CBSE Class 11 Physics Chapter 12 Kinetic Theory – NCERT Solutions 2025-26

Kinetic Theory of Gases

1. Assumptions of Kinetic Theory of Gas

(i) The molecules of an ideal gas are in a state of never stopping random motion. They move in all directions with different speeds., (of the order of 500 m/s) and obey Newton’s laws of motion.

(ii) The molecules of an ideal gas do not exert any force of attraction or repulsion on each other, except during collision.

(iii) The molecules of an ideal gas moves along a straight line between two successive collisions and the average straight distance covered between two successive collisions is called the mean free path of the molecules.

(iv) The size of ideal gas molecules is very small as compared to the distance between them.

(v) The molecules of an ideal do not exert any force of attraction or repulsion on each other.

2. Pressure of an ideal gas is given by, $P=\frac{1}{3}\rho _{gas}v_{rms}^{2}$

3. Ideal Gas: That gas which strictly obeys the gas laws, (such as Boyle’s Law, Charles’, Gay Lussac’s Law etc.)

4. Real Gas: All gases are referred to as real gases. All real gases showes the ideal gas behavior at low pressures and temperatures high enough, where they cannot be liquified

5. Gay Lussac’s Law/Boyle’s Law:  PV = constant for given mass of gas at constant temperature.

6. Charle’s Law: $PV\propto T$ If P is constant, $V\propto T$.

7. Constant Volume Law:$PV\propto T$, If V is constant, $P\propto T$

8. Avogadro’s Law: Equal volumes of all ideal gases existing under the same conditions of temperature and pressure contain equal number of molecules.

9. Ideal Gas Equation: $PV=Nk_{b}T$

10. Speed of Gas Molecules: 

11. Degrees of Freedom: The number of degrees of freedom of a dynamical system is defined as the total number of co-ordinates or independent quantities required to describe completely the position and configuration of the system.

12. Law of Equipartition of Energy: According to this law, for any dynamical system in thermal equilibrium, the total energy is distributed equally amongst all the degrees of freedom, and the energy associated with each molecule per degree of freedom is  $\frac{1}{2}k_{B}T$ where kB is Boltzman constant, and T is temperature of the system.

13. Mean Free Path: The path traversed by a molecule between two successive collisions with another molecule is called the mean free path

$1\vec{} = \frac{\text{Total distance travelled by a molecule}}{\text{No. of collisions it makes with other molecules}}$

Thermodynamics

• The Zeroth Law: The zeroth law of thermodynamics states that ‘two systems in thermal equilibrium with a third system separately are in thermal equilibrium with each other’. 

• Internal energy: Internal energy of a system is the sum of kinetic energies and potential energies of the molecular constituents of the system. 

• First law of thermodynamics: This is the general law of conservation of energy applied to any system in which energy transfer from or to the surroundings (through heat and work) is taken into account. It states that

$\Delta Q=\Delta U+\Delta W$

Where, $\Delta Q$ is the heat supplied to the system.

$\Delta W$ is the work done by the system and $\Delta U$ is the change in internal energy of the system.

• Specific heat capacity:  Specific heat capacity of a substance is defined by $s=\frac{1}{m}\frac{\Delta Q}{\Delta T}$

Where m is the mass of the substance and $\Delta Q$ is the heat required to change its temperature by $\Delta T$. 

• Molar specific heat capacity: Molar specific heat capacity of a substance is defined by $C=\frac{1}{\mu }\frac{\Delta Q}{\Delta T}$

Where, $\mu$ is the number of moles of the substance.

• Mayer’s Law:  For an ideal gas, the molar specific heat capacities at constant pressure (Cp) and at constant volume (Cv) satisfy the relation. 

Cp- Cv = R

where R is the universal gas constant. 

• State Variables: Equilibrium states of a thermodynamic system are described by state variables.

The value of a state variable depends only on the particular state, not on the path used to arrive at thestate. Examples of state variables are pressure (p), volume (V), temperature (T), and mass (m). 

• Isothermal Process: In an isothermal expansion of an ideal gas from volume V1 to V2 at temperature T the heat absorbed (Q) equals the work done (W) by the gas, each given by 

$Q=W=\mu RT\left ( \frac{V_{2}}{V_{1}} \right )$

• Adiabatic Process: In an adiabatic process of an ideal gas $PV^{\gamma }=Constant$

Where $\gamma=\frac{C_{p}}{C_{v}}$

Work done by an ideal gas in an adiabatic change of state from(P1,V1,T1) to (P1,V2,T2)  is 

$W=\frac{\mu RT\left ( T_{1} +T_{2}\right )}{\gamma -1}$

• Second law of thermodynamics: Second law of thermodynamics disallows some processes consistent with the First Law of Thermodynamics. 

Kelvin -Planck statement: No process is possible whose sole result is the absorption of heat from a reservoir and complete conversion of the heat into work.

Clausius statement: No process is possible whose sole result the transfer of heat from a colder object to a hotter object.

The second Law implies that no heat engine can have efficiency η equal to 1 or no refrigerator can have co-efficient of performance α equal to infinity. 

• Reversible Process: A process is reversible if it can be reversed such that both the system and the surroundings return to their original states, with no other change anywhere else in the universe. Spontaneous processes of nature are irreversible. The idealized reversible process is a quasi-static process with no dissipative factors such as friction, viscosity, etc. 

• Carnot engine: Carnot engine is a reversible engine operating between two temperatures T1 (source) and T2 (sink). The Carnot cycle consists of two isothermal processes connected by two adiabatic processes. The efficiency of a Carnot engine is given by 

$\eta = 1 - \frac{T_{2}}{T_{1}}$

No engine operating between two temperatures can have efficiency greater than that of the Carnot engine. 

NCERT Exercise

1. Calculate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Consider taking the diameter of an oxygen molecule to be $3\mathop A\limits^ \circ  $

Ans: The diameter of an oxygen molecule is given as: $d = 3\mathop A\limits^ \circ  $

Radius, $r = \frac{d}{2} = \frac{3}{2} = 1.5\mathop A\limits^ \circ   = 1.5 \times {10^{ - 8}}cm$

At STP, the actual volume occupied by 1 mole of oxygen gas is given as: $22400{\kern 1pt} {\kern 1pt} c{m^3}$

The molecular volume of oxygen gas is given as: $V = \frac{4}{3}\pi {r^3}.{N_A}$

Where, ${N_A}$ is Avogadro’s number: \[6.023 \times {10^{23}}molecules/mole\]. Hence:

$V = \frac{4}{3}\pi {r^3}.{N_A}$

$ \Rightarrow \frac{4}{3} \times 3.14 \times {\left( {1.5 \times {{10}^{ - 8}}} \right)^3}6.023 \times {10^{23}}$

$ \Rightarrow 8.51{\kern 1pt} c{m^3}$

Therefore, the molecular volume of one mole of oxygen gas will be $8.51{\kern 1pt} c{m^3}$.

Now, the ratio of the molecular volume to the actual volume of oxygen can be given as:

$\frac{{{V_{molar}}}}{{{V_{actual}}}} = \frac{{8.51}}{{22400}} = 3.8 \times {10^{ - 4}}$

2. The volume which is occupied by 1 mole of any (ideal) gas at standard temperature and pressure (STP: 1 atmospheric pressure, ${0^0}C$ ) is molar volume. Show that it is 22.4 liters.

Ans: The ideal gas equation is:

$PV = nRT$

R is the universal gas constant, $R = 8.314Jmo{l^{ - 1}}{K^{ - 1}}$

n is the number of moles, $n = 1$

T is standard temperature, $T = 273K$

P is standard pressure, $P = 1atm = 1.013 \times {10^5}N{m^{ - 2}}$

$\therefore V = \frac{{nRT}}{P}$

$V = \frac{{1 \times 8.314 \times 273}}{{1.013 \times {{10}^5}}} = 0.0224{m^3} = 22.4litres$

So, we can say that the molar volume of a gas is 22.4 liters at STP.

3. The diagram below shows a plot of ${}^{PV}/{}_{T}$versus P for $1.00 \times {10^{ - 3}}Kg$ oxygen gas at two different temperatures.

1. What does the dotted plot signify?

Ans: 

In the graph, the dotted plot signifies the ideal behaviour of the gas, i.e., the ratio$\frac{{PV}}{T}$is equal. $\mu R$  is a constant quality.

$\mu $ is the number of moles 

 R is the universal gas constant

It is independent on the pressure of the gas.

2. Which is true: ${T_1} > {T_2}$ or ${T_1} < {T_2}$?

Ans: In the given graph, the dotted plot represents an ideal gas. At temperature ${T_1}$ , the curve of the gas is very closer to the dotted plot than for the curve of the gas at temperature ${T_2}$. The behavior of a real gas approaches ideal gas when its temperature increases. Therefore, ${T_1} > {T_2}$is true.

3. $\frac{{PV}}{T}$ value, where the curves meet on the y-axis is?

Ans: The ratio $\frac{{PV}}{T}$for the meeting of two curves is $\mu R$. So, the ideal gas equation is,

$PV = \mu RT$

Where P is the pressure

T is the temperature

V is the volume

$\mu $is the number of moles

R is the universal constant

The molecular mass of oxygen=32.0g

Mass of oxygen$ = 1 \times {10^{ - 3}}kg = 1g$

$R = 8.314Jmo{l^{ - 1}}{K^{ - 1}}$

$\therefore \frac{{PV}}{T} = \frac{1}{{32}} \times 8.314 = 0.26J{K^{ - 1}}$

The value of the ratio So, the value of the ratio $\frac{{PV}}{T}$, where the curves meet on the y-axis, is $0.26J{K^{ - 1}}$

4. Will we be getting the same value of $\frac{PV}{T}$at the point where the curves meet on the y-axis, if for $1.00 \times {10^{ - 3}}Kg$ of hydrogen we get similar plots? Mass of hydrogen that produces the same value of $\frac {PV}{T}$ (for a low-pressure high-temperature region of the plot) if it is not the case? (Molecular mass of ${H_2} = 2.02u$,${O_2} = 32.0u$, and $R = 8.314Jmo{l^{ - 1}}{K^{ - 1}}$ )

Ans:

If a similar plot for $1.00 \times {{10}^{ - 3}}Kg$ of hydrogen, then we won’t get the same value of  $\frac {PV}{T}$ at the point where the curves meet the y-axis. Since the molecular mass of hydrogen (2.02 u) is not the same as that of oxygen (32.0 u).

We have:

$\therefore \frac{{PV}}{T} = 0.26J{K^{ - 1}}$

$R = 8.314Jmo{l^{ - 1}}{K^{ - 1}}$

Molecular mass M of ${H_2} = 2.02u$

$PV = \mu RT$at constant temperature

$\mu  = \frac{m}{M}$

m is the mass of ${H_2}$

$m = \frac{{PV}}{T} \times \frac{M}{R} = \frac{{0.26 \times 2.02}}{{8.31}} = 6.3 \times {10^{ - 2}}g = 6.3 \times {10^{ - 5}}kg$

Hence, $6.3 \times {10^{ - 2}}g$of ${H_2}$will get the same value of $\frac{{PV}}{T}$

4. A 30 liters oxygen cylinder has an initial gauge pressure of 15 atm and a temperature of ${27^0}C$ . The gauge pressure drops to 11 atm, and its temperature drops to ${17^0}C$ when some oxygen is withdrawn from the cylinder. Estimate the mass of oxygen taken out of the cylinder ( $R = 8.314Jmo{l^{ - 1}}{K^{ - 1}}$, the molecular mass of ${O_2} = 32u$).

Ans:  The volume of oxygen, ${V_1} = 30litres = 30 \times {10^{ - 3}}{m^3}$

Gauge pressure, ${P_1} = 15atm = 15 \times 1.013 \times {10^5}Pa$

Temperature, ${T_1} = {27^0}C = 300K$

Universal gas constant, $R = 8.314Jmo{l^{ - 1}}{K^{ - 1}}$

Consider the initial number of moles of oxygen gas in the cylinder be ${n_1}$

The gas equation is given as:

${{P_1}{V_1}} = {n_1}R{T_1}$  

$\therefore {n_1} = \frac{{{P_1}V}}{{R{T_1}}} = \frac{{15.195 \times {{10}^5} \times 30 \times {{10}^{ - 3}}}}{{8.314 \times 300}} = 18.276$

But ${n_1} = \frac{{{m_1}}}{M}$

Where,

${m_1} = $the initial mass of oxygen

$M = $The molecular mass of oxygen=32g

$\therefore {m_1} = {n_1}M = 18.276 \times 32 = 584.84g$

The pressure and temperature reduce after some oxygen is withdrawn from the cylinder.

Volume, ${V_2} = 30litres = 30 \times {10^{ - 3}}{m^3}$

Gauge pressure, ${P_2} = 11atm = 11 \times 1.013 \times {10^5}Pa$

Temperature, ${T_2} = {17^0}C = 290K$

Let consider${n_2}$, the number of moles of oxygen left in the cylinder.

The gas equation is given as:

${P_2}{V_2} = {n_2}R{T_2}$

$\therefore {n_2} = \frac{{{P_2}{V_2}}}{{R{T_2}}} = \frac{{11.143 \times {{10}^5} \times 30 \times {{10}^{ - 3}}}}{{8.314 \times 290}} = 13.86$

But, ${n_2} = \frac{{{m_2}}}{M}$

Where,

The remaining mass of oxygen in the cylinder is ${m_2}$

$\therefore {m_2} = {n_2}M = 13.86 \times 32 = 443.52g$

So, the mass of oxygen taken out is:

The initial mass of oxygen in the cylinder – Final mass of oxygen in the cylinder

$ \Rightarrow {m_1} - {m_2} = 584.84 - 443.522 = 141.32g = 0.141kg$

0.141kg of oxygen is hence taken out of the cylinder.

5. An air bubble which is having a volume $1.0c{m^3}$rises from the bottom of a lake 40 m deep at a temperature of ${12^0}C$. When it reaches the surface, which is at a temperature of ${35^0}C$, to what volume does it grow?

Ans: The volume of the air bubble,${V_1} = 1.0c{m^3} = 1.0 \times {10^{ - 6}}{m^3}$

The bubble rises to height, $d = 40m$

The temperature at a depth of 40m, ${T_1} = {12^0}C = 285K$

The temperature is ${T_2} = {35^0}C = 308K$ , at the surface of the lake

On the surface of the lake the pressure,

${P_2} = 1atm = 1 \times 1.013 \times {10^5}Pa$

The pressure at the depth of 40m, ${P_1} = 1atm + d\rho g$

Where,

$\rho $is the density of water$ = {10^3}kg{m^{ - 3}}$

$g$is the acceleration due to gravity$ = 9.8m{s^{ - 1}}$

$\therefore {P_1} = 1.013 \times {10^5} + 40 \times {10^3} \times 9.8 = 493300Pa$

We have: $\frac{{{P_1}{V_1}}}{{{T_1}}} = \frac{{{P_2}{V_2}}}{{{T_2}}}$

 ${V_2}$is the air bubbles volume when it reaches the surface

${V_2} = \frac{{{P_1}{V_1}{T_2}}}{{{T_1}{P_2}}} = \frac{{493300 \times 1.0 \times {{10}^{ - 6}} \times 308}}{{285 \times 1.013 \times {{10}^5}}} = 5.263 \times {10^{ - 6}}{m^3} = 5.263c{m^3}$

The volume of air bubble becomes$5.263c{m^3}$when it reaches the surface.

6. Determine the total number of air molecules ( that includes oxygen, nitrogen, water vapor, and other constituents) in a room of capacity $25.0{m^3}$ at a temperature of ${27^0}C$and 1atm pressure.

Ans:   The volume of the room, $V = 25.0{m^3}$

The temperature of the room, $T = {27^0}C = 300K$

Pressure in the room, $P = 1atm = 1 \times 1.1013 \times {10^5}Pa$

The ideal gas equation:

$PV = {K_B}NT$

Where,

${K_B}$ is Boltzmann constant, ${K_B} = 1.38 \times {10^{ - 23}}{m^2}kg{s^{ - 2}}{K^{ - 1}}$

Number of air molecules in the room be N.

$N = \frac{{PV}}{{{k_B}T}} = \frac{{1.013 \times {{10}^5} \times 25}}{{1.38 \times {{10}^{ - 23}} \times 300}} = 6.11 \times {10^{26}}molecules$

The total number of air molecules is $6.11 \times {10^{26}}$ 

7. Find out the average thermal energy of a helium atom at the following cases:

1. Room temperature$\left( {{{27}^0}C} \right)$

Ans:

At room temperature, $T = {27^0}C = 300K$

Average thermal energy$ = \frac{3}{2}kT$

Where k is Boltzmann constant$ = 1.38 \times {10^{ - 23}}{m^2}kg{s^{ - 2}}{K^{ - 1}}$

$\therefore \frac{3}{2}kT = \frac{3}{2} \times 1.38 \times {10^{ - 38}} \times 300 = 6.21 \times {10^{ - 21}}J$

So, the average thermal energy is $\left( {{{27}^0}C} \right)$is $6.21 \times {10^{ - 21}}J$

2. The temperature on the sun’s surface $\left( {6000K} \right)$

Ans:

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

Average thermal energy$ = \frac{3}{2}kT = \frac{3}{2} \times 1.38 \times {10^{ - 38}} \times 6000 = 1.241 \times {10^{ - 19}}J$

Hence, the average thermal energy is $1.241 \times {10^{ - 19}}J$

3. At a temperature of 10 million kelvin ( the typical core temperature in the case of a star).

Ans:

At temperature, $T = {10^7}K$

Average thermal energy$ = \frac{3}{2}kT = \frac{3}{2} \times 1.38 \times {10^{ - 23}} \times {10^7} = 2.07 \times {10^{ - 16}}J$

Hence, the average thermal energy is $2.07 \times {10^{ - 16}}J$

8. Three vessels all of the same capacity have gases at the same pressure and temperature. It consists of neon which is monatomic, in the first one, the second contains diatomic chlorine, and the third contains uranium hexafluoride (polyatomic).

1. Do you think all the vessels contain an equal number of respective molecules?

Ans:

Yes. The same number of the respective molecules is there in all the vessels.

They have the same volume since the three vessels have the same capacity.

All gases are of same pressure, volume, and temperature.

Avogadro’s law states the three vessels consist of an equal number of molecules. This equals Avogadro’s number, $N = 6.023 \times {10^{23}}$

2. Is in all three cases, the root mean square speed of molecules the same? If it is not the case, in which case is ${v_{rms}}$the largest?

Ans:

No. Neon has the largest root-mean-square speed.

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

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

Where k is Boltzmann constant

k and T are constants for the given gases.

 ${v_{rms}}$ only depends on the mass of the atoms, i.e.,

${v_{rms}}\alpha \sqrt {\frac{l}{m}} $ 

So, in the three cases, the root-means-square speed of the molecules is not the same.

The mass of neon is the smallest among neon, chlorine, and uranium hexafluoride and so possesses the largest root mean square speed.

9. Calculate the temperature at which the root mean square speed of an argon atom in a gas cylinder is equal to the RMS speed of a helium gas atom at $ - {20^0}C$? (atomic mass of Ar = 39.9 u, of He = 4.0 u)

Ans: The temperature of the helium atom, ${T_{He}} =  - {20^0}C = 253K$

The atomic mass of argon, ${M_{Ar}} = 39.9u$

The atomic mass of helium, ${M_{He}} = 4.0u$

Let, ${\left( {{v_{rms}}} \right)_{Ar}}$be the rms speed of argon.

Let, ${\left( {{v_{rms}}} \right)_{He}}$be the rms speed of helium.

Argon as an rms speed of,

${\left( {{v_{rms}}} \right)_{Ar}} = \sqrt {\frac{{3R{T_{Ar}}}}{{{M_{Ar}}}}} \,\,\,\,\,\,\,......\left( i \right)$

Where,

R is the universal gas constant

${T_{Ar}}$is the temperature of argon gas

Helium has an rms speed of,

${\left( {{v_{rms}}} \right)_{He}} = \sqrt {\frac{{3R{T_{He}}}}{{{M_{He}}}}} \,\,\,\,\,.....(ii)$

It is given that:

${\left( {{v_{rms}}} \right)_{Ar}} = {\left( {{v_{rms}}} \right)_{He}}$

$\sqrt {\frac{{3R{T_{Ar}}}}{{{M_{Ar}}}}}  = \sqrt {\frac{{3R{T_{He}}}}{{{M_{He}}}}} $

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

${T_{Ar}} = \frac{{{T_{He}}}}{{{M_{He}}}} \times {M_{Ar}} = \frac{{253}}{4} \times 39.9 = 2523.675 = 2.52 \times {10^3}K$

Argon atom is at a temperature of $2.52 \times {10^3}K$

10. Find out the collision frequency and also the mean free path of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature ${17^0}C$. The nitrogen molecule has a radius of   roughly$1.0\mathop A\limits^0 $ . How collision time is related with the time the molecule moves freely between two successive collisions (Molecular mass of ${N_2} = 28.0u$).

Ans:  Mean free path\[ = 1.11 \times {10^{ - 7}}m\]

Collision frequency$ = 4.58 \times {10^9}{s^{ - 1}}$

Successive collision time$ \approx 500 \times collision\,time$

The pressure inside the cylinder containing nitrogen, $P = 2.0atm = 2.026 \times {10^5}Pa$

Temperature inside the cylinder, $T = {17^0}C = 290K$

The radius of nitrogen molecule, $r = 1.0\mathop A\limits^0  = 1 \times {10^{10}}m$

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

Molecular mass of nitrogen, $M = 28.0g = 28 \times {10^{ - 3}}kg$

For the nitrogen, root mean square speed is,

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

Where,

$R = 8.314mol{e^{ - 1}}{K^{ - 1}}$, is universal gas constant

$\therefore {v_{rms}} = \sqrt {\frac{{3 \times 8.314 \times 290}}{{28 \times {{10}^{ - 3}}}}}  = 508.26m{s^{ - 1}}$

The mean free path (l) is,

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

Where,

$k = 1.38 \times {10^{ - 23}}kg{m^2}{s^{ - 2}}{K^{ - 1}}$ is the Boltzmann constant

$\therefore l = \frac{{1.38 \times {{10}^{ - 23}} \times 290}}{{\sqrt 2  \times 3.14 \times {{\left( {2 \times {{10}^{ - 10}}} \right)}^2} \times 2.026 \times {{10}^5}}} = 1.11 \times {10^{ - 7}}m$

Collision frequency$ = \frac{{{v_{rms}}}}{l} = \frac{{508.26}}{{1.11 \times {{10}^{ - 7}}}} = 4.58 \times {10^9}{s^{ - 1}}$

The collision time is given as:

$T = \frac{d}{{{v_{rms}}}} = \frac{{2 \times {{10}^{ - 10}}}}{{508.26}} = 3.93 \times {10^{ - 13}}s$

Between successive collisions, the time taken is

$T' = \frac{l}{{{v_{rms}}}} = \frac{{1.11 \times {{10}^{ - 7}}m}}{{508.26m{s^{ - 1}}}} = 2.18 \times {10^{ - 10}}s$

$\therefore \frac{{T'}}{T} = \frac{{2.18 \times {{10}^{ - 10}}}}{{3.93 \times {{10}^{ - 13}}}} = 500$

For successive collisions, the time taken is 500 times the time taken for a collision. 

11. A 1-meter narrow bore that is kept horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. Suppose the tube is kept vertically with its open end at the bottom, what will happens?

Ans:  Length of the narrow bore, L=1m=100cm

Length of the mercury thread, l=76cm

Air column’s length between mercury and the closed-end, ${l_a} = 15cm$

The air space occupied by mercury length: 100 – (76 + 15) = 9 cm

Hence, the total length of the air column=15+9=24cm

Let us consider as a result of atmospheric pressure, h cm of mercury flow out.

In the bore, length of air column =24+h cm

Mercury column’s length = 76 -h cm

Initial pressure, ${P_1} = 76cm$of mercury

Initial volume, ${V_1} = 15c{m^3}$

Final pressure, ${P_2} = 76 - \left( {76 - h} \right) = h\,cm$of mercury

Final volume, ${V_2} = \left( {24 + h} \right)c{m^3}$

Temperature here is constant

$\therefore {P_1}{V_1} = {P_2}{V_2}$  

$76 \times 15 = h\left( {24 + h} \right)$

$ \Rightarrow {h^2} + 24h - 1140 = 0$

$\therefore h = \frac{{ - 24 \pm \sqrt {{{24}^2} + \left( {4 \times 1 \times 1140} \right)} }}{{2 \times 1}} = 23.8cm\,\,or\,\, - 47.8cm$

Height cannot be negative. 

So, 23.8 cm of mercury will flow out.

52.2cm of mercury will remain in the bore.

The length is,

24 + 23.8 = 47.8 cm

12. The diffusion rate of hydrogen has an average value of $28.7c{m^3}{s^{ - 1}}$ from a certain apparatus. Under the same condition, the diffusion of another gas is measured to have an average rate of$7.2c{m^3}{s^{ - 1}}$. Identify the gas.

(Hint: Use Graham’s law of diffusion${{{R}_{1}}}/{{{R}_{2}}={{\left( {{{M}_{2}}}/{{{M}_{1}}}\; \right)}^{{1}/{2}\;}}}\;$, where ${R_1},{R_2}$are diffusion rates of gases 1 and 2, and ${M_1}\,\,and\,\,{M_2}$their respective molecular masses.)

Ans:Rate of diffusion of hydrogen, ${R_1} = 28.7c{m^3}{s^{ - 1}}$

Rate of diffusion of another gas, ${R_2} = 7.2c{m^3}{s^{ - 1}}$

From Graham’s Law, we have:

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

Where,

${M_1}$is the molecular mass of hydrogen=2.020g

${M_2}$is the unknown gas’s molecular mass

$\therefore {M_2} = {M_1}{\left( {\frac{{{R_1}}}{{{R_2}}}} \right)^2} = 2.02{\left( {\frac{{28.27}}{{7.2}}} \right)^2} = 32.09g$

Since 32g is the molecular mass, the unknown gas is oxygen.

13. Gas in equilibrium will have uniform density and pressure throughout its volume A gas column under gravity, for example, does not have a uniform density (and pressure). The density decreases with height. The precise dependence is given by the law of atmospheres

${n_2} = {n_1}\exp \left[ { - mg\left( {{h_2} - {h_1}} \right)/kBT} \right]\,\,$

Where ${n_2},{n_1}$is number density at heights ${h_2},{h_1}$respectively. 

The sedimentation equilibrium,

${n_2} = {n_1}\exp \left[ { - mg{N_4}\left( {\rho  - \rho '} \right)\left( {{h_2} - {h_1}} \right)/\rho RT} \right]\,\,$

Where$\rho $is the density of the suspended particle, and $\rho '$that of the surrounding medium. (${N_A}$is Avagadro’s number, and R the universal gas constant)

( Hint: Apparent weight can be found by using Archimedes principle)

Ans:

According to the law of atmosphere, we have:

${n_2} = {n_1}\exp \left[ { - mg\left( {{h_2} - {h_1}} \right)/{k_B}T} \right]\,\,\,\,\,\,....(i)$

${n_2},{n_1}$is number density at heights ${h_2},{h_1}$respectively.

The weight of the particle in the gas column is mg

Density of the medium=$\rho '$

Density of the suspended particle =$\rho $

Mass of one suspended particle = $m'$

Mass of the medium displaced = $m$

Volume of a suspended particle= V

The weight of the suspended particle is given from Archimedes’ principle as:

Displaced medium’s weight-Suspended particle’s weight =$mg - m'g$

$\Rightarrow mg - m'g$

$\Rightarrow mg = V\rho 'g = mg\left( {\frac{m}{\rho }} \right)\rho 'g$

$mg\left( {1 - \frac{{\rho '}}{\rho }} \right)\,\,\,\,\,......\left( {ii} \right)$

Gas constant, $R = {k_B}N$

${k_B} = \frac{R}{N}\,\,\,\,.....(iii)$

substituting in the equations we get:

${n_2} = {n_1}\exp \left[ { - mg\left( {{h_2} - {h_1}} \right)/{k_B}T} \right]$

$\Rightarrow {n_1}\exp \left[ { - mg\left( {1 - \frac{{\rho '}}{\rho }} \right)\left( {{h_2} - {h_1}} \right)\frac{N}{{RT}}} \right] $

$ \Rightarrow {n_1}\exp \left[ { - mg\left( {\rho  - \rho '} \right)\left( {{h_2} - {h_1}} \right)\frac{N}{{RT\rho }}} \right]$

14. Observe the below table showing the densities of some solids and liquids. Determine the size of their atoms:

(Hint: Atoms are tightly packed in a solid or liquid phase. Use the known value of Avogadro’s number. You shouldn’t take the actual numbers you obtain for various atomic sizes too literally. Due to the crudeness of the tight packing approximation, the results only indicate that atomic sizes are in the range of a few $\mathop A\limits^0 $ ).

Ans:

The atomic mass of a substance=M

Density of the substance=$\rho $

Avogadro’s number$ = N = 6.023 \times {10^{23}}$

Volume of each atom $ = \frac{4}{3}\pi {r^3}$

Volume of N number of molecules $ = \frac{4}{3}\pi {r^3}N\,\,\,\,\,.....(i)$

Volume of one mole of a substance$ = \frac{M}{\rho }\,\,\,....(ii)$

$\frac{4}{3}\pi {r^3}N = \frac{M}{\rho }\,\,\,\,$

$\therefore r = \sqrt[3]{{\frac{{3M}}{{4\pi \rho N}}}}$

For Carbon:

$M = 12.01 \times {10^{ - 3}}kg$  

$\rho  = 2.22 \times {10^3}kg{m^{ - 3}}$

Radius is,

$\therefore r = {\left( {\frac{{3 \times 12.01 \times {{10}^{ - 3}}}}{{4\pi  \times 2.22 \times {{10}^3} \times 6.023 \times {{10}^{23}}}}} \right)^{\frac{1}{3}}} = 1.29{\overset{0} A}$

For Gold:

$M = 197.00 \times {10^{ - 3}}kg$  

$\rho  = 19.32 \times {10^3}kg{m^{ - 3}}$

Radius is,

$\therefore r = {\left( {\frac{{3 \times 197 \times {{10}^{ - 3}}}}{{4\pi  \times 19.32 \times {{10}^3} \times 6.023 \times {{10}^{23}}}}} \right)^{\frac{1}{3}}} = 1.59{$\mathop A\limits^ \circ $}$

For Liquid Nitrogen:

$M = 14.01 \times {10^{ - 3}}kg$  

$\rho  = 1.00 \times {10^3}kg{m^{ - 3}}$

Radius is,

\[\therefore r = {\left( {\frac{{3 \times 14.01 \times {{10}^{ - 3}}}}{{4\pi  \times 1.00 \times {{10}^3} \times 6.023 \times {{10}^{23}}}}} \right)^{\frac{1}{3}}} = 1.77{\overset{0} A}\]

For Lithium:

$M = 6.94 \times {10^{ - 3}}kg$  

$\rho  = 0.53 \times {10^3}kg{m^{ - 3}}$

Radius is,

$\therefore r = {\left( {\frac{{3 \times 6.94 \times {{10}^{ - 3}}}}{{4\pi  \times 0.53 \times {{10}^3} \times 6.023 \times {{10}^{23}}}}} \right)^{\frac{1}{3}}} = 1.73{\overset{0} A}$

For Liquid Fluorine:

$M = 19.00 \times {10^{ - 3}}kg$

$\rho  = 1.14 \times {10^3}kg{m^{ - 3}}$

Radius is,

$\therefore r = {\left( {\frac{{3 \times 19 \times {{10}^{ - 3}}}}{{4\pi  \times 1.14 \times {{10}^3} \times 6.023 \times {{10}^{23}}}}} \right)^{\frac{1}{3}}} = 1.88{\overset{0} A}$

NCERT Solutions For Class 11 Physics Chapter 12 – Free PDF Download

NCERT Solutions for Class 11 Physics Chapter Kinetic Theory will not only help students to form effective answers but also help them practice better. These solutions have been prepared adhering strictly to the CBSE guidelines and come with suggested questions for your final and entrance exams. To avail these study materials offline, download NCERT Solutions for Class 11 Physics Chapter 12 PDF from Vedantu and enjoy uninterrupted learning.

NCERT Solutions for Class 11 Physics Chapter 12 Sub - Topics

For a precise idea of the concepts taught in Kinetic Theory of Gases Class 11, you can go through the brief overview of all seven topics covered in the chapter provided below.

Part 1: Introduction

The first part lays a basic idea of the Kinetic Theory of gases and how it came into formulation, before jumping into the intricacies. Students will learn about the structure and constituents of gas molecules, their properties as compared to liquids and solids, and their thermodynamic behaviour. In NCERT Solutions Class 11 Physics Kinetic Theory, questions related to terms like conduction, viscosity, diffusion and specific heat capacities have been explained with care and detailed outline.

Part 2: Molecular Nature of Matter

This section delves into the details of the molecular behaviour of gas. It comprises the Atomic Theory and its laws. The importance of interatomic spaces, mean free path and dynamic equilibrium of gases has been stressed upon. Students will be required to understand Gay Loussac’s Law and Avogadro’s Law and their connection.

Part 3: Behaviour of Gases

Here, students are introduced to the title concept of Kinetic Theory of Gases and its postulates. This is a vital section in NCERT Solutions for Class 11 Physics Chapter 12 and covers key terms and concepts like Boltzmann constant and Avogadro’s number, and their use in deriving relationships between temperature, pressure, and volume of gases. You will also be required to solve problems based on mole number and molar mass.

Part 4: Kinetic Theory of an Ideal Gas

This part is further divided into two subsections, where the already established concepts and relations are further discussed according to specific conditions.

• Pressure of an Ideal Gas

With the fixed velocities of gas particles, their momentum and eventually their pressure is deduced for calculation. A detailed explanation of Pascal’s Law is provided.

• Kinetic Interpretation of Temperature

Incorporating ideal gas equation with previous derivations, the connection between kinetic energy, pressure and temperature is derived with the aid of Boltzmann constant.

Chapter 12 Physics Class 11 NCERT Solutions include examples and exercises relating to the above-mentioned concepts to reinforce the equations and derivations for efficient learning.

Part 5: Law of Equipartition of Energy

This segment spans over some of the important concepts of vibrational energy of molecules, rotational vibration, moment of inertia, and degrees of movement of gas particles are discussed before defining the Law of equipartition of energy.

Part 6: Specific Heat Capacity

The Law of equipartition energy is used in the calculations of molar specific heat capacities of gases, solids and water at constant volume (cv) and constant pressure (cp). Tips are also provided on avoiding discrepancies in predicted and experimental values of specific heat capacity. Kinetic Theory of Gases Class 11 NCERT conveys the importance of quantum physics and its application in solving numerical.

Part 7: Mean Free Path

The last subtopic in Kinetic Theory of Gases Class 11 Physics defines the term mean free path and explains its course with the help of illustrative real-life examples. The formula to calculate it is also derived in relation to the size, density, and number of gas molecules. Students will find the perfectly explained guide to numerical based on the formula in Kinetic Theory of Gases Class 11 Physics NCERT Solutions that will help them improvise on their problem-solving skills.

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