#### Chapters

Chapter 2: Units and Measurements

Chapter 3: Motion in a Straight Line

Chapter 4: Motion in a Plane

Chapter 5: Laws of Motion

Chapter 6: Work, Energy and Power

Chapter 7: System of Particles and Rotational Motion

Chapter 8: Gravitation

Chapter 9: Mechanical Properties of Solids

Chapter 10: Mechanical Properties of Fluids

Chapter 11: Thermal Properties of Matter

Chapter 12: Thermodynamics

Chapter 13: Kinetic Theory

Chapter 14: Oscillations

Chapter 15: Waves

#### NCERT Physics Class 11

## Chapter 13: Kinetic Theory

#### Chapter 13: Kinetic Theory solutions [Pages 333 - 335]

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 3Å.

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

The figure shows the plot of *PV*/*T *versus Pfor 1.00×10^{–3} kg of oxygen gas at two different temperatures.

(a) What does the dotted plot signify?

(b) Which is true: *T*_{1 }> *T*_{2} 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 ×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 u, of O_{2} = 32.0 u, *R *= 8.31 J mo1^{–1} K^{–1}.)

An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder (*R *= 8.31 J mol^{–1} K^{–1}, molecular mass of O_{2} = 32 u)

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?

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 27 °C and 1 atm pressure

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).

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 the 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?

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).

Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 °C. Take the radius of a nitrogen molecule to be roughly 1.0 Å. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N_{2} = 28.0 u).

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?

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

[Hint: Use Graham’s law of diffusion: R_{1}/R_{2} = (M_{2}/M_{1})^{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. The law is a simple consequence of kinetic theory.]

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 the 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 [-mg (h_{2 }– h_{1})/ k_{B}T]

Where n_{2}, n_{1} refer to number density at heights h_{2} and h_{1} respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column:

n_{2} = n_{1} exp [-mg N_{A}(ρ - P′) (h_{2} –h_{1})/ (ρRT)]

Where ρ is the density of the suspended particle, and ρ’ that of surrounding medium. [N_{A} is Avogadro’s number, and R the universal gas constant.] [Hint: Use Archimedes principle to find the apparent weight of the suspended particle.]

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 | 197.00 | 19.32 |

Nitrogen (liquid) | 14.01 | 1.00 |

Lithium | 6.94 | 0.53 |

Fluorine (liquid) | 19.00 | 1.14 |

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

## Chapter 13: Kinetic Theory

#### NCERT Physics Class 11

#### Textbook solutions for Class 11

## NCERT solutions for Class 11 Physics chapter 13 - Kinetic Theory

NCERT solutions for Class 11 Physics chapter 13 (Kinetic Theory) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CBSE Physics Textbook for Class 11 solutions in a manner that help students grasp basic concepts better and faster.

Further, we at Shaalaa.com are providing such solutions so that students can prepare for written exams. NCERT textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.

Concepts covered in Class 11 Physics chapter 13 Kinetic Theory are Kinetic Theory of Gases - Concept of Pressure, Mean Free Path, Specific Heat Capacities - Gases, Kinetic Interpretation of Temperature, Introduction of Kinetic Theory of an Ideal Gas, Work Done in Compressing a Gas, Equation of State of a Perfect Gas, Behaviour of Gases, Law of Equipartition of Energy, Molecular Nature of Matter, Avogadro's Number, Degrees of Freedom, rms Speed of Gas Molecules, Kinetic Theory of Gases- Assumptions.

Using NCERT Class 11 solutions Kinetic Theory exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in NCERT Solutions are important questions that can be asked in the final exam. Maximum students of CBSE Class 11 prefer NCERT Textbook Solutions to score more in exam.

Get the free view of chapter 13 Kinetic Theory Class 11 extra questions for Physics and can use Shaalaa.com to keep it handy for your exam preparation