas.

Example:
A container contains nitrogen at 2 atm and oxygen at 1 atm. Total pressure = 2 + 1 = 3 atm

8. Absolute Density of a Gas

Density is the mass of a substance per unit volume.

d = Mass / Volume

For an ideal gas:

d = PM / RT

SI Unit: kg/m³ Another commonly used unit is g/L.

If 10 g of a gas occupies 5 L: d = 10/5 d = 2 g/L

9. Vapour Density

Vapour density is the ratio of the mass of a given volume of gas to the mass of an equal volume of hydrogen under the same temperature and pressure.

VD = Molecular Mass / 2

Therefore:

Molecular Mass = 2 × VD

If the vapour density of a gas is 16: Molecular mass = 2 × 16 = 32 g/mol

10. Molecular Speed of Gas Molecules

Gas molecules move continuously with different speeds. Three important types of molecular speeds are:

A. Average Speed

u(avg) = √(8RT/πM)

B. Root Mean Square Speed

u(rms) = √(3RT/M)

C. Most Probable Speed

u(mp) = √(2RT/M)

Where:

• R = Gas constant
• T = Temperature in Kelvin
• M = Molar mass in kg/mol
• π = Mathematical constant (approximately 3.1416)

u(mp) : u(avg) : u(rms) = 1 : 1.128 : 1.224

Important: Most probable speed is the speed possessed by the maximum number of molecules. RMS speed is always greater than average speed for a Maxwellian gas distribution.

11. Kinetic Equation of Gases

The kinetic theory explains the relationship between gas pressure and the motion of molecules.

PV = (1/3)Nm u(rms)²

Where:

• P = Pressure
• V = Volume
• N = Number of molecules
• m = Mass of one molecule
• u(rms) = Root mean square speed

This equation shows that gas pressure is related to the kinetic energy of its molecules.

Do not confuse N (number of molecules) with n (number of moles).

12. Kinetic Energy of Gas Molecules

The average translational kinetic energy of one mole of an ideal gas is:

KE per mole = (3/2)RT

Average kinetic energy per molecule:

KE per molecule = (3/2)kT

Where:

• R = Universal gas constant
• k = Boltzmann constant
• T = Absolute temperature

Boltzmann constant: k = 1.380649 × 10⁻²³ J/K

Important Concept:
The average translational kinetic energy of gas molecules depends only on absolute temperature, not on the nature or mass of the gas.

13. Van der Waals Equation

Real gases do not behave ideally under all conditions because gas molecules have finite volume and intermolecular forces. Van der Waals introduced two correction factors. For one mole:

(P + a/V²)(V - b) = RT

For n moles:

(P + an²/V²)(V - nb) = nRT

Meaning of Constants:

• a = Correction for intermolecular attraction.
• b = Correction for the actual volume occupied by gas molecules.

Important: A larger value of a indicates stronger intermolecular attraction. A larger value of b indicates a larger excluded volume per mole.

The ideal gas equation is a good approximation at low pressure and high temperature.

14. Boyle Temperature

Boyle temperature is the temperature at which a real gas behaves approximately like an ideal gas over a range of low pressures.

T(B) = a / (Rb)

Where:

• a = Van der Waals attraction constant
• b = Van der Waals volume constant
• R = Universal gas constant

15. Critical Constants

Critical constants are the temperature, pressure and volume associated with the critical point of a substance. For a Van der Waals gas:

Critical Temperature

Tc = 8a / 27Rb

Critical Pressure

Pc = a / 27b²

Critical Molar Volume

Vc = 3b

Where:

• Tc = Critical temperature
• Pc = Critical pressure
• Vc = Critical molar volume

Important Concept: Above the critical temperature, a gas cannot be liquefied by applying pressure alone.

16. Inversion Temperature

Inversion temperature is the temperature at which the Joule-Thomson coefficient changes sign. For a Van der Waals gas, the maximum inversion temperature is:

Ti = 2a / (Rb)

Important: Below the inversion temperature, a gas generally cools during Joule-Thomson expansion. Above it, the gas generally warms.

17. Law of Corresponding States

The law of corresponding states states that different gases show approximately similar behaviour when compared at the same reduced temperature, reduced pressure and reduced volume. Reduced properties:

Pr = P/Pc

Tr = T/Tc

Vr = V/Vc

Van der Waals equation in reduced form:

(Pr + 3/Vr²)(3Vr - 1) = 8Tr

Where:

• Pr = Reduced pressure
• Tr = Reduced temperature
• Vr = Reduced molar volume

18. Mean Free Path

Mean free path is the average distance travelled by a gas molecule between two successive collisions.

λ = 1 / (√2 π σ² n*)

Where:

• λ = Mean free path
• σ = Molecular diameter
• n* = Number of molecules per unit volume
• π = 3.1416 approximately

Since n* = N/V:

λ = V / (√2 π σ² N)

SI Unit: Metre (m)

Mean free path decreases when pressure increases, provided temperature and molecular diameter remain constant.

19. Collision Frequency

Collision frequency is the average number of collisions made by a molecule per unit time.

Collision Frequency = u(rms) / λ

Where:

• u(rms) = Root mean square speed
• λ = Mean free path

SI Unit: s⁻¹

If RMS speed = 500 m/s and mean free path = 0.0001 m: Collision frequency = 500/0.0001 = 5 × 10⁶ s⁻¹

20. Specific Heat and Molar Heat Capacity

Heat capacity is the amount of heat required to raise the temperature of a substance by one degree. Two important heat capacities of gases are:

A. Heat Capacity at Constant Pressure

Cp = Molar heat capacity at constant pressure.

B. Heat Capacity at Constant Volume

Cv = Molar heat capacity at constant volume. For one mole of an ideal gas:

Cp - Cv = R

For specific heat capacities per unit mass:

cp - cv = R/M

Ratio of heat capacities:

γ = Cp/Cv = cp/cv

Where:

• Cp = Molar heat capacity at constant pressure
• Cv = Molar heat capacity at constant volume
• cp = Specific heat at constant pressure
• cv = Specific heat at constant volume
• γ = Heat capacity ratio

Important Concept: For an ideal gas, Cp is always greater than Cv. This is because a gas expands and performs work when heated at constant pressure.

IMPORTANT FORMULA REVISION TABLE

Topic	Important Formula
Boyle's Law	PV = Constant
Charles's Law	V/T = Constant
Pressure Law	P/T = Constant
Avogadro's Law	V ∝ n
Ideal Gas	PV = nRT
Graham's Law	r₁/r₂ = √(M₂/M₁)
Dalton's Law	Pt = ΣPi
Density	d = PM/RT
Vapour Density	M = 2 × VD
RMS Speed	√(3RT/M)
Average KE	(3/2)RT per mole
Van der Waals	(P+a/V²)(V-b)=RT
Boyle Temperature	a/Rb
Critical Temperature	8a/27Rb
Critical Pressure	a/27b²
Critical Volume	3b
Mean Free Path	1/(√2πσ²n*)
Collision Frequency	u(rms)/λ
Heat Capacity	Cp - Cv = R

NEET / JEE PRACTICE QUESTIONS

Q1. According to Boyle's law, at constant temperature, pressure is:
A. Directly proportional to volume
B. Inversely proportional to volume
C. Equal to volume
D. Independent of volume

Correct Answer: B

Q2. The ideal gas equation is:
A. PV = RT
B. PV = nRT
C. P = VT
D. V = PT

Correct Answer: B

Q3. Which gas diffuses faster?
A. Oxygen
B. Carbon dioxide
C. Hydrogen
D. Nitrogen

Correct Answer: C

Q4. The RMS speed of gas molecules is proportional to:
A. T
B. √T
C. 1/T
D. T²

Correct Answer: B

Q5. The average translational kinetic energy per mole of an ideal gas is:
A. RT
B. 2RT
C. 3RT/2
D. RT/2

Correct Answer: C

Q6. For one mole of an ideal gas, the value of Cp - Cv is:
A. Zero
B. R
C. 2R
D. 3R

Correct Answer: B

Q7. The critical volume of one mole of a Van der Waals gas is:
A. b
B. 2b
C. 3b
D. 4b

Correct Answer: C

Q8. Mean free path is inversely proportional to:
A. Temperature
B. Molecular number density
C. Molecular speed
D. Molecular mass

Correct Answer: B

FINAL LEARNING CHECKLIST

• Understand all four fundamental gas laws.
• Learn the ideal gas equation and its applications.
• Understand diffusion and partial pressure.
• Memorize molecular speed formulas.
• Understand kinetic energy and kinetic theory.
• Study Van der Waals equation and critical constants.
• Practise mean free path and collision frequency problems.
• Learn the relationship between Cp and Cv.

Conclusion

The Gaseous State chapter is an important part of Physical Chemistry. A clear understanding of gas laws, molecular motion, kinetic energy and real gas behaviour provides a strong foundation for solving numerical problems in NEET and JEE examinations. Students should understand the concepts first and then practise numerical problems regularly.

End of Lesson – Gaseous State