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<a href=”/”>eduPhysics</a> <span>/</span> <a href=”/notes/”>Notes</a> <span>/</span> <a href=”/notes/class-xi/”>Class XI</a> <span>/</span> <a href=”/notes/class-xi/kinetic-theory/”>Kinetic Theory</a> <span>/</span> <span style=”color:var(–ink-dim);”>Previous Year Questions</span>
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<div class=”edup-ch-header”>
<div>
<span class=”edup-ch-num”>Class XI · Chapter 12</span>
<h1 class=”edup-ch-title”>Kinetic Theory — Exam Practice</h1>
<p class=”edup-ch-sub”>Practice questions in CBSE, NEET, and JEE Main style covering gas laws, kinetic theory pressure, rms speed, and specific heats.</p>
<div class=”edup-meta-row”>
<span class=”edup-chip”>📝 10 questions</span>
<span class=”edup-chip”>🎯 CBSE · NEET · JEE Main</span>
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<a href=”/notes/class-xi/kinetic-theory/” class=”edup-btn edup-btn-outline”>← Back to Notes</a>
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<div class=”edup-notice”>
<strong>A note on these questions:</strong> these are original practice questions, written to match the exact style, format, and difficulty of CBSE Board, NEET, and JEE Main papers — not verbatim reproductions of specific past papers.
</div>
<div style=”max-width:800px;”>
<section class=”edup-section” id=”cbse-questions”>
<h2><span class=”n”>1</span>CBSE Board Exam Style</h2>
<div class=”edup-pyq-card”>
<div class=”edup-pyq-meta”><span class=”edup-exam-tag cbse”>CBSE · 1 Mark</span></div>
<div class=”edup-pyq-q”>Q1. Why does the pressure of a gas increase when it is heated at constant volume?</div>
<div class=”edup-pyq-solution”>
<span class=”lbl”>Answer</span>
Increased temperature raises the average kinetic energy (and hence speed) of gas molecules, causing more frequent and more forceful collisions with the container walls — this increased rate of momentum transfer is exactly what pressure measures.
</div>
</div>
<div class=”edup-pyq-card”>
<div class=”edup-pyq-meta”><span class=”edup-exam-tag cbse”>CBSE · 3 Marks</span></div>
<div class=”edup-pyq-q”>Derive the expression relating gas pressure to the mean square speed of its molecules, based on the kinetic theory.</div>
<div class=”edup-pyq-solution”>
<span class=”lbl”>Answer</span>
Considering molecular collisions with a container wall and the resulting momentum transfer, kinetic theory analysis (summed over all molecules and averaged over all directions) gives:
<br><br>
<strong>P = (1/3) ρ (v²)<sub>avg</sub></strong>, where ρ is gas density and (v²)<sub>avg</sub> is the mean square speed of the molecules — showing pressure emerges directly from molecular motion, not as a separate assumption.
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<div class=”edup-pyq-card”>
<div class=”edup-pyq-meta”><span class=”edup-exam-tag cbse”>CBSE · 5 Marks</span></div>
<div class=”edup-pyq-q”>Q3. (a) State the assumptions of the kinetic theory of gases. (b) Find the rms speed of nitrogen molecules at 300 K. (Molar mass of N₂ = 28×10⁻³ kg/mol, R = 8.314 J/mol·K)</div>
<div class=”edup-pyq-solution”>
<span class=”lbl”>Answer</span>
<strong>(a)</strong> Molecules are in continuous random motion; collisions are perfectly elastic; molecules exert no force on each other except during collision; the total volume of the molecules is negligible compared to the container; molecules obey Newton’s laws.
<br><br>
<strong>(b)</strong> v<sub>rms</sub> = √(3RT/M) = √[(3×8.314×300)/0.028] = √267,236 ≈ <strong>517 m/s</strong>.
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</section>
<section class=”edup-section” id=”neet-questions”>
<h2><span class=”n”>2</span>NEET Style (Single-Correct MCQ)</h2>
<div class=”edup-pyq-card”>
<div class=”edup-pyq-meta”><span class=”edup-exam-tag neet”>NEET Style</span></div>
<div class=”edup-pyq-q”>Q4. The average kinetic energy of a gas molecule depends only on:</div>
<div class=”edup-pyq-options”>
<div class=”edup-pyq-opt”>Pressure</div>
<div class=”edup-pyq-opt correct”>✓ Absolute temperature</div>
<div class=”edup-pyq-opt”>Volume</div>
<div class=”edup-pyq-opt”>Molar mass</div>
</div>
<div class=”edup-pyq-solution”>
<span class=”lbl”>Solution</span>
KE<sub>avg</sub> = (3/2)kT — this depends only on temperature, not on the gas’s identity, pressure, or volume.
</div>
</div>
<div class=”edup-pyq-card”>
<div class=”edup-pyq-meta”><span class=”edup-exam-tag neet”>NEET Style</span></div>
<div class=”edup-pyq-q”>Q5. At the same temperature, lighter gas molecules have:</div>
<div class=”edup-pyq-options”>
<div class=”edup-pyq-opt”>Lower rms speed</div>
<div class=”edup-pyq-opt correct”>✓ Higher rms speed</div>
<div class=”edup-pyq-opt”>The same rms speed as heavier molecules</div>
<div class=”edup-pyq-opt”>Zero rms speed</div>
</div>
<div class=”edup-pyq-solution”>
<span class=”lbl”>Solution</span>
v<sub>rms</sub> = √(3RT/M) — for the same average kinetic energy at a given temperature, lighter molecules (smaller M) must move faster.
</div>
</div>
<div class=”edup-pyq-card”>
<div class=”edup-pyq-meta”><span class=”edup-exam-tag neet”>NEET Style</span></div>
<div class=”edup-pyq-q”>Q6. The number of degrees of freedom for a monatomic gas molecule is:</div>
<div class=”edup-pyq-options”>
<div class=”edup-pyq-opt”>2</div>
<div class=”edup-pyq-opt correct”>✓ 3</div>
<div class=”edup-pyq-opt”>5</div>
<div class=”edup-pyq-opt”>6</div>
</div>
<div class=”edup-pyq-solution”>
<span class=”lbl”>Solution</span>
A monatomic molecule can only translate in three independent directions (x, y, z) — no rotational or vibrational modes are relevant at ordinary temperatures.
</div>
</div>
<div class=”edup-pyq-card”>
<div class=”edup-pyq-meta”><span class=”edup-exam-tag neet”>NEET Style</span></div>
<div class=”edup-pyq-q”>Q7. The mean free path of gas molecules decreases with an increase in:</div>
<div class=”edup-pyq-options”>
<div class=”edup-pyq-opt”>Temperature at constant pressure</div>
<div class=”edup-pyq-opt correct”>✓ Density (or pressure at constant temperature)</div>
<div class=”edup-pyq-opt”>Molecular speed alone</div>
<div class=”edup-pyq-opt”>Container volume, at constant density</div>
</div>
<div class=”edup-pyq-solution”>
<span class=”lbl”>Solution</span>
λ = 1/(√2πd²n) — a higher number density n (more molecules per unit volume, from higher density/pressure) means more frequent collisions and a shorter mean free path.
</div>
</div>
</section>
<section class=”edup-section” id=”jee-questions”>
<h2><span class=”n”>3</span>JEE Main Style</h2>
<div class=”edup-pyq-card”>
<div class=”edup-pyq-meta”><span class=”edup-exam-tag jee”>JEE Main Style · Numerical</span></div>
<div class=”edup-pyq-q”>Q8. Find the average translational kinetic energy of one mole of an ideal gas at 300 K. (R = 8.314 J/mol·K)</div>
<div class=”edup-pyq-solution”>
<span class=”lbl”>Solution</span>
KE (per mole) = (3/2)RT = (3/2) × 8.314 × 300 ≈ <strong>3741 J</strong>.
</div>
</div>
<div class=”edup-pyq-card”>
<div class=”edup-pyq-meta”><span class=”edup-exam-tag jee”>JEE Main Style · MCQ</span></div>
<div class=”edup-pyq-q”>Q9. The ratio of specific heats (Cp/Cv) for a diatomic gas is:</div>
<div class=”edup-pyq-options”>
<div class=”edup-pyq-opt”>5/3</div>
<div class=”edup-pyq-opt correct”>✓ 7/5</div>
<div class=”edup-pyq-opt”>3/2</div>
<div class=”edup-pyq-opt”>1</div>
</div>
<div class=”edup-pyq-solution”>
<span class=”lbl”>Solution</span>
For a diatomic gas, Cv = (5/2)R, Cp = (7/2)R, giving γ = Cp/Cv = 7/5 = 1.4.
</div>
</div>
<div class=”edup-pyq-card”>
<div class=”edup-pyq-meta”><span class=”edup-exam-tag jee”>JEE Main Style · Numerical</span></div>
<div class=”edup-pyq-q”>Q10. At the same temperature, find the ratio of rms speeds of hydrogen (M=2 g/mol) to oxygen (M=32 g/mol).</div>
<div class=”edup-pyq-solution”>
<span class=”lbl”>Solution</span>
v<sub>rms</sub> ∝ 1/√M, so ratio (H₂/O₂) = √(M<sub>O₂</sub>/M<sub>H₂</sub>) = √(32/2) = √16 = <strong>4</strong> — hydrogen molecules move 4× faster on average.
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<div class=”lbl”>← Previous · PYQ Set 11</div>
<h4>Thermodynamics</h4>
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<div class=”lbl”>Up Next · PYQ Set 13</div>
<h4>Oscillations</h4>
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