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NCERT Solutions for Class 11 Physics Chapter 12 – Kinetic Theory

Kinetic Theory is a conceptual and scoring chapter in Class 11 Physics that explains the microscopic behavior of gases. This chapter covers important topics such as kinetic theory of gases, assumptions of the model, pressure of a gas, kinetic energy of molecules, degrees of freedom, law of equipartition of energy, specific heat capacities, and mean free path, which are frequently tested in school exams and competitive exams like JEE and NEET.

NCERT Solutions for Class 11 Physics Chapter 12 – Kinetic Theory are prepared strictly according to the latest CBSE syllabus and exam pattern. The solutions are written in simple, step-by-step language with clear derivations, graphs, and solved numericals, helping students develop strong conceptual clarity and score well in Class 11 examinations.

NCERT Solutions for Class 11 Physics Chapter 12 – Kinetic Theory

Kinetic Theory

Medium

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 3A.

Kinetic Theory

Medium

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

Kinetic Theory

Medium

Figure 13.8 shows plot of PV/T versus P for 1.00×10^–3 kg of oxygen gas at two different temperatures.
(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 ×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 of PV/T (for low pressure high temperature region of the plot)? (Molecular mass of H2 = 2.02 u, of O2 = 32.0 u, R = 8.31 J mo1^–1K^–1.)

Kinetic Theory

Medium

An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27°C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17°C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31J mol^–1 K^–1, molecular mass of O₂= 32 u).

Kinetic Theory

Medium

An air bubble of volume 1.0 cm³ rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grow when it reaches the surface, which is at a temperature of 35°C?

Kinetic Theory

Medium

Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m³ at a temperature of 27 °C and 1 atm pressure.

Kinetic Theory

Medium

Estimate the average thermal energy of a helium atom at
(i) room temperature (27°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).

Kinetic Theory

Medium

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 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?

Kinetic Theory

Medium

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°C? (Atomic mass of Ar = 39.9 u, of He = 4.0 u).

Kinetic Theory

Difficult

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

Kinetic Theory

Difficult

A meter long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom?

Kinetic Theory

Difficult

From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm³ s^–1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm³ s^–1. Identify the gas.
[Hint: Use Graham’s law of diffusion: R₁/R₂ = (M₂/M₁)^1/2, where R₁, R₂ are diffusion rates of gases 1 and 2, and M₁ and M₂ their respective molecular masses. The law is a simple consequence of kinetic theory.]

Kinetic Theory

Difficult

A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have
uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called law of atmospheres
n₂ = n₁ exp [-mg (h₂ – h₁)/ k_BT]
Where n₂, n₁ refer to number density at heights h₂ and h₁ respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column:
n₂ = n₁ exp [-mg N_A(ρ - P′) (h₂ –h₁)/ (ρRT)]
Where ρ is the density of the suspended particle, and ρ’ that of surrounding medium.
[N_A is Avogadro’s number, and R the universal gas constant.] [Hint: Use Archimedes principle to find the apparent weight of the suspended particle.]

Kinetic Theory

Difficult

Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms:

Substance	Atomic Mass(u)	Density (10³ Kg m^–3)
Carbon (diamond)	12.01	2.22
Gold	197.00	19.32
Nitrogen (liquid)	14.01	1.00
Lithium	6.94	0.53
Fluorine (liquid)	19.00	1.14

[Hint: Assume the atoms to be ‘tightly packed’ in a solid or liquid phase, and use the known value of Avogadro’s number. You should, however, not take the actual numbers you obtain for various atomic sizes too literally. Because of the crudeness of the tight packing approximation, the results only indicate that atomic sizes are in the range of a few A].

NCERT Solutions for Class 11 Physics Chapter 12 – Kinetic Theory

1) Q. 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 Å.

Answer:
Diameter

$d = 3\ \text{Å} \Rightarrow r = 1.5\times10^{-10}\ \text{m}$

$d = 3 A ˚ \Rightarrow r = 1.5 \times 1 0^{-10} m$
Volume of one molecule:

$v=\frac{4}{3}\pi r^3$

$v = 34 π r^{3}$

Volume per molecule at STP:

$V_\text{per}=\frac{22.4\times10^{-3}}{N_A}$

$V_{per} = N ^{A} 22.4 \times 10 ^{-3}$

Fraction:

$\frac{v}{V_\text{per}} \approx 3.8\times10^{-4}$

$V ^{per} v \approx 3.8 \times 1 0^{-4}$

2) Q. Molar volume is the volume occupied by 1 mol of any (ideal) gas at STP. Show that it is 22.4 litres.

Answer:
At STP:

$P=1.013\times10^5\ \text{Pa},\ T=273\ \text{K},\ n=1$

$P = 1.013 \times 1 0^{5} Pa, T = 273 K, n = 1$

$PV=nRT \Rightarrow V=\frac{RT}{P}$

$P V = n RT \Rightarrow V = P RT$

$V=\frac{8.31\times273}{1.013\times10^5}\approx 2.24\times10^{-2}\ \text{m}^3 = 22.4\ \text{L}$

$V = 1.013 \times 10 ^{5} 8.31 \times 273 \approx 2.24 \times 1 0^{-2} m^{3} = 22.4 L$

3) Q. Figure 13.8 shows plot of PV/T versus P for $1.00\times10^{-3}$

$1.00 \times 1 0^{-3}$ kg of oxygen gas at two different temperatures.

(a) What does the dotted plot signify?
Answer: The dotted line represents ideal gas behaviour where

$PV/T$

$P V / T$ is constant (independent of

$P$

$P$ ).

(b) Which is true:

$T_1>T_2$

$T_{1} > T_{2}$ or

$T_1<T_2$

$T_{1} < T_{2}$ ?
Answer: The curve closer to the dotted (ideal) line corresponds to the higher temperature. Hence,

$T_1>T_2$

$T_{1} > T_{2}$ if curve

$T_1$

$T_{1}$ is nearer to ideal behaviour.

(c) What is the value of

$PV/T$

$P V / T$ where the curves meet on the y-axis?
Answer: At very low pressure, gas is ideal:

$\frac{PV}{T}=nR$

$T P V = n R$

For oxygen:

$m=10^{-3}\ \text{kg},\ M=0.032\ \text{kg mol}^{-1}\Rightarrow n=0.03125\ \text{mol}$

$m = 1 0^{-3} kg, M = 0.032 kg mol^{-1} \Rightarrow n = 0.03125 mol$

$\frac{PV}{T}=nR=0.03125\times 8.31 \approx 0.26\ \text{J K}^{-1}$

$T P V = n R = 0.03125 \times 8.31 \approx 0.26 J K^{-1}$

(d) For

$1.00\times10^{-3}$

$1.00 \times 1 0^{-3}$ kg hydrogen, will we get same

$PV/T$

$P V / T$ at y-axis? If not, what mass of H

$_2$

$_{2}$ gives same

$PV/T$

$P V / T$ ?
Answer: No, because

$PV/T=nR$

$P V / T = n R$ depends on number of moles.
To match oxygen’s

$n=0.03125$

$n = 0.03125$ mol:

$m_{H_2}=nM_{H_2}=0.03125\times0.00202 \approx 6.3\times10^{-5}\ \text{kg}$

$m_{H 2} = n M_{H 2} = 0.03125 \times 0.00202 \approx 6.3 \times 1 0^{-5} kg$

$m_{H_2}\approx 0.063\ \text{g}$

$m_{H 2} \approx 0.063 g$

4) Q. An oxygen cylinder of volume 30 litres has initial gauge pressure 15 atm at 27°C. After some oxygen is withdrawn, gauge pressure becomes 11 atm and temperature 17°C. Estimate mass of oxygen taken out.

Answer:
Use absolute pressure:

$P_\text{abs}=P_\text{gauge}+1$

$P_{abs} = P_{gauge} + 1$

$P_1=16\ \text{atm},\ T_1=300\ \text{K}$

$P_{1} = 16 atm, T_{1} = 300 K$

$P_2=12\ \text{atm},\ T_2=290\ \text{K}$

$P_{2} = 12 atm, T_{2} = 290 K$

$V=30\ \text{L}=0.03\ \text{m}^3$

$V = 30 L = 0.03 m^{3}$

$n=\frac{PV}{RT}$

$n = RT P V$

$n_1\approx 19.50,\quad n_2\approx 15.13 \Rightarrow \Delta n \approx 4.37\ \text{mol}$

$n_{1} \approx 19.50, n_{2} \approx 15.13 \Rightarrow Δ n \approx 4.37 mol$

Mass removed:

$m=\Delta n \times 0.032 \approx 0.14\ \text{kg}\approx 140\ \text{g}$

$m = Δ n \times 0.032 \approx 0.14 kg \approx 140 g$

5) Q. An air bubble of volume 1.0 cm³ rises from bottom of a lake 40 m deep at 12°C. To what volume does it grow at surface at 35°C?

Answer:
Bottom pressure:

$P_1=P_0+\rho gh$

$P_{1} = P_{0} + ρ g h$

$P_1=1.013\times10^5+1000\times9.8\times40 \approx 4.93\times10^5\ \text{Pa}$

$P_{1} = 1.013 \times 1 0^{5} + 1000 \times 9.8 \times 40 \approx 4.93 \times 1 0^{5} Pa$

Surface:

$P_2=P_0$

$P_{2} = P_{0}$

$T_1=285\ \text{K},\ T_2=308\ \text{K}$

$T_{1} = 285 K, T_{2} = 308 K$

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} \Rightarrow V_2=V_1\frac{P_1}{P_2}\frac{T_2}{T_1}$

$T ^{1} P ^{1} V ^{1} = T ^{2} P ^{2} V ^{2} \Rightarrow V_{2} = V_{1} P ^{2} P ^{1} T ^{1} T ^{2}$

$V_2 \approx 5.26\ \text{cm}^3$

$V_{2} \approx 5.26 cm^{3}$

6) Q. Estimate total number of air molecules in a room of capacity 25.0 m³ at 27°C and 1 atm.

Answer:

$n=\frac{PV}{RT}=\frac{(1.013\times10^5)\times25}{8.31\times300}\approx 1.02\times10^3\ \text{mol}$

$n = RT P V = 8.31 \times 300 ( 1.013 \times 10 ^{5} ) \times 25 \approx 1.02 \times 1 0^{3} mol$

$N=nN_A\approx (1.02\times10^3)(6.022\times10^{23}) \approx 6.1\times10^{26}\ \text{molecules}$

$N = n N_{A} \approx (1.02 \times 1 0^{3}) (6.022 \times 1 0^{23}) \approx 6.1 \times 1 0^{26} molecules$

7) Q. Estimate average thermal energy of a helium atom at (i) 27°C (ii) 6000 K (iii) 10 million K.

Answer: (per atom)

$\langle E\rangle=\frac{3}{2}kT$

$⟨ E ⟩ = 23 k T$

(i)

$T=300\ \text{K}\Rightarrow E\approx 6.2\times10^{-21}\ \text{J}$

$T = 300 K \Rightarrow E \approx 6.2 \times 1 0^{-21} J$
(ii)

$T=6000\ \text{K}\Rightarrow E\approx 1.24\times10^{-19}\ \text{J}$

$T = 6000 K \Rightarrow E \approx 1.24 \times 1 0^{-19} J$
(iii)

$T=10^7\ \text{K}\Rightarrow E\approx 2.07\times10^{-16}\ \text{J}$

$T = 1 0^{7} K \Rightarrow E \approx 2.07 \times 1 0^{-16} J$

8) Q. Three vessels of equal capacity have gases at same T and P: neon, chlorine, uranium hexafluoride. Do they contain equal number of molecules? Is $v_{rms}$

$v_{r m s}$ same? If not, where is it largest?

Answer:

Yes, they contain equal number of molecules (Avogadro’s law: same

$P,V,T\Rightarrow$ $P, V, T \Rightarrow$ same

$n$ $n$ ).
$v_{rms}$ $v_{r m s}$ is not same:

$v_{rms}=\sqrt{\frac{3RT}{M}}$

$v_{r m s} = M 3 RT$

Largest

$v_{rms}$ $v_{r m s}$ for smallest molar mass → Neon.

9) Q. At what temperature is rms speed of argon equal to rms speed of helium at –20°C?

Answer:

$\sqrt{\frac{T_{Ar}}{M_{Ar}}}=\sqrt{\frac{T_{He}}{M_{He}}} \Rightarrow T_{Ar}=T_{He}\frac{M_{Ar}}{M_{He}}$

$M ^{A r} T ^{A r} = M ^{He} T ^{He} \Rightarrow T_{A r} = T_{He} M ^{He} M ^{A r}$

$T_{He}=253\ \text{K},\ M_{Ar}=39.9,\ M_{He}=4.0$

$T_{He} = 253 K, M_{A r} = 39.9, M_{He} = 4.0$

$T_{Ar}\approx 253\times\frac{39.9}{4.0}\approx 2.52\times10^3\ \text{K}$

$T_{A r} \approx 253 \times 4.039.9 \approx 2.52 \times 1 0^{3} K$

10) Q. Estimate mean free path and collision frequency of N₂ at 2.0 atm and 17°C. Radius ≈ 1 Å. Compare collision time with free-flight time.

Answer:
Diameter

$d=2r=2\ \text{Å}=2\times10^{-10}\ \text{m}$

$d = 2 r = 2 A ˚ = 2 \times 1 0^{-10} m$
Number density:

$n=\frac{P}{kT}$

$n = k T P$

Mean free path:

$\lambda=\frac{1}{\sqrt{2}\pi d^2 n}\approx 1.1\times10^{-7}\ \text{m}$

$λ = 2 π d ^{2} n 1 \approx 1.1 \times 1 0^{-7} m$

Mean speed

$\bar v \approx 4.7\times10^2\ \text{m s}^{-1}$

$v ˉ \approx 4.7 \times 1 0^{2} m s^{-1}$
Collision frequency:

$f\approx \frac{\bar v}{\lambda}\approx 4.2\times10^9\ \text{s}^{-1}$

$f \approx λ v ˉ \approx 4.2 \times 1 0^{9} s^{-1}$

Time between collisions:

$\tau \approx \frac{1}{f}\approx 2.4\times10^{-10}\ \text{s}$

$τ \approx f 1 \approx 2.4 \times 1 0^{-10} s$

Collision duration is typically much smaller (~

$10^{-12}$

$1 0^{-12}$ s), so collision time

$\ll$

$≪$ free-flight time.

11) Q. A 1 m narrow tube (closed at one end) contains a 76 cm mercury thread trapping 15 cm air. What happens if tube is held vertical with open end at bottom?

Answer:
When vertical with open end at bottom, the mercury falls down and some mercury comes out, so the remaining mercury height becomes less than 76 cm. Trapped air expands until equilibrium:

$P_{air}=P_{atm}-\rho g h \quad (\text{in cm Hg: } P_{air}=76-h)$

$P_{ai r} = P_{a t m} - ρ g h (in cm Hg: P_{ai r} = 76 - h)$

If mercury height becomes

$h$

$h$ cm, air length becomes

$100-h$

$100 - h$ cm, and Boyle’s law gives:

$(76-h)(100-h)=76\times15$

$(76 - h) (100 - h) = 76 \times 15$

Solving:

$h\approx 52.2\ \text{cm}$

$h \approx 52.2 cm$

So trapped air length

$\approx 47.8$

$\approx 47.8$ cm and about

$76-52.2\approx 23.8$

$76 - 52.2 \approx 23.8$ cm of mercury flows out.

12) Q. Diffusion rate of H₂ is 28.7 cm³/s. Another gas diffuses at 7.2 cm³/s. Identify the gas.

Answer:
Graham’s law:

$\frac{R_1}{R_2}=\sqrt{\frac{M_2}{M_1}}$

$R ^{2} R ^{1} = M ^{1} M ^{2}$

$\frac{28.7}{7.2}\approx 3.99 \Rightarrow \left(\frac{28.7}{7.2}\right)^2\approx 15.9$

$7.228.7 \approx 3.99 \Rightarrow (7.228.7)^{2} \approx 15.9$

For H₂,

$M_1=2\Rightarrow M_2\approx 31.8\approx 32$

$M_{1} = 2 \Rightarrow M_{2} \approx 31.8 \approx 32$
Hence the gas is O₂ (oxygen).

13) Q. Use law of atmospheres to derive sedimentation equilibrium formula.

Answer (outline):
Law of atmospheres:

$n_2=n_1\exp\left[-\frac{mg(h_2-h_1)}{k_BT}\right]$

$n_{2} = n_{1} exp [- k ^{B} T m g ( h ^{2} - h ^{1} )]$

For a particle in liquid, effective (apparent) weight:

$W_{eff}=mg\left(1-\frac{\rho'}{\rho}\right)=mg\frac{(\rho-\rho')}{\rho}$

$W_{e ff} = m g (1 - ρ ρ ^{'}) = m g ρ ( ρ - ρ ^{'} )$

Replace

$m$

$m$ by molar mass form using

$k_B=\frac{R}{N_A}$

$k_{B} = N ^{A} R$ , giving:

$n_2=n_1\exp\left[-\frac{mgN_A(\rho-\rho')(h_2-h_1)}{\rho RT}\right]$

$n_{2} = n_{1} exp [- ρRT m g N ^{A} ( ρ - ρ ^{'} ) ( h ^{2} - h ^{1} )]$

which is the required expression.

14) Q. Given densities, estimate atomic sizes (tight packing approximation).

Answer (rough estimate):
Volume per atom

$\approx \frac{M}{\rho N_A}$

$\approx ρ N ^{A} M$ , atomic size

$\sim (\text{volume})^{1/3}$

$\sim (volume)^{1/3}$ .
Approx atomic sizes (order of Å):

Carbon (diamond): ~ 2.1 Å
Gold: ~ 2.6 Å
Nitrogen (liquid): ~ 2.9 Å
Lithium: ~ 2.8 Å
Fluorine (liquid): ~ 3.0 Å

So atomic sizes are in the range of a few Å.

FAQs: Class 11 Physics Chapter 12 – Kinetic Theory

Q1. Is Kinetic Theory important for exams?
Yes, it is a high-weightage chapter in thermal physics.

Q2. Which topics are most important in this chapter?
Kinetic theory of gases, equipartition of energy, and degrees of freedom.

Q3. Are numericals asked from this chapter?
Yes, RMS speed, pressure, and energy-based numericals are common.

Q4. Are derivations important here?
Yes, derivations related to pressure of gas and kinetic energy are frequently asked.

Q5. How do NCERT Solutions help?
They provide NCERT-aligned, exam-ready explanations with solved numericals.

NCERT Solutions for Class 11 Physics Chapter 12 – Kinetic Theory

NCERT Solutions for Class 11 Physics Chapter 12 – Kinetic Theory

NCERT Solutions for Class 11 Physics Chapter 12 – Kinetic Theory

1) Q. 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 Å.

2) Q. Molar volume is the volume occupied by 1 mol of any (ideal) gas at STP. Show that it is 22.4 litres.

3) Q. Figure 13.8 shows plot of PV/T versus P for $1.00\times10^{-3}$

$1.00 \times 1 0^{-3}$ kg of oxygen gas at two different temperatures.

4) Q. An oxygen cylinder of volume 30 litres has initial gauge pressure 15 atm at 27°C. After some oxygen is withdrawn, gauge pressure becomes 11 atm and temperature 17°C. Estimate mass of oxygen taken out.

5) Q. An air bubble of volume 1.0 cm³ rises from bottom of a lake 40 m deep at 12°C. To what volume does it grow at surface at 35°C?

6) Q. Estimate total number of air molecules in a room of capacity 25.0 m³ at 27°C and 1 atm.

7) Q. Estimate average thermal energy of a helium atom at (i) 27°C (ii) 6000 K (iii) 10 million K.

8) Q. Three vessels of equal capacity have gases at same T and P: neon, chlorine, uranium hexafluoride. Do they contain equal number of molecules? Is $v_{rms}$

$v_{r m s}$ same? If not, where is it largest?

9) Q. At what temperature is rms speed of argon equal to rms speed of helium at –20°C?

10) Q. Estimate mean free path and collision frequency of N₂ at 2.0 atm and 17°C. Radius ≈ 1 Å. Compare collision time with free-flight time.

11) Q. A 1 m narrow tube (closed at one end) contains a 76 cm mercury thread trapping 15 cm air. What happens if tube is held vertical with open end at bottom?

12) Q. Diffusion rate of H₂ is 28.7 cm³/s. Another gas diffuses at 7.2 cm³/s. Identify the gas.

13) Q. Use law of atmospheres to derive sedimentation equilibrium formula.

14) Q. Given densities, estimate atomic sizes (tight packing approximation).

FAQs: Class 11 Physics Chapter 12 – Kinetic Theory

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NCERT Solutions for Class 11 Physics Chapter 12 – Kinetic Theory

NCERT Solutions for Class 11 Physics Chapter 12 – Kinetic Theory

NCERT Solutions for Class 11 Physics Chapter 12 – Kinetic Theory

1) Q. 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 Å.

2) Q. Molar volume is the volume occupied by 1 mol of any (ideal) gas at STP. Show that it is 22.4 litres.

3) Q. Figure 13.8 shows plot of PV/T versus P for 1.00×10−31.00\times10^{-3}

1.00×10−3 kg of oxygen gas at two different temperatures.

4) Q. An oxygen cylinder of volume 30 litres has initial gauge pressure 15 atm at 27°C. After some oxygen is withdrawn, gauge pressure becomes 11 atm and temperature 17°C. Estimate mass of oxygen taken out.

5) Q. An air bubble of volume 1.0 cm³ rises from bottom of a lake 40 m deep at 12°C. To what volume does it grow at surface at 35°C?

6) Q. Estimate total number of air molecules in a room of capacity 25.0 m³ at 27°C and 1 atm.

7) Q. Estimate average thermal energy of a helium atom at (i) 27°C (ii) 6000 K (iii) 10 million K.

8) Q. Three vessels of equal capacity have gases at same T and P: neon, chlorine, uranium hexafluoride. Do they contain equal number of molecules? Is vrmsv_{rms}

vrms​ same? If not, where is it largest?

9) Q. At what temperature is rms speed of argon equal to rms speed of helium at –20°C?

10) Q. Estimate mean free path and collision frequency of N₂ at 2.0 atm and 17°C. Radius ≈ 1 Å. Compare collision time with free-flight time.

11) Q. A 1 m narrow tube (closed at one end) contains a 76 cm mercury thread trapping 15 cm air. What happens if tube is held vertical with open end at bottom?

12) Q. Diffusion rate of H₂ is 28.7 cm³/s. Another gas diffuses at 7.2 cm³/s. Identify the gas.

13) Q. Use law of atmospheres to derive sedimentation equilibrium formula.

14) Q. Given densities, estimate atomic sizes (tight packing approximation).

FAQs: Class 11 Physics Chapter 12 – Kinetic Theory

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3) Q. Figure 13.8 shows plot of PV/T versus P for $1.00\times10^{-3}$

$1.00 \times 1 0^{-3}$ kg of oxygen gas at two different temperatures.

8) Q. Three vessels of equal capacity have gases at same T and P: neon, chlorine, uranium hexafluoride. Do they contain equal number of molecules? Is $v_{rms}$

$v_{r m s}$ same? If not, where is it largest?