Exercises — Kinetic Theory

1. 12.13 marksNCERT Cl-11 Physics Part II, Ch12 Exercises, Q12.1

   Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3 angstrom.

   Hint. Find the volume of one molecule as a sphere of the given diameter, multiply by the number of molecules in a mole (Avogadro's number), and compare that to the molar volume at STP (22.4 L).

   Step 1 — Volume of one molecule. Radius r = 1.5 angstrom = 1.5 x 10^-10 m. Volume = (4/3) pi r^3 = (4/3) x 3.1416 x (1.5x10^-10)^3 = 1.414 x 10^-29 m^3.

   Step 2 — Total molecular volume in one mole. Molecular volume = N_A x volume of one molecule = 6.02x10^23 x 1.414x10^-29 = 8.51 x 10^-6 m^3.

   Step 3 — Compare to the actual (molar) volume at STP. Actual volume = 22.4 L = 2.24 x 10^-2 m^3. Fraction = (8.51x10^-6) / (2.24x10^-2) = 3.8 x 10^-4.

   ✦ Molecules occupy only about 3.8 x 10^-4, or roughly 0.04%, of the gas's actual volume, since at STP the molecules of a gas sit far apart compared to their own size — the same fact that lets kinetic theory treat gases as free particles.

2. 12.23 marksNCERT Cl-11 Physics Part II, Ch12 Exercises, Q12.2

   Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP: 1 atmospheric pressure, 0 degrees C). Show that it is 22.4 litres.

   Hint. Apply the ideal gas equation directly with mu = 1 mol, and STP values for P and T.

   Step 1 — Write the ideal gas equation for 1 mole. PV = RT, so V = RT/P.

   Step 2 — Substitute STP values. T = 273 K, P = 1 atm = 1.013 x 10^5 Pa, R = 8.314 J/mol/K. V = (8.314 x 273) / (1.013x10^5) = 2269.7 / 101300 = 2.240 x 10^-2 m^3.

   Step 3 — Convert to litres. V = 2.240 x 10^-2 m^3 = 22.4 litres (since 1 m^3 = 1000 L).

   ✦ One mole of any ideal gas occupies 22.4 L at STP, since this result depends only on R, T, and P, none of which involve which particular gas is being measured.

3. 12.35 marksNCERT Cl-11 Physics Part II, Ch12 Exercises, Q12.3

   Figure 12.8 shows a plot of PV/T versus P for 1.00 x 10^-3 kg of oxygen gas at two different temperatures T1 and T2, as two curves that dip below a common dotted horizontal line at low-to-moderate P before rising again, with the curve labelled T1 dipping lower than the curve labelled T2. (a) What does the dotted plot signify? (b) Which is true: T1 > T2 or T1 < T2? (c) What is the value of PV/T where the curves meet on the y-axis? (d) If we obtained similar plots for 1.00 x 10^-3 kg of hydrogen, would we get the same value of PV/T at the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value? (Molecular mass of H2 = 2.02 u, of O2 = 32.0 u, R = 8.31 J/mol/K.)

   Hint. The dotted line is the ideal-gas value of PV/T, which real gases only approach at low pressure. The curve that deviates more strongly from this line belongs to the lower temperature, since real-gas non-ideality is stronger when a gas is closer to the temperature at which it would liquefy.

   (a) The dotted plot signifies the value of PV/T predicted by the ideal gas equation, mu R — constant and independent of P. It is the value both real curves approach only in the limit of low pressure, where the gas behaves most ideally.

   (b) T1 < T2. A real gas deviates from ideal behaviour more strongly at a lower temperature, since it is then relatively closer to the conditions under which it would liquefy; the curve dipping further below the dotted line is therefore the lower-temperature curve.

   (c) At the y-axis (P to 0), the curves meet at PV/T = mu R. Moles of oxygen: mu = mass/molar mass = (1.00x10^-3 kg) / (32.0x10^-3 kg/mol) = 0.03125 mol. PV/T = 0.03125 x 8.31 = 0.26 J/K.

   (d) No — mu R depends on how many moles are present, and 1.00x10^-3 kg of hydrogen is a different number of moles than 1.00x10^-3 kg of oxygen, so its curve would meet the y-axis at a different value. To get the same PV/T = 0.26 J/K, hydrogen needs the same mu = 0.03125 mol. Mass of hydrogen = mu x molar mass = 0.03125 x 2.02x10^-3 kg = 6.3 x 10^-5 kg (about 0.063 g).

   ✦ Since PV/T equals mu R at the y-axis, matching the value across two different gases is entirely about matching the number of moles, not the mass, which is why the required hydrogen mass comes out so much smaller than oxygen's.

4. 12.44 marksNCERT Cl-11 Physics Part II, Ch12 Exercises, Q12.4

   An oxygen cylinder of volume 30 litre has an initial gauge pressure of 15 atm and a temperature of 27 degrees C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 degrees C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J/mol/K, molecular mass of O2 = 32 u).

   Hint. Find the number of moles present before and after withdrawal separately using PV = mu RT, then convert the difference in moles to a mass using the molar mass of oxygen.

   Step 1 — Moles present initially. P1 = 15 atm = 15 x 1.013x10^5 Pa, V = 0.030 m^3, T1 = 300 K. mu1 = P1V/(RT1) = (15x1.013x10^5 x 0.030) / (8.31 x 300) = 45585 / 2493 = 18.29 mol.

   Step 2 — Moles present after withdrawal. P2 = 11 atm = 11 x 1.013x10^5 Pa, T2 = 290 K. mu2 = P2V/(RT2) = (11x1.013x10^5 x 0.030) / (8.31 x 290) = 33429 / 2409.9 = 13.87 mol.

   Step 3 — Convert the mole difference to mass. Delta-mu = 18.29 - 13.87 = 4.41 mol. Mass withdrawn = 4.41 x 0.032 kg/mol = 0.141 kg.

   ✦ About 0.141 kg (141 g) of oxygen was withdrawn, since the number of moles remaining in a fixed-volume cylinder is what actually determines the gauge pressure reading, not the mass alone.

5. 12.54 marksNCERT Cl-11 Physics Part II, Ch12 Exercises, Q12.5

   An air bubble of volume 1.0 cm^3 rises from the bottom of a lake 40 m deep at a temperature of 12 degrees C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 degrees C? (Take atmospheric pressure = 1.013 x 10^5 Pa, density of water = 1000 kg/m^3, g = 9.8 m/s^2.)

   Hint. Find the absolute pressure at the bottom of the lake by adding the water-column pressure to atmospheric pressure, then apply the combined gas law between the bottom and surface states.

   Step 1 — Pressure at the bottom. P1 = P_atm + rho g h = 1.013x10^5 + (1000 x 9.8 x 40) = 1.013x10^5 + 3.92x10^5 = 4.933 x 10^5 Pa.

   Step 2 — Pressure at the surface. P2 = P_atm = 1.013 x 10^5 Pa.

   Step 3 — Apply P1V1/T1 = P2V2/T2. T1 = 285 K, T2 = 308 K. V2 = V1 x (P1/P2) x (T2/T1) = 1.0 x (4.933x10^5/1.013x10^5) x (308/285) = 1.0 x 4.870 x 1.081 = 5.26 cm^3.

   ✦ The bubble grows to about 5.3 cm^3, since both the falling pressure and the rising temperature act in the same direction — expanding the bubble — as it rises to the surface.

6. 12.63 marksNCERT Cl-11 Physics Part II, Ch12 Exercises, Q12.6

   Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m^3 at a temperature of 27 degrees C and 1 atm pressure.

   Hint. Use PV = N kB T directly to find the total molecule count N — this counts every molecule in the mixture regardless of which gas it belongs to.

   Step 1 — Write PV = N kB T and solve for N. N = PV / (kB T).

   Step 2 — Substitute values. P = 1.013 x 10^5 Pa, V = 25.0 m^3, kB = 1.38 x 10^-23 J/K, T = 300 K. N = (1.013x10^5 x 25.0) / (1.38x10^-23 x 300) = 2.5325x10^6 / 4.14x10^-21 = 6.12 x 10^26.

   ✦ The room holds about 6.12 x 10^26 molecules, since PV = N kB T applies to the total gas mixture exactly as it would to any single ideal gas — the molecules do not need to be identical for this count to hold.

7. 12.73 marksNCERT Cl-11 Physics Part II, Ch12 Exercises, Q12.7

   Estimate the average thermal energy of a helium atom at (i) room temperature (27 degrees C), (ii) the temperature on the surface of the Sun (6000 K), (iii) the temperature of 10 million kelvin (the typical core temperature in the case of a star).

   Hint. Average thermal (translational kinetic) energy per molecule is (3/2) kB T regardless of which gas it is — just substitute each temperature in turn.

   Using E = (3/2) kB T with kB = 1.38 x 10^-23 J/K:

   (i) T = 300 K: E = 1.5 x 1.38x10^-23 x 300 = 6.21 x 10^-21 J.

   (ii) T = 6000 K: E = 1.5 x 1.38x10^-23 x 6000 = 1.242 x 10^-19 J.

   (iii) T = 1.0x10^7 K: E = 1.5 x 1.38x10^-23 x 1.0x10^7 = 2.07 x 10^-16 J.

   ✦ Average thermal energy scales linearly with absolute temperature, so the core of a star gives a helium atom roughly 33,000 times the thermal energy it has at room temperature, since 10^7 K is about 33,000 times 300 K.

8. 12.83 marksNCERT Cl-11 Physics Part II, Ch12 Exercises, Q12.8

   Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain an equal number of respective molecules? Is the root mean square speed of molecules the same in the three cases? If not, in which case is v_rms the largest?

   Hint. Number of molecules follows directly from the ideal gas equation and does not depend on molecular mass. rms speed, by contrast, depends explicitly on molecular mass.

   Step 1 — Number of molecules. At the same T, P, and V, the ideal gas equation PV = N kB T gives the same N for every gas, regardless of its identity. So yes, all three vessels contain an equal number of molecules — this is exactly Avogadro's law.

   Step 2 — rms speed comparison. v_rms = sqrt(3 kB T / m). At the same T, v_rms depends only on molecular mass m, so the three gases do NOT have the same v_rms.

   Step 3 — Identify the largest. Neon (about 20 u) is far lighter than chlorine (about 71 u) or uranium hexafluoride (about 352 u), so neon has the smallest m and therefore the largest v_rms.

   ✦ Molecule count is equal in all three vessels since it depends only on T, P, and V, but v_rms is largest for neon, since it is the lightest of the three gases and lighter molecules move faster to carry the same average kinetic energy.

9. 12.93 marksNCERT Cl-11 Physics Part II, Ch12 Exercises, Q12.9

   At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at -20 degrees C? (Atomic mass of Ar = 39.9 u, of He = 4.0 u.)

   Hint. Set the two v_rms expressions equal and solve for the unknown temperature — the masses do not cancel, since they are different gases.

   Step 1 — Write both rms speeds and set them equal. sqrt(3RT_Ar/M_Ar) = sqrt(3RT_He/M_He), so T_Ar/M_Ar = T_He/M_He.

   Step 2 — Solve for T_Ar. T_Ar = T_He x (M_Ar/M_He).

   Step 3 — Substitute values. T_He = -20 + 273 = 253 K. T_Ar = 253 x (39.9/4.0) = 253 x 9.975 = 2523.7 K.

   ✦ Argon needs to be at about 2523.7 K, nearly 10 times helium's temperature, since matching rms speeds across two gases of very different mass requires the heavier gas to be proportionally hotter.

10. 12.105 marksNCERT Cl-11 Physics Part II, Ch12 Exercises, Q12.10

    Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 degrees C. Take the radius of a nitrogen molecule to be roughly 1.0 angstrom. Compare the collision time with the time the molecule moves freely between two successive collisions (molecular mass of N2 = 28.0 u).

    Hint. Find the number density n from P = n kB T first, then use the mean free path formula. For the comparison, estimate the actual collision duration as roughly (molecular diameter)/(speed), and compare it to the free-flight time between collisions.

    Step 1 — Number density. n = P/(kB T) = (2.0 x 1.013x10^5) / (1.38x10^-23 x 290) = 2.026x10^5 / 4.002x10^-21 = 5.06 x 10^25 m^-3.

    Step 2 — Mean free path. d = 2 x 1.0 angstrom = 2.0 x 10^-10 m. l = 1/(sqrt(2) n pi d^2) = 1/(1.4142 x 5.06x10^25 x 3.1416 x 4.0x10^-20) = 1/(8.996x10^6) = 1.11 x 10^-7 m.

    Step 3 — rms speed and collision (free) time. v_rms = sqrt(3RT/M) = sqrt(3 x 8.31 x 290 / 0.028) = sqrt(258010) = 508 m/s. tau (free time between collisions) = l/v_rms = 1.11x10^-7 / 508 = 2.19 x 10^-10 s. Collision frequency = 1/tau = 4.57 x 10^9 collisions per second.

    Step 4 — Compare with actual collision duration. Collision duration (time within interaction range d of another molecule) is roughly d/v_rms = 2.0x10^-10 / 508 = 3.94 x 10^-13 s. Ratio = tau / (collision duration) = 2.19x10^-10 / 3.94x10^-13 = 556.

    ✦ The molecule spends about 556 times longer moving freely between collisions than it spends actually colliding, which is exactly why kinetic theory can treat collisions as effectively instantaneous events without affecting the pressure derivation.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 11 Physics Part II textbook, Reprint 2026-27 (keph205.pdf, 15 pages, "CHAPTER TWELVE"), cross-checked against the official CBSE curriculum 2026-27, Subject Code 042, Unit IX. One end-of-chapter Exercises set (10 questions, 12.1-12.10). One orphaned finding: CBSE's Unit IX line reads "equation of state of a perfect gas, work done in compressing a gas" but this chapter never derives a work-done formula — that belongs to Chapter 11 (Thermodynamics); this chapter supplies only the gamma value needed to use it.. Questions are referenced from the NCERT textbook for identification.