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CBSE · Class 11 · Physics

Kinetic Theory

Standard difficulty · 20 questions · 40 marks

Name: Date: Score: ____ / 40

Show your working for Sections B and C. Answers are on the last page — try every question first!

Section A — Choose the correct answer8 × 1 = 8 marks
1.
How many moles of gas are in a 21 L container at 2 × 105 N/m2 and 430 K? (R = 8.31 J/mol K)
  1. 0.12
  2. 11.75
  3. 2.35
  4. 1.18
2.
Nitrogen is heated from 410 K to 6560 K. Its rms speed becomes how many times?
  1. 16
  2. 8
  3. 5
  4. 4
3.
The average kinetic energy of a gas molecule at 390 K is (k = 1.38 × 10−23 J/K):
  1. 2.69 × 10−21 J
  2. 5.38 × 10−21 J
  3. 8.07 × 10−21 J
  4. 1.61 × 10−20 J
4.
For a polyatomic (non-linear) ideal gas (rigid molecules), the molar heat capacity at constant pressure, Cp, is:
  1. 4R
  2. 9R/2
  3. 5R/2
  4. 7R/2
5.
The ideal gas equation is:
  1. pT = nRV
  2. pV = RT/n
  3. p/V = nRT
  4. pV = nRT
6.
A gas is at 380 K. To what temperature must it be heated so that the rms speed of its molecules becomes 4 times?
  1. 6,080 K
  2. 24,320 K
  3. 1,520 K
  4. 6,353 K
7.
5 L of gas at 400 K is cooled to 200 K at constant pressure. Its new volume is:
  1. 2.5 L
  2. 5 L
  3. 10 L
  4. 1.34 L
8.
Assertion (A): If the temperature of a gas rises from 810 K to 1620 K, the average kinetic energy of its molecules becomes 4 times.
Reason (R): The average kinetic energy of a molecule is proportional to the absolute temperature.
  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.
Section B — Short answer8 × 2 = 16 marks
9.
A gas occupies 18 L at 5 atm and 400 K. What is its volume at 1 atm and 520 K? (2 decimal places where needed.)
10.
How many molecules are there in 49 L of a gas at 9 × 105 N/m2 and 460 K? (k = 1.38 × 10−23 J/K)
11.
A gas at a pressure of 2.0 × 105 N/m2 has a density of 2.4 kg/m3. Find the rms speed of its molecules.
12.
For a polyatomic (non-linear) ideal gas with rigid molecules, find the degrees of freedom, Cv, Cp and γ (in terms of R and in J/mol K), and the heat capacity at constant volume of 3 mol of it. (R = 8.31 J/mol K)
13.
Find the total translational kinetic energy of 4 mol of an ideal gas at 570 K. (R = 8.31 J/mol K) (2 decimal places where needed.)
14.
The average kinetic energy of the molecules of a gas is 20 × 10−21 J. Find the temperature of the gas. (k = 1.38 × 10−23 J/K) (2 decimal places where needed.)
15.
A tyre holds air at 150 kPa at 280 K. After a long drive the temperature of the air is 320 K. Assuming the volume does not change, find the new pressure. (2 decimal places where needed.)
16.
At a certain temperature, the rms speed of oxygen molecules is 2000 m/s. Find the rms speed of hydrogen molecules at the same temperature.
Section C — Word problems4 × 4 = 16 marks
17.
A cylinder of volume 7 L contains argon (molar mass 40 g/mol) at 10 × 105 N/m2 and 330 K. Find (a) the number of moles, (b) the mass of gas, (c) the number of molecules. (R = 8.31 J/mol K, Nₐ = 6.02 × 1023) (2 decimal places where needed.)
18.
A gas at 1.0 × 105 N/m2 and 280 K has density 7.5 kg/m3. (a) Find the rms speed of its molecules. (b) What is the rms speed at 2520 K? (c) What happens to the average kinetic energy?
19.
2 mol of a diatomic ideal gas is heated through 20 K (a) at constant volume, (b) at constant pressure. Find the heat needed in each case and explain why they differ. (R = 8.31 J/mol K) (2 decimal places where needed.)
20.
A student records the pressure and volume of a fixed mass of gas at constant temperature: p = 5 kPa, V = 100 L; p = 10 kPa, V = 50 L; p = 20 kPa, V = ? L; p = 25 kPa, V = 20 L. (a) Show that the data obey Boyle's law. (b) Find the missing volume. (c) Sketch the p–V graph.
AmbifulWorksheets
Practice that never repeats
CBSE · Class 11 · Physics

Answer key

Kinetic Theory · AW-11P-KTG-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.

QAnswerWorking
1(d) 1.18n = pV/RT
2(d) 4v_rms ∝ √T
3(c) 8.07 × 10−21 J(3/2)kT
4(a) 4Rfrom the degrees of freedom: Cv = (f/2)R, Cp = Cv + R
5(d) pV = nRT—
6(a) 6,080 Kv_rms ∝ √T
7(a) 2.5 LV/T is constant (Charles's law)
8(d) A is false, but R is true.A is false: KE = (3/2)kT, so it becomes 2 times. R is true.
9117 Lp1V1/T1 = p2V2/T2
106.95 × 1024N = pV/(kT)
11500 m/sp = ⅓ρv2, so v = √(3p/ρ)
12f = 6; Cv = (3)R = 24.93 J/mol K; Cp = (4)R = 33.24 J/mol K; γ = 4/3; 74.79 J/KCv = (f/2)R; Cp = Cv + R
1328,420.2 JE = (3/2)nRT
14966.18 KT = 2E/(3k)
15171.43 kPap/T is constant at constant volume
168,000 m/sv ∝ 1/√M: 2000 × √(32/2)
17(a) 2.55 mol; (b) 102.10 g; (c) 1.54 × 1024n = pV/RT
18(a) 200 m/s; (b) 600 m/s; (c) it becomes 9 timesv = √(3p/ρ); v ∝ √T; KE ∝ T
19(a) 831 J; (b) 1,163.4 J. At constant pressure the gas also does work (nRΔT = 332.4 J) as it expands.Cv = (f/2)R; Cp = Cv + R
20(a) pV = 500 kPa L each time. (b) 25 L. (c) A curve (rectangular hyperbola): as p rises, V falls.pV = constant at constant temperature