Chapter - Gaseous State
1. Introduction to Gaseous State
Gaseous State: The gaseous state is one of the fundamental states of matter characterized by particles that are far apart, moving randomly with high kinetic energy, and having negligible intermolecular forces.
Characteristics of Gases:
- Gases have no definite shape or volume
- They completely fill the container in which they are placed
- Gases are highly compressible
- They have very low density compared to solids and liquids
- Gases mix evenly and completely in all proportions
- They exert pressure equally in all directions
Note: The behavior of gases can be explained using kinetic molecular theory and various gas laws.
2. Measurable Properties of Gases
2.1 Pressure (P)
Pressure: The force exerted by gas molecules per unit area on the walls of the container.
Pressure = Force / Area
P = F/A
Units of Pressure:
| Unit |
Symbol |
Relation with atm |
| Atmosphere |
atm |
1 atm |
| Pascal |
Pa |
1 atm = 101325 Pa |
| Torr |
Torr |
1 atm = 760 Torr |
| Bar |
bar |
1 atm = 1.01325 bar |
| mmHg |
mmHg |
1 atm = 760 mmHg |
2.2 Volume (V)
Volume: The space occupied by gas molecules.
Units: Litre (L), millilitre (mL), cubic meter (m³), cubic centimeter (cm³)
1 L = 1000 mL = 1 dm³ = 10⁻³ m³
2.3 Temperature (T)
Temperature: A measure of the average kinetic energy of gas molecules.
Units: Kelvin (K), Celsius (°C), Fahrenheit (°F)
K = °C + 273.15
°C = (°F - 32) × 5/9
2.4 Amount of Gas (n)
Amount of Gas: The quantity of gas expressed in moles.
n = m/M
where m = mass, M = molar mass
3. Gas Laws
3.1 Boyle's Law
Boyle's Law: At constant temperature, the pressure of a fixed amount of gas is inversely proportional to its volume.
P ∝ 1/V (at constant T and n)
PV = constant
P₁V₁ = P₂V₂
Example: A gas occupies 2.0 L at 1.5 atm pressure. What volume will it occupy at 3.0 atm at the same temperature?
Solution:
Given: V₁ = 2.0 L, P₁ = 1.5 atm, P₂ = 3.0 atm
Using Boyle's Law: P₁V₁ = P₂V₂
V₂ = P₁V₁/P₂ = (1.5 × 2.0)/3.0 = 1.0 L
3.2 Charles's Law
Charles's Law: At constant pressure, the volume of a fixed amount of gas is directly proportional to its absolute temperature.
V ∝ T (at constant P and n)
V/T = constant
V₁/T₁ = V₂/T₂
Example: A balloon has a volume of 2.5 L at 27°C. What will be its volume at 127°C at constant pressure?
Solution:
Given: V₁ = 2.5 L, T₁ = 27 + 273 = 300 K, T₂ = 127 + 273 = 400 K
Using Charles's Law: V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁ = (2.5 × 400)/300 = 3.33 L
3.3 Gay-Lussac's Law (Pressure Law)
Gay-Lussac's Law: At constant volume, the pressure of a fixed amount of gas is directly proportional to its absolute temperature.
P ∝ T (at constant V and n)
P/T = constant
P₁/T₁ = P₂/T₂
3.4 Avogadro's Law
Avogadro's Law: At constant temperature and pressure, the volume of a gas is directly proportional to the number of moles of gas.
V ∝ n (at constant T and P)
V/n = constant
V₁/n₁ = V₂/n₂
Avogadro's Number: 6.022 × 10²³ particles per mole
Molar Volume: At STP, 1 mole of any gas occupies 22.4 L
4. Ideal Gas Equation
Ideal Gas Equation: Combining all gas laws, we get the ideal gas equation.
PV = nRT
Where:
- P = Pressure (atm, Pa, bar)
- V = Volume (L, m³)
- n = Number of moles (mol)
- R = Universal gas constant
- T = Absolute temperature (K)
4.1 Values of Gas Constant R
| Value |
Units |
| 0.0821 |
L atm mol⁻¹ K⁻¹ |
| 8.314 |
J mol⁻¹ K⁻¹ |
| 1.987 |
cal mol⁻¹ K⁻¹ |
| 8.314 × 10⁻² |
L bar mol⁻¹ K⁻¹ |
4.2 Other Forms of Ideal Gas Equation
PV = (m/M)RT
Density form: PM = ρRT
where ρ = density = m/V
For mixture of gases: PtotalV = ntotalRT
Example: Calculate the pressure exerted by 2.0 moles of CO₂ in a 10.0 L container at 27°C.
Solution:
Given: n = 2.0 mol, V = 10.0 L, T = 27 + 273 = 300 K
Using PV = nRT
P = nRT/V = (2.0 × 0.0821 × 300)/10.0 = 4.926 atm
5. Dalton's Law of Partial Pressures
Dalton's Law: The total pressure exerted by a mixture of non-reacting gases is equal to the sum of partial pressures of individual gases.
Ptotal = P₁ + P₂ + P₃ + ... + Pn
Partial Pressure: The pressure that would be exerted by a component gas if it alone occupied the entire volume at the same temperature.
P₁ = n₁RT/V = χ₁ × Ptotal
where χ₁ = mole fraction = n₁/ntotal
Example: A mixture contains 2.0 mol N₂ and 3.0 mol O₂ at 1 atm total pressure. Calculate partial pressures.
Solution:
ntotal = 2.0 + 3.0 = 5.0 mol
χN₂ = 2.0/5.0 = 0.4
χO₂ = 3.0/5.0 = 0.6
PN₂ = 0.4 × 1 = 0.4 atm
PO₂ = 0.6 × 1 = 0.6 atm
6. Graham's Law of Diffusion and Effusion
6.1 Diffusion and Effusion
Diffusion: The process of spontaneous spreading of gases to fill the available space.
Effusion: The process of escape of gas molecules through a tiny hole into vacuum.
Graham's Law: The rate of diffusion or effusion of gases is inversely proportional to the square root of their molar masses.
r₁/r₂ = √(M₂/M₁)
For effusion time: t₁/t₂ = √(M₁/M₂)
Example: Compare the rates of effusion of H₂ and O₂ gases.
Solution:
MH₂ = 2 g/mol, MO₂ = 32 g/mol
rH₂/rO₂ = √(32/2) = √16 = 4
Therefore, H₂ effuses 4 times faster than O₂.
7. Kinetic Molecular Theory
Kinetic Molecular Theory: A theoretical model that explains the behavior of gases based on the motion of gas molecules.
Postulates of Kinetic Molecular Theory:
- Gas molecules are in constant, random motion
- The volume of gas molecules is negligible compared to the container volume
- Gas molecules exert no intermolecular forces except during collisions
- Collisions between gas molecules are perfectly elastic
- The average kinetic energy is directly proportional to absolute temperature
7.1 Mathematical Relations from KMT
Average Kinetic Energy: KEavg = (3/2)RT
Root Mean Square Velocity: urms = √(3RT/M)
Most Probable Velocity: ump = √(2RT/M)
Average Velocity: uavg = √(8RT/πM)
Note: urms > uavg > ump
Example: Calculate the rms velocity of O₂ at 27°C.
Solution:
Given: M = 32 g/mol = 0.032 kg/mol, T = 300 K
urms = √(3RT/M) = √(3 × 8.314 × 300/0.032) = 483.7 m/s
8. Real Gases and Deviations from Ideal Behavior
Real Gases: Gases that do not obey ideal gas law perfectly due to finite molecular size and intermolecular forces.
8.1 Conditions for Maximum Deviation
- High pressure (molecules close together)
- Low temperature (low kinetic energy)
- Large molecular size
- Strong intermolecular forces
8.2 Compressibility Factor (Z)
Z = PV/nRT
For ideal gas: Z = 1
For real gas: Z ≠ 1
8.3 Van der Waals Equation
Van der Waals Equation: A modified form of ideal gas equation that accounts for molecular size and intermolecular forces.
(P + a/V²)(V - b) = RT
For n moles: (P + an²/V²)(V - nb) = nRT
Where:
- a: Van der Waals constant for intermolecular attraction (atm L² mol⁻²)
- b: Van der Waals constant for molecular volume (L mol⁻¹)
Note: Larger 'a' values indicate stronger intermolecular forces. Larger 'b' values indicate larger molecular size.
9. Liquefaction of Gases
Liquefaction: The process of converting gas into liquid state by cooling and/or applying pressure.
9.1 Critical Constants
Critical Temperature (Tc): The temperature above which a gas cannot be liquefied regardless of pressure applied.
Critical Pressure (Pc): The minimum pressure required to liquefy a gas at its critical temperature.
Critical Volume (Vc): The volume occupied by one mole of gas at critical temperature and pressure.
Tc = 8a/27Rb
Pc = a/27b²
Vc = 3b
9.2 Methods of Liquefaction
- Linde's Method: Cooling by Joule-Thomson effect
- Claude's Method: Cooling by adiabatic expansion doing external work
- Cascade Process: Using series of refrigerants
10. Important Formulas and Constants
10.1 Key Formulas Summary
| Law/Concept |
Formula |
Conditions |
| Boyle's Law |
P₁V₁ = P₂V₂ |
Constant T, n |
| Charles's Law |
V₁/T₁ = V₂/T₂ |
Constant P, n |
| Gay-Lussac's Law |
P₁/T₁ = P₂/T₂ |
Constant V, n |
| Avogadro's Law |
V₁/n₁ = V₂/n₂ |
Constant T, P |
| Ideal Gas Equation |
PV = nRT |
All variables |
| Density Form |
PM = ρRT |
ρ = density |
| Dalton's Law |
Ptotal = ΣPi |
Non-reacting gases |
| Graham's Law |
r₁/r₂ = √(M₂/M₁) |
Same T, P |
| RMS Velocity |
urms = √(3RT/M) |
Kinetic theory |
| Van der Waals |
(P + a/V²)(V - b) = RT |
Real gases |
10.2 Important Constants
| Constant |
Symbol |
Value |
Units |
| Avogadro's Number |
NA |
6.022 × 10²³ |
particles/mol |
| Gas Constant |
R |
8.314 |
J mol⁻¹ K⁻¹ |
| Gas Constant |
R |
0.0821 |
L atm mol⁻¹ K⁻¹ |
| Molar Volume (STP) |
Vm |
22.4 |
L/mol |
| Standard Pressure |
P° |
1 |
atm = 760 mmHg |
| Standard Temperature |
T° |
273.15 |
K = 0°C |
11. Numerical Problems and Solutions
Problem 1: A gas at 27°C and 1 atm pressure occupies 500 mL. What volume will it occupy at 127°C and 2 atm pressure?
Solution:
Given: T₁ = 27 + 273 = 300 K, P₁ = 1 atm, V₁ = 500 mL
T₂ = 127 + 273 = 400 K, P₂ = 2 atm, V₂ = ?
Using combined gas law: P₁V₁/T₁ = P₂V₂/T₂
V₂ = P₁V₁T₂/(P₂T₁) = (1 × 500 × 400)/(2 × 300) = 333.33 mL
Problem 2: Calculate the density of CO₂ at STP.
Solution:
At STP: P = 1 atm, T = 273 K
Molar mass of CO₂ = 12 + (2 × 16) = 44 g/mol
Using PM = ρRT
ρ = PM/RT = (1 × 44)/(0.0821 × 273) = 1.96 g/L
Problem 3: A mixture contains 0.5 mol N₂ and 1.5 mol O₂ at 2 atm total pressure. Calculate partial pressures.
Solution:
Total moles = 0.5 + 1.5 = 2.0 mol
Mole fraction of N₂ = 0.5/2.0 = 0.25
Mole fraction of O₂ = 1.5/2.0 = 0.75
PN₂ = 0.25 × 2 = 0.5 atm
PO₂ = 0.75 × 2 = 1.5 atm
Problem 4: Calculate the rms velocity of H₂ gas at 25°C.
Solution:
Given: T = 25 + 273 = 298 K
Molar mass of H₂ = 2 g/mol = 0.002 kg/mol
urms = √(3RT/M) = √(3 × 8.314 × 298/0.002)
urms = √(3,712,266) = 1926 m/s
Problem 5: If H₂ effuses through a hole in 10 minutes, how long will it take for O₂ to effuse under similar conditions?
Solution:
Given: tH₂ = 10 min, MH₂ = 2 g/mol, MO₂ = 32 g/mol
Using Graham's Law: t₁/t₂ = √(M₁/M₂)
tO₂/tH₂ = √(MO₂/MH₂) = √(32/2) = 4
tO₂ = 4 × 10 = 40 minutes
Multiple Choice Questions
Additional Practice Questions
1. At STP, 22.4 L of any gas contains:
(A)
6.022 × 10²³ molecules
(B)
6.022 × 10²² molecules
(C)
3.011 × 10²³ molecules
(D)
12.044 × 10²³ molecules
2. Which gas law is represented by PV = constant?
(A)
Charles's Law
(B)
Boyle's Law
(C)
Gay-Lussac's Law
(D)
Avogadro's Law
3. The value of universal gas constant R in SI units is:
(A)
0.0821 L atm mol⁻¹ K⁻¹
(B)
8.314 J mol⁻¹ K⁻¹
(C)
1.987 cal mol⁻¹ K⁻¹
(D)
62.36 L Torr mol⁻¹ K⁻¹
4. According to Graham's law, the rate of effusion is:
(A)
Directly proportional to molar mass
(B)
Inversely proportional to molar mass
(C)
Inversely proportional to square root of molar mass
(D)
Directly proportional to square root of molar mass
5. In van der Waals equation, the constant 'a' represents:
(A)
Molecular volume
(B)
Intermolecular forces
(C)
Temperature correction
(D)
Pressure correction
6. The compressibility factor (Z) for an ideal gas is:
(A)
Greater than 1
(B)
Less than 1
(C)
Equal to 1
(D)
Equal to 0
7. At constant temperature and pressure, 2 L of CO₂ contains the same number of molecules as:
(A)
2 L of O₂
(B)
1 L of N₂
(C)
4 L of H₂
(D)
0.5 L of CH₄
8. Which of the following relationships is correct for root mean square velocity?
(A)
urms = √(2RT/M)
(B)
urms = √(3RT/M)
(C)
urms = √(8RT/πM)
(D)
urms = √(RT/M)
9. The critical temperature of CO₂ is 31°C. This means:
(A)
CO₂ cannot exist as liquid above 31°C
(B)
CO₂ cannot exist as gas above 31°C
(C)
CO₂ freezes at 31°C
(D)
CO₂ boils at 31°C
10. Which gas deviates maximum from ideal behavior?
(A)
H₂
(B)
He
(C)
NH₃
(D)
Ne
11. The ratio of rms velocities of H₂ and O₂ at the same temperature is:
(A)
1:4
(B)
4:1
(C)
1:16
(D)
16:1
12. At what temperature will the rms velocity of SO₂ be equal to that of O₂ at 300 K?
(A)
600 K
(B)
150 K
(C)
450 K
(D)
900 K
13. The van der Waals constant 'b' has units:
(A)
L mol⁻¹
(B)
atm L² mol⁻²
(C)
atm L mol⁻¹
(D)
L² mol⁻²
14. According to kinetic molecular theory, gas pressure is due to:
(A)
Intermolecular attraction
(B)
Molecular collisions with container walls
(C)
Molecular volume
(D)
Temperature of gas
15. In the van der Waals equation, the term an²/V² accounts for:
(A)
Finite molecular size
(B)
Intermolecular forces of attraction
(C)
High pressure
(D)
Low temperature
12. Answer Key and Explanations
| Question |
Answer |
Explanation |
| 1 |
(A) |
At STP, 1 mole = 22.4 L contains NA = 6.022 × 10²³ molecules |
| 2 |
(B) |
Boyle's Law states PV = constant at constant temperature |
| 3 |
(B) |
R = 8.314 J mol⁻¹ K⁻¹ is the SI unit value |
| 4 |
(C) |
Graham's Law: r ∝ 1/√M |
| 5 |
(B) |
Constant 'a' represents intermolecular forces of attraction |
| 6 |
(C) |
For ideal gas, Z = PV/nRT = 1 |
| 7 |
(A) |
Avogadro's Law: Equal volumes contain equal molecules at same T, P |
| 8 |
(B) |
RMS velocity formula: urms = √(3RT/M) |
| 9 |
(A) |
Above critical temperature, gas cannot be liquefied |
| 10 |
(C) |
NH₃ has strong intermolecular forces (hydrogen bonding) |
| 11 |
(B) |
urms ∝ 1/√M, ratio = √(32/2) = 4:1 |
| 12 |
(A) |
urms ∝ √(T/M), TSO₂ = 300 × (64/32) = 600 K |
| 13 |
(A) |
Constant 'b' represents excluded volume per mole |
| 14 |
(B) |
Pressure results from molecular collisions with walls |
| 15 |
(B) |
The term an²/V² corrects for intermolecular attractions |
Summary Points
- Gases have no definite shape or volume and completely fill their containers
- Gas behavior is described by measurable properties: pressure, volume, temperature, and amount
- Gas laws relate these properties under specific conditions
- The ideal gas equation PV = nRT combines all gas laws
- Real gases deviate from ideal behavior due to molecular size and intermolecular forces
- Van der Waals equation accounts for real gas behavior
- Kinetic molecular theory explains gas behavior based on molecular motion
- Critical constants define the conditions for gas liquefaction
Remember: Always convert temperature to Kelvin when using gas law equations!