NCERT Solutions for Class 11 Physics Chapter 13Â 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
where is the number of moles
N is the number of molecules
R and k _{ B } are universal constants.
- The pressure of an ideal gas
where n is the number density, m is the mass of the molecule and is the mean of the squared speed.
- The RMS velocity of the gas molecule
Where T is the temperature.
- The translational kinetic energy is
- The mean free path
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 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
Q13.3 Figure 13.8 shows plot of versus P for kg of oxygen gas at two different temperatures.
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
kg of hydrogen we would not get the same
value of
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
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
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 since it is monoatomic
(i)
(ii)
(iii)
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
As we can see v _{ rms } 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.
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 ,
The molecular mass of N _{ 2 } = 28 u
The molar mass of N _{ 2 } = 28 g
From 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 is given as
The root mean square velocity v _{ rms } is given as
The time between collisions T is given as
Collision time T' is equal 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 time between collisions is much larger than the collision time.
NCERT solutions for class 11 physics chapter 13 kinetic theory additional exercise
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 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.
Q13.13 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
Answer:
Let the suspended particles be spherical and have 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 have molar mass M
Let us assume the atoms to be spherical
Avogadro's number is
For Carbon
For gold
For Nitrogen
For Lithium
For Fluorine
NCERT solutions for class 11 physics chapter wise
NCERT Solutions for Class 11 Subject wise
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.