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Kinetic Theory: Class 11 Physics Chapter 12 is one of the important chapters in Physics of Class 11 and candidates appearing on CBSE Board Examination, JEE, NEET, and other tests are advised to go through this Topics. We have developed NCERT Solutions - Class 11 Physics Chapter 12 Kinetic Theory that will help at the time of exams. The experts of the subject have elaborated the explanation to illustrate the concept to the students which are difficult to understand in Class 11 NCERT Solutions. These NCERT solutions are in accordance with the new CBSE only curriculum and would enable the students to confidently endeavor both the theoretical and numerical questions.
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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.
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The NCERT Solutions to Class 11 Chapter 12 Kinetic Theory includes step-by-step solutions of the textbook questions, which makes the students to have a clear picture of the behavior of gases at molecular scale. One can access these solutions online and download as a PDF solution without any payment making practice and revision convenient.
Answer:
The diameter of an oxygen molecule, d = 3 Å.
The actual volume of a mole of oxygen molecules V actual is
The volume occupied by a mole of oxygen gas at STP is V molar = 22.4 litres
Answer:
As per the ideal gas equation
For one mole of a gas at STP we have
Q12.3 Figure 13.8 shows plot of
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
The molar mass of oxygen = 32 g
R = 8.314
(d) If we obtained similar plots for
Molar Mass of Hydrogen M = 2 g
mass of hydrogen
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
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
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
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
Initial Pressure is P 1
Depth of the bottom of the lake = 40 m
Final Temperature, T 2 = 35 o C = 35 + 273 = 308 K
Final Pressure = Atmospheric Pressure
Let the final volume be V 2
As the number of moles inside the bubble remains constant, we have
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
Avogadro's Number,
Number of molecules inside the room is N
Answer:
The average energy of a Helium atom is given as
(i)
(ii)
(iii)
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
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.
Answer:
As we know root mean square velocity is given as
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
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
The above tells us about the number of moles per unit volume; the number of molecules per unit volume would be given as
The mean free path
The root mean square velocity v rms is given as
The time between collisions T is given as
Collision time T' is equal to the average time taken by a molecule to travel a distance equal to its diameter
The ratio of the average time between collisions to the collision time is
Thus, we can see that the time between collisions is much larger than the collision time.
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
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
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
In the given question
R 1 = 28.7 cm 3 s -1
R 2 = 7.2 cm 3 s -1
M 1 = 2 g
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
Answer:
Let the suspended particles be spherical and have a radius r
The gravitational force acting on the suspended particles would be
The buoyant force acting on them would be
The net force acting on the particles become
Replacing mg in equation (i) with the above equation, we get
The above is the equation to be derived
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
Let us assume the atoms to be spherical
Avogadro's number is
For Carbon
For gold
For Nitrogen
For Lithium
For Fluorine
Q1:
Two moles of an ideal gas with
Answer:
For ideal gas:-
For first case:-
For second case:-
Q2:
If temperature of the atmosphere varies with height as
Answer:
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:
Q4:
The temperature of an open room of volume 30 m3 increases from 17oC to 27oC due to the sunshine. The atmospheric pressure in the room remains 1×105 Pa. If
Answer:
PV = nRT
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:
According to the question,
Chapter 12: Kinetic Theory in Class 11 Physics explains the microscopic behavior of gases based on the motion of their molecules. It covers concepts like the kinetic theory of gases, pressure-temperature relationship, and degrees of freedom.
12.1 Introduction
12.2 Molecular Nature Of Matter
12.3 Behaviour Of Gases
12.4 Kinetic Theory Of An Ideal Gas
12.4.1 Pressure Of An Ideal Gas
12.4.2 Kinetic Interpretation Of Temperature
12.5 Law Of Equipartition Of Energy
12.6 Specific Heat Capacity
12.6.1 Monatomic Gases
12.6.2 Diatomic Gases
12.6.3 Polyatomic Gases
12.6.4 Specific Heat Capacity Of Solids
12.7 Mean Free Path
The important formulae from this chapter that help in solving numerical problems related to molecular speeds, kinetic energy, and gas laws, making them essential for board exams and competitive tests like JEE and NEET are listed below:
Where:
Where
Where
Where:
Where:
-
-
-
Where
Take some time reading the question or cross it out, and underline the available data (such as temperature, pressure, volume, the number of moles or mass of gas).
See what the question actually asks (e.g. the speed, the kinetic energy, internal energy, pressure).
Express everything in terms of SI units:
Temperature in Kelvin
Cube volume in m3
Mass in kg
Molar Mass in kilogram per mol
NCERT textbooks are almost a must have in studying and preparing competitive exams such as JEE and NEET, but are definitely lacking in preparing the numericals, tricky applications and deeper conceptual advancement. This is a list of what NCERT does not cover and what additional students are expected to learn in the chapter Kinetic Theory of Gases in preparation of JEE/NEET:
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 .
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.
It’s like sharing energy equally! In a gas, molecules divide energy between movement types—sliding, spinning, and vibrating—all getting a fair share.
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!
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