Kinetic Theory Class 11 Chapter 13 Questions And Answers

NCERT Solutions for Class 11 Physics Physics Chapter 13 Kinetic Theory includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Physics exams must go through NCERT Solutions for Class 11 Physics Physics Chapter 13 Kinetic Theory. Going through the solutions provided on this page will help you to know how to approach and solve the problems.

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Class 11 Physics Chapter 13 Kinetic Theory NCERT Solutions

Kinetic Theory NCERT Solutions

NCERT Solutions for Class 11 Physics Physics Chapter 13 Kinetic Theory are been solved by expert teachers of CBSETuts.com. All the solutions given in this page are solved based on CBSE Syllabus and NCERT guidelines.

NCERT Exercises

Question 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:
Here, the diameter of an oxygen molecule
d = 3 A = 3 x 10^-10 m
The radius of an oxygen molecule

Question 2.
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 liters.
Answer:
Here, R = 8.31 J mol^-1 K ^-1
T = 0°C = 273 K; n= 1 mole
P = 1 atm = 0.76 of Hg column
= 0.76 x 13.6 x 10³ x 9.8 = 1.013 x 10⁵ N m²

Question 3.
Figure shows plot of PV/T versus P for 1.00 x 10^-3 kg of oxygen gas at two different temperatures.

(a) What does the dotted plot signify?
(b) Which is true: T, > T₂ or T, < T₂?
(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 x 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 H₂ = 2.02 u, of 0₂ = 32.0 u, ff = 8.31 J mol^-1 MK^-1.)
Answer:
(a) Since the dotted plot is parallel to P-axis, it tells that value of \(\cfrac { pv }{ T } (=nR)\) remains constant even when P is changed.
(b) Curve at temperature T, is more close to the dotted plot than the curve at the temperature T₂. Since the behaviour of a real gas approaches the perfect gas behaviour, as the temperature is increased therefore T₁ > T₂.
(c) If the amount of gas under consideration is 1 mole, then the value of PV/T, where the curves meet PV/T- axis will be R (=8.31 J mole^-1 K^-1). For oxygen, molecular mass is 32.0 g i.e. 32.0 x 10^-3 kg. Since mass of oxygen gas under considerations is 1.0 ^x 10^-3 kg, the value of PV/T, where the curves meet the PV/T-axis is given by
\(\cfrac { pv }{ T } =\cfrac { 8.31 }{ 32.0 } \times { 10 }^{ -3 }=0.26j{ k }^{ -1 }\)
(d) If we obtained similar plots for 1.0 x 10^-3 kg of hydrogen, the value of PV/T, where the curves meet the y-axis will not be the same. Since molecular mass of hydrogen is 2.02 x 10^-3 kg, the mass of hydrogen that will yield same value of PV/T i.e., 0.26 J K^-1 will be
\(\cfrac { 2.02\times { 10 }^{ -3 } }{ 8.31 } \times 0.26=6.32\times { 10 }^{ -5 }kg\)

Question 4.
An oxygen cylinder of volume 30 liters 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 0₂ = 32 u).
Answer:
Here, initially in the oxygen cylinder
Volume V₂ = 30 liters = 30 x 10^-3 m³
Pressure P₁= 15 atm = 15 x 1.013 x 10⁵ Pa
Temperature = 27 + 273 = 300 K
If the cylinder contains,let n₂ moles of the oxygen gas then P₁V₁ = n₁RT,

Initial mass of oxygen in the cylinder
m_x = n₁ molecular weight of 0₂
= 18.253 x 32 = 584.1 g
Finally in the oxygen cylinder, let n₂ moles of oxygen be left
Final volume V₂ = 30 x 10^-3 m³
Final pressure P₂ = 11 atm = 11 x 1.013 x 10⁵ Pa
Final temperature T₂ = 17 + 273 = 290 K

Question 5.
An air bubble of volume 1.0 cm³ 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?
Answer:
When the air bubble is at depth of 40 m
Volume of air bubble V₂ = 1.0 cm³ = 1.0 x 10^-6 m³
Pressure of air bubble P₁ = 1 atm + 40 m depth of water = 1 atm + h₁ρg
= 1.013 x 10⁵ + 40 x 10³ x 9.8 = 4.93 x 10⁵ Pa
Temperature of air bubble T₁ = 12 + 273 = 285 K
When the air bubbles reaches at the surface of lake, pressure of air bubbles
P₂ = 1 atm = 1.013 x 10⁵ Pa
Temperature of air bubble T₂ = 35°C = 35 + 273 = 308 K
Volume of air bubbles = ?

Question 6.
Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m³ at a temperature of 27°C and 1 atm pressure.
Answer:
Here, volume of air molecules, V = 25.0 m³
Temperature of air molecules
T = 27°C = 27 + 273 = 300 K
Pressure of air molecules
P = 1 atm = 1.013 x 10^s Pa
Now, PV = nRT

Question 7.
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 case of a star).
Answer:
The average kinetic energy of the gas at a temperature T

Question 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 v_ms the largest?
Answer:
All the three vessels (at the same temperature and pressure) have same volume. So, in accordance with the Avogadro’s law, the three vessels will contain equal number  of respective molecules, being equal to Avogadro’s number N = 6.023 x 10²³.

at a given temperature therefore, rms speed of molecules will not be the same in the three cases. As neon has the smallest mass, therefore, rms speed will be the largest in case of neon.

Question 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°C? (atomic mass of Ar = 39.9 u, of He = 4.0 u).
Answer:
Here, atomic mass of argon M₁= 39.9
Atomic mass of helium He = 4.0
Let C₁ and C₂ be the rms velocities of argon and helium at temperature T₁K and T₂ K respectively.
We know that rms speed is given by

Question 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°C. Take the radius of a nitrogen molecule to be roughly 1.0 A. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N₂ = 28.0 u).
Answer:
Here, Pressure of nitrogen = 2 atm
= 2 x 1.013 x 10⁵ N m-² = 2.026 x 10⁵ Nm²
Temperature of nitrogen = 17° C
= 17 + 273 = 290 K
Molecular diameter d = 2 x 1= 2Å= 2x 10^-10 m

Question 11.
A meter 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:
When the tube is held horizontally, the mercury thread of length 76 cm traps a length of air = 15 cm. A length of 9 cm of the tube will be left at the open end, figure
(a) The pressure of air enclosed in tube will be atmospheric pressure. Let area of cross-section of the tube be 1 sq. cm.

∴ P₁ = 76 cm and V₁ = 15 cm³
When the tube is held vertically, 15 cm air gets another 9 cm of air filled in the right hand side (in the horizontal position) and let h cm of mercury flows out to balance the atmospheric pressure, figure
(b) Then the heights of air column and mercury column are (24 + h) cm and (76 – h) cm respectively.
The pressure of air = 76 – (76 – h) = h cm of mercury
∴ V₂ = (24 + h) cm³ and P₂ = h cm.

Since h cannot be negative (because more mercury cannot flow into the tube), therefore h = 23.8 cm. Thus, in the vertical

Question 12.
From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm³ s^-1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm³ s^-1. Identify the gas.
Answer:
According to Graham’s law of diffusion

Question 13.
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
n₂ = n₁ exp [-mg (h₂ – h₁)/k_BT]
where n₂,n₁ refer to number density at heights h₂ and /i, respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column:
n₂ = n₁ exp [-mg N_A (ρ – ρ’) [h₂– h₁)/(ρRT)]
where ρ is the density of the suspended particle, and ρ’ that of surrounding medium.
[N_A is Avogadro’s number, and R is the universal gas constant.]
Answer:
According to the law of atmospheres,

where n₂, n₂ refer to number density of particles at heights h₂ and h₁ , respectively.
If we consider the sedimentation equilibrium of suspended particles in a liquid, then in place of mg, we will have to take effective weight of the suspended particles.
Let V = average volume of a suspended particle, r = density of suspended particle, r’ = density of liquid, m = mass of equal volume of liquid displaced.
According to Archimede’s principle, effective weight of one suspended particle = actual weight – weight of liquid displaced = mg- m’g

Question 14.
Given below are the densities of some solids and liquids. Give rough estimates of the sizes of their atoms

Substance	Atomic Mass (u)	Density (103km-3)
Carbon(diamond)	12.01	2.22
Gold	197.0	19.32
Nitrogen (liquid)	14.01	1.00
Lithium	6.94	0.53
Fluorine (liquid)	19.00	1.14

Answer:

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