PiTheory

PiTheory JEE Physics Practice & Revision Pack · Class 11 · Chapter 12 · JEE Main

Kinetic Theory

Chapter-wise practice and revision for JEE Main: concise notes on the ideal gas law, pressure and molecular speeds, degrees of freedom and specific heats, and mean free path, concept figures, and 46 selected previous-year JEE Main questions covering the major concepts and question patterns, each with a step-by-step solution.

46 selected questions4 topics5 concept figuresPreview · first 3 of 46 questions

How to study this chapter

  1. Look at the topic's figures, then read its notes (5 to 10 minutes). They give the major equations and methods the questions use.
  2. Solve the topic's questions with a timer: 2 minutes for Easy, 3 to 4 for Medium and Hard.
  3. Check the solution. Read the Point to note line even if you were right: it is the idea you will reuse.
  4. Mark each question Got it or Revise. Come back to the Revise list after two days.

Where the marks come from

Main idea of each of the 139 previous-year JEE questions in the PiTheory bank

Molecular speeds48
Specific heats44
Ideal gas33
Mean free path14

Contents

  1. 1Ideal gas equation and gas lawsQ1–Q12
  2. 2Pressure, temperature and molecular speedsQ13–Q25
  3. 3Degrees of freedom, equipartition and specific heatsQ26–Q40
  4. 4Mean free path and collision frequencyQ41–Q46
  5. ✓Last-day checklist
01

Topic 1 · Q1–Q12

Ideal gas equation and gas laws

33 of the 139 JEE PYQs in the bank test this as their main idea

Look first

N molecules, volume VPV = nRT= NkBTn = m / Mρ = PM / RT
The ideal gas law ties the macroscopic state to the number of molecules.

Quick notes

$$PV=nRT=Nk_BT,\qquad n=\frac{m}{M}=\frac{N}{N_A},\qquad \rho=\frac{PM}{RT}$$

$T$ always in kelvin. $R=8.314$ J mol$^{-1}$ K$^{-1}$ $=2$ cal mol$^{-1}$ K$^{-1}$ (approx.), $k_B=R/N_A=1.38\times10^{-23}$ J K$^{-1}$. Number density $n_v=\dfrac{N}{V}=\dfrac{P}{k_BT}$.

LawHeld fixedResult
Boyle$n$, $T$$PV$ const
Charles$n$, $P$$V\propto T$
Gay-Lussac$n$, $V$$P\propto T$
Avogadro$P$, $T$$V\propto n$
Daltonmixture$P=\sum P_i$, $P_i=\dfrac{n_iRT}{V}$
  • Connected vessels: total moles conserved, $\sum\dfrac{P_iV_i}{T_i}=P\sum\dfrac{V_i}{T_i'}$; final pressure is common.
  • Insulated mixing (no work, no heat): total $U$ conserved. Same type of gas: $\sum n_iT_i=(\sum n_i)T$, equivalently $\sum P_iV_i=PV$.
  • Dissociation $X_2\to2X$: a fraction $\alpha$ dissociated multiplies moles by $(1+\alpha)$.
  • Process given as $P(V)$: $T=\dfrac{PV}{nR}$; for maximum $T$ set $\dfrac{d(PV)}{dV}=0$.
  • Expansion coefficient: if $V\propto T^k$ along the process, $\gamma_V=\dfrac1V\dfrac{dV}{dT}=\dfrac kT$ (at constant $P$: $\dfrac1T$).
  • Piston in a vertical cylinder: $(P_{\text{below}}-P_{\text{above}})A=mg$.
Traps. Celsius in $PV=nRT$. Assuming a common temperature when vessels are kept at different temperatures. Forgetting that the number of moles changes on dissociation or leakage.

Practice: first 3 of 12 questions

Q1JEE Main 2026 (6 Apr, Shift 1)Medium

Two closed vessels of same volume are joined through a narrow tube and both vessels are filled with air of pressure $90$ kPa and temperature $400$ K. Keeping the temperature of one vessel constant at $400$ K the second vessel temperature is raised to $500$ K. The final pressure in the vessels is ______ kPa.

  1. A$100$
  2. B$120$
  3. C$90$
  4. D$105$
Answer and solution

Answer: Option A: $100$

Concept. The vessels are joined, so the gas can move but its total amount stays fixed, and the final pressure is the same in both. Count moles with $n=\dfrac{PV}{RT}$ in each vessel before and after.

Step 1. Before (both at $400$ K): $n=\dfrac{90V}{R}\left(\dfrac1{400}+\dfrac1{400}\right)$.

Step 2. After (pressure $P$, temperatures $400$ and $500$ K): $n=\dfrac{PV}{R}\left(\dfrac1{400}+\dfrac1{500}\right)$.

Step 3. Equate. $90\times\dfrac{2}{400}=P\times\dfrac{9}{2000}\Rightarrow P=\dfrac{0.45}{0.0045}=100$ kPa.

Answer. Option A.

Point to note. Moles conserved: $\sum V/T$ weighted by $P$.

Q2JEE Main 2025 (4 Apr, Shift 2)Medium

There are two vessels filled with an ideal gas where the volume of one is double the volume of the other. The larger vessel contains the gas at $8\ \text{kPa}$ at $1000\ \text{K}$ while the smaller vessel contains the gas at $7\ \text{kPa}$ at $500\ \text{K}$. If the vessels are connected to each other by a thin tube allowing the gas to flow and the temperature of both vessels is maintained at $600\ \text{K}$, at steady state the pressure in the vessels will be (in kPa):

  1. A$4.4$
  2. B$6$
  3. C$24$
  4. D$18$
Answer and solution

Answer: Option B: $6$

Concept. Joining the vessels conserves the total amount of gas; afterwards both are at the same pressure and temperature. Count $n\propto\dfrac{PV}{T}$.

Step 1. Before (volumes $2V$ and $V$): $\dfrac{8(2V)}{1000}+\dfrac{7V}{500}=0.016V+0.014V=0.030V$.

Step 2. After. $\dfrac{P(3V)}{600}=0.005PV$.

Step 3. Equate. $P=6$ kPa.

Answer. Option B.

Point to note. Same method, unequal volumes.

Q3JEE Main 2026 (6 Apr, Shift 2)Easy

If $2$ mole of an ideal monoatomic gas at temperature $T$, is mixed with $6$ mole of another ideal monoatomic gas at temperature $2T$ then the temperature of mixture is:

  1. A$\dfrac{5}{2}T$
  2. B$\dfrac{5}{4}T$
  3. C$\dfrac{7}{2}T$
  4. D$\dfrac{7}{4}T$
Answer and solution

Answer: Option D: $\dfrac{7}{4}T$

Concept. When gases mix in isolation, the total internal energy is conserved. For monatomic gases $U=\tfrac32nRT$, so $\sum n_iT_i=(\sum n_i)T$.

Step 1. Energy balance. $2T+6(2T)=8T_f$.

Step 2. Solve. $T_f=\dfrac{14T}{8}=\dfrac74T$.

Answer. Option D.

Point to note. $U$ conserved: $\sum n_iT_i$.

43 more questions and 3 more topics in the full chapter

  1. 2Pressure, temperature and molecular speeds13 questions with solutions
  2. 3Degrees of freedom, equipartition and specific heats15 questions with solutions
  3. 4Mean free path and collision frequency6 questions with solutions
  4. ✓Last-day checklist

The complete Kinetic Theory chapter is in the Class 11 Pack and the Complete Pack, as a PDF on WhatsApp. Want to see a whole chapter first? Laws of Motion and Current Electricity are free in full.

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