NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory

 

NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory: Maxwell, Boltzmann, and others developed the kinetic theory of gas in the 19th century. Before going to solutions of NCERT for class 11 physics chapter 13 kinetic theory one must be familiar with ideal gas equations and gas laws. The kinetic theory interprets the pressure and temperature at the molecular level. Kinetic theory is consistent with gas laws and it correctly explains specific heat and relates measurable properties of gases such as viscosity, diffusion, etc. The CBSE NCERT solutions for class 11 physics chapter 13 kinetic theory gives an idea about how to apply the formulas studied in the chapter in numerical problems. NCERT solutions explain all the exercise and additional exercise questions.

A brief summary of formulas required for NCERT solutions for class 11 physics chapter 13 kinetic theory are given below.

  • The ideal gas equation is given by

 PV=\mu RT=K_BNT

where \mu is the number of moles

N is the number of molecules

R and kB are universal constants.

  • The pressure of an ideal gas

            P=\frac{1}{3}nm\bar v^2

where n is the number density, m is the mass of the molecule and \bar v^2 is the mean of the squared speed. 

  • The RMS velocity of the gas molecule

           v_{rms}=\sqrt{\bar v^2}=\sqrt{\frac{3K_BT}{m}} Where T is the temperature.

  • The translational kinetic energy is

            \frac{3K_BNT}{2}

  • The mean free path

           l=\frac{1}{\sqrt{2}n\pi d^2} 

     where n is the number density and d is the diameter of the molecule

The main topics of  NCERT Class 11 Physics Chapter 13 are listed below.

13.1 Introduction

13.2 Molecular nature of matter

13.3 Behaviour of gases

13.4 Kinetic theory of an ideal gas

13.5 Law of equipartition of energy

13.6 Specific heat capacity

13.7 Mean free path

NCERT solutions for class 11 physics chapter 13 kinetic theory exercise

Q13.1 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 Vactual 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 Vmolar = 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}

Q13.3 Figure 13.8 shows plot of PV/T  versus P for 1.00\times 10^{-3}  kg  of oxygen gas at two different temperatures.

(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 for1.00\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 T1 is closer to the horizontal line that the one corresponding to T2 we conclude T1 is greater than T2.

(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 for1.00\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

Q13.4  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, V1 = Volume of Cylinder = 30 l

Initial Pressure P1 = 15 atm

Initial Temperature T1 = 27 oC = 300 K

The initial number of moles n1 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, V2 = Volume of Cylinder = 30 l

Final Pressure P2 = 11 atm

Final Temperature T2 = 17 oC = 290 K

Final number of moles n2 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 = n2 -n1 = 18.28 - 13.86 = 4.42

Mass of oxygen taken out of the cylinder m is

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

Q13.5 An air bubble of volume1.0 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, V1 = 1.0 cm3

Initial temperature, T1 = 12 oC = 273 + 12 = 285 K

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

Initial Pressure is P1

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, T2 = 35 oC = 35 + 273 = 308 K

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

Let the final volume be V2

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}

Q13.6 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 m3

Temperature of the room, T = 27oC = 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}

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}

Q13.8  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 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 vrms the largest?

Answer:

As per Avogadro's Hypothesis under similar conditions of temperature and pressure equal volumes of gases contain equal number of molecules. Since the volume of the vessels are the same and all vessels are kept at the same conditions of pressure and temperature they would contain 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 case of Neon as its molar mass is the least.

Q13.9 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

Q13.10 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 oC

The radius of the Nitrogen molecule , r=1\ \AA

The molecular mass of  N= 28 u

The molar mass of N= 28 g

From 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 vrms 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 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

\\\frac{T}{T'}=\frac{2.18\times 10^{-10}}{3.935\times 10^{-13}}\\ \frac{T}{T'}=554

Thus we can see time between collisions is much larger than the collision time.

NCERT solutions for class 11 physics chapter 13 kinetic theory additional exercise

Q13.11 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, P1 = 1 atm = 76 cm of Mercury

Let the crossectional area of the tube be x cm2

The initial volume of the air column, V1 = 15x cm3

Let's assume 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= 76 - (76 - y) cm of Mercury = y cm of mercury

Final volume of air column V2 = (24 + y)x cm3

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 13.12 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 M1 and M2 diffuse with rates R1 and R2 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

R1 = 28.7 cm3 s-1

R2 = 7.2 cm3 s-1 

M1 = 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.

Q13.14 Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms :

Substance Atomic Mass (u) Density (10Kg m3)
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\AA

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\AA

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\AA

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\AA

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\AA

NCERT solutions for class 11 physics chapter wise

Chapter 1

NCERT solutions for class 11 physics chapter 1 Physical world

Chapter 2

Solutions of NCERT for class 11 physics chapter 2 Units and Measurement

Chapter 3

CBSE NCERT solutions for class 11 physics chapter 3 physics Motion in a straight line

Chapter 4

NCERT solutions for class 11 physics chapter 4 Motion in a Plane

Chapter 5

Solutions of NCERT for class 11 physics chapter 5 Laws of Motion

Chapter 6

CBSE NCERT solutions for class 11 physics chapter 6 Work, Energy and Power

Chapter 7

NCERT solutions for class 11 physics chapter 7 System of Particles and Rotational motion

Chapter 8

Solutions of NCERT for class 11 physics chapter 8 Gravitation

Chapter 9

CBSE NCERT solutions for class 11 physics chapter 9 Mechanical Properties of Solids

Chapter 10

NCERT solutions for class 11 physics chapter 10 Mechanical Properties of Fluids

Chapter 11

Solutions of NCERT for class 11 physics chapter 11 Thermal Properties of Matter

Chapter 12

CBSE NCERT solutions for class 11 physics chapter 12 Thermodynamics

Chapter 13

NCERT solutions for class 11 physics chapter 13 Kinetic Theory

Chapter 14

Solutions of NCERT for class 11 physics chapter 14 Oscillations

Chapter 15

CBSE NCERT solutions for class 11 physics chapter 15 Waves

NCERT Solutions for Class 11 Subject wise

CBSE NCERT solutions for class 11 biology

NCERT solutions for class 11 maths

CBSE NCERT solutions for class 11 chemistry

NCERT solutions for class 11 physics

Use of NCERT solutions for class 11 physics chapter 13 kinetic theory:

  • To familiarise with the concepts studied in the chapter, it is important to study the CBSE NCERT solutions for class 11 physics chapter 13 kinetic theory.
  • All the equations studied in the chapter are utilized in the solutions of NCERT class 11 physics chapter 13 kinetic theory.
  • The chapter is important for class exams and competitive exams. For exams like NEET and JEE Mains one question can be expected from the chapter and the questions can be solved easily if you are familiar with the concepts and formulas of the chapter.
 

Recently Asked Questions

 

Related Articles

Exams
Articles
Questions