**Definition:** Matter is composed of tiny indivisible particles called atoms and molecules that are in perpetual motion, attracting each other at moderate distances and repelling when compressed.
**Key Historical Development:**
**Avogadro's Number (NA):** 6.02 × 10²³ molecules/mole — the number of molecules in 22.4 litres of any gas at STP (273 K, 1 atm).
**Important Spacing and Distances:**
**Intermolecular Forces:**
**Exam Key Point:** Static appearance of gases is misleading; equilibrium is dynamic with constant molecular collisions and velocity changes, maintaining constant average properties.
---
**Ideal Gas Equation (Fundamental Relation):**
For a given sample of gas at low pressures and high temperatures:
PV = KT (where K varies with amount of gas)
Introducing number of molecules N:
**PV = NkBT**
where kB is **Boltzmann constant** = 1.38 × 10⁻²³ J K⁻¹
**For µ moles of gas:**
**PV = µRT** ... (12.3)
where:
**Alternative Forms of Ideal Gas Equation:**
1) **P = kBnT** where n = N/V = number density (molecules per unit volume)
2) **P = (ρ/M₀)RT** where ρ = mass density, M₀ = molar mass
**Definition of Ideal Gas:** A gas that satisfies PV = µRT exactly at all pressures and temperatures. No real gas is truly ideal; real gases approach ideal behaviour at low pressures and high temperatures (when molecules are far apart and interactions negligible).
**Derivation of Boltzmann Constant from Gas Laws:**
Starting from ideal gas equation for two different gases:
(P₁V₁)/(N₁T₁) = (P₂V₂)/(N₂T₂) = constant = kB
This constant is same for ALL gases (Avogadro's hypothesis justification).
**Boyle's Law:** For fixed µ and T: **PV = constant**
**Charles' Law:** For fixed P: **V ∝ T**
**Dalton's Law of Partial Pressures:**
For mixture of non-interacting ideal gases in same vessel:
**P = P₁ + P₂ + P₃ + ...**
where **Pi = (µiRT)/V** is partial pressure of gas i (pressure gas i would exert alone at same V and T)
**Derivation:** PV = (µ₁ + µ₂ + ...)RT shows total pressure is sum of individual contributions.
**Example Problem:** A vessel contains neon (monatomic) and oxygen (diatomic) with partial pressure ratio 3:2. Atomic mass Ne = 20.2 u, O₂ = 32.0 u. Find ratio of number of molecules and mass densities.
**Solution:**
**Estimation of Molecular Volumes:**
**Example:** Water density = 1000 kg m⁻³; water vapour at 100°C, 1 atm = 0.6 kg m⁻³
Volume increase ratio = 1000/0.6 = 1/(6 × 10⁻⁴)
Fractional molecular volume in vapour = **6 × 10⁻⁴** (ratio of molecular volume to total volume)
**Volume of Single Water Molecule:**
**Average Interatomic Distance in Water Vapour:**
**Exam Key Points:**
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**Basic Postulates:**
1. Gas consists of large number of molecules (order 10²³)
2. Molecules in incessant random motion
3. Average distance between molecules >> molecular size (factor of 10 or more)
4. Intermolecular forces negligible except during collision
5. Collisions are elastic (kinetic energy conserved)
6. Both molecular momentum and total gas momentum conserved
**Isotropic Distribution:** No preferred direction; by symmetry: **vₓ² = vᵧ² = vᵤ² = (1/3)v²**
**Derivation of Pressure Formula:**
Consider gas in cube of side l. Molecule with velocity component vx hits wall of area A = l².
**Step 1 - Momentum Change:**
**Step 2 - Number of Collisions:**
Only molecules with velocity vx approaching wall will collide. In time Δt, all molecules within distance vxΔt reach wall.
Number hitting wall in time Δt = (1/2) × n × A × vx × Δt
where n = number density (molecules per unit volume); factor 1/2 because only half molecules move towards wall.
**Step 3 - Total Momentum Transfer:**
Q = (2mvx) × (1/2 × n × A × vx × Δt) = n × m × A × vx² × Δt
**Step 4 - Pressure (Force per Unit Area):**
P = Q/(A·Δt) = **nmvx²** ... (12.11)
**Step 5 - Summing Over All Velocities:**
For molecules with different vx values:
**P = nm⟨vx²⟩** ... (12.12)
where ⟨vx²⟩ denotes average of vx².
**Step 6 - Using Isotropy:**
Since gas is isotropic (no preferred direction):
⟨vx²⟩ = ⟨vᵧ²⟩ = ⟨vᵤ²⟩
And v² = vx² + vᵧ² + vᵤ² means:
⟨v²⟩ = ⟨vx²⟩ + ⟨vᵧ²⟩ + ⟨vᵤ²⟩ = 3⟨vx²⟩
Therefore: **⟨vx²⟩ = (1/3)⟨v²⟩** ... (12.13)
**Final Result:**
**P = (1/3)nm⟨v²⟩** ... (12.14)
Or: **PV = (1/3)Nm⟨v²⟩**
**Remarks on Derivation:**
**Relating Pressure Equation to Ideal Gas Equation:**
From kinetic theory: P = (1/3)nm⟨v²⟩
From ideal gas: P = nkBT
Equating both expressions:
(1/3)nm⟨v²⟩ = nkBT
**Average Kinetic Energy per Molecule:**
**(1/2)m⟨v²⟩ = (3/2)kBT** ... (12.15)
**Key Interpretation:**
**Root Mean Square (RMS) Speed:**
From (1/2)m⟨v²⟩ = (3/2)kBT:
⟨v²⟩ = 3kBT/m
**vrms = √⟨v²⟩ = √(3kBT/m)** ... (12.16)
**Alternative Form:**
**vrms = √(3RT/M)** where M = molar mass = NAm
**Example:** Find RMS speed of nitrogen molecules at 300 K.
**Temperature and Molecular Kinetic Energy (Fundamental Result):**
Average total kinetic energy per molecule:
**⟨KE⟩ = (3/2)kBT**
Total kinetic energy for N molecules:
**E = (3/2)NkBT = (3/2)µRT**
**Physical Interpretation:**
Combining P = (1/3)nm⟨v²⟩ and (1/2)m⟨v²⟩ = (3/2)kBT:
**P = (2/3)n × (average kinetic energy per molecule)**
Or: **PV = (2/3) × (total kinetic energy)** ... (12.17)
This shows **pressure directly related to random kinetic energy of molecules**.
---
**Statement:** In a gas in thermal equilibrium, energy is equally distributed among all degrees of freedom, with each degree of freedom contributing (1/2)kBT per molecule (or (1/2)RT per mole).
**Degrees of Freedom:**
**For Monatomic Gas (He, Ne, Ar):**
**For Diatomic Gas at Room Temperature (H₂, N₂, O₂):**
**For Polyatomic Gas at Room Temperature (CO₂, H₂O non-linear):**
**Why Vibrational Modes Inactive at Room Temperature:**
**Derivation of Equipartition Principle (Qualitative):**
Consider component of velocity (say vx). From Maxwell-Boltzmann distribution, average kinetic energy associated with this direction:
⟨(1/2)mvx²⟩ = (1/2)kBT
By symmetry, each translational degree of freedom contributes (1/2)kBT to average kinetic energy. Same applies to rotational energy due to rotational inertia.
**Mathematical Form:**
For each degree of freedom with quadratic contribution to energy:
**Average energy = (1/2)kBT per molecule or (1/2)RT per mole**
---
**Definition:** Heat capacity C is heat required to raise temperature of substance by 1 K.
**Specific heat capacity c:** heat required per unit mass per unit K = C/m
**Molar heat capacity CM:** heat required per mole per unit K = C/µ
**Relation:** CM = M × c (where M is molar mass)
**Definition:** Heat supplied per mole at constant volume per unit temperature change.
**From First Law of Thermodynamics:**
dQ = dU + PdV
At constant volume (dV = 0):
dQ = dU
Therefore: **CV = (dU/dT)V = (1/µ)(dE/dT)**
**For Monatomic Ideal Gas:**
**For Diatomic Ideal Gas (5 degrees of freedom):**
**For Polyatomic Ideal Gas (6 degrees of freedom):**
**General Formula:** **CV = (f/2)R** where f = number of degrees of freedom
**Definition:** Heat supplied per mole at constant pressure per unit temperature change.
From first law: dQ = dU + PdV
At constant P: dQ = dU + PdV = dU + d(PV) = dU + d(µRT) = dU + µRdT
Therefore: **CP = (dU/dT)P + µR = CV + R**
**Mayer's Relation:** **CP = CV + R** ... (12.18)
This derives from thermodynamic relations and ideal gas law; universal for ideal gases.
**For Monatomic Gas:**
**For Diatomic Gas:**
**For Polyatomic Gas:**
**Definition:** **γ = CP/CV** (ratio of molar heat capacities)
**For Monatomic Gas:**
γ = (5/2)R / (3/2)R = **5/3 ≈ 1.67**
**For Diatomic Gas:**
γ = (7/2)R / (5/2)R = **7/5 = 1.40**
**For Polyatomic Gas:**
γ = 4R / 3R = **4/3 ≈ 1.33**
**General Relation:** **γ = (f+2)/f** where f = degrees of freedom
**Derivation:**
**Experimental Verification:**
Measured values of γ for real gases agree well with kinetic theory predictions:
**Temperature Dependence of CV:**
At low temperatures: only translational modes active → CV ≈ (3/2)R
As T increases: rotational modes activated
At very high T: vibrational modes activate → CV increases further
**Physical Interpretation:**
**Numerical Example:** Find CV, CP, and γ for diatomic oxygen at room temperature.
---
**Definition:** Average distance a molecule travels in a straight line before colliding with another molecule.
**Physical Concept:**
**Step 1 - Molecular Collision Cross-Section:**
Consider molecule of diameter d (radius r = d/2). It will collide with any other molecule whose center comes within distance d from its center. Effective collision cross-section:
σ = πd² = 4πr²
**Step 2 - Collision Rate per Molecule:**
In time dt, molecule moves distance vrmsdt. Volume swept = σ × vrmsdt
Number of molecules in this volume = n × σ × vrmsdt (where n = number density)
Collision rate (collisions per unit time): Z = n σ vrms
**However, considering relative motion** (other molecules also moving):
**Step 3 - Mean Free Path:**
Distance traveled in time interval = vrms × (1/Z)
**Mean free path λ = vrms/Z = vrms/(√2 × n π d² vrms)**
**λ = 1/(√2 × π d² × n)** ... (12.20)
Or: **λ = kBT/(√2 × π d² × P)** (using P = nkBT)
**Larger d → smaller λ:** Bigger molecules collide more frequently
**Higher n → smaller λ:** More molecules to collide with
**Higher T → larger λ:** Higher pressure increases n, decreases λ (at fixed V, higher T decreases n; at fixed P, higher T increases V, decreases n)
**Numerical Values:**
At STP (P = 1 atm, T = 273 K):
This is much larger than atomic size (~Å) but much smaller than typical container dimensions.
**Temperature and Pressure Dependence:**
Using ideal gas law n = P/(kBT):
λ = kBT/(√2 π d² P)
**Physical Significance:**
Mean free path determines:
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| Concept | Formula | SI Units | Remarks |
|---------|---------|----------|---------|
| Ideal Gas Equation | PV = μRT | Pa·m³ = J | R = 8.314 J mol⁻¹ K⁻¹ |
| Boltzmann Constant | kB = R/NA | 1.38 × 10⁻²³ J K⁻¹ | Universal constant |
| Pressure (Kinetic) | P = (1/3)nm⟨v²⟩ | Pa | n = number density |
| RMS Speed | vrms = √(3kBT/m) | m/s | Related to T directly |
| Kinetic Energy/Molecule | ⟨KE⟩ = (3/2)kBT | J | Monatomic; 5 degrees f |
| Equipartition | E = (f/2)RT per mole | J mol⁻¹ | f = degrees of freedom |
| CV (Monatomic) | CV = (3/2)R | 12.47 J mol⁻¹ K⁻¹ | 3 translational |
| CV (Diatomic) | CV = (5/2)R | 20.79 J mol⁻¹ K⁻¹ | 3 trans + 2 rot |
| Mayer's Relation | CP = CV + R | — | Always valid for ideal gas |
| Heat Capacity Ratio | γ = (f+2)/f | — | γ = CP/CV |
| Mean Free Path | λ = 1/(√2πd²n) | m | d = molecular diameter |
---
1. **Ideal gas equation applies only at low pressures and high temperatures** where intermolecular forces are negligible.
2. **Temperature is measure of average kinetic energy:** (3/2)kBT per molecule for translational motion alone.
3. **Equipartition theorem:** Each degree of freedom contributes (1/2)kBT to average molecular energy.
4. **Heat capacity depends on degrees of freedom:** CV = (f/2)R; doubling f doubles CV (approximately).
5. **Mayer relation CP - CV = R is universal** for all ideal gases regardless of molecular complexity.
6. **Heat capacity ratio γ varies with molecule type:** Monatomic (5/3), Diatomic (7/5), Polyatomic (4/3).
7. **Mean free path inversely proportional to pressure** at fixed temperature: λ ∝ 1/P.
8. **Pressure derives from molecular collisions:** Each collision transfers momentum 2mvx to wall.
9. **Avogadro's hypothesis justified by kinetic theory:** Equal P, V, T → equal N for all gases.
10. **Boltzmann constant kB = 1.38 × 10⁻²³ J K⁻¹** is fundamental constant relating energy and temperature at molecular scale.
11. **RMS speed increases with temperature:** vrms = √(3RT/M); doubles when T quadruples.
12. **In gas-wall collisions:** elastic collision assumed; component perpendicular to wall reverses; parallel components unchanged.
13. **Isotropic distribution essential:** ⟨vx²⟩ = ⟨vy²⟩ = ⟨vz²⟩ = (1/3)⟨v²⟩ allows simplification to P = (1/3)nm⟨v²⟩.
14. **Partial pressures sum for gas mixtures:** Pi = (μiRT)/V for each component; total P = ΣPi (Dalton's law).
15. **Molecular volume fraction in vapour:** Ratio of actual molecular volume to container volume is ~10⁻⁴ to 10⁻⁵ at STP.
Q1. The size of an atom is approximately:
Answer: B — One angstrom (Å) equals 10⁻¹⁰ m, which is the standard atomic diameter.
Q2. In the Vaiseshika school of ancient India, how many kinds of atoms were postulated?
Answer: C — Kanada's Vaiseshika school proposed four atomic types: earth, water, fire, and air; Akasa (space) was thought continuous, not atomic.
Q3. Which scientist explained Gay-Lussac's law by combining Dalton's theory with a molecular hypothesis?
Answer: B — Avogadro's hypothesis (equal volumes of gases at same T and P contain equal molecules) combined with Dalton's atomic theory explains why gas volumes mix in simple integer ratios.
Q4. The mean free path of molecules in a gas is approximately:
Answer: C — Mean free path in gases is much larger than atomic separation due to low density; molecules travel thousands of angstroms before collision.
Q5. Why are interatomic forces negligible in gases but important in solids and liquids?
Answer: B — Gas molecules are separated by tens of angstroms while solids/liquids have atoms only ~2 angstroms apart; interatomic forces are short-range and only matter at close distances.
Q6. A gas sample at STP contains 2 litres of oxygen and another contains 2 litres of nitrogen. Which statement is true according to Avogadro's hypothesis?
Answer: C — Avogadro's hypothesis states equal volumes of all gases at equal temperature and pressure contain the same number of molecules, regardless of identity or density.
Q7. Why did ancient atomic ideas of Kanada and Democritus fail to evolve into modern science despite being ingenious?
Answer: B — The text explicitly states these ideas lacked quantitative experimental support, which is the hallmark of modern science needed for development.
Q8. Which pair of statements is correct according to kinetic theory? (A) Gas is in static equilibrium; pressure is continuous force on walls. (B) Gas is in dynamic equilibrium; pressure results from molecular collisions.
Answer: B — Static appearance of gas is misleading—it is in dynamic equilibrium with molecules continuously colliding, and gas pressure arises from these molecular impacts on container walls, not continuous force.
Q9. If atomic separation in solids is 2 angstroms and in gases is tens of angstroms, approximately how many times farther apart are gas molecules compared to solid atoms?
Answer: B — Tens of angstroms divided by 2 angstroms gives a factor of 5–50; the lower estimate of 'tens' (e.g. 10 Å) yields approximately 5 times, while higher tens (e.g. 30 Å) yield ~15 times, placing the answer in the 5–10 range for typical values.
Q10. Modern kinetic theory successfully predicts which of the following properties? (I) Specific heat capacities of gases (II) Viscosity and thermal conduction of gases (III) Molecular sizes and masses (IV) Colour and taste of liquids
Answer: B — Kinetic theory successfully explains specific heat, viscosity, conduction, diffusion, and yields estimates of molecular parameters; colour and taste are chemical properties outside kinetic theory scope.
What is the Atomic Hypothesis in simple terms?
All matter consists of tiny atoms in perpetual motion that attract at distance and repel when squeezed together.
Why can we neglect interatomic forces in gases but not in solids?
In gases, atoms are separated by tens of angstroms so are far apart; in solids, atoms are only 2 angstroms apart making forces critical.
What is the mean free path of a molecule?
The average distance a molecule travels without colliding with another molecule, typically thousands of angstroms in gases.
State Avogadro's hypothesis.
Equal volumes of all gases at equal temperature and pressure contain the same number of molecules.
How do ancient Indian and Greek ideas about atoms differ from Dalton's theory?
Ancient ideas were intuitive conjectures unsupported by quantitative experiments, while Dalton's theory used experimental evidence to explain chemical laws.
What did Democritus suggest about the shape of water atoms?
Water atoms were smooth and round, unable to hook onto each other, explaining why liquid flows easily.
Why is the equilibrium of a gas dynamic, not static?
Molecules continuously collide and change speeds while only average properties remain constant.
What is the typical size of an atom in metres?
Approximately one angstrom, which equals 10⁻¹⁰ m.
How does Gay-Lussac's law relate to Avogadro's hypothesis?
Avogadro's hypothesis combined with Dalton's atomic theory explains why gas volumes mix in simple integer ratios.
Name two famous scientists who developed kinetic theory in the 19th century.
Maxwell and Boltzmann were the principal developers of kinetic theory.
Define the mean free path of a gas molecule. Why is it much larger in gases than in solids? [2 marks]
Mean free path = average distance before collision. Compare atomic separation distances in gases (tens Å) vs solids (2 Å); larger separation → fewer collisions → longer path.
Explain how Avogadro's hypothesis combined with Dalton's atomic theory explains Gay-Lussac's law of combining volumes. Provide an example with two gases. [5 marks]
Use Dalton: each element has fixed atoms forming molecules. Apply Avogadro: equal volumes at same T, P have equal molecules. Show: if 1 volume H₂ + 1 volume Cl₂ → 2 volumes HCl (ratio 1:1:2), this must hold because molecule numbers are equal; demonstrate with nitrogen and oxygen combination or hydrogen and chlorine.
Derive why kinetic theory of gases neglects interatomic forces, whereas such forces are crucial in solids and liquids. In your answer, compare (i) atomic separation distances, (ii) nature of intermolecular forces, and (iii) the resulting physical state properties. Also explain what 'dynamic equilibrium' means in the context of a gas. [6 marks]
Show atomic separations: solids ~2 Å, gases ~tens of Å. State intermolecular forces are short-range (effective within ~2–3 Å). Conclude: in gases, average separation exceeds force range, making forces negligible; in solids, separations within force range make them critical. Define dynamic equilibrium: molecules collide continuously, exchanging speed and direction, yet average properties (P, T) remain constant over time.
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