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Kinetic TheoryNEET Physics · Class 11 · NCERT Chapter 12

32 NEET previous-year questions on Kinetic Theory, each with the correct answer and a step-by-step solution. Filter by topic and expand any question to see how to solve it.

PYQ frequency · topic × year

17
18
19
20
21
22
23
24
Ideal gas law
1
1
1
Pressure formula
1
1
1
1
1
Kinetic temperature
1
1
1
1
1
Speeds (rms / avg / vp)
1
1
1
1
1
1
Equipartition
1
1
1
1
Cp / Cv from f
1
1
1
1
Mean free path
2
1
Maxwell distribution
1
1

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All (32)
Ideal gas law (3)
Pressure formula (5)
Kinetic temperature (5)
Speeds (rms / avg / vp) (6)
Equipartition (4)
Cp / Cv from f (4)
Mean free path (3)
Maxwell distribution (2)

A

B

C

D

Solution

From , .

A

B

C

D

Solution

Three translational degrees of freedom: each carries , so total .

A

B

C

D

Solution

At fixed T, . , . Ratio . So .

A

B

C

D

Solution

Standard kinetic theory result: .

A

times

B

times

C

times

D

unchanged

Solution

. Doubling multiplies by .

A

B

C

D

Solution

.

A

B

C

D

Solution

.

A

B

C

D

Solution

, , . So .

A

241 K

B

362 K

C

483 K

D

724 K

Solution

.

A

11.2 L

B

22.4 L

C

44.8 L

D

depends on the gas

Solution

At STP (273 K, 101325 Pa), . Same for any ideal gas.

A

Only the temperature

B

Only the density

C

Both density and v_rms

D

Only the volume

Solution

. So pressure depends on both density (rho = m/V) and v_rms (which depends on T).

A

B

C

D

Solution

Half per quadratic degree of freedom per molecule. This is the equipartition theorem.

A

B

C

D

Solution

Monoatomic: , so .

A

B

C

D

Solution

.

A

The same average speed

B

The same average translational kinetic energy

C

The same density

D

The same molar mass

Solution

Avg translational KE depends only on T, not on mass. So both have the same avg KE; the heavier gas just moves slower.

A

Number density only

B

Molecular diameter only

C

Both number density and molecular diameter

D

Temperature only

Solution

. Depends on both molecular size and number density .

A

0.5 atm

B

1 atm

C

2 atm

D

4 atm

Solution

At constant V, = const. . (Gay-Lussac's law)

A

Half

B

Twice

C

Four times

D

Same

Solution

Boyle's law at constant T: = const. Half V doubles P.

A

Is the same as v_rms

B

Is greater than v_rms

C

Is less than both v_avg and v_rms

D

Is the average of all molecular speeds

Solution

The order is for any Maxwell distribution. v_p is the speed at the peak of f(v), the smallest of the three.

A

B

C

D

Solution

Diatomic at room T: , .

A

150 K

B

300 K

C

450 K

D

600 K

Solution

. To double KE, double T: .

A

1.87 kJ

B

3.74 kJ

C

7.48 kJ

D

12.5 kJ

Solution

.

A

Halves

B

Doubles

C

Quadruples

D

Stays the same

Solution

. Halve P, double .

A

Doubles

B

Triples

C

Quadruples

D

Stays the same

Solution

at fixed density. Doubling v_rms quadruples P.

A

B

C

D

Solution

.

A

B

C

D

Solution

, so . Lighter gas, faster molecules.

A

B

C

D

Solution

, so .

A

Shifts the peak left, narrows the curve

B

Shifts the peak right, broadens the curve

C

Shifts the peak right, narrows the curve

D

Has no effect on the shape

Solution

Higher T means higher v_p (peak shifts right) and a flatter, broader distribution.

A

B

C

D

Solution

At STP, 22.4 L contains 1 mole, which is molecules. (Avogadro's number)

A

3

B

6

C

9

D

27

Solution

. Tripling v_rms multiplies P by .

A

B

C

D

Solution

Each vibrational mode = 2 quadratic dof (kinetic + potential). So energy = .

A

0.4 nm

B

40 nm

C

70 nm

D

700 nm

Solution

Standard textbook result: lambda for air at STP is about 70 nm, several hundred molecular diameters.

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