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Chapter 12 · Kinetic Theory of Gases — NCERT Class 11 Physics

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25 MCQ Questions 15 Numerical Problems Moderate Level Show Solutions NCERT Aligned
Topics Covered
Molecular Nature of Matter Ideal Gas Equation PV = μRT Boltzmann Constant k_B Avogadro's Hypothesis Dalton's Law of Partial Pressures Boyle's Law & Charles' Law Kinetic Theory Pressure Derivation RMS Speed v_rms Kinetic Interpretation of Temperature Law of Equipartition of Energy Degrees of Freedom Specific Heat Capacity Cv & Cp Monatomic, Diatomic, Polyatomic Gases Mean Free Path Graham's Law of Diffusion
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Section A — Multiple Choice Questions
Q.1–25  ·  1 mark each
Q1
The Boltzmann constant k_B relates: Easy
(A)Pressure and volume of a gas
(B)Average kinetic energy of a molecule and absolute temperature
(C)Number of moles and pressure
(D)Molar mass and rms speed
Solution
✓ Correct: (B)
Average KE per molecule = (3/2)k_B T k_B = 1.38 × 10⁻²³ J K⁻¹ = R/N_A

k_B (Boltzmann constant) is the bridge between macroscopic (temperature T) and microscopic (molecular KE) domains. R = N_A × k_B where R = 8.314 J mol⁻¹K⁻¹ and N_A = 6.022 × 10²³ mol⁻¹.

Q2
Avogadro's hypothesis states that equal volumes of all gases at the same temperature and pressure have: Easy
(A)Same mass
(B)Same number of molecules
(C)Same density
(D)Same molar mass
Solution
✓ Correct: (B)

Avogadro's hypothesis: Equal volumes of all gases at same T and P contain the same number of molecules. At STP, 22.4 L of any gas contains N_A = 6.022 × 10²³ molecules. This is consistent with the ideal gas equation P₁V₁/N₁T₁ = P₂V₂/N₂T₂ = k_B.

Q3
The pressure of an ideal gas according to kinetic theory is: Moderate
(A)P = nmv²
(B)P = (1/3)nm⟨v²⟩
(C)P = (2/3)nm⟨v²⟩
(D)P = (1/2)nm⟨v²⟩
Solution
✓ Correct: (B)
P = (1/3)nm⟨v²⟩ where n = number density, m = molecular mass, ⟨v²⟩ = mean square speed

The factor 1/3 arises from isotropy — molecules move equally in all 3 directions, so ⟨vₓ²⟩ = ⟨vy²⟩ = ⟨vz²⟩ = (1/3)⟨v²⟩. This leads to PV = (2/3)E (translational KE).

Q4
The average kinetic energy per molecule of an ideal gas at temperature T is: Easy
(A)(1/2)k_BT
(B)k_BT
(C)(3/2)k_BT
(D)(5/2)k_BT
Solution
✓ Correct: (C)
(1/2)m⟨v²⟩ = (3/2)k_BT

This is the kinetic interpretation of temperature. Average translational KE per molecule = (3/2)k_BT regardless of the nature of the gas (mono, di, polyatomic). Temperature is a measure of average molecular KE.

Q5
The root mean square (rms) speed of gas molecules is given by: Easy
(A)v_rms = √(2RT/M)
(B)v_rms = √(3RT/M)
(C)v_rms = √(8RT/πM)
(D)v_rms = √(RT/M)
Solution
✓ Correct: (B)
v_rms = √(3RT/M) = √(3k_BT/m) v_avg = √(8RT/πM) [option C is v_avg] v_mp = √(2RT/M) [option A is most probable speed]

Three important molecular speeds: v_mp < v_avg < v_rms. All ∝ √(T/M) — higher temperature or lower molar mass → faster molecules.

Q6
The law of equipartition of energy states that each degree of freedom contributes an average energy of: Moderate
(A)k_BT
(B)(1/2)k_BT
(C)(3/2)k_BT
(D)2k_BT
Solution
✓ Correct: (B)
Each translational/rotational DOF: (1/2)k_BT Each vibrational mode (2 DOF): k_BT (kinetic + potential)

Law of equipartition: Energy is equally distributed among all available modes. A vibrational mode contributes 2 × (1/2)k_BT = k_BT because it has both kinetic AND potential energy quadratic terms.

Q7
Number of degrees of freedom for a diatomic molecule (treated as rigid rotator) is: Moderate
(A)3
(B)5
(C)6
(D)7
Solution
✓ Correct: (B)
Rigid diatomic: 3 translational + 2 rotational = 5 DOF Non-rigid diatomic: 5 + 2 vibrational = 7 DOF (each vib = 2 DOF) Monatomic: 3 translational only = 3 DOF

Rotation along the molecular axis has negligible moment of inertia (quantum reasons) and doesn't contribute. So diatomic has only 2 (not 3) rotational DOF.

Q8
For a monatomic ideal gas, the ratio γ = Cp/Cv is: Moderate
(A)1.0
(B)1.4
(C)1.67
(D)1.33
Solution
✓ Correct: (C)
Monatomic: Cv = 3R/2, Cp = 5R/2, γ = 5/3 ≈ 1.67 Diatomic (rigid): Cv = 5R/2, Cp = 7R/2, γ = 7/5 = 1.40 Triatomic: γ ≈ 4/3 ≈ 1.33

More degrees of freedom → higher Cv → lower γ. γ = (f+2)/f where f = number of DOF (translational + rotational only, at moderate temperatures).

Q9
Dalton's Law of Partial Pressures states that for a mixture of ideal gases: Easy
(A)Total pressure = product of partial pressures
(B)Total pressure = sum of partial pressures
(C)Total pressure = average of partial pressures
(D)Each gas exerts equal pressure
Solution
✓ Correct: (B)
P_total = P₁ + P₂ + P₃ + ... where Pᵢ = μᵢRT/V (pressure each gas would exert alone)

Valid for non-reacting ideal gases. Each gas contributes its partial pressure independently. In kinetic theory terms: P = (n₁ + n₂ + ...)k_BT since all molecules at same T have equal average KE.

Q10
At the same temperature, if we compare rms speeds of H₂ and O₂ molecules: Moderate
(A)v_rms(H₂) = v_rms(O₂)
(B)v_rms(H₂) < v_rms(O₂)
(C)v_rms(H₂) > v_rms(O₂)
(D)v_rms depends on pressure, not temperature
Solution
✓ Correct: (C)
v_rms = √(3RT/M) ∝ 1/√M M(H₂) = 2, M(O₂) = 32 v_rms(H₂)/v_rms(O₂) = √(32/2) = √16 = 4

H₂ molecules move 4 times faster than O₂ at the same temperature. Lighter molecules move faster — at fixed T, both have same average KE but different speeds.

Q11
The mean free path l of gas molecules (diameter d, number density n) is: Moderate
(A)l = 1/(nπd²)
(B)l = 1/(√2 nπd²)
(C)l = √2 nπd²
(D)l = 1/(2nπd²)
Solution
✓ Correct: (B)
l = 1/(√2 nπd²) The √2 factor accounts for relative motion of all molecules

At STP for air (d ≈ 2Å): l ≈ 2.9 × 10⁻⁷ m ≈ 1500d. Mean free path increases with lower pressure (fewer molecules) and decreases with larger molecular size.

Q12
Molar specific heat at constant volume for a rigid diatomic ideal gas is: Moderate
(A)3R/2
(B)5R/2
(C)7R/2
(D)R
Solution
✓ Correct: (B)
Rigid diatomic: f = 5 DOF (3T + 2R) U = (5/2)RT per mole Cv = dU/dT = 5R/2 ≈ 20.8 J/mol·K Cp = Cv + R = 7R/2 ≈ 29.1 J/mol·K

These values match experimental data for N₂, O₂, H₂ very well. Note: Cp − Cv = R always holds for any ideal gas (Mayer's relation).

Q13
According to kinetic theory, collisions between gas molecules are assumed to be: Easy
(A)Perfectly inelastic
(B)Partially elastic
(C)Perfectly elastic
(D)Non-existent in ideal gas
Solution
✓ Correct: (C)

Kinetic theory assumes perfectly elastic collisions — total kinetic energy and total momentum are both conserved. This ensures the distribution of molecular speeds remains steady (dynamic equilibrium). No energy is lost to heat (no internal molecular excitations).

Q14
Graham's law of diffusion states that rate of diffusion of a gas is: Moderate
(A)Directly proportional to √M
(B)Inversely proportional to √M
(C)Independent of molecular mass
(D)Directly proportional to M
Solution
✓ Correct: (B)
Rate of diffusion ∝ v_rms ∝ 1/√M r₁/r₂ = √(M₂/M₁)

This is the basis of uranium enrichment: ²³⁵UF₆ (M=349) diffuses 0.44% faster than ²³⁸UF₆ (M=352). H₂ diffuses 4× faster than O₂ at the same temperature.

Q15
The molar specific heat of a solid (Dulong-Petit law) predicted by equipartition is approximately: Moderate
(A)R
(B)2R
(C)3R
(D)5R/2
Solution
✓ Correct: (C)
Solid: each atom vibrates in 3D 3 vibrational modes × 2 DOF each = 6 DOF U = 6 × (1/2)k_BT × N_A = 3RT C = dU/dT = 3R ≈ 25 J/mol·K

Dulong-Petit law (C ≈ 3R ≈ 25 J/mol·K) works well for metals at room temperature. Carbon (diamond) is an exception — quantum effects make its C much lower at room temperature.

Q16
At STP, the number of molecules in 22.4 litres of any ideal gas is: Easy
(A)6.022 × 10²⁰
(B)6.022 × 10²³
(C)6.022 × 10²⁶
(D)Depends on the type of gas
Solution
✓ Correct: (B)

Avogadro's number N_A = 6.022 × 10²³. This is the number of molecules in one mole (22.4 L at STP) of any ideal gas. It is universal — same for He, O₂, CO₂, etc. (Avogadro's hypothesis + kinetic theory).

Q17
Kinetic theory of gases assumes that gas molecules: Easy
(A)Are in random motion and mutually attractive
(B)Occupy significant volume compared to container
(C)Move in random straight lines with negligible intermolecular forces (except during collisions)
(D)Have uniform speed distribution
Solution
✓ Correct: (C)

Key assumptions of kinetic theory: (1) Molecules are in incessant random motion. (2) Molecular volume is negligible compared to container. (3) Intermolecular forces negligible except during brief collisions. (4) Collisions are perfectly elastic. (5) Distribution of speeds follows Maxwell-Boltzmann distribution.

Q18
The total translational kinetic energy of N molecules of an ideal gas at temperature T is: Moderate
(A)E = Nk_BT
(B)E = (3/2)Nk_BT
(C)E = (1/2)Nk_BT
(D)E = 3Nk_BT
Solution
✓ Correct: (B)
E = N × (3/2)k_BT = (3/2)Nk_BT = (3/2)μRT Also: PV = (2/3)E → E = (3/2)PV

This relates macroscopic (P, V, T) and microscopic (KE, N) quantities. Internal energy of ideal gas depends only on temperature, not P or V.

Q19
As temperature increases, mean free path of gas molecules: Moderate
(A)Increases (if pressure constant)
(B)Decreases (if pressure constant)
(C)Remains unchanged
(D)Depends only on molecular diameter
Solution
✓ Correct: (A)
l = 1/(√2 nπd²) At constant P: n = P/(k_BT) → n ∝ 1/T So l ∝ T (increases with temperature at constant P)

Higher temperature → lower number density (gas expands at const P) → fewer collisions → longer mean free path. At constant volume (n fixed), l is unchanged.

Q20
In a mixture of argon (monatomic) and oxygen (diatomic) at the same temperature, which has greater average kinetic energy per molecule? Moderate
(A)Argon
(B)Oxygen
(C)Both have equal translational KE
(D)Depends on their masses
Solution
✓ Correct: (C)

Average translational KE per molecule = (3/2)k_BT for all gases, regardless of monatomic, diatomic or polyatomic nature. However, total KE per molecule differs: Ar has 3 DOF, O₂ has 5 DOF (rigid) → O₂ has higher total energy.

Q21
The ideal gas equation PV = μRT is most accurately obeyed at: Easy
(A)High pressure and low temperature
(B)Low pressure and high temperature
(C)High pressure and high temperature
(D)Low pressure and low temperature
Solution
✓ Correct: (B)

At low pressure: molecules are far apart → intermolecular forces negligible → behaves ideally. At high temperature: thermal KE >> intermolecular PE → interaction effects negligible. Real gases deviate at high pressure and low temperature (near liquefaction).

Q22
If the temperature of a gas is doubled (in Kelvin), the rms speed of its molecules: Moderate
(A)Doubles
(B)Becomes √2 times
(C)Becomes 4 times
(D)Remains unchanged
Solution
✓ Correct: (B)
v_rms = √(3RT/M) ∝ √T If T → 2T: v_rms → √(2T) = √2 × v_rms

v_rms ∝ √T (not T). Doubling T → v_rms increases by factor √2 ≈ 1.414. For KE ∝ T, so doubling T doubles the average KE per molecule.

Q23
The relation PV = (2/3)E, where E is translational kinetic energy, shows: Moderate
(A)Pressure equals kinetic energy
(B)Pressure depends only on temperature
(C)Pressure × volume = (2/3) of total translational KE
(D)Kinetic energy depends on pressure only
Solution
✓ Correct: (C)
PV = (2/3)E where E = (3/2)Nk_BT So PV = (2/3)(3/2)Nk_BT = Nk_BT = μRT ✓

This is consistent with the ideal gas equation — a remarkable result showing that macroscopic pressure arises from molecular kinetic energy. The 2/3 factor comes from the 3D isotropy (only one direction contributes to wall pressure at a time).

Q24
The mean free path in a gas depends on: Moderate
(A)Temperature only
(B)Number density and molecular diameter
(C)Molecular mass only
(D)Volume of the container only
Solution
✓ Correct: (B)
l = 1/(√2 nπd²) l ∝ 1/n (smaller n → longer l) l ∝ 1/d² (larger molecules → shorter l)

Mean free path is inversely proportional to number density n and square of molecular diameter d. In a highly evacuated tube (small n), mean free path can be as large as the tube length itself.

Q25
Which of the following relationships between molecular speeds is correct? Moderate
(A)v_mp > v_avg > v_rms
(B)v_rms > v_avg > v_mp
(C)v_avg > v_rms > v_mp
(D)All three are equal
Solution
✓ Correct: (B)
v_mp = √(2RT/M) ← most probable speed v_avg = √(8RT/πM) ← average speed v_rms = √(3RT/M) ← root mean square speed Ratio: v_mp : v_avg : v_rms = √2 : √(8/π) : √3 ≈ 1.414 : 1.596 : 1.732 So v_rms > v_avg > v_mp
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Section B — Numerical Problems
N1–N15  ·  3–5 marks each
N1
Calculate the rms speed of nitrogen molecules at 27°C. (M_N₂ = 28 g/mol, R = 8.31 J/mol·K) Easy
Solution
T = 27 + 273 = 300 K M = 28 × 10⁻³ kg/mol v_rms = √(3RT/M) = √(3 × 8.31 × 300 / 28 × 10⁻³) = √(7479 / 0.028) = √(267,107) = 517 m/s ≈ 517 m/s

About 517 m/s — close to the speed of sound in air (~340 m/s). This explains why sound can propagate through air effectively.

N2
Estimate the average kinetic energy of a helium atom at (a) 27°C, (b) 6000 K (Sun's surface), (c) 10⁷ K (stellar core). (k_B = 1.38 × 10⁻²³ J/K) Easy
Solution
Average KE = (3/2)k_BT (a) T = 300 K: KE = (3/2)(1.38 × 10⁻²³)(300) = 6.21 × 10⁻²¹ J ≈ 0.039 eV (b) T = 6000 K: KE = (3/2)(1.38 × 10⁻²³)(6000) = 1.24 × 10⁻¹⁹ J ≈ 0.78 eV (c) T = 10⁷ K: KE = (3/2)(1.38 × 10⁻²³)(10⁷) = 2.07 × 10⁻¹⁶ J ≈ 1293 eV ≈ 1.3 keV

At stellar core temperatures (~10⁷ K), molecules have enough energy to overcome nuclear Coulomb barriers → nuclear fusion occurs!

N3
An oxygen cylinder (V = 30 L) has gauge pressure 15 atm at 27°C. After withdrawing oxygen, gauge pressure drops to 11 atm at 17°C. Find mass of oxygen withdrawn. (R = 8.31 J/mol·K, M_O₂ = 32 g/mol, 1 atm = 1.013 × 10⁵ Pa) Moderate
Solution
Absolute pressures: P₁ = (15 + 1) atm = 16 atm, T₁ = 300 K P₂ = (11 + 1) atm = 12 atm, T₂ = 290 K V = 30 L = 30 × 10⁻³ m³ Moles initially: μ₁ = P₁V/(RT₁) = (16 × 1.013 × 10⁵ × 30 × 10⁻³) / (8.31 × 300) = 48624 / 2493 = 19.50 mol Moles finally: μ₂ = P₂V/(RT₂) = (12 × 1.013 × 10⁵ × 30 × 10⁻³) / (8.31 × 290) = 36468 / 2409.9 = 15.13 mol Moles withdrawn: Δμ = 19.50 − 15.13 = 4.37 mol Mass withdrawn = 4.37 × 32 = 139.8 g ≈ 141 g
N4
An air bubble (V₁ = 1.0 cm³) rises from a lake bottom at depth 40 m (T₁ = 12°C) to the surface (T₂ = 35°C). Find the volume at the surface. (ρ_water = 1000 kg/m³, g = 10 m/s², P_atm = 1.013 × 10⁵ Pa) Moderate
Solution
P₁ = P_atm + ρgh = 1.013 × 10⁵ + 1000 × 10 × 40 = 1.013 × 10⁵ + 4 × 10⁵ = 5.013 × 10⁵ Pa P₂ = P_atm = 1.013 × 10⁵ Pa T₁ = 285 K, T₂ = 308 K V₁ = 1.0 cm³ Using P₁V₁/T₁ = P₂V₂/T₂: V₂ = V₁ × (P₁/P₂) × (T₂/T₁) = 1.0 × (5.013/1.013) × (308/285) = 1.0 × 4.949 × 1.081 = 5.35 cm³

The bubble expands to about 5.35 cm³ — more than 5 times its original volume as it rises due to decreasing pressure.

N5
Find the ratio of rms speeds of argon (Ar) and helium (He) atoms at the same temperature. (M_Ar = 40, M_He = 4 u) Easy
Solution
v_rms ∝ 1/√M (at constant T) v_rms(Ar)/v_rms(He) = √(M_He/M_Ar) = √(4/40) = √(1/10) = 1/√10 ≈ 0.316 OR v_rms(He)/v_rms(Ar) = √(M_Ar/M_He) = √(40/4) = √10 ≈ 3.16

Helium atoms move about 3.16 times faster than argon atoms at the same temperature. This is why helium escapes Earth's atmosphere more readily than heavier gases.

N6
A flask contains argon (M = 40 u) and chlorine (M = 71 u) in ratio 2:1 by mass at 27°C. Find (a) ratio of average KE per molecule, (b) ratio of v_rms. Moderate
Solution
(a) Average KE per molecule = (3/2)k_BT (same for all gases at same T) Ratio = 1:1 (mass composition is irrelevant!) (b) v_rms(Ar)/v_rms(Cl₂) = √(M_Cl₂/M_Ar) = √(71/40) = √1.775 = 1.332 So v_rms(Ar) : v_rms(Cl₂) = 1.332 : 1

Note: The 2:1 mass ratio doesn't affect the answer at all! Any mass ratio would give the same result, since KE and v_rms depend only on temperature and molecular mass.

N7
At what temperature will argon atoms have the same rms speed as helium atoms at −20°C? (M_Ar = 39.9 u, M_He = 4.0 u) Moderate
Solution
T_He = −20 + 273 = 253 K For same v_rms: √(3RT_Ar/M_Ar) = √(3RT_He/M_He) T_Ar/M_Ar = T_He/M_He T_Ar = T_He × (M_Ar/M_He) = 253 × (39.9/4.0) = 253 × 9.975 = 2523.7 K ≈ 2524 K = 2251°C

Argon needs to be at ~2524 K for its atoms to have the same rms speed as helium at 253 K. Heavier atoms need much higher temperature to achieve the same speed.

N8
Estimate the total number of air molecules in a room of 25.0 m³ at 27°C and 1 atm. (k_B = 1.38 × 10⁻²³ J/K, 1 atm = 1.013 × 10⁵ Pa) Easy
Solution
PV = Nk_BT → N = PV/(k_BT) N = (1.013 × 10⁵ × 25.0) / (1.38 × 10⁻²³ × 300) = 2.5325 × 10⁶ / (4.14 × 10⁻²¹) = 6.12 × 10²⁶ molecules

Over 6 × 10²⁶ air molecules in a typical room — about 1000 times Avogadro's number! Despite their enormous number, they occupy negligible volume compared to the room.

N9
A cylinder of fixed capacity 44.8 litres contains helium at STP. What is the heat needed to raise the temperature by 15°C? (R = 8.31 J/mol·K) Moderate
Solution
At STP, 1 mole occupies 22.4 L So 44.8 L contains μ = 44.8/22.4 = 2 moles of He He is monatomic: Cv = (3/2)R (volume is fixed → use Cv) Q = μ × Cv × ΔT = 2 × (3/2)R × 15 = 2 × 1.5 × 8.31 × 15 = 2 × 1.5 × 124.65 = 374 J

Since volume is fixed (rigid cylinder), we use Cv. No work is done, all heat goes to internal energy.

N10
Uranium hexafluoride gas contains two isotopes: ²³⁵UF₆ (M=349) and ²³⁸UF₆ (M=352). Find the percentage difference in their rms speeds. (Atomic mass F = 19) Moderate
Solution
M₁ = 235 + 6×19 = 235 + 114 = 349 M₂ = 238 + 6×19 = 238 + 114 = 352 v₁/v₂ = √(M₂/M₁) = √(352/349) = √1.00860 = 1.00429 % difference = (v₁−v₂)/v₂ × 100 = (1.00429 − 1) × 100 ≈ 0.43%

Only 0.43% speed difference! This tiny difference requires the gas diffusion to be repeated many thousands of times to separate enough ²³⁵U for nuclear fuel. This is done in uranium enrichment plants.

N11
A vessel contains two non-reacting gases: neon (M = 20.2 u) and oxygen (M = 32 u). Ratio of partial pressures P_Ne:P_O₂ = 3:2. Find (a) ratio of number of molecules, (b) ratio of mass densities. Moderate
Solution
(a) From ideal gas: P₁V = μ₁RT, P₂V = μ₂RT P₁/P₂ = μ₁/μ₂ = N₁/N₂ = 3/2 N_Ne : N_O₂ = 3 : 2 (b) ρ₁/ρ₂ = (m₁/V)/(m₂/V) = m₁/m₂ = (μ₁M₁)/(μ₂M₂) = (3/2) × (20.2/32.0) = (3 × 20.2)/(2 × 32.0) = 60.6/64.0 = 0.947

ρ_Ne/ρ_O₂ = 0.947 — neon is slightly less dense than oxygen in this mixture despite more neon molecules, because neon atoms are lighter.

N12
Estimate the mean free path for nitrogen molecules at 2 atm and 17°C. Radius of N₂ molecule ≈ 1.0 Å. (R = 8.31 J/mol·K, N_A = 6.02 × 10²³) Moderate
Solution
d = 2 × r = 2 × 10⁻¹⁰ m T = 290 K, P = 2 atm = 2.026 × 10⁵ Pa Number density: n = P/(k_BT) = 2.026 × 10⁵/(1.38 × 10⁻²³ × 290) = 2.026 × 10⁵ / (4.002 × 10⁻²¹) = 5.06 × 10²⁵ m⁻³ Mean free path: l = 1/(√2 nπd²) = 1/(√2 × 5.06 × 10²⁵ × π × (2 × 10⁻¹⁰)²) = 1/(√2 × 5.06 × 10²⁵ × 1.257 × 10⁻¹⁹) = 1/(8.99 × 10⁶ × √2) = 1/(1.272 × 10⁷) l ≈ 7.9 × 10⁻⁸ m ≈ 79 nm
N13
Calculate heat required to raise temperature of 2 moles of N₂ (diatomic rigid) by 30°C at (a) constant volume, (b) constant pressure. (R = 8.31 J/mol·K) Moderate
Solution
N₂ is diatomic rigid: Cv = 5R/2, Cp = 7R/2 (a) At constant volume: Q_v = μ × Cv × ΔT = 2 × (5/2) × 8.31 × 30 = 2 × 2.5 × 8.31 × 30 = 1246.5 J ≈ 1247 J (b) At constant pressure: Q_p = μ × Cp × ΔT = 2 × (7/2) × 8.31 × 30 = 2 × 3.5 × 8.31 × 30 = 1745.1 J ≈ 1745 J Difference = Q_p − Q_v = μRΔT = 2 × 8.31 × 30 = 498.6 J (This extra energy does P-V work)
N14
Estimate the fraction of molecular volume to total volume occupied by oxygen gas at STP. Take diameter of O₂ molecule = 3 Å. Moderate
Solution
Volume of one O₂ molecule: V_mol = (4/3)π(d/2)³ = (4/3)π(1.5 × 10⁻¹⁰)³ = (4/3)π × 3.375 × 10⁻³⁰ = 1.414 × 10⁻²⁹ m³ 1 mole at STP occupies V_gas = 22.4 L = 22.4 × 10⁻³ m³ Total molecular volume for 1 mole: V_total_mol = N_A × V_mol = 6.022 × 10²³ × 1.414 × 10⁻²⁹ = 8.516 × 10⁻⁶ m³ Fraction = V_total_mol / V_gas = 8.516 × 10⁻⁶ / 22.4 × 10⁻³ ≈ 3.8 × 10⁻⁴

Molecular volume is only ~0.04% of total volume — confirming that gas molecules are extremely far apart compared to their size. This validates the ideal gas assumption.

N15
Three vessels of equal volume have gases at same T and P: (1) Ne (monatomic, M=20), (2) Cl₂ (diatomic, M=71), (3) UF₆ (polyatomic, M=352). (a) Do they have equal number of molecules? (b) Is v_rms same? Which has largest v_rms? Moderate
Solution
(a) From PV = Nk_BT: at same P, V, T → same N YES, all three vessels contain equal number of molecules (Avogadro's hypothesis) (b) v_rms = √(3RT/M) ∝ 1/√M Ne (M=20): v_rms ∝ 1/√20 = 0.224 (relative) Cl₂ (M=71): v_rms ∝ 1/√71 = 0.119 (relative) UF₆ (M=352): v_rms ∝ 1/√352 = 0.053 (relative) v_rms is NOT the same. NEON has the LARGEST v_rms. Ratio: v_Ne : v_Cl₂ : v_UF₆ = 1/√20 : 1/√71 : 1/√352 = 4.21 : 2.23 : 1

Despite having equal numbers of molecules, lighter molecules move fastest. Neon's rms speed is 4.21 times that of UF₆ at the same temperature!