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.
where is the number of moles
N is the number of molecules
R and k_{B} are universal constants.
where n is the number density, m is the mass of the molecule and is the mean of the squared speed.
Where T is the temperature.
where n is the number density and d is the diameter of the molecule
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
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
As per the ideal gas equation
For one mole of a gas at STP we have
(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
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
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
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
The average energy of a Helium atom is given as since it is monoatomic
(i)
(ii)
(iii)
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.
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
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.
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
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
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 |
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
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
where n2, n1 refer to number density at heights h2 and h1 respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column:
where is the density of the suspended particle, and , that of the surrounding medium. [ s Avogadro’s number, and R the universal gas constant.] [Hint: Use Archimedes principle to find the apparent weight of the suspended particle.]
Q13.6 Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity at a temperature of and pressure.
Q13.4 An oxygen cylinder of volume litres has an initial gauge pressure of atm and a temperature of . After some oxygen is withdrawn from the cylinder, the gauge pressure drops to atm and its temperature drops to . Estimate the mass of
oxygen taken out of the cylinder (, molecular mass of ).