68 free MCQs on Kinetic Theory with worked answers and explanations. Kinetic theory of gases connects molecular motion to pressure and temperature, explaining gas laws, specific heats, and molecular speeds.
Take the timed Kinetic Theory chapterwise test →Below are 68 practice questions on Kinetic Theory, sorted Easy → Hard. Tap “Show answer & explanation” under any question to check yourself. Want the full theory first? Read the Kinetic Theory notes.
Not all gas molecules move at the same speed - the Maxwell-Boltzmann distribution shows the spread, with the most probable speed vp slightly less than the average vavg, which is slightly less than the root-mean-square speed vrms used in the pressure formula.
Q1.
According to kinetic theory, gas molecules are assumed to undergo collisions that are:
Answer: B. Perfectly elastic
Why: Kinetic theory assumes molecular collisions (with each other and with walls) are perfectly elastic, conserving total kinetic energy.
Q2.
The pressure exerted by an ideal gas, according to kinetic theory, is given by P =
Answer: A. (1/3)ρv<sub>rms</sub>²
Why: Kinetic theory gives the result P = (1/3)ρv<sub>rms</sub>², where ρ is the density of the gas and v<sub>rms</sub> its root mean square speed.
Q3.
The average kinetic energy of a gas molecule is directly proportional to:
Answer: C. Absolute temperature of the gas
Why: Average KE per molecule = (3/2)k<sub>BT</sub>, showing it depends only on absolute temperature, not on the type of gas.
Q4.
The number of degrees of freedom for a monatomic gas molecule is:
Answer: B. 3
Why: A monatomic gas molecule (like helium) has only 3 translational degrees of freedom and no rotational or vibrational modes.
Q5.
For a diatomic gas at moderate temperatures (ignoring vibration), the number of degrees of freedom is:
Answer: B. 5
Why: A diatomic molecule has 3 translational + 2 rotational degrees of freedom at moderate temperature, giving f = 5.
Q6.
The law of equipartition of energy states that each degree of freedom contributes an average energy of:
Answer: B. (1/2)k<sub>BT</sub>
Why: According to the law of equipartition of energy, each quadratic degree of freedom contributes (1/2)k<sub>BT</sub> of average energy per molecule.
Q7.
Mean free path of gas molecules is the:
Answer: B. Average distance travelled between two successive collisions
Why: Mean free path is the average distance a gas molecule travels between two successive collisions with other molecules.
Q8.
Which of the three characteristic molecular speeds (rms, average, most probable) has the highest value?
Answer: C. RMS speed
Why: The order is v<sub>p</sub> < v<sub>avg</sub> < v<sub>rms</sub>; the root mean square speed is always the highest of the three.
Q9.
The root mean square speed of gas molecules is given by:
Answer: B. v<sub>rms</sub> = √(3RT/M)
Why: v<sub>rms</sub> = √(3RT/M), where R is the gas constant, T is absolute temperature, and M is the molar mass.
Q10.
At absolute zero temperature, the kinetic energy of gas molecules (classically) is:
Answer: B. Zero
Why: Classically, at 0 K, translational kinetic energy of gas molecules becomes zero since average KE = (3/2)k<sub>BT</sub>.
Q11.
For a monatomic ideal gas, the ratio of specific heats γ = Cp/Cv is:
Answer: B. 5/3
Why: For monatomic gases, Cv=(3/2)R and Cp=(5/2)R, giving γ = 5/3 ≈ 1.67.
Q12.
For a diatomic ideal gas (no vibration), the ratio of specific heats γ = Cp/Cv is:
Answer: B. 7/5
Why: For diatomic gases without vibration, Cv=(5/2)R and Cp=(7/2)R, giving γ = 7/5 = 1.4.
Q13.
Kinetic theory assumes that the actual volume occupied by gas molecules is:
Answer: B. Negligible compared to the volume of the container
Why: One of the key postulates of kinetic theory is that the size of gas molecules is negligible compared to the average distance between them and the volume of the container.
Q14.
The kinetic theory of gases assumes that gas molecules are in ___ motion:
Answer: A. constant random
Why: Gas molecules move continuously and randomly in all directions.
Q15.
The pressure exerted by a gas arises from molecular collisions with the:
Answer: A. the container walls
Why: Gas pressure is caused by molecules striking the container walls.
Q16.
In kinetic theory, the actual volume of the gas molecules themselves is taken to be ___ the container volume:
Answer: A. negligible beside
Why: The molecules’ own volume is assumed negligible relative to the space they occupy.
Q17.
The average kinetic energy of gas molecules depends only on the:
Answer: A. temperature
Why: Average molecular kinetic energy is proportional to the absolute temperature.
Q18.
At absolute zero, the kinetic energy of gas molecules is theoretically:
Answer: A. zero
Why: At 0 K molecular motion (and hence kinetic energy) theoretically ceases.
Q19.
In the kinetic theory, molecular collisions are assumed to be perfectly:
Answer: A. elastic
Why: Collisions are assumed perfectly elastic, conserving kinetic energy.
Q20.
The temperature scale used in the kinetic theory of gases is the:
Answer: A. kelvin scale
Why: Absolute temperature in kelvin is used throughout kinetic theory.
Q21.
If the absolute temperature of an ideal gas is increased 4 times, the rms speed of its molecules becomes:
Answer: A. 2 times
Why: v<sub>rms</sub> ∝ √T, so increasing T by 4 times increases v<sub>rms</sub> by √4 = 2 times.
Q22.
Equal volumes of two different ideal gases at the same temperature and pressure have:
Answer: B. The same number of molecules (Avogadro hypothesis)
Why: Equal volumes of any ideal gas at the same temperature and pressure contain the same number of molecules, consistent with Avogadros hypothesis and the ideal gas law.
Q23.
Two gases, hydrogen (M=2 g/mol) and oxygen (M=32 g/mol), are at the same temperature. The ratio of their rms speeds (v<sub>H</sub><sub>2</sub> / v<sub>O</sub><sub>2</sub>) is:
Answer: A. 4
Why: v<sub>rms</sub> ∝ 1/√M, so ratio = √(M<sub>O</sub><sub>2</sub>/M<sub>H</sub><sub>2</sub>) = √(32/2) = √16 = 4.
Q24.
An ideal gas is heated at constant volume. The heat supplied increases:
Answer: B. Only the kinetic energy of molecules (and hence temperature)
Why: At constant volume, all heat supplied goes into increasing the internal (kinetic) energy of the gas molecules, raising its temperature, and consequently the pressure rises too (since V constant).
Q25.
If the pressure of an ideal gas is doubled while keeping temperature constant, the rms speed of the molecules:
Answer: C. Remains the same
Why: v<sub>rms</sub> depends only on temperature and molar mass (v<sub>rms</sub> = √(3RT/M)), not on pressure, so it remains unchanged at constant T.
Q26.
For a gas mixture of monatomic and diatomic molecules with degrees of freedom f<sub>1</sub> and f<sub>2</sub> respectively in equal number of moles, the average degrees of freedom of the mixture is:
Answer: B. (f<sub>1</sub>+f<sub>2</sub>)/2
Why: For an equimolar mixture, the average degrees of freedom is the simple average of the individual values: (f<sub>1</sub>+f<sub>2</sub>)/2.
Q27.
The mean free path of gas molecules is inversely proportional to which factor (for fixed temperature)?
Answer: B. Number density of molecules
Why: λ = 1/(√2·n·π·d²), so mean free path decreases as the number density n of molecules increases (e.g., at higher pressure).
Q28.
A vessel contains a mixture of two ideal gases at the same temperature. Which statement is true regarding their molecules?
Answer: B. Both gases have the same average kinetic energy per molecule
Why: At a given temperature, average KE per molecule = (3/2)k<sub>BT</sub> is the same for all ideal gases regardless of molar mass; only their speeds differ (lighter molecules move faster).
Q29.
If a gas obeys PV = (2/3)E where E is the total translational kinetic energy of the gas, this kinetic theory result combined with PV = nRT shows that E equals:
Answer: C. (3/2)nRT
Why: Equating (2/3)E = nRT gives E = (3/2)nRT, the total translational kinetic energy of n moles of an ideal gas.
Q30.
Why does a balloon filled with helium gas leak (deflate) faster than one filled with air, through the same tiny porous wall?
Answer: B. Helium has a higher rms speed (lower molar mass) so it diffuses out faster
Why: Since v<sub>rms</sub> ∝ 1/√M, lighter helium molecules move faster than air molecules at the same temperature, leading to faster effusion/diffusion through small pores.
Q31.
For a polyatomic gas with 3 translational, 3 rotational, and 2 vibrational degrees of freedom, the total degrees of freedom f is:
Answer: C. 8
Why: f = 3 (translational) + 3 (rotational) + 2 (vibrational, each vibrational mode contributes 2 due to KE and PE) = 8.
Q32.
For an ideal gas, internal energy depends only on:
Answer: C. Temperature
Why: For an ideal gas, internal energy is purely a function of temperature, since intermolecular potential energy is neglected (no intermolecular forces assumed).
Q33.
The rms speed of gas molecules, v<sub>rms</sub> = √(3RT/M), is proportional to:
Answer: A. √T
Why: v<sub>rms</sub> varies as the square root of the absolute temperature.
Q34.
The average translational kinetic energy per molecule of a gas is:
Answer: A. (3/2)kT
Why: Average translational KE per molecule = (3/2)kT.
Q35.
As the temperature of a gas increases, the rms speed of its molecules:
Answer: A. increases
Why: Since v<sub>rms</sub> ∝ √T, a higher temperature gives a higher rms speed.
Q36.
In the relation PV = (1/3)Nm·v²_rms, the pressure is linked to the ___ of the molecules:
Answer: A. mean square speed
Why: Pressure depends on the mean square speed of the molecules.
Q37.
At the same temperature, a lighter gas has a ___ rms speed than a heavier gas:
Answer: A. higher
Why: Since v<sub>rms</sub> ∝ 1/√M, lighter molecules move faster at the same temperature.
Q38.
The number of degrees of freedom of a monatomic gas molecule is:
Answer: A. 3
Why: A monatomic molecule has only three translational degrees of freedom.
Q39.
The number of degrees of freedom of a diatomic gas molecule at moderate temperature is:
Answer: A. 5
Why: A diatomic molecule has 3 translational plus 2 rotational degrees of freedom = 5.
Q40.
The mean free path is the average distance a molecule travels between successive:
Answer: A. collisions
Why: Mean free path is the average distance between one collision and the next.
Q41.
A gas of N molecules is enclosed in a container. If the container volume is suddenly doubled at constant temperature (free expansion, no heat exchange), the rms speed of the molecules:
Answer: C. Remains the same since temperature is unchanged
Why: In free expansion of an ideal gas at constant temperature, since v<sub>rms</sub> depends only on T and M, and T is unchanged (ideal gas, no work done against external pressure in free expansion), v<sub>rms</sub> remains the same.
Q42.
Using kinetic theory, derive the relationship: if a gas has n moles and the total translational KE is (3/2)nRT, then for a diatomic gas with rotational energy also included, the total internal energy U at temperature T (f=5) is:
Answer: B. (5/2)nRT
Why: Total internal energy U = (f/2)nRT. For diatomic gas with f=5 (3 translational + 2 rotational), U = (5/2)nRT.
Q43.
A vessel contains N<sub>2</sub> gas at temperature T. If the temperature is raised such that the gas molecules begin to dissociate into individual N atoms at very high T, the degrees of freedom of the system effectively:
Answer: B. Decreases from 5 to 3 per resulting particle, since atoms only have translational degrees of freedom
Why: Diatomic N<sub>2</sub> molecules have f=5 (translational+rotational). After dissociation into individual N atoms, each atom (monatomic) only has translational freedom, f=3 per atom.
Q44.
Two ideal gas samples, A (monatomic) and B (diatomic), have equal moles and are at the same temperature. The ratio of their total internal energies U<sub>A</sub> : U<sub>B</sub> is:
Answer: A. 3:5
Why: U = (f/2)nRT. For equal n and T, ratio U<sub>A</sub>:U<sub>B</sub> = f<sub>A</sub>:f<sub>B</sub> = 3:5 (monatomic f=3, diatomic f=5).
Q45.
If the most probable speed of gas molecules is v<sub>p</sub> = √(2RT/M), and the rms speed is v<sub>rms</sub> = √(3RT/M), the ratio v<sub>rms</sub>/v<sub>p</sub> is:
Answer: A. √(3/2)
Why: v<sub>rms</sub>/v<sub>p</sub> = √(3RT/M)/√(2RT/M) = √(3/2) ≈ 1.22, showing v<sub>rms</sub> is always greater than v<sub>p</sub>.
Q46.
A closed rigid container has an ideal monatomic gas. Heat Q is supplied at constant volume, raising the temperature by ΔT. The fraction of heat that goes into increasing the rotational kinetic energy of the molecules is:
Answer: A. 0, since monatomic gas molecules have no rotational degrees of freedom
Why: Monatomic gas molecules (treated as point masses) have no rotational degrees of freedom, so all the heat supplied at constant volume goes entirely into translational kinetic energy.
Q47.
The pressure of an ideal gas is given by P = (1/3)nm(v<sub>rms</sub>)², where n is number density and m is the mass of each molecule. If the gas is compressed isothermally to half its volume, the number density n doubles. The new pressure compared to the original is:
Answer: B. Double
Why: At constant temperature, v<sub>rms</sub> is unchanged. Since P ∝ n and n doubles when volume is halved (n=N/V), the new pressure is double the original, consistent with Boyles law.
Q48.
Mean free path of a gas molecule is found to be λ at pressure P and temperature T. If the pressure is doubled at constant temperature, the new mean free path becomes:
Answer: B. λ/2
Why: At constant T, doubling pressure doubles the number density n (since PV=nRT, n=P/kT at fixed T). Since λ ∝ 1/n, the mean free path becomes λ/2.
Q49.
According to the Maxwell speed distribution, as temperature increases for a fixed gas, the distribution curve of molecular speeds:
Answer: B. Becomes broader and shifts to higher speeds, with the peak height decreasing
Why: As temperature increases, the Maxwell-Boltzmann speed distribution broadens and its peak shifts toward higher speeds, while the peak height decreases since the area under the curve (total probability) stays equal to 1.
Q50.
An ideal gas undergoes a process in which its pressure and volume are related by PV² = constant. Starting from an initial temperature T, if the volume of the gas doubles, the final temperature of the gas is:
Answer: C. T/2
Why: From the ideal gas law, PV = nRT, and PV² = constant means P ∝ V⁻². Substituting, T = PV/(nR) ∝ V⁻¹, so doubling V halves T, giving a final temperature of T/2.
Q51.
By the law of equipartition of energy, each degree of freedom contributes ___ per molecule:
Answer: A. (1/2)kT
Why: Each degree of freedom carries an average energy of (1/2)kT.
Q52.
At the same temperature, the rms speed of oxygen (M = 32) compared with hydrogen (M = 2) is:
Answer: A. 1/4 times
Why: v<sub>rms</sub> ∝ 1/√M, so v<sub>O</sub>/v<sub>H</sub> = √(2/32) = 1/4.
Q53.
If the absolute temperature of a gas is quadrupled, its rms speed becomes:
Answer: A. double
Why: Since v<sub>rms</sub> ∝ √T, quadrupling T multiplies the speed by √4 = 2.
Q54.
The ratio of specific heats γ = Cp/Cv for a monatomic ideal gas is:
Answer: A. 5/3
Why: For a monatomic gas, γ = 5/3 ≈ 1.67.
Q55.
The ratio γ = Cp/Cv for a diatomic ideal gas is:
Answer: A. 7/5
Why: For a diatomic gas, γ = 7/5 = 1.4.
Q56.
For one mole of an ideal gas, Cp − Cv equals:
Answer: A. R
Why: Mayer’s relation gives Cp − Cv = R for one mole of an ideal gas.
Q57.
The mean free path of a gas is ___ proportional to the number density of molecules:
Answer: A. inversely
Why: A higher molecular density means more frequent collisions, so a shorter mean free path.
Q58.
The internal energy of an ideal gas depends only on its:
Answer: A. temperature
Why: For an ideal gas, internal energy is a function of temperature alone.
Q59.
At the same temperature, all ideal gases have the same average:
Answer: A. translational kinetic energy
Why: Average translational KE = (3/2)kT depends only on temperature, so it is equal for all gases.
Q60.
The kinetic theory explains the gas laws by modelling a gas as a very large number of tiny:
Answer: A. moving molecules
Why: A gas is treated as many small molecules in constant random motion.
Q61.
At the same temperature, the ratio of rms speeds of hydrogen (M = 2) to oxygen (M = 32) molecules is:
Answer: B. 4:1
Why: v<sub>rms</sub> ∝ 1/√M, so ratio = √(32/2) = √16 = 4:1.
Q62.
At a fixed temperature, the average translational kinetic energy of a molecule depends on:
Answer: A. temperature only
Why: Average translational KE = (3/2)kT, which depends only on absolute temperature.
Q63.
The number of degrees of freedom of a rigid diatomic molecule at ordinary temperature is:
Answer: B. 5
Why: A rigid diatomic molecule has 3 translational + 2 rotational = 5 degrees of freedom.
Q64.
At constant temperature, the pressure of a gas is doubled. Its mean free path becomes:
Answer: B. halved
Why: Mean free path ∝ 1/n; doubling pressure at constant T doubles n, so the mean free path is halved.
Q65.
The rms speed of gas molecules at 27°C is v. At 927°C, the rms speed becomes:
Answer: B. 2v
Why: v<sub>rms</sub> ∝ √T; T goes from 300 K to 1200 K (4×), so speed becomes √4 = 2v.
Q66.
The internal energy of 2 moles of a monatomic ideal gas at 300 K (R = 8.31 J/mol·K) is about:
Answer: C. 7.48 kJ
Why: U = (3/2)nRT = 1.5·2·8.31·300 ≈ 7480 J ≈ 7.48 kJ.
Q67.
For a monatomic ideal gas, the ratio Cp/Cv equals:
Answer: D. 1.67
Why: Cv = 3R/2, Cp = 5R/2, so γ = 5/3 ≈ 1.67.
Q68.
From kinetic theory, P = (1/3)ρ v²_rms. The rms speed in terms of pressure P and density ρ is:
Answer: A. √(3P/ρ)
Why: Rearranging gives v<sub>rms</sub> = √(3P/ρ).