Kinetic Theory - Vardaan Master Notes

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References

Compiled from: NCERT Physics Part-2 (Kinetic Theory) • H.C. Verma — Concepts of Physics Vol-2 (Ch. 24: Kinetic Theory of Gases) • D.C. Pandey — Understanding Physics: Waves & Thermodynamics • Pradeep's Fundamental Physics XI • S.L. Arora • Cengage (B.M. Sharma) • Errorless Physics. Target exams: Boards • JEE Main • JEE Advanced • NEET • NDA.

1. Molecular Nature of Matter & the Ideal Gas

Kinetic theory explains the bulk behaviour of gases (pressure, temperature) from the random motion of molecules. Historical anchors: Dalton's atomic theory, Avogadro's hypothesis (equal volumes of all gases at the same T and P contain equal numbers of molecules), and Brownian motion as the visible evidence of molecular agitation. Avogadro number $N_A = 6.022\times10^{23}$ per mole; 22.4 L at STP contains 1 mole.

Ideal Gas

An ideal gas has point-sized molecules with no intermolecular forces, so it obeys PV = nRT exactly at all T and P. Real gases approach ideal behaviour at low pressure and high temperature (molecules far apart, forces negligible) — the standard 1-marker.

1.1 The Gas Laws (All of Them)

Law	Constant	Statement
Boyle's law	T, n	$PV = $ constant ⟹ P ∝ 1/V (isotherm: rectangular hyperbola on P–V)
Charles' law	P, n	$V/T = $ constant; V–T graph is a straight line through the origin (absolute T!)
Gay-Lussac's law	V, n	$P/T = $ constant (pressure law)
Avogadro's law	P, T	V ∝ n — equal volumes hold equal molecules
Dalton's law	—	Total pressure of a non-reacting mixture = sum of partial pressures: $P = P_1 + P_2 + \dots$

Ideal Gas Equation — All Forms $$PV = nRT = \frac{m}{M}RT = Nk_BT \qquad P = \frac{\rho RT}{M} \qquad PM = \rho RT$$

R = 8.314 J mol⁻¹K⁻¹ (universal); $k_B = R/N_A = 1.38\times10^{-23}$ J/K (Boltzmann constant — gas constant per molecule). Combined-law working form for one gas sample: $\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}$. Always use absolute temperature (kelvin).

Image Placeholder — Fig. 1

Gas-law graph gallery: isotherms, Charles' line, and the P–T pressure law (Landscape • Background #ffffff)

AI IMAGE PROMPT: Create a three-panel physics textbook graph gallery in a row, LANDSCAPE orientation (16:9), pure white background (#ffffff). PANEL 1 titled "Boyle's law (T constant)": axes "V" (horizontal) and "P" (vertical); TWO smooth rectangular-hyperbola curves in blue (#1e88e5), the upper one labelled "T₂ > T₁" and the lower "T₁"; caption "PV = constant — isotherms never intersect". PANEL 2 titled "Charles' law (P constant)": axes "T (K)" and "V"; a straight blue line through the ORIGIN labelled "V ∝ T"; a dashed grey extension of the line back to the origin with the origin labelled "T = 0 K (absolute zero, V → 0)"; a second steeper line labelled "lower P". PANEL 3 titled "Gay-Lussac's law (V constant)": axes "T (K)" and "P"; a straight blue line through the origin labelled "P ∝ T"; caption "pressure law". All three panels: thin black axes with arrowheads, small clean sans-serif labels. Style: flat minimalist vector textbook graphs, blue curves/lines, dashed grey extrapolations, no gradients, no shadows, pure white background.

Graph identification (hyperbola vs origin-line) is a repeat MCQ in NEET and NDA.

✍ IN-TEXT PRACTICE 1.1 — Boards / NDA (Combined gas law)

Q. A gas at 27°C and 1 atm is compressed to half its volume while being heated to 127°C. Find the final pressure.

Kelvin first: T₁ = 300 K, T₂ = 400 K.

Combined law: $P_2 = P_1\cdot\dfrac{V_1}{V_2}\cdot \dfrac{T_2}{T_1} = 1\times2\times\dfrac{400}{300} = \mathbf{\dfrac{8}{3} \approx 2.67 \text{ atm}}$

✍ IN-TEXT PRACTICE 1.2 — NEET (PM = ρRT)

Q. Find the density of oxygen (M = 32 g/mol) at 2 atm and 27°C. (R = 8.3 J mol⁻¹K⁻¹, 1 atm ≈ 10⁵ Pa)

ρ = PM/RT: $\rho = \dfrac{2\times10^5\times0.032}{8.3\times300} \approx \mathbf{2.57 \text{ kg/m}^3}$

2. Postulates of the Kinetic Theory of Gases

The Assumptions (quote in Board answers)

A gas is a huge number of identical molecules in incessant random motion in all directions with all possible speeds.
Molecular size is negligible compared with intermolecular separation (point masses); total molecular volume ≪ container volume.
No intermolecular forces except during collisions ⟹ between collisions molecules move in straight lines with constant velocity; PE = 0, all energy is kinetic.
Collisions (molecule–molecule and molecule–wall) are perfectly elastic and of negligible duration compared with the time between collisions.
Molecular density is uniform; velocity distribution is isotropic: $\overline{v_x^2} = \overline{v_y^2} = \overline{v_z^2} = \tfrac13\overline{v^2}$.

3. Pressure of an Ideal Gas — Full Derivation

Setup: N molecules, each of mass m, in a cube of side L (volume V = L³). Consider one molecule with x-velocity $v_x$ hitting the wall perpendicular to x.

Step 1 — momentum transfer per hit: elastic bounce reverses $v_x$: $\Delta p = 2mv_x$.

Step 2 — hits per second: round trip 2L → frequency $= v_x/2L$.

Step 3 — force by one molecule: $f = \dfrac{2mv_x \cdot v_x}{2L} = \dfrac{mv_x^2}{L}$.

Step 4 — all molecules: $F = \dfrac{m}{L}\sum v_{x}^2 = \dfrac{mN\overline{v_x^2}}{L}$; pressure $P = F/L^2 = \dfrac{mN\overline{v_x^2}}{V}$.

Step 5 — isotropy: $\overline{v_x^2} = \tfrac13\overline{v^2}$:

Pressure of a Gas $$\boxed{\,P = \frac{1}{3}\frac{mN}{V}\overline{v^{2}} = \frac{1}{3}\rho\,\overline{v^{2}} = \frac{1}{3}\rho\,v_{rms}^{2}\,}$$ $$\text{Equivalently: } PV = \tfrac{2}{3}E_{trans}\ \ (\text{E = total translational KE})$$

Pressure is two-thirds of the translational kinetic energy per unit volume — the bridge between mechanics and thermodynamics.

Image Placeholder — Fig. 2

The pressure derivation: molecule bouncing in a cube (Landscape • Background #ffffff)

AI IMAGE PROMPT: Create a professional physics textbook diagram, LANDSCAPE orientation (16:9), pure white background (#ffffff). LEFT TWO-THIRDS: draw a cube in light 3-D perspective with thin black edges, side length labelled "L" on one edge, and coordinate axes x, y, z drawn at one corner with small arrows. Inside the cube scatter about ten tiny grey dots (molecules) with short random grey velocity arrows in varied directions. Highlight ONE molecule as a larger blue (#1e88e5) dot near the shaded right wall (the wall perpendicular to x, lightly tinted #e3f2fd): draw a green (#43a047) arrow toward the wall labelled "+v_x (before)" and a red (#d32f2f) arrow bouncing away labelled "−v_x (after, elastic)"; an annotation with a leader line: "momentum change per hit Δp = 2mv_x; hits per second = v_x/2L". RIGHT THIRD: a boxed derivation ladder of four short lines of text stacked vertically: "f (one molecule) = mv_x²/L", "F (all) = Nm v̄x²/L", "isotropy: v̄x² = v̄²/3", "⟹ P = ⅓ ρ v_rms² (PV = ⅔ E_trans)". Style: flat minimalist vector textbook figure, thin black cube edges, lightly tinted target wall, blue/green/red highlighted molecule arrows, grey background molecules, clean sans-serif labels, no gradients, no heavy shading, pure white background.

The four-line ladder on the right mirrors Steps 1–5 of the derivation — memorise it as a picture.

4. Kinetic Interpretation of Temperature & RMS Speed

Combine $PV = \tfrac13 Nm\overline{v^2}$ with $PV = Nk_BT$:

$$\tfrac13 m\overline{v^2} = k_BT \;\Rightarrow\; \tfrac12m\overline{v^2} = \tfrac32k_BT$$

Temperature = Molecular KE $$\boxed{\,\overline{KE}_{trans}\text{ per molecule} = \frac{3}{2}k_BT\,} \qquad \text{per mole} = \frac{3}{2}RT$$ $$v_{rms} = \sqrt{\overline{v^2}} = \sqrt{\frac{3k_BT}{m}} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3P}{\rho}}$$

Temperature IS average translational KE — independent of the gas! At the same T, every gas has the same KE per molecule, but lighter molecules move faster ($v_{rms} \propto 1/\sqrt{M}$). At absolute zero, molecular motion (classically) ceases. Note $v_{rms} \propto \sqrt{T}$: to double v_rms, T must become 4×.

⭐ CLASSIC COMPARISONS (NEET / NDA REGULARS)

H₂ vs O₂ at the same T: $\dfrac{v_{H_2}}{v_{O_2}} = \sqrt{\dfrac{32}{2}} = 4$ — hydrogen is 4× faster (and escapes the atmosphere!).
Same gas: $v_{rms}$ at 927°C vs 27°C → $\sqrt{1200/300} = 2×$.
Mixture of gases: each has the same average KE, NOT the same speed.
At what T do O₂ molecules match H₂'s speed at 27°C? $T = 300\times16 = 4800$ K — scale by the mass ratio.

✍ IN-TEXT PRACTICE 4.1 — NEET (v_rms numeric)

Q. Find v_rms of nitrogen (M = 28 g/mol) at 27°C. (R = 8.3 J mol⁻¹K⁻¹)

Formula: $v_{rms} = \sqrt{\dfrac{3RT}{M}} = \sqrt{\dfrac{3\times8.3\times300}{0.028}}$

$= \sqrt{2.67\times10^5} \approx \mathbf{517 \text{ m/s}}$ — gas molecules move at bullet speeds even at room temperature!

5. Molecular Speeds & the Maxwell Distribution

The Three Speeds (memorise the trio) $$v_{p} = \sqrt{\frac{2RT}{M}} \qquad \bar{v} = \sqrt{\frac{8RT}{\pi M}} \qquad v_{rms} = \sqrt{\frac{3RT}{M}}$$ $$v_p : \bar{v} : v_{rms} = \sqrt2 : \sqrt{8/\pi} : \sqrt3 \approx 1.41 : 1.60 : 1.73$$

Order (always): $\boxed{v_p < \bar{v} < v_{rms}}$ — mnemonic "PAR": Probable < Average < RMS.

Maxwell Speed Distribution (qualitative — what exams ask)

The number of molecules vs speed curve starts at zero, peaks at the most probable speed v_p, and falls with a long high-speed tail.
Raising T: the peak shifts RIGHT (higher v_p) and gets LOWER/broader — but the total area (= total number of molecules) stays the same.
Heavier gas at the same T: peak shifts LEFT (slower) and taller.

Image Placeholder — Fig. 3

Maxwell speed distribution at two temperatures with the three marked speeds (Landscape • Background #ffffff)

AI IMAGE PROMPT: Create a single large physics graph, LANDSCAPE orientation (16:9), pure white background (#ffffff). Axes: horizontal "molecular speed v" with arrowhead, vertical "number of molecules dN/dv" with arrowhead, origin "O". Draw TWO smooth Maxwell-distribution curves both starting at the origin: a TALLER, narrower blue (#1e88e5) curve peaking at moderate speed labelled "T₁ (lower temperature)", and a LOWER, broader orange (#d9480f) curve peaking farther right labelled "T₂ > T₁ — peak shifts right, curve flattens; area under both curves EQUAL (= N)". On the BLUE curve mark three speeds with dashed vertical lines to the axis, left to right: "v_p = √(2RT/M) (peak)", "v̄ = √(8RT/πM)", "v_rms = √(3RT/M)", with a bracket note above them "v_p < v̄ < v_rms (ratio ≈ 1.41 : 1.60 : 1.73)". Add a short note near the right tail: "long high-speed tail". Style: flat minimalist vector textbook graph, thin black axes, smooth blue and orange curves, dashed grey marker lines, clean sans-serif labels, no gradients, no shadows, pure white background.

Every Maxwell-curve MCQ (peak shift, area invariance, speed ordering) is answered by this one graph.

✍ IN-TEXT PRACTICE 5.1 — JEE Main (Speed trio)

Q. For a gas at temperature T, calculate the ratio v_p : v̄ : v_rms, and find the temperature at which v_rms of the gas equals its v_p at 600 K.

Ratio: $\sqrt2 : \sqrt{8/\pi} : \sqrt3 \approx \mathbf{1.41 : 1.60 : 1.73}$

Match: $\sqrt{\dfrac{3RT}{M}} = \sqrt{\dfrac{2R(600)}{M}} \Rightarrow 3T = 1200 \Rightarrow T = \mathbf{400 \text{ K}}$

6. Degrees of Freedom & the Law of Equipartition

Degrees of Freedom (f)

The number of independent ways a molecule can store energy (independent coordinates needed to specify its configuration/motion). General count: $f = 3N_{atoms} - (\text{constraints})$.

Gas type	Translational	Rotational	Total f (ordinary T)	Example
Monatomic	3	0	3	He, Ne, Ar
Diatomic (rigid)	3	2	5	H₂, O₂, N₂, CO
Diatomic (vibrating, high T)	3	2 (+2 vibrational)	7	H₂ at very high T
Polyatomic (non-linear)	3	3	6	CO₂*, CH₄, NH₃, H₂O

*CO₂ is linear — strictly f = 5 like a diatomic at ordinary temperatures; exams usually treat "polyatomic" as f = 6. A vibrational mode counts TWICE (KE + PE).

Law of Equipartition of Energy $$\text{Energy per molecule per degree of freedom} = \frac{1}{2}k_BT$$ $$U_{per\ molecule} = \frac{f}{2}k_BT \qquad U_{per\ mole} = \frac{f}{2}RT$$

In thermal equilibrium, energy is shared equally among ALL degrees of freedom. This single law generates all the specific heats below.

7. Specific Heats of Gases (C_V, C_P, γ) & Mayer's Relation

From equipartition: $U = \dfrac{f}{2}RT$ per mole ⟹

$$C_V = \frac{dU}{dT} = \frac{f}{2}R \qquad C_P = C_V + R = \left(\frac{f}{2}+1\right)R \qquad \gamma = \frac{C_P}{C_V} = 1 + \frac{2}{f}$$

Mayer's relation C_P − C_V = R (derivation): at constant volume all heat raises U: $dQ_V = C_VdT = dU$. At constant pressure the gas also does work $PdV = RdT$ (from PV = RT): $dQ_P = dU + RdT = (C_V + R)dT$ ⟹ $C_P = C_V + R$. (In mass units: $c_p - c_v = R/M$.) C_P > C_V because constant-pressure heating must also pay for expansion work.

Gas	f	U per mole	C_V	C_P	γ
Monatomic	3	$\frac32RT$	$\frac32R$	$\frac52R$	5/3 ≈ 1.67
Diatomic (rigid)	5	$\frac52RT$	$\frac52R$	$\frac72R$	7/5 = 1.40
Diatomic (vibrating)	7	$\frac72RT$	$\frac72R$	$\frac92R$	9/7 ≈ 1.29
Polyatomic	6	$3RT$	$3R$	$4R$	4/3 ≈ 1.33

⭐ γ DECODER + MIXTURES

More atoms → larger f → γ closer to 1. Read γ backwards to identify the gas: 1.67 ⟹ monatomic; 1.40 ⟹ diatomic; 1.33 ⟹ polyatomic (a direct NEET question). Solids: by equipartition U = 3RT ⟹ C ≈ 3R (Dulong–Petit); water's high heat capacity ≈ 9R.

Mixture of gases (JEE): $C_{V,mix} = \dfrac{n_1C_{V1} + n_2C_{V2}}{n_1+n_2}$ (same for C_P), then $\gamma_{mix} = C_{P,mix}/C_{V,mix}$ — never average γ directly!

✍ IN-TEXT PRACTICE 7.1 — JEE Main (Mixture γ)

Q. One mole of helium is mixed with one mole of oxygen. Find γ of the mixture.

C_V mix: $\dfrac{1(\tfrac32R) + 1(\tfrac52R)}{2} = 2R$; C_P mix: $2R + R = 3R$

γ: $\dfrac{3R}{2R} = \mathbf{1.5}$ (note: NOT the average of 1.67 and 1.40)

✍ IN-TEXT PRACTICE 7.2 — NEET (Heat at constant P vs V)

Q. 2 moles of a diatomic gas are heated from 300 K to 400 K (a) at constant volume, (b) at constant pressure. Find the heat required in each case. (R = 8.3 J mol⁻¹K⁻¹)

(a) $Q_V = nC_V\Delta T = 2\times\tfrac52(8.3)\times100 = \mathbf{4150 \text{ J}}$

(b) $Q_P = nC_P\Delta T = 2\times\tfrac72(8.3)\times100 = \mathbf{5810 \text{ J}}$ — the extra 1660 J = nRΔT went into expansion work.

8. Mean Free Path

Definition

The mean free path λ is the average distance a molecule travels between two successive collisions. A molecule's actual path is a rapid zig-zag of straight segments (this is what makes diffusion slow even though molecular speeds are ~500 m/s — the answer to NCERT's famous "why does a gas take time to spread" question).

Mean Free Path $$\boxed{\,\lambda = \frac{1}{\sqrt{2}\,\pi n d^{2}}\, = \frac{k_BT}{\sqrt{2}\,\pi d^{2}P}}$$

n = number density (molecules/m³), d = molecular diameter. Dependence: λ ∝ 1/n ∝ 1/P (at constant T) and λ ∝ T (at constant P); λ ∝ 1/d². Typical air value at STP: λ ≈ 10⁻⁷ m ≈ 1000 × molecular size, with ~10⁹ collisions per second. Collision frequency $= v_{rms}/\lambda$.

Image Placeholder — Fig. 4

Zig-zag molecular path and the collision cylinder behind the λ formula (Landscape • Background #ffffff)

AI IMAGE PROMPT: Create a two-panel physics textbook diagram side by side, LANDSCAPE orientation (16:9), pure white background (#ffffff). LEFT PANEL titled "Random walk of a molecule": a square region outlined in thin black scattered with about fifteen small grey circles (other molecules); one blue (#1e88e5) molecule traces a zig-zag path drawn as connected straight blue segments with small direction changes at each grey molecule it grazes, arrowheads showing direction; THREE of the straight segments marked with dimension labels "λ₁", "λ₂", "λ₃" and a caption "mean free path λ = average of the free segments ≈ 10⁻⁷ m in air (≈ 10⁹ collisions/s)". RIGHT PANEL titled "The collision cylinder": a horizontal dashed cylinder (outline only) of length labelled "v̄t" and diameter labelled "2d (collision cross-section π d²)"; a blue molecule at the left end moving right with a green (#43a047) arrow "v̄"; two grey molecules drawn INSIDE the cylinder (will be hit) and two OUTSIDE it (missed); below the cylinder the boxed derivation chain: "collisions in time t = n·(πd²)(v̄t) ⟹ λ = 1/(πnd²) → with relative motion: λ = 1/(√2 π n d²)". Style: flat minimalist vector textbook figure, thin black outlines, blue tracked molecule, grey field molecules, dashed cylinder, clean sans-serif labels, no gradients, no shadows, pure white background.

The √2 comes from relative motion of the targets — a one-line JEE viva point shown in the chain.

✍ IN-TEXT PRACTICE 8.1 — JEE Main (λ scaling)

Q. The pressure of a gas is doubled at constant temperature. What happens to the mean free path and the collision frequency?

λ ∝ 1/P: λ becomes half.

Frequency = v_rms/λ: v_rms unchanged (same T) → frequency doubles.

9. Solved Examples — Every Question Type in the Chapter

One worked model of each distinct question type asked from this chapter, tagged by exam. Attempt each yourself first, then check.

✍ TYPE 1 — Boards · State, Derive & Explain

Q. State the postulates of kinetic theory and derive P = ⅓ρv²_rms (5 marks). / Show that average KE per molecule = 3/2 k_BT (3 marks). / State the law of equipartition and use it to find γ for mono- and diatomic gases (5 marks). / Derive Mayer's relation C_P − C_V = R (3 marks).

Where to answer from: Section 2 + Section 3 with Fig. 2 (quote all five postulates first), Section 4, Sections 6–7 (equipartition → C_V → C_P → γ chain), Section 7 (Mayer proof).

✍ TYPE 2 — NEET · KE per Molecule / per Mole

Q. Find the average translational KE of a molecule of any ideal gas at 27°C, and the total translational KE of one mole. (k_B = 1.38 × 10⁻²³ J/K, R = 8.3 J mol⁻¹K⁻¹)

Per molecule: $\tfrac32k_BT = 1.5\times1.38\times10^{-23}\times300 = \mathbf{6.21\times10^{-21} \text{ J}}$ — same for every gas at 300 K.

Per mole: $\tfrac32RT = 1.5\times8.3\times300 = \mathbf{3735 \text{ J}}$

✍ TYPE 3 — NEET / NDA · v_rms Ratio of Two Gases

Q. Compare the rms speeds of helium (M = 4) and methane (M = 16) at the same temperature. At what temperature will methane have the same v_rms as helium at 300 K?

Ratio: $\dfrac{v_{He}}{v_{CH_4}} = \sqrt{\dfrac{16}{4}} = \mathbf{2}$

Match: $\dfrac{T}{16} = \dfrac{300}{4} \Rightarrow T = \mathbf{1200 \text{ K}}$ (scale T by the mass ratio).

✍ TYPE 4 — JEE Main · Dalton's Law / Mixing Two Containers

Q. A 2 L vessel at 3 atm and a 3 L vessel at 2 atm (same gas, same T) are connected. Find the final common pressure.

Moles conserved (T constant): $P(V_1+V_2) = P_1V_1 + P_2V_2$

$P = \dfrac{3\times2 + 2\times3}{5} = \dfrac{12}{5} = \mathbf{2.4 \text{ atm}}$

✍ TYPE 5 — JEE · Gas Leak / Molecules Escaping

Q. A vessel at 27°C and pressure P loses half its gas molecules while being heated. If the pressure stays the same, find the final temperature.

PV = Nk_BT with P, V fixed: $NT = $ constant ⟹ $T_2 = \dfrac{N_1}{N_2}T_1 = 2\times300 = \mathbf{600 \text{ K} = 327°C}$

✍ TYPE 6 — NEET · Internal Energy of a Sample

Q. Find the internal energy of 2 moles of oxygen (rigid diatomic) at 300 K, and the energy stored in its rotational modes alone. (R = 8.3 J mol⁻¹K⁻¹)

Total: $U = n\tfrac{f}{2}RT = 2\times\tfrac52\times8.3\times300 = \mathbf{12450 \text{ J}}$

Rotational (2 of the 5): $\dfrac{2}{5}\times12450 = \mathbf{4980 \text{ J}}$ (equipartition shares equally).

✍ TYPE 7 — JEE Main · γ from f, and f from γ

Q. A gas has γ = 1.5. Find its degrees of freedom, C_V and C_P. Which real situation could this represent?

f: $\gamma = 1 + \dfrac2f \Rightarrow f = \dfrac{2}{0.5} = \mathbf{4}$

Heats: $C_V = 2R$, $C_P = 3R$. No simple pure gas has f = 4 — but a 1:1 He–O₂ mixture behaves exactly so (see Practice 7.1): γ questions often hide mixtures.

✍ TYPE 8 — Boards / NEET · Maxwell Curve Reading

Q. The speed-distribution curves of the same gas at temperatures T₁ and T₂ show the second peak at a higher speed but lower height. (i) Which temperature is greater? (ii) How do the areas under the curves compare? (iii) Mark v_p, v̄, v_rms in order.

(i) T₂ > T₁ (peak right-shifts with T). (ii) Areas equal — both equal the total number of molecules N. (iii) v_p < v̄ < v_rms (1.41 : 1.60 : 1.73) — see Fig. 3.

✍ TYPE 9 — NDA / Boards · Pressure from Molecular Data

Q. 10²² molecules, each of mass 5 × 10⁻²⁶ kg, are in a 1-litre box with v_rms = 500 m/s. Find the pressure.

P = ⅓ (mN/V)v²_rms: $P = \dfrac{1}{3}\cdot\dfrac{5\times10^{-26}\times10^{22}}{10^{-3}}\times(500)^2$

$= \dfrac13\times0.5\times2.5\times10^5 \approx \mathbf{4.2\times10^4 \text{ Pa}}$

✍ TYPE 10 — JEE Advanced Level · Mean Free Path Numeric

Q. Estimate the mean free path of air molecules at STP. (d ≈ 2 × 10⁻¹⁰ m, n ≈ 2.7 × 10²⁵ m⁻³) Also estimate the collision frequency if v̄ ≈ 480 m/s.

λ: $\lambda = \dfrac{1}{\sqrt2\pi nd^2} = \dfrac{1}{1.414\times3.14\times2.7\times10^{25}\times4\times10^{-20}} \approx \mathbf{2\times10^{-7} \text{ m}}$

Frequency: $\dfrac{\bar v}{\lambda} = \dfrac{480}{2\times10^{-7}} \approx \mathbf{2.4\times10^{9} \text{ collisions/s}}$ — billions per second, which is why diffusion is slow despite huge molecular speeds.

10. Common Mistakes & Misconceptions — Final Checklist

Night Before Exam

Read this the night before the exam:

ALWAYS convert to kelvin before any gas-law or speed formula (27°C = 300 K, not 27!).
Real gases → ideal at LOW pressure and HIGH temperature (not the reverse).
In v_rms = √(3RT/M), M is in kg/mol when R = 8.31 J mol⁻¹K⁻¹ (0.032, not 32, for O₂).
Average KE per molecule (3/2 k_BT) is the SAME for all gases at the same T — speeds differ, energies don't.
3/2 k_BT counts only TRANSLATIONAL KE; total energy per molecule is (f/2)k_BT.
v_rms ∝ √T: doubling the speed needs 4× the absolute temperature.
Speed order is fixed: v_p < v̄ < v_rms (1.41 : 1.60 : 1.73); the Maxwell peak marks v_p, not v_rms.
Raising T shifts the Maxwell peak right and lowers it, but the AREA (total N) never changes.
C_P > C_V always — constant-pressure heating also pays for expansion work; the difference is exactly R (per mole).
For gas mixtures, average C_V and C_P mole-weighted, then divide — never average γ directly.
Each vibrational mode contributes k_BT (two half-shares: KE + PE), not ½k_BT.
γ decodes the gas: 1.67 mono, 1.40 diatomic, 1.33 polyatomic; more atoms → γ closer to 1.
Mean free path: λ ∝ T/P and ∝ 1/d²; doubling P halves λ but leaves v_rms unchanged (same T).
Diffusion is slow because of ~10⁹ collisions per second, not because molecules are slow.
At absolute zero the ideal-gas KE → 0 (classically); V → 0 on the Charles' extrapolation — that's how −273.15°C is defined.

11. Rapid Revision — One-Page Formula Sheet

Concept	Formula	Condition / Note
Ideal gas equation	$PV = nRT = Nk_BT$	$PM = \rho RT$; T in kelvin
Combined law	$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$	Fixed amount of gas
Boltzmann constant	$k_B = R/N_A = 1.38\times10^{-23}$ J/K	Gas constant per molecule
Dalton / two vessels	$P(V_1+V_2) = P_1V_1 + P_2V_2$	Same T, connected
Pressure (KTG)	$P = \tfrac13\rho v_{rms}^2$	$PV = \tfrac23E_{trans}$
KE per molecule	$\tfrac32k_BT$	Same for all gases at same T
RMS speed	$v_{rms} = \sqrt{3RT/M} = \sqrt{3P/\rho}$	∝ √T, ∝ 1/√M
Average speed	$\bar v = \sqrt{8RT/\pi M}$	$v_p : \bar v : v_{rms} = 1.41 : 1.60 : 1.73$
Most probable speed	$v_p = \sqrt{2RT/M}$	$v_p : \bar v : v_{rms} = 1.41 : 1.60 : 1.73$
Equipartition	$\tfrac12k_BT$ per degree of freedom	$U = \tfrac{f}{2}k_BT$ per molecule
Degrees of freedom	mono 3, diatomic 5, poly 6	Vibrating diatomic: 7
Specific heats	$C_V = \tfrac{f}{2}R$, $C_P = C_V + R$	$\gamma = 1 + 2/f$
γ values	5/3, 7/5, 4/3	mono / diatomic / polyatomic
Mayer's relation	$C_P - C_V = R$	Per mole; per kg: $c_p - c_v = R/M$
Mixture heats	$C_{V,mix} = \frac{n_1C_{V1}+n_2C_{V2}}{n_1+n_2}$	Then γ_mix = C_P,mix/C_V,mix
Mean free path	$\lambda = \frac{1}{\sqrt2\pi nd^2} = \frac{k_BT}{\sqrt2\pi d^2P}$	∝ T/P, ∝ 1/d²; air: ~10⁻⁷ m
Collision frequency	$v_{rms}/\lambda$	~10⁹ s⁻¹ at STP
Dulong–Petit (solids)	$C \approx 3R$	From equipartition (3 vibr. modes × k_BT)

Study Plan

How to use these notes: Day 1: Sections 1–2 (gas laws + postulates) + Fig 1; solve five combined-law numericals. Day 2: Sections 3–4 (pressure derivation + temperature) + Fig 2; write the five-step derivation unaided. Day 3: Sections 5–7 (speeds, equipartition, specific heats) + Fig 3; reproduce the C_V/C_P/γ table and the Mayer proof from memory. Day 4: Section 8 (mean free path) + Fig 4, then all ten Types of Section 9 without looking, followed by the mistakes checklist. Finish every session by writing the formula sheet from memory.