Chapter - 12 Kinetic Theory

Frequently Asked Questions

Join ALLEN!

(Session 2026 - 27)

Name

Mobile Number

Class

Choose class

Goal

Choose your goal

Preferred Programs

Preferred Mode

State

Choose State

I agree to

I authorise ALLEN Career Institute Pvt Ltd to send me regular updates via Phone calls, Whatsapp, SMS, Robocalls (Automated Calls), Emails, or on postal address.

Frequently Asked Questions

Join ALLEN!

(Session 2026 - 27)

Name

Mobile Number

Class

Choose class

Goal

Choose your goal

Preferred Programs

Preferred Mode

State

Choose State

I agree to

I authorise ALLEN Career Institute Pvt Ltd to send me regular updates via Phone calls, Whatsapp, SMS, Robocalls (Automated Calls), Emails, or on postal address.

NCERT Solutions Class 11 Physics Chapter 12 – Kinetic Theory

NCERT Solutions for Class 11 Physics Chapter 12 by ALLEN provide clear, step-by-step explanations of Kinetic Theory and the molecular behaviour of gases. The solutions cover important concepts such as pressure of an ideal gas, RMS speed, degrees of freedom, law of equipartition of energy, and mean free path.

Students can understand the derivations and apply the concepts to numerical problems, including finding the RMS speed of oxygen molecules and calculating the mean free path of gas molecules. The solutions are prepared according to the latest NCERT syllabus and are useful for CBSE, JEE, and NEET preparation.

1.0Class 11 Physics Chapter 12 Kinetic Theory: Key Concepts

This chapter focuses on the postulates of the kinetic theory of gases and their mathematical consequences. Key lessons include:

Molecular Nature of Matter: Understanding the historical context from Dalton’s atomic theory to Avogadro’s hypothesis.
Ideal Gas Laws: Revisiting Boyle’s Law, Charles’s Law, and the Ideal Gas Equation (PV = nRT).
Kinetic Theory Postulates: The assumptions made about gas molecules (e.g., they are point masses, undergo elastic collisions, and have no intermolecular forces).
Pressure of an Ideal Gas: Deriving the formula for pressure based on the momentum transfer of molecules hitting the container walls:
$P = \frac{1}{3} ρ v_{r m s}^{2}$
Kinetic Interpretation of Temperature: Proving that the average kinetic energy of a molecule is proportional to the absolute temperature:
$E = \frac{3}{2} k_{B} T$
Law of Equipartition of Energy: The principle that total energy is shared equally among all degrees of freedom ( $1/2 k_{B}$ T per degree of freedom).
Degrees of Freedom: * Monatomic gases: 3 translational degrees.
- Diatomic gases: 3 translational + 2 rotational (+ 2 vibrational at high temp).
Specific Heat Capacity: Using the Law of Equipartition to calculate $C_{p}$ and $C_{v}$ for different gases and the ratio \gamma.
Mean Free Path ( $λ$ ): The average distance a molecule travels between two successive collisions:
$λ = \frac{1}{2 π d ^{2} n}$

2.0Detailed NCERT Textbook Solutions : Class 11 Physics Chapter 12

SOLVED EXAMPLES

The density of water is $1000 kg m^{- 3}$ . The density of water vapour at $10 0^{\circ} C$ and 1 atm pressure is $0.6 kg m^{- 3}$ . The volume of a molecule multiplied by the total number gives, what is called, molecular volume. Estimate the ratio (or fraction) of the molecular volume to the total volume occupied by the water vapour under the above conditions of temperature and pressure. Sol. For a given mass of water molecules, the density is less if volume is large. So the volume of the vapour is $1000/0.6 = 1/ (6 \times 1 0^{- 4})$ times larger. If densities of bulk water and water molecules are same, then the fraction of molecular volume to the total volume in liquid state is 1. As volume in vapour state has increased, the fractional volume is less by the same amount, i.e., $6 \times 1 0^{- 4}$ .
Estimate the volume of a water molecule using the data in Example 1. Sol. In the liquid (or solid) phase, the molecules of water are quite closely packed. The density of water molecule may therefore, be regarded as roughly equal to the density of bulk water $= 1000 kg m^{- 3}$ . To estimate the volume of a water molecule, we need to know the mass of a single water molecule. We know that 1 mole of water has a mass approximately equal to $(2 + 16) g = 18 g = 0.018 kg$ Since 1 mole contains about $6 \times 1 0^{23}$ molecules (Avogadro's number), the mass of a molecule of water is $(0.018) / (6 \times 1 0^{23}) kg = 3 \times 1 0^{- 26} kg$ . Therefore, a rough estimate of the volume of a water molecule is as follows : Volume of a water molecule

$= (3 \times 1 0^{- 26} kg) / (1000 kg m^{- 3}) = 3 \times 1 0^{- 29} m^{3} = (4/3) π (Radius)^{3}$

Hence, Radius $\approx 2 \times 1 0^{- 10} m = 2 \overset{˚}{A}$

What is the average distance between atoms (interatomic distance) in water? Use the data given in Example 1 & 2. Sol. A given mass of water in vapour state has $1.67 \times 1 0^{3}$ times the volume of the same mass of water in liquid state (Ex.-1). This is also the increase in the amount of volume available for each molecule of water. When volume increases by $1 0^{3}$ times the radius increases by $V^{1/3}$ or 10 times, i.e., $10 \times 2 \overset{˚}{A} = 20 \overset{˚}{A}$ . So the average distance is $2 \times 20 = 40 \overset{˚}{A}$ .
A vessel contains two non-reactive gases : neon (monoatomic) and oxygen (diatomic). The ratio of their partial pressures is 3 : 2. Estimate the ratio of (i) number of molecules and (ii) mass density of neon and oxygen in the vessel. Atomic mass of $Ne = 20.2 u$ , molecular mass of $O_{2} = 32.0 u$ . Sol. Partial pressure of a gas in a mixture is the pressure it would have for the same volume and temperature if it alone occupied the vessel. (The total pressure of a mixture of non-reactive gases is the sum of partial pressures due to its constituent gases.) Each gas (assumed ideal) obeys the gas law. Since V and T are common to the two gases, we have $P_{1} V = μ_{1} RT$ and $P_{2} V = μ_{2} RT$ , i.e. $(P_{1} / P_{2}) = (μ_{1} / μ_{2})$ . Here 1 and 2 refer to neon and oxygen respectively. Since $(P_{1} / P_{2}) = (3/2)$ (given), $(μ_{1} / μ_{2}) = 3/2$ . (i) By definition $μ_{1} = (N_{1} / N_{A})$ and $μ_{2} = (N_{2} / N_{A})$ where $N_{1}$ and $N_{2}$ are the number of molecules of 1 and 2, and $N_{A}$ is the Avogadro's number. Therefore, $(N_{1} / N_{2}) = (μ_{1} / μ_{2}) = 3/2$ .

(ii) We can also write $μ_{1} = (m_{1} / M_{1})$ and $μ_{2} = (m_{2} / M_{2})$ where $m_{1}$ and $m_{2}$ are the masses of 1 and 2; and $M_{1}$ and $M_{2}$ are their molecular masses. (Both $m_{1}$ and $M_{1}$ ; as well as $m_{2}$ an $M_{2}$ should be expressed in the same units). If $ρ_{1}$ and $ρ_{2}$ are the mass densities of 1 and 2 respectively, we have

$\frac{ρ _{1}}{ρ _{2}} = \frac{m _{1} / V}{m _{2} / V} = \frac{m _{1}}{m _{2}} = \frac{μ _{1}}{μ _{2}} \times (\frac{M _{1}}{M _{2}}) = \frac{3}{2} \times \frac{20.2}{32.0} = 0.947$

A flask contains argon and chlorine in the ratio of 2 : 1 by mass. The temperature of the mixture is $2 7^{\circ} C$ . Obtain the ratio of (i) average kinetic energy per molecule and (ii) root mean square speed $v_{rms}$ of the molecules of the two gases. Atomic mass of argon = 39.9 u; Molecular mass of chlorine $= 70.9 u$ . Sol. The important point to remember is that the average kinetic energy (per molecule) of any (ideal) gas (be it monatomic like argon, diatomic like chlorine or polyatomic) is always equal to (3/2) $k_{B} T$ . It depends only on temperature, and is independent of the nature of the gas. (i) Since argon and chlorine both have the same temperature in the flask, the ratio of average kinetic energy (per molecule) of the two gases is 1 : 1. (ii) Now $\frac{1}{2} m v_{rms}^{2} =$ average kinetic energy per molecule $= (\frac{3}{2}) k_{B} T$ where m is the mass of a molecule of the gas. Therefore.

$\frac{( v _{rms}^{2} ) _{Ar}}{( v _{rms}^{2} ) _{Cl}} = \frac{( m ) _{Cl}}{( m ) _{Ar}} = \frac{( M ) _{Cl}}{( M ) _{Ar}} = \frac{70.9}{39.9} = 1.77$

where M denotes the molecular mass of the gas. (For argon, a molecule is just an atom of argon.) Taking square root of both sides, $\frac{( v _{rms} ) _{Ar}}{( v _{rms} ) _{Cl}} = 1.33$ You should note that the composition of the mixture by mass is quite irrelevant to the above calculation. Any other proportion by mass of argon and chlorine would give the same answers to (i) and (ii), provided the temperature remains unaltered.

Uranium has two isotopes of masses 235 and 238 units. If both are present in Uranium hexafluoride gas which would have the larger average speed? If atomic mass of fluorine is 19 units, estimate the percentage difference in speeds at any temperature. Sol. At a fixed temperature the average energy $= \frac{1}{2}$ $m < v^{2} >$ is constant. So smaller the mass of the molecule, faster will be the speed. The ratio of speeds is inversely proportional to the square root of the ratio of the masses. The masses are 349 and 352 units. So

$\frac{v _{349}}{v _{352}} = (\frac{352}{349})^{1/2} = 1.0044$

Hence difference $\frac{Δ V}{V} = 0.44%$ $^{235} U$ is the isotope needed for nuclear fission. To separate it from the more abundant isotope $^{238} U$ , the mixture is surrounded by a porous cylinder. The porous cylinder must be thick and narrow, so that the molecule wanders through individually, colliding with the walls of the long pore. The faster molecule will leak out more than the slower one and so there is more of the lighter molecule (enrichment) outside the porous cylinder. The method is not very efficient and has to be repeated several times for sufficient enrichment.

(a) When a molecule (or an elastic ball) hits a (massive) wall, it rebounds with same speed. When a ball hits a massive bat held firmly, the same thing happens. However, when the bat is moving towards the ball, the ball rebounds with a different speed. Does the ball move faster or slower ? (Ch. 5 will refresh your memory on elastic collisions.) (b) When gas in a cylinder is compressed by pushing in a piston, its temperature rises. Guess at an explanation of this in terms of kinetic theory using (a) above. (c) What happens when a compressed gas pushes a piston out and expands. What would you observe? (d) Sachin Tendulkar used a heavy cricket bat while playing. Did it help him in anyway? Sol. (a) Let the speed of the ball be u relative to the wicket behind the bat. If the bat is moving towards the ball with a speed V relative to the wicket, then the relative speed of the ball to bat is $V + u$ towards the bat. When the ball rebounds (after hitting the massive bat) its speed, relative to bat, is $V + u$ moving away from the bat. So relative to the wicket the speed of the rebounding ball is V+ $(V + u) = 2 V + u$ , moving away from the wicket. So the ball speeds up after the collision with the bat. The rebound speed will be less than $u$ if the bat is not massive. For a molecule this would imply an increase in temperature. You should be able to answer (b) (c) and (d) based on the answer to (a). (Hint: Note the correspondence, piston → bat, cylinder → wicket, molecule → ball)
A cylinder of fixed capacity 44.8 litres contains helium gas at standard temperature and pressure. What is the amount of heat needed to raise the temperature of the gas in the cylinder by 15.0°C? $(R = 8.31 J mol^{- 1} K^{- 1})$ . Sol. Using the gas law $PV = μ RT$ , you can easily show that 1 mol of any (ideal) gas at standard temperature (273 K) and pressure ( $1 atm = 1.01 \times 1 0^{5} Pa$ ) occupies a volume of 22.4 litres. This the cylinder in this example contains 2 mol of helium. Further, since helium is monatomic, its predicted (and observed) molar specific heat at constant volume, $C_{V} = (3/2) R$ , and molar specific heat at constant pressure, $C_{P} = (3/2) R + R$ $= (5/2) R$ . Since the volume of the cylinder is fixed, the heat required is determined by $C_{v}$ . Therefore, Heat required = no. of moles × molar specific heat rise in temperature

$= 2 \times 1.5 R \times 15.0 = 45 R = 45 \times 8.31 = 374 J$

Estimate the mean free path for a water molecule in water vapour at 373 K . Use information from Exercises 12.1 and Eq. above. Sol. The d for water vapour is same as that of air. The number density is inversely proportional to absolute temperature.

$So n = 2.7 \times 1 0^{25} \times \frac{273}{373} = 2 \times 1 0^{25} m^{- 3}$

Hence, mean free path $l = 4 \times 1 0^{- 7} m$ Note that the mean free path is 100 times the interatomic distance $\sim 40 \overset{˚}{A} = 4 \times 1 0^{- 9} m$ calculated earlier. It is this large value of mean free path that leads to the typical gaseous behaviour. Gases cannot be confined without a container.

EXERCISE QUESTIONS WITH SOLUTIONS

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 $3 \overset{˚}{A}$ . Sol. Diameter of an oxygen molecule, $d = 3 \overset{˚}{A}$ .

$Radius, r = \frac{d}{2} = \frac{3}{2} = 15 \overset{˚}{A} = 1.5 \times 1 0^{- 8} cm$

Actual volume occupied by 1 mole of oxygen gas at STP $= 22400 cm^{3}$

Molecular volume of oxygen gas,

$V = \frac{4}{3} π r^{3} \cdot N$

Where, N is Avogadro's number

$= 6.023 \times 1 0^{23} molecules / mole ∴ V = \frac{4}{3} \times 3.14 \times (1.5 \times 1 0^{- 8})^{3} \times 6.023 \times 1 0^{23} = 8.51 cm^{3}$

Ratio of the molecular volume to the actual of

$oxygen = \frac{8.51}{22400} = 3.8 \times 1 0^{- 4}$

Molar volume is the volume occupied by 1 mole of any (ideal) gas at standard temperature and pressure (STP : 1 atmospheric pressure, 0°C). Show that it is 22.4 litres. Sol. The ideal gas equation relating pressure (P), volume (V), and absolute temperature (T) is given as : $PV = nRT$ Where, R is the universal gas constant $= 8.314 J mol^{- 1}$ $K^{- 1}$

$n = Number of moles = 1 T = Standard temperature = 273 K P = Standard pressure = 1 atm = 1.013 \times 1 0^{5} Nm^{- 2} ∴ V = \frac{nRT}{P} = \frac{1 \times 8.314 \times 273}{1.013 \times 1 0 ^{5}} = 0.0224 m^{3} = 22.4 litres$

Hence, the molar volume of a gas at STP is 22.4 litres.

Figure shows plot of PV/T versus P for $1.00 \times 1 0^{- 3} kg$ of oxygen gas at two different temperatures.

exercise-ques-3-class-11-chap-12-physics

(a) What does the dotted plot signify? (b) Which is true : $T_{1} > T_{2}$ or $T_{1} < T_{2}$ ? (c) What is the value of PV/T where the curves meet on the y-axis? (d) If we obtained similar plots for $1.00 \times 1 0^{- 3} kg$ 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 $H_{2} = 2.02 u$ , of $O_{2} = 32.0 u, R = 8.31 J mol^{- 1} K^{- 1}$ ) Sol. (a) The dotted plot in the graph signifies the ideal behaviour of the gas, i.e., the ratio $\frac{PV}{T}$ is equal. $μ R$ ( $μ$ is the number of moles and R is the universal gas constant) is a constant quality. It is not dependent on the pressure of the gas. (b) The dotted plot in the given graph represents an ideal gas. The curve of the gas at temperature $T_{1}$ is closer to the dotted plot than the curve of the gas at temperature $T_{2}$ . A real gas approaches the behaviour of an ideal gas when its temperature increases. Therefore, $T_{1} > T_{2}$ is true for the given plot. (c) The value of the ratio PV/T, where the two curves meet, is $μ R$ . This is because the ideal gas equation is given as : $PV = μ RT$ $\frac{PV}{T} = μ R$ Where, P is the pressure T is the temperature V is the volume $μ$ is the number of moles R is the universal constant

Molecular mass of oxygen $= 32.0 g$ Mass of oxygen $= 1 \times 1 0^{- 3} kg = 1 g$ $R = 8.314 J mole^{- 1} K^{- 1}$ $∴ \frac{PV}{T} = \frac{1}{32} \times 8.314 = 0.26 JK^{- 1}$ Therefore, the value of the ratio PV/T, where the curves meet on the y-axis, is $0.26 JK^{- 1}$ . (d) If we obtain similar plots for $1.00 \times 1 0^{- 3} kg$ of hydrogen, then we will not get the same value of PV/T at the point where the curves meet the y-axis. This is because the molecular mass of hydrogen (2.02 u) is different from that of oxygen (32.0 u). We have :

$∴ \frac{PV}{T} = 0.26 JK^{- 1} R = 8.314 J mol^{- 1} K^{- 1}$

Molecular mass (M) of $H_{2} = 2.02 u$ $\frac{PV}{T} = μ R$ at constant temperature Where, $μ = \frac{m}{M}$

$m = Mass of H_{2} m = \frac{PV}{T} \times \frac{M}{R} = \frac{0.26 \times 2.02}{8.31} = 6.3 \times 1 0^{- 2} g = 6.3 \times 1 0^{- 5} kg$

Hence, $6.3 \times 1 0^{- 5} kg$ of $H_{2}$ will yield the same value of PV/T

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 $1 7^{\circ} C$ . Estimate the mass of oxygen taken out of the cylinder $(R = 8.31 J mol^{- 1} K^{- 1}$ , molecular mass of $O_{2} = 32 u$ ) Sol. Volume of oxygen,

$V_{1} = 30 litres = 30 \times 1 0^{- 3} m^{3}$

Gauge pressure,

$P_{1} = 15 atm = 15 \times 1.013 \times 1 0^{5} Pa$

Temperature, $T_{1} = 2 7^{\circ} C = 300 K$ Universal gas constant, $R = 8.314 J mol^{- 1} K^{- 1}$ Let the initial no. of moles of oxygen gas in the cylinder be $n_{1}$ . The gas equation is given as :

$P_{1} V_{1} = n_{1} R T_{1} ∴ n_{1} = \frac{P _{1} V _{1}}{R T _{1}} = \frac{15.195 \times 1 0 ^{5} \times 30 \times 1 0 ^{- 3}}{( 8.314 ) \times 300} = 18.276$

But $n_{1} = \frac{m _{1}}{M}$ Where,

$m_{1} = Initial no. mass of oxygen M = Molecular mass of oxygen = 32 g ∴ m_{1} = n_{1} M = 18.276 \times 32 = 584.84 g$

After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces. Volume, $V_{2} = 30$ litres $= 30 \times 1 0^{- 3} m^{3}$ Gauge pressure,

$P_{2} = 11 atm = 11 \times 1.013 \times 1 0^{5} Pa Temperature, T_{2} = 1 7^{\circ} C = 290 K$

Let $n_{2}$ be the number of moles of oxygen left in the cylinder. The gas equation is given as :

$P_{2} V_{2} = n_{2} RT_{2} ∴ n_{2} = \frac{P _{2} V _{2}}{RT _{2}} = \frac{11.143 \times 1 0 ^{5} \times 30 \times 1 0 ^{- 3}}{8.314 \times 290} = 13.86$

But, $n_{2} = \frac{m _{2}}{M}$ Where, $m_{2}$ is the mass of oxygen remaining in the cylinder

$∴ m_{2} = n_{2} M = 13.86 \times 32 = 453.1 g$

The mass of oxygen taken out of the cylinder is given by the relation : Initial mass of oxygen in the cylinder - Final mass of oxygen in the cylinder

$= m_{1} - m_{2} = 584.84 g - 453.1 g$

$= 131.74 g = 0.131 kg$

Therefore, 0.131 kg of oxygen is taken out of the cylinder.

An air bubble of volume $1.0 cm^{3}$ 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? Sol. Volume of the air bubble,

$V_{1} = 1.0 cm^{3} = 1.0 \times 1 0^{- 6} m^{3}$

Bubble rises to height, $d = 40 m$ Temperature at a depth of 40 m ,

$T_{1} = 1 2^{\circ} C = 285 K$

Temperature at the surface of the lake,

$T_{2} = 3 5^{\circ} C = 308 K$

The pressure on the surface of the lake :

$P_{2} = 1 atm = 1 \times 1.013 \times 1 0^{5} Pa$

The pressure at the depth of 40 m:

$P_{1} = 1 atm + d ρ g$

Where,

$ρ is density of water = 1 0^{3} kg / m^{3}$

g is the acceleration due to gravity $= 9.8 m / s^{2}$

$∴ P_{1} = 1.013 \times 1 0^{5} + 40 \times 1 0^{3} \times 9.8 = 493300 Pa$

We have : $\frac{P _{1} V _{1}}{T _{1}} = \frac{P _{2} V _{2}}{T _{2}}$ Where, $V_{2}$ is the volume of the air bubble when it reaches the surface $V_{2} = \frac{P _{1} V _{1} T _{2}}{T _{1} P _{2}}$

$= \frac{( 493300 ) ( 1.0 \times 1 0 ^{- 6} ) 308}{285 \times 1.013 \times 1 0 ^{5}} = 5.263 \times 1 0^{- 6} m^{3} or 5.263 cm^{3}$

Therefore, when the air bubble reaches the surface, its volume becomes $5.263 cm^{3}$ .

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 $2 7^{\circ} C$ and 1 atm pressure. Sol. Volume of the room,

$V = 25.0 m^{3}$

Temperature of the room,

$T = 2 7^{\circ} C = 300 K$

Pressure in the room,

$P = 1 atm = 1 \times 1.013 \times 1 0^{5} Pa$

The ideal gas equation relating pressure (P), Volume (V), and absolute temperature (T) can be written as :

$PV = k_{B} NT$

Where, $k_{B}$ is Boltzmann constant

$= 1.38 \times 1 0^{- 23} m^{2} kgs^{- 2} K^{- 1}$

N is the number of air molecules in the room

$N = \frac{PV}{k _{B} T} = \frac{1.013 \times 1 0 ^{5} \times 25}{1.38 \times 1 0 ^{- 23} \times 300} = 6.11 \times 1 0^{26} molecules$

Therefore, the total number of air molecules in the given room is $6.11 \times 1 0^{26}$ .

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). Sol. At room temperature,

$T = 2 7^{\circ} C = 300 K$

Average thermal energy $= \frac{3}{2} kT$ Where k is Boltzmann constant

$= 1.38 \times 1 0^{- 23} m^{2} kg s^{- 2} K^{- 1} ∴ \frac{3}{2} kT = \frac{3}{2} \times 1.38 \times 1 0^{- 38} \times 300 = 6.21 \times 1 0^{- 21} J$

Hence, the average thermal energy of a helium atom at room temperature (27°C) is

$6.21 \times 1 0^{- 21} J$

On the surface of the Sun, $T = 6000 K$

$Average thermal energy = \frac{3}{2} kT = \frac{3}{2} \times 1.38 \times 1 0^{- 38} \times 6000 = 1.241 \times 1 0^{- 19} J$

Hence, the average thermal energy of a helium atom on the surface of the Sun is

$1.241 \times 1 0^{- 19} J$

At temperature, $T = 1 0^{7} K$

$Average thermal energy = \frac{3}{2} kT = \frac{3}{2} \times 1.38 \times 1 0^{- 23} \times 1 0^{7} = 2.07 \times 1 0^{- 16} J$

Hence, the average thermal energy of a helium atom at the core of a star is $2.07 \times 1 0^{- 16} J$ .

Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (mono atomic), 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? Sol. Yes. All contain the same number of the respective molecules. No. The root mean square speed of neon is the largest. Since the three vessels have the same capacity, they have the same volume.

Hence, each gas has the same pressure, volume, and temperature. According to Avogadro's law the three vessels will contain an equal number of the respectively molecules. This number is equal to Avogadro's number. $N = 6.023 \times 1 0^{23}$ .

The root mean square speed $(v_{rms})$ of a gas of mass m , and temperature T , is given by the relation :

$v_{rms} = \frac{3 kT}{m}$

Where, k is Boltzmann constant For the given gases, k and T are constants. Hence $v_{rms}$ depends only on the mass of the atoms i.e.,

$v_{rms} \propto \frac{1}{m}$

Therefore, the root mean square speed of the molecules in the three cases is not the same. Among neon, chlorine, and uranium hexafluoride, the mass of neon is the smallest. Hence, neon has the largest root mean square speed among the given gases.

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 $He = 4.0 u$ ).

Sol. Temperature of the helium atom,

$T_{He} = - 2 0^{\circ} C = 253 K$

Atomic mass of argon, $M_{Ar} = 39.9 u$ Atomic mass of helium, $M_{He} = 4.0 u$ Let $(v_{rms})_{Ar}$ be the rms speed of argon. Let $(v_{rms})_{He}$ be the rms speed of helium. The rms speed of argon is given by :

$(v_{rms})_{Ar} = \frac{3 RT _{Ar}}{M _{Ar}}$

Where, R is the universal gas constant $T_{Ar}$ is temperature of argon gas The rms speed of helium is given by:

$(v_{rms})_{He} = \frac{3 RT _{He}}{M _{He}}$

It is given that :

$(v_{rms})_{Ar} = (v_{rms})_{He}$

$\frac{3 R T _{Ar}}{M _{Ar}} = \frac{3 R T _{He}}{M _{He}} \frac{T _{Ar}}{M _{Ar}} = \frac{T _{He}}{M _{He}} T_{Ar} = \frac{T _{He}}{M _{He}} \times M_{Ar} = \frac{253}{4} \times 39.9 = 2523.675 = 2.52 \times 1 0^{3} K$

Therefore, the temperature of the argon atom is $2.52 \times 1 0^{3} K$ .

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 $1.0 \overset{˚}{A}$ . Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of $N_{2} = 28.0 u$ ). Sol. [Mean free path $= 1.11 \times 1 0^{- 7} m$ Collision frequency $= 4.58 \times 1 0^{9} s^{- 1}$ Successive collision time $\approx 500 \times$ (Collision time)] Pressure inside the cylinder containing nitrogen,

$P = 2.0 atm = 2.026 \times 1 0^{5} Pa$

Temperature inside the cylinder,

$T = 1 7^{\circ} C = 290 K$

Radius of a nitrogen molecule,

$r = 1.0 \overset{˚}{A} = 1 \times 1 0^{10} m$

Diameter, $d = 2 \times 1 \times 1 0^{10}$

$= 2 \times 1 0^{10} m$

Molecular mass of nitrogen,

$M = 28.0 g = 28 \times 1 0^{- 3} kg$

The root mean square speed of nitrogen is given by the relation :

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

Where, R is the universal gas constant

$= 8.314 J mole^{- 1} K^{- 1} ∴ v_{rms} = \frac{3 \times 8.314 \times 290}{28 \times 1 0 ^{- 3}} = 508.26 m / s$

The mean free path ( $l$ ) is given by the relation :

$l = \frac{kT}{2 \times d ^{2} \times P}$

Where, k is the Boltzmann constant

$= 1.38 \times 1 0^{- 23} kg m^{2} s^{- 2} K^{- 1} ∴ l = \frac{1.38 \times 1 0 ^{- 23} \times 290}{2 \times 2.14 \times ( 2 \times 1 0 ^{- 10} ) ^{2} \times 2.026 \times 1 0 ^{5}} = 1.11 \times 1 0^{- 7} m$

Collision frequency $= \frac{v _{rms}}{l}$

$= \frac{508.26}{1.11 \times 1 0 ^{- 7}} = 4.58 \times 1 0^{9} s^{- 1}$

Collision time is given as :

$T = \frac{d}{v _{rms}} = \frac{2 \times 1 0 ^{- 10}}{508.26} = 3.93 \times 1 0^{- 13} s$

Time taken between successive collisions :

$T^{'} = \frac{l}{v _{rms}} = \frac{1.11 \times 1 0 ^{- 7} m}{508.26 m / s} = 2.18 \times 1 0^{- 10} S ∴ \frac{T ^{'}}{T} = \frac{2.18 \times 1 0 ^{- 10}}{3.93 \times 1 0 ^{- 13}} = 500$

Hence, the time taken between successive collision is 500 times the time taken for a collision.

3.0NCERT Class 11 Physics – Chapter-Wise Solutions & Explanations

Study NCERT Solutions for Class 11 Physics chapter-wise with simple explanations for every textbook question. Understand important concepts, follow clear solution steps, and strengthen your Physics fundamentals across all chapters.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Units and Measurements
Chapter 2	Motion in a Straight Line
Chapter 3	Motion in a Plane
Chapter 4	Laws of Motion
Chapter 5	Work, Energy and Power
Chapter 6	System of Particles and Rotational Motion
Chapter 7	Gravitation
Chapter 8	Mechanical Properties of Solids
Chapter 9	Mechanical Properties of Fluids
Chapter 10	Thermal Properties of Matter
Chapter 11	Thermodynamics
Chapter 13	Oscillations
Chapter 14	Waves

4.0Key Features of NCERT Solutions for Class 11 Physics Chapter 12

RMS Velocity Explained: Understand why Root Mean Square velocity is used to describe the motion of gas molecules and how it differs from average velocity.
Degrees of Freedom and Internal Energy: Learn how degrees of freedom determine the internal energy of monatomic, diatomic, and polyatomic gases.
Molar Heat Capacities: Understand the relation between $C_{p}$ , $C_{V}$ and $R$ and learn the derivation of Mayer’s formula $C_{p} - C_{v} = R$
Step-by-Step Numerical Solutions: Solve NCERT problems based on the number of molecules, molecular motion, mean free path, pressure, and temperature of gases with clear calculations and units.
Prepared by ALLEN Subject Experts: Get accurate and detailed NCERT Solutions for Class 11 Physics Chapter 12, prepared according to the latest NCERT syllabus.
Simple and Concept-Based Explanations: Understand Kinetic Theory, gas molecules, molecular motion, degrees of freedom, specific heat capacities, and mean free path through clear explanations and step-by-step solutions.
Complete NCERT Exercise Coverage: Get detailed solutions to NCERT Class 11 Physics Chapter 12 examples and exercise questions, covering important concepts for CBSE, JEE, and NEET preparation.

Frequently Asked Questions

What makes NCERT Solutions for Class 11 Physics Chapter 12 Kinetic Theory crucial for learners?

How do Class 11 Physics NCERT Solutions for Chapter 12 help in Board, JEE, and NEET preparation?

Why do we call it an "Ideal Gas"?

What is RMS speed in Kinetic Theory?

What are the degrees of freedom of a gas molecule?

Join ALLEN!

Frequently Asked Questions

What makes NCERT Solutions for Class 11 Physics Chapter 12 Kinetic Theory crucial for learners?

How do Class 11 Physics NCERT Solutions for Chapter 12 help in Board, JEE, and NEET preparation?

Why do we call it an "Ideal Gas"?

What is RMS speed in Kinetic Theory?

What are the degrees of freedom of a gas molecule?

Join ALLEN!

NCERT Solutions Class 11 Physics Chapter 12 – Kinetic Theory

1.0Class 11 Physics Chapter 12 Kinetic Theory: Key Concepts

2.0Detailed NCERT Textbook Solutions : Class 11 Physics Chapter 12

SOLVED EXAMPLES

EXERCISE QUESTIONS WITH SOLUTIONS

3.0NCERT Class 11 Physics – Chapter-Wise Solutions & Explanations

4.0Key Features of NCERT Solutions for Class 11 Physics Chapter 12

Table of Contents