NCERT Solutions for Class 11 Physics Chapter 12 Kinetic Theory

NCERT Solutions for Class 11 Physics Chapter 12 Kinetic Theory

Edited By Vishal kumar | Updated on Apr 29, 2025 04:34 PM IST

Did you ever consider why gases seem to expand and how they do apply pressure? The Kinetic Theory of Gases explains the processes by making assumptions that gases consist of small particles that move very rapidly. Boyle, Newton, Maxwell, and Boltzmann were among the prominent scientists who developed the theory, which predicts the actions of gases, how they will respond to pressure and temperature alterations, and their ability to flow and mix.

This Story also Contains
  1. NCERT Solutions for Class 11 Physics Chapter 12: Download PDF
  2. Additional Questions
  3. Class 11 physics NCERT Chapter 12: Higher Order Thinking Skills (HOTS) Questions
  4. Approach to Solve Questions of Kinetic Theory of Gases
  5. What Extra Should Students Study Beyond NCERT for JEE/NEET?
  6. NCERT Solutions for Class 11 Physics Chapter-Wise
  7. NCERT Solutions for Class 11 Subject Wise
  8. Subject wise NCERT Exemplar solutions

NCERT solutions for important chapter kinetic theory class 11 chapter 12 physics are designed by subject matter experts to give proper and accurate solutions to each NCERT exercise problem. Students can build a firm concept by solving Kinetic Theory NCERT Class 11 Physics solutions. Simple step-by-step solutions make it easy to learn and are discussed below.

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NCERT Solutions for Class 11 Physics Chapter 12: Download PDF

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NCERT Solutions for Class 11 Physics Chapter 12 Kinetic Theory - Exercise Questions

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 Å.

Answer:

The diameter of an oxygen molecule, d = 3 Å.

The actual volume of a mole of oxygen molecules V actual is

$\\V_{actual}=N_{A}\frac{4}{3}\pi \left ( \frac{d}{2} \right )^{3}$ $ \\ $
$V_{actual}=6.023\times 10^{23}\times \frac{4}{3}\pi \times \left ( \frac{3\times 10^{-10}}{2} \right )^{3}\\$
$ V_{actual}=8.51\times 10^{-6}m^{3}\\ $
$V_{actual}=8.51\times 10^{-3}litres$

The volume occupied by a mole of oxygen gas at STP is V molar = 22.4 litres

$\\\frac{V_{actual}}{V_{molar}}=\frac{8.51\times 10^{-3}}{22.4}\\ $

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

Q12.2 Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP: 1 atmospheric pressure, $0^{0}C$ ). Show that it is $22.4 Litres$ .

Answer:

As per the ideal gas equation

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

For one mole of a gas at STP we have

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

$V=22.4\ litres$

Q12.3 Figure 13.8 shows plot of $PV/T$ versus P for $1.00\times 10^{-3}$ kg of oxygen gas at two different temperatures.

PV/T versus P graph

(a) What does the dotted plot signify?
(b) Which is true: $T_{1}> T_{2}\: \: orT_{1}< T_{2}$ ?
(c) What is the value of $PV/T$ where the curves meet on the y-axis?
(d) If we obtained similar plots for $1.00\times 10^{-3}$ 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 $H_{2}=2.02\mu$ , of $O_{2}=32.0\mu$ ,
$R=8.31J mol^{-1}K^{-1}$ .)

Answer:

(a) The dotted plot corresponds to the ideal gas behaviour.

(b) We know the behaviour of a real gas tends close to that of ideal gas as its temperature increases and since the plot corresponding to temperature T 1 is closer to the horizontal line that the one corresponding to T 2 we conclude T 1 is greater than T 2.

(c) As per the ideal gas equation

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

The molar mass of oxygen = 32 g

$n=\frac{1}{32}$

R = 8.314

$\\nR=\frac{1}{32}\times 8.314\\ nR=0.256JK^{-1}$

(d) If we obtained similar plots for $1.00\times 10^{-3}$ kg of hydrogen we would not get the same
value of $PV/T$ at the point where the curves meet on the y-axis as 1 g of Hydrogen would contain more moles than 1 g of Oxygen because of having smaller molar mass.

Molar Mass of Hydrogen M = 2 g

mass of hydrogen

$m=\frac{PV}{T} \frac{M}{R}={0.256}\times \frac{2}{8.314}=5.48\times10^{-5}Kg$

Q12.4 An oxygen cylinder of volume $30$ litres has an initial gauge pressure of $15$ atm and a temperature of $27^{0}C$ . After some oxygen is withdrawn from the cylinder, the gauge pressure drops to $11$ atm and its temperature drops to $17^{0}C$ . Estimate the mass of
oxygen taken out of the cylinder ( $R=8.31Jmol^{-1}K^{-1}$ , molecular mass of $O_{2}=32\mu$ ).

Answer:

Initial volume, V 1 = Volume of Cylinder = 30 l

Initial Pressure P 1 = 15 atm

Initial Temperature T 1 = 27 o C = 300 K

The initial number of moles n 1 inside the cylinder is

$\\n_{1}=\frac{P_{1}V_{1}}{RT_{1}}\\ n_{1}=\frac{15\times 1.013\times 10^{5}\times 30\times 10^{-3}}{8.314\times 300}\\ n_{1}=18.28$

Final volume, V 2 = Volume of Cylinder = 30 l

Final Pressure P 2 = 11 atm

Final Temperature T 2 = 17 o C = 290 K

The final number of moles n 2 inside the cylinder is

$\\n_{2}=\frac{P_{2}V_{2}}{RT_{2}}\\ n_{2}=\frac{11\times 1.013\times 10^{5}\times 30\times 10^{-3}}{8.314\times 290}\\ n_{2}=13.86$

Moles of oxygen taken out of the cylinder = n 2 -n 1 = 18.28 - 13.86 = 4.42

Mass of oxygen taken out of the cylinder m is

$\\m=4.42\times 32\\ m=141.44g$

Q12.5 An air bubble of volume $1.0 cm^{3}$ rises from the bottom of a lake $40m$ deep at a temperature of $12^{0}C$ . To what volume does it grow when it reaches the surface, which is at a temperature of $35^{0}C$ ?

Answer:

Initial Volume of the bubble, V 1 = 1.0 cm 3

Initial temperature, T 1 = 12 o C = 273 + 12 = 285 K

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

Initial Pressure is P 1

Depth of the bottom of the lake = 40 m

$\\P_{1}=Atmospheric\ Pressure+Pressure\ due\ to\ water\\ P_{1}=P_{atm}+\rho _{w}gh\\ P_{1}=1.013\times 10^{5}+10^{3}\times 9.8\times 40\\ P_{1}=4.93\times 10^{5}Pa$

Final Temperature, T 2 = 35 o C = 35 + 273 = 308 K

Final Pressure = Atmospheric Pressure $=1.013\times 10^{5}Pa$

Let the final volume be V 2

As the number of moles inside the bubble remains constant, we have

$\\\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}\\$

$ V_{2}=\frac{P_{1}T_{2}V_{1}}{P_{2}T_{1}}\\ $

$V_{2}=\frac{4.93\times 10^{5}\times 308\times 1}{1.013\times 10^{5}\times 285}\\ $

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

Q12.6 Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity $25.0m^{3}$ at a temperature of $27^{0}C$ and $1atm$ pressure.

Answer:

The volume of the room, V = 25.0 m 3

Temperature of the room, T = 27 o C = 300 K

The pressure inside the room, P = 1 atm

Let the number of moles of air molecules inside the room be n

$\\n=\frac{PV}{RT}\\ n=\frac{1.013\times 10^{5}\times 25}{8.314\times 300}\\ n=1015.35$

Avogadro's Number, $N_{A}=6.022\times 10^{23}$

The number of molecules inside the room is N

$\\N=nN_{A}\\ N=1015.35\times 6.022\times 10^{23}\\ N=6.114\times 10^{26}$

Q12.7 Estimate the average thermal energy of a helium atom at (i) room temperature ( $27^{0}C$ ), (ii) the temperature on the surface of the Sun ( $6000K$ ), (iii) the temperature of $10$ million kelvin (the typical core temperature in the case of a star).

Answer:

The average energy of a Helium atom is given as $\frac{3kT}{2}$ since it is monoatomic

(i)

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

$E=\frac{3\times 1.38\times 10^{-23}\times 300}{2}\\ $

$E=6.21\times 10^{-21}\ J$

(ii)

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

$E=\frac{3\times 1.38\times 10^{-23}\times 6000}{2}\\ $

$E=1.242\times 10^{-19}\ J$

(iii)

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

$ E=\frac{3\times 1.38\times 10^{-23}\times10^{7}}{2}\\ $

$E=2.07\times 10^{-16}\ J$

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?

Answer:

As per Avogadro's Hypothesis, under similar conditions of temperature and pressure, equal volumes of gases contain an equal number of molecules. Since the volume of the vessels is the same and all vessels are kept at the same conditions of pressure and temperature, they would contain an equal number of molecules.

Root mean square velocity is given as

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

As we can see, vrms is inversely proportional to the square root of the molar mass; the root mean square velocity will be maximum in the case of Neon as its molar mass is the least.

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 $-20^{0}C$ ? (atomic mass of $Ar=39.9\mu$ , of $He=4.0\mu$ ).

Answer:

As we know root mean square velocity is given as $v_{rms}=\sqrt{\frac{3RT}{M}}$

Let at temperature T the root mean square speed of an atom in an argon cylinder equal to the rms speed of a helium gas atom at $-20^{0}C$

$\\\sqrt{\frac{3R\times T}{39.9}}=\sqrt{\frac{3R\times 253}{4}}\\ T=2523.7\ K$

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 $17^{0}C$ . Take the radius of a nitrogen molecule to be roughly $1.0A$ . Compare the collision time with the time the
molecule moves freely between two successive collisions (Molecular mass of $N_{2}=28.0\mu$ ).

Answer:

Pressure, P = 2atm

Temperature, T = 17 o C

The radius of the Nitrogen molecule, r=1 Å.

The molecular mass of N 2 = 28 u

The molar mass of N 2 = 28 g

From the ideal gas equation

$\\PV=nRT\\ $

$\frac{n}{V}=\frac{P}{RT}\\$

The above tells us about the number of moles per unit volume; the number of molecules per unit volume would be given as

$\\n'=\frac{N_{A}n}{V}=\frac{6.022\times 10^{23}\times 2\times 1.013\times 10^{5}}{8.314\times (17+273)}\\ $

$n'=5.06\times 10^{25}$

The mean free path $\lambda$ is given as

$\\\lambda =\frac{1}{\sqrt{2}\pi n'd^{2}}\\$

$ \lambda =\frac{1}{\sqrt{2}\times \pi \times 5.06\times 10^{25}\times (2\times 1\times 10^{-10})^{2}}\\ $

$\lambda =1.11\times 10^{-7}\ m$

The root mean square velocity v rms is given as

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

$v_{rms}=\sqrt{\frac{3\times 8.314\times 290}{28\times 10^{-3}}}\\ $

$v_{rms}=508.26\ m\ s^{-1}$

The time between collisions T is given as

$\\T=\frac{1}{Collision\ Frequency}\\ $

$T=\frac{1}{\nu }\\$

$ T=\frac{\lambda }{v_{rms}}\\ $

$T=\frac{1.11\times 10^{-7}}{508.26}\\ $

$T=2.18\times 10^{-10}s$

Collision time T' is equal to the average time taken by a molecule to travel a distance equal to its diameter

$\\T'=\frac{d}{v_{rms}}\\ $

$T'=\frac{2\times 1\times 10^{-10}}{508.26}\\ $

$T'=3.935\times 10^{-13}s$

The ratio of the average time between collisions to the collision time is

$
\begin{aligned}
\frac{T}{T^{\prime}} & =\frac{2.18 \times 10^{-10}}{3.935 \times 10^{-13}} \\
\frac{T}{T^{\prime}} & =554
\end{aligned}
$
Thus, we can see that the time between collisions is much larger than the collision time.

Additional Questions

Q 1) A metre-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?

Answer:

Initially, the pressure of the 15 cm long air column is equal to the atmospheric pressure, P 1 = 1 atm = 76 cm of Mercury

Let the crossectional area of the tube be x cm 2

The initial volume of the air column, V 1 = 15x cm 3

Let's assume that once the tube is held vertical, y cm of Mercury flows out of it.

The pressure of the air column after y cm of Mercury has flown out of the column P 2 = 76 - (76 - y) cm of Mercury = y cm of mercury

Final volume of air column V 2 = (24 + y)x cm 3

Since the temperature of the air column does not change

$\\P_{1}V_{1}=P_{2}V_{2}\\ 76\times 15x=y\times (24+y)x\\ 1140=y^{2}+24y\\ y^{2}+24y-1140=0$

Solving the above quadratic equation, we get y = 23.8 cm or y = -47.8 cm

Since a negative amount of mercury cannot flow out of the column, y cannot be negative. Therefore, y = 23.8 cm.

Length of the air column = y + 24 = 47.8 cm.

Therefore, once the tube is held vertically, 23.8 cm of Mercury flows out of it, and the length of the air column becomes 47.8 cm

Q 2) From a certain apparatus, the diffusion rate of hydrogen has an average value of $28.7cm^{3}s^{-1}$. The diffusion of another gas under the same conditions is measured to have an average rate of $7.2cm^{3}s^{-1}$. Identify the gas.

Answer:

As per Graham's Law of diffusion, if two gases of Molar Mass M 1 and M 2 diffuse with rates R 1 and R 2 respectively, their diffusion rates are related by the following equation

$\frac{R_{1}}{R_{2}}=\sqrt{\frac{M_{2}}{M_{1}}}$

In the given question

R 1 = 28.7 cm 3 s -1

R 2 = 7.2 cm 3 s -1

M 1 = 2 g

$\\M_{2}=M_{1}\left ( \frac{R_{1}}{R_{2}} \right )^{2}\\ M_{2}=2\times \left ( \frac{28.7}{7.2} \right )^{2}\\ M_{2}=31.78g$

The above Molar Mass is close to 32, therefore, the gas is Oxygen.

Q 3) 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 $n_{2}=n_{1}exp\left [ -mg(h_{2}-h_{1}) /k_{b}T\right ]$

where n2, n1 refer to number density at heights h2 and h1 respectively. Use this relation to derive the equation for the sedimentation equilibrium of a suspension in a liquid column:

$n_{2}=n_{1}exp\left [ -mg N_{A}(\rho -\rho ^{'})(h_{2}-h_{1})/ (\rho RT)\right ]$
where $\rho$ is the density of the suspended particle, and $\rho ^{'}$ , that of the surrounding medium. [ $N_{A}$ s Avogadro’s number, and R the universal gas constant.] [Hint: Use Archimedes' principle to find the apparent weight of the suspended particle.]

Answer:

$n_{2}=n_{1}exp\left [ -mg(h_{2}-h_{1}) /k_{b}T\right ]$ $(i)$

Let the suspended particles be spherical and have a radius r

The gravitational force acting on the suspended particles would be

$F_{G}=\frac{4}{3}\pi r^{3}\rho g$

The buoyant force acting on them would be

$F_{B}=\frac{4}{3}\pi r^{3}\rho' g$

The net force acting on the particles become

$\\F_{net}=F_{G}-F_{B}\\ F_{net}=\frac{4}{3}\pi r^{3}\rho g-\frac{4}{3}\pi r^{3}\rho' g\\F_{net}=\frac{4}{3}\pi r^{3}g(\rho -\rho ')$

Replacing mg in equation (i) with the above equation, we get

$\\n_{2}=n_{1}exp\left [ -\frac{4}{3}\pi r^{3}g(\rho -\rho ')(h_{2}-h_{1}) /k_{b}T\right ]\\ n_{2}=n_{1}exp\left [ \frac{-\frac{4}{3}\pi r^{3}g(\rho -\rho ')(h_{2}-h_{1})}{\frac{RT}{N_{A}}} \right ]\\ n_{2}=n_{1}exp\left [ \frac{-mgN_{A}(\rho -\rho ')(h_{2}-h_{1})}{RT\rho '} \right ]$

The above is the equation to be derived

Q 4) Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms :

Substance
Atomic Mass (u)
Density (10 3 Kg m 3 )
Carbon (diamond)
12.01
2.22
Gold
197
19.32
Nitrogen (liquid)
14.01
1
Lithium
6.94
0.53
Fluorine
19
1.14

Answer:

Let one mole of a substance of atomic radius r and density $\rho$ have molar mass M

Let us assume the atoms to be spherical

Avogadro's number is $N_{A}=6.022\times 10^{23}$

$\\N_{A}\frac{4}{3}\pi r^{3}\rho =M\\ r=\left ( \frac{3M}{4N_{A}\pi \rho } \right )^{\frac{1}{3}}$

For Carbon

$\\N_{A}\frac{4}{3}\pi r^{3}\rho =M\\ r=\left ( \frac{3\times 12.01}{4\times 6.022\times 10^{23}\times \pi\times 2.22\times 10^{3} } \right )^{\frac{1}{3}}\\ r=1.29$ Å.

For gold

$\\N_{A}\frac{4}{3}\pi r^{3}\rho =M\\ r=\left ( \frac{3\times 197.00}{4\times 6.022\times 10^{23}\times \pi\times 19.32\times 10^{3} } \right )^{\frac{1}{3}}\\ r=1.59$ Å.

For Nitrogen

$\\N_{A}\frac{4}{3}\pi r^{3}\rho =M\\ r=\left ( \frac{3\times 14.01}{4\times 6.022\times 10^{23}\times \pi\times 1.00\times 10^{3} } \right )^{\frac{1}{3}}\\ r=1.77$ Å.

For Lithium

$\\N_{A}\frac{4}{3}\pi r^{3}\rho =M\\ r=\left ( \frac{3\times 6.94}{4\times 6.022\times 10^{23}\times \pi\times 0.53\times 10^{3} } \right )^{\frac{1}{3}}\\ r=1.73$ Å.

For Fluorine

$\\N_{A}\frac{4}{3}\pi r^{3}\rho =M\\ r=\left ( \frac{3\times19.00}{4\times 6.022\times 10^{23}\times \pi\times 1.14\times 10^{3} } \right )^{\frac{1}{3}}\\ r=1.88 $ Å.

Class 11 Physics Chapter 12 NCERT solutions help students understand how tiny particles in matter behave. This chapter explains gases and their thermal properties, linking to thermodynamics. Studying it helps in exams and builds a strong foundation in physics.


Class 11 physics NCERT Chapter 12: Higher Order Thinking Skills (HOTS) Questions

Q1:

Two moles of an ideal gas with $\frac{C_p}{C_v}=\frac{5}{3}$ are mixed 3 moles of another ideal gas with $\frac{C_p}{C_v}=\frac{4}{3}$ The value of $\frac{C_p}{C_v}$ for the mixture is:-

Answer:

For ideal gas:- $C_p-C_v=R$
For first case:-

$
\frac{C_{p 1}}{C_{v 1}}=\frac{5}{3} \text { and } C_{p 1}-C_{v 1}=R
$


$
C_{p 1}=\frac{5}{3} C_{v 1} \text { and } \frac{5}{3} C_{v 1}-C_{v 1}=R \Rightarrow \frac{2}{3} C_{v 1}=R \Rightarrow C_{v 1}=\frac{3}{2} R
$

${ }_{\text {So, }} C_{p 1}=\frac{5}{2} R$
For second case:-

$
\begin{aligned}
& \frac{C_{p 2}}{C_{v 2}}=\frac{4}{3} \text { and } C_{p 2}-C_{v 2}=R \\
& C_{p 2}=\frac{4}{3} C_{v 2} \text { and } \frac{4}{3} C_{v 2}-C_{v 2}=R \Rightarrow C_{v 2}=3 R{ }_{\text {and }} C_{p 2}=4 R \\
& \qquad Y_{\text {mix }}=\frac{n_1 C_{p 1}+n_2 C_{p 2}}{n_1 C_{v 1}+n_2 C_{v 2}}=\frac{2 \times \frac{5}{2} R+3 \times 4 R}{2 \times \frac{3}{2} R+3 \times 3 R}=1.417=1.42
\end{aligned}
$


Q2:

If temperature of the atmosphere varies with height as $\mathrm{T}=\left(\mathrm{T}_0-\mathrm{ah}\right)$, where a and $\mathrm{T}_0$ are positive constants, then the pressure as a function of height $h$ is (assume atmospheric pressure at sea level ( $\mathrm{h}=0$ ) is ${P_0}$ and molecule mass M of the air and acceleration due to gravity g be constant)

Answer:

$
\frac{\mathrm{dp}}{\mathrm{dh}}=-\rho \mathrm{g}=-\left(\frac{\mathrm{pM}}{\mathrm{RT}}\right) \mathrm{g}=-\left[\frac{\mathrm{pM}}{\mathrm{R}\left(\mathrm{~T}_0-\mathrm{ah}\right)}\right] g
$


$
\frac{\mathrm{dp}}{\mathrm{p}}=-\left(\frac{\mathrm{Mg}}{\mathrm{R}}\right) \frac{\mathrm{dh}}{\mathrm{~T}_0-\mathrm{ah}}
$


$
\int_{\rho_0}^{\mathrm{p}} \frac{\mathrm{dp}}{\mathrm{p}}=-\left(\frac{\mathrm{Mg}}{\mathrm{R}}\right) \int_0^{\mathrm{h}} \frac{\mathrm{dh}}{\left(\mathrm{~T}_0-\mathrm{ah}\right)}
$


$
\mathrm{p}=\mathrm{p}_0\left(\frac{\mathrm{~T}_0-\mathrm{ah}}{\mathrm{~T}_0}\right)^{\frac{\mathrm{M}_{\mathrm{s}}}{\mathrm{Ra}}}
$


Q3:

In two jars A and B, the pressure, volume and temperature in jar A are respectively P, V, and T and that of B are 2P, V/4 and 2T. Then the ratio of the number of molecules in jar A and B will be,

Answer:

$\begin{aligned} & \frac{N_A}{N_B}=\frac{P_A V_A}{P_B P_B} \times \frac{T_B}{T_A} \\ & \frac{N_A}{N_B}=\frac{P \times V \times(2 T)}{2 P \times \frac{V}{4} \times T}=\frac{4}{1}\end{aligned}$


Q4:

The temperature of an open room of volume 30 m3 increases from 170C to 270C due to the sunshine. The atmospheric pressure in the room remains 1×105 Pa. If $n_i$ and $n_f$ are the number of molecules in the room before and after heating, then $n_f-n_i$ will be :

Answer:

PV = nRT

$\begin{aligned} & \Rightarrow n_i=\frac{P V}{R T_i}, \quad n_f=\frac{P V}{R T_f} \\ & \Rightarrow n_f-n_i=\frac{P V}{R}\left(\frac{1}{T_f}-\frac{1}{T_i}\right)=\frac{10^5 \times 30}{8.31} \times\left(\frac{1}{300}-\frac{1}{290}\right) \\ & n_f-n_i=\frac{10^5 \times 30}{8.31} \times \frac{-10}{300 \times 290}=\frac{-10^5}{290 \times 8.31} \\ & \text { change in the Number of molecules }=\frac{-10^5 \times 6.023 \times 10^{23}}{290 \times 8.31}=-2.5 \times 10^{25}\end{aligned}$


Q5:

A gas mixture contains 3 moles of oxygen and x mole of monoatomic gas at temperature T Considering only translational and rotational but not vibrational modes, the total energy of the system is 15 RT then the value of x is.

Answer:

$U=\frac{n f}{2} R T$

According to the question,

$\begin{aligned} 15 R T & =3 \times \frac{5}{2} R T+x \times \frac{3}{2} R T \\ 15 & =\frac{15}{2}+\frac{3 x}{2} \\ x & =5\end{aligned}$


Approach to Solve Questions of Kinetic Theory of Gases

  • Understand basic assumptions – Gas is made up of very small particles in random continuous motion; collisions are elastic.
  • Learn the important words – Pressure, temperature, volume, kinetic energy, mean free path.
  • Master key formulas –
  1. Average kinetic energy: $K E=\frac{3}{2} k T$
  2. Total KE of $n$ molecules: $K E_{\text {total }}=\frac{3}{2} n R T$
  3. Pressure: $P=\frac{1}{3} \rho \bar{c}^2(\rho=$ density, $\bar{c}=r m s$ speed $)$
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  • Apply gas laws -
  1. Boyle's Law: $P V=$ constant
  2. Charles's Law: $\frac{V}{T}=$ constant
  3. Avogadro's Law: $V \propto n$
  • Learn universal gas equation – PV=nRT
  • Apply kinetic theory relations – Represent pressure, volume, and temperature in terms of molecules.
  • Understand formula for rms speed – $\bar{v}=\sqrt{\frac{3 k T}{m}}$ or $\bar{v}=\sqrt{\frac{3 R T}{M}}$
  • Keep units consistent –
  1. $\mathrm{R}=8.314 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{~K}$

  2. T in Kelvin

  3. Mass in kg or g as per requirement

  • Numerals solve step by step – Write the given, select suitable formula, change units, solve neatly. Revisit derivations and fundamentals – Certain theory questions are questions of assumptions or postulates.

What Extra Should Students Study Beyond NCERT for JEE/NEET?

NCERT Solutions for Class 11 Physics Chapter-Wise

NCERT Solutions for Class 11 Subject Wise

Also, check NCERT Books and NCERT Syllabus here

Subject wise NCERT Exemplar solutions

Frequently Asked Questions (FAQs)

1. What is the molecular nature of matter?

Everything wheather it is solids, liquids, and gases are made of tiny moving particles called atoms and molecules. Their movement decides whether something is solid liquid or gas .

2. How do gases behave in kinetic theory?

Gas molecules move randomly bouncing off each other and their container. This explains why gases expand and fill spaces following rules like Boyle’s and Charles’ laws.

3. What is the law of equipartition of energy?

It’s like sharing energy equally! In a gas, molecules divide energy between movement types—sliding, spinning, and vibrating—all getting a fair share.

4. Why do gases have two specific heat capacities?

Gases absorb heat differently depending on conditions:

Cₚ (at constant pressure) when the gas expands.

Cᵥ (at constant volume) when it stays in the same space.

The extra energy in Cₚ goes into expansion!

What is the mean free path?

It’s how far a gas molecule travels before colliding with another. Like walking in a crowd, some go farther before bumping into someone!

Articles

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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