Kinetic Theory of Gases & Thermodynamics — Complete Revision (JEE 2026)

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State eqn (ideal)
$PV = nRT = NkT$

KTG pressure
$P=\dfrac{1}{3}nm\langle v^2\rangle$

RMS / mpd / avg
$v_{rms}=\sqrt{\dfrac{3kT}{m}}$ , $v_{mp}=\sqrt{\dfrac{2kT}{m}}$

Part A — Kinetic Theory of Gases (KTG)

1. Basic assumptions (ideal gas)

Molecules are point masses; collisions elastic.
No long-range intermolecular forces (except during collisions).
Random motion; Newtonian mechanics holds between collisions.
Time of collision ≪ time between collisions.

2. Pressure from microscopic motion (derivation sketch)

For N molecules in volume V, number density $n=N/V$. For one molecule with velocity component $v_x$, momentum change on elastic collision with wall gives impulse $2mv_x$. Rate of collisions and averaging over all molecules yields: \[ P = \frac{1}{3} n m \langle v^2\rangle. \] Combining with ideal gas $PV=NkT$ gives $ \tfrac{1}{2}m\langle v^2\rangle = \tfrac{3}{2}kT$.

3. Characteristic speeds

For molecule mass $m$ (kg) and molar mass $M$ (kg·mol⁻¹), and temperature $T$:

Most probable speed: \[ v_{mp} = \sqrt{\frac{2kT}{m}} = \sqrt{\frac{2RT}{M}} \]
Mean speed: \[ v_{avg} = \sqrt{\frac{8kT}{\pi m}} = \sqrt{\frac{8RT}{\pi M}} \]
Root-mean-square: \[ v_{rms} = \sqrt{\langle v^2\rangle} = \sqrt{\frac{3kT}{m}} = \sqrt{\frac{3RT}{M}} \]

Ordering: $v_{mp} < v_{avg} < v_{rms}$ (always).

4. Energy per molecule & equipartition

Each degree of freedom contributes $\tfrac{1}{2}kT$ per molecule to average energy. For a molecule with $f$ degrees of freedom: \[ \text{avg energy per molecule} = \frac{f}{2}kT,\qquad U_{\text{(per mole)}} = \frac{f}{2}RT. \] Typical $f$: monoatomic $=3$; diatomic (trans.+rot.) $\approx5$ (vibrational adds 2 per mode at high T).

5. Maxwell–Boltzmann distributions

Velocity (speed) distribution in 3D: \[ f(v)=4\pi\left(\frac{m}{2\pi kT}\right)^{3/2} v^{2} e^{-mv^{2}/2kT}. \] Kinetic-energy distribution (3D): \[ f_E(E)=\frac{2}{\sqrt{\pi}} \frac{1}{(kT)^{3/2}}\sqrt{E}\,e^{-E/kT},\qquad E=\tfrac{1}{2}mv^{2}. \] (General Boltzmann factor: probability ∝ $e^{-E/kT}$; normalize by partition function for full pdf.)

6. Mean free path & collision frequency

Effective molecular diameter $d$ and density $n$: \[ \lambda = \frac{1}{\sqrt{2}\,\pi d^{2} n}. \] Mean collision frequency for one molecule (approx): \[ \nu = \frac{\bar v}{\lambda} = \sqrt{2}\,\pi d^{2} n \bar v, \] where $\bar v$ is mean speed.

7. Transport & real-gas correction (mention)

Real gases deviate: van der Waals equation (molar volume $V_m$): \[ \left(P+\frac{a}{V_m^2}\right)(V_m - b) = RT, \] with critical constants: $T_c=\dfrac{8a}{27Rb},\ P_c=\dfrac{a}{27b^2},\ V_c=3b$.

Part B — Thermodynamics (Core Formula Bank & Insights)

1. Laws — quick

Zeroth: thermal equilibrium defines temperature.
First: $\Delta Q = \Delta U + W$ (energy conservation).
Second: entropy of isolated system does not decrease; reversible processes maximize efficiency.
Third: entropy → 0 as $T\to0$ (ideal limit).

2. Internal energy & heat capacities

For ideal gas (per mole): \[ U = \frac{f}{2} RT,\quad C_v = \frac{f}{2}R,\quad C_p = C_v + R,\quad \gamma=\frac{C_p}{C_v}. \] Example: monoatomic $f=3\Rightarrow C_v=\tfrac{3}{2}R,\ C_p=\tfrac{5}{2}R,\ \gamma=\tfrac{5}{3}$.

3. Useful differential identities

For an ideal gas: $dU = C_v dT$, $dH = C_p dT$, with $H=U+PV$ (enthalpy).

4. Standard processes & work

Isothermal (T const): $PV=\text{const}$, $\Delta U=0$, \[ W = nRT\ln\frac{V_2}{V_1}. \]
Adiabatic (reversible, no heat): $PV^\gamma=\text{const}$, \[ TV^{\gamma-1}=\text{const},\quad W=\frac{P_1V_1-P_2V_2}{\gamma-1}. \]
Isochoric: $V$ const → $W=0$, $Q=nC_v\Delta T$.
Isobaric: $P$ const → $W=P\Delta V$, $Q=nC_p\Delta T$.

5. Entropy (reversible paths)

Differential: $dS=\dfrac{dQ_{rev}}{T}$. For ideal gas between states: \[ \Delta S = nC_v\ln\frac{T_2}{T_1} + nR\ln\frac{V_2}{V_1} \] (valid for reversible change).

6. Heat engines & Carnot

Efficiency: \[ \eta=\frac{W}{Q_h}=1-\frac{Q_c}{Q_h}. \] Carnot (ideal reversible): \[ \eta_{\text{Carnot}} = 1 - \frac{T_c}{T_h}. \]

7. PV diagram quick rules

Area under curve = work done by system (integral $P\,dV$).
Isotherm slope gentler than adiabatic (adiabatic steeper).
Clockwise cycle → net work output; anticlockwise → net work input.

Worked Examples (Quick & Accurate)

Ex 1 — RMS change: If $T$ doubles, how does $v_{rms}$ change?

Solution: $v_{rms}\propto\sqrt{T}$ → factor $\sqrt{2}$. So $v_{rms}\to\sqrt{2}\,v_{rms}$.

Ex 2 — Isothermal work: One mole ideal gas expands isothermally at $T=300\,$K from $V_1$ to $V_2=2V_1$. Find $W$.

Solution: $W=nRT\ln(V_2/V_1)=1\times R\times300\times\ln2\approx (8.314)(300)\ln2\approx1728\ \text{J (approx).}$

Ex 3 — Adiabatic work: For adiabatic compression, relate $T$ and $V$ via $TV^{\gamma-1}=\text{const}$ and compute $W$ via endpoints.

Solution idea: use $P= nRT/V$ and integrate or use $W=(P_1V_1-P_2V_2)/(\gamma-1)$ after computing $P_2$ from adiabatic relation.

High-Yield JEE Traps & How to Avoid Them

Units: use SI: $m$ in kg, $M$ in kg·mol⁻¹, $k=1.38\times10^{-23}$ J·K⁻¹, $R=8.314\,$J·mol⁻¹K⁻¹.
Degrees of freedom: vibrational modes activate at high T — check context.
Sign convention: Work by system positive (expansion). Many errors stem from sign slips.
Isothermal ≠ adiabatic: do not mix formulas; check whether heat exchange occurs.
Entropy formula validity: only for reversible path or state function independent of path.
Mean free path caveat: uses hard-sphere approx; real molecules differ slightly.

Practice Problems (Try First)

Derive $P=\dfrac{1}{3}nm\langle v^2\rangle$ for an ideal monatomic gas (short derivation).
Compute mean kinetic energy per molecule at $T=300\,$K (in eV and J).
One mole ideal gas expands isothermally from 10 L to 20 L at 300 K. Calculate $W$, $\Delta U$, $Q$.
Find mean free path for air at STP assuming effective diameter $d=3.7\times10^{-10}\,$m.
For van der Waals gas with given a and b compute critical constants $T_c,P_c,V_c$.
Show that for monoatomic ideal gas, $C_v=\tfrac{3}{2}R$ and $\gamma=5/3$.

Show Answers & Hints

Standard derivation: consider momentum transfer to wall, use average $v_x^2=\\langle v^2\\rangle/3$, count collisions per unit time.
Mean KE = $\\tfrac{3}{2}kT$. At 300K: $\\tfrac{3}{2}\\times1.38\\times10^{-23}\\times300\\approx6.21\\times10^{-21}$ J ≈ $3.88\\times10^{-2}$ eV.
$W=nRT\\ln2\\approx(8.314)(300)\\ln2\\approx1728$ J; $\\Delta U=0$; $Q=W$ for isothermal.
$n\\approx2.69\\times10^{25}$ m⁻3 at STP → $\\lambda\\approx 1/(\\sqrt2\\pi d^2 n)\\approx 6.7\\times10^{-8}$ m (order of 10⁻⁷ m).
Use $T_c=8a/(27Rb)$ etc. (plug numbers to compute).
Equipartition gives $C_v=3R/2$; then $C_p=C_v+R=5R/2$ → $\\gamma=5/3$.

Final One-line Takeaways

KTG links microscopic motion to macroscopic $P,V,T$ — remember $P=\\tfrac{1}{3}nm\\langle v^2\\rangle$ and $\\langle E\\rangle=\\tfrac{3}{2}kT$ for monoatomic.
Use polarities: isothermal keeps T constant (ΔU=0); adiabatic keeps Q=0 (PV^γ const).
Watch units, DOF activation, and sign conventions — these cost marks in JEE.

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