Kinetic Theory
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Consider a rectangular block of wood moving with a velocity v0 in a gas at temperature T and mass density ρ. Assume the velocity is along x-axis and the area of cross-section of the block perpendicular to v0 is A. Show that the drag force on the block is 4ρ 0 kT Av m , where m is the mass of the gas molecule
Consider a rectangular block of wood moving with a velocity v0 in a gas at temperature T and mass density ρ. Assume the velocity is along x-axis and the area of cross-section of the block perpendicular to v0 is A. Show that the drag force on the block is 4ρ 0 kT Av m , where m is the mass of the gas molecule
This is a long answer type question as classified in NCERT Exemplar
Consider the diagram

let n =number of molecules per unit volume
Vrms= rms speed of gas molecule
When block is moving with speed vo, relative speed of molecules w.r.t front face =v+vo
Coming head on, momentum transferred to block per coll
A box of 1.00m3 is filled with nitrogen at 1.50 atm at 300K. The box has a hole of an area 0.010 mm2. How much time is required for the pressure to reduce by 0.10 atm, if the pressure outside is 1 atm.
A box of 1.00m3 is filled with nitrogen at 1.50 atm at 300K. The box has a hole of an area 0.010 mm2. How much time is required for the pressure to reduce by 0.10 atm, if the pressure outside is 1 atm.
This is a long answer type question as classified in NCERT Exemplar
Given volume V = 1m3
area = 0.01mm2

= 8.01 m2= m2
Temperature both inside and outside are equal
So initial pressure inside the box = 1.50atm
Final pressure inside the box= 0.1atm
Assuming Vx= speed of nitrogen molecule in x direction
ni
Ten small planes are flying at a speed of 150 km/h in total darkness in an air space that is 20 × 20 × 1.5 km3 in volume. You are in one of the planes, flying at random within this space with no way of knowing where the other planes are. On the average about how long a time will elapse between near collision with your plane. Assume for this rough computation that a safety region around the plane can be approximated by a sphere of radius 10m.
Ten small planes are flying at a speed of 150 km/h in total darkness in an air space that is 20 × 20 × 1.5 km3 in volume. You are in one of the planes, flying at random within this space with no way of knowing where the other planes are. On the average about how long a time will elapse between near collision with your plane. Assume for this rough computation that a safety region around the plane can be approximated by a sphere of radius 10m.
This is a long answer type question as classified in NCERT Exemplar
Time require to avoid the collision T= l/v where l = mean free path =1/
Where n = N/V
n=number of aeroplanes/volume
= -3
T=
T= =
Consider an ideal gas with the following distribution of speeds

(a) Calculate Vrms and hence T(m=3 )
(b) If all the molecules with speed 1000m/s escape from the system , calculate new Vrms and hence T.
Consider an ideal gas with the following distribution of speeds

(a) Calculate Vrms and hence T(m=3 )
(b) If all the molecules with speed 1000m/s escape from the system , calculate new Vrms and hence T.
This is a long answer type question as classified in NCERT Exemplar

Explain why
(a) There is no atmosphere on moon.
(b) There is fall in temperature with altitude. 13.28 Consider an ideal gas with following distribution of speeds
(i) Calculate Vrms and hence T. 26 ( 3.0 10 kg) m − = ×
(ii) If all the molecules with speed 1000 m/s escape from the system, calculate new Vrms and hence T.
Explain why
(a) There is no atmosphere on moon.
(b) There is fall in temperature with altitude. 13.28 Consider an ideal gas with following distribution of speeds
(i) Calculate Vrms and hence T. 26 ( 3.0 10 kg) m − = ×
(ii) If all the molecules with speed 1000 m/s escape from the system, calculate new Vrms and hence T.
This is a long answer type question as classified in NCERT Exemplar
(a) The moon has small gravitational force and hence, the escape velocity is small .
As the moon in the proximity of the earth as seen from the sun, the moonhas the same amount of heat per unit area as that of the earth.
The air molec
13.14 Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms:
Substance
Atomic Mass (u)
Density (
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].
13.14 Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms:
|
Substance |
Atomic Mass (u) |
Density ( |
|
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].
13.14 Let the atomic mass of a substance be = M and the density of the substance be =
Avogadro's number, N = 6.023
Volume of N number of molecules = ……. (i)
Volume of one mole of a substance = …… (ii)
Equating (i) and (ii), we get

13.13 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)/ kBT]
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:
n2 = n1 exp [ -mg NA ( - ) (h2 –h1)/ ( RT)]
where is the density of the suspended particle, and that of surrounding medium. [NA is Avogadro’s number, and R the universal gas constant.] [Hint : Use Archimedes principle to find the apparent weight of the suspended particle.]
13.13 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)/ kBT]
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:
n2 = n1 exp [ -mg NA ( - ) (h2 –h1)/ ( RT)]
where is the density of the suspended particle, and that of surrounding medium. [NA is Avogadro’s number, and R the universal gas constant.] [Hint : Use Archimedes principle to find the apparent weight of the suspended particle.]
13.13 According to law of atmospheres, we have
n2 = n1 exp [ -mg (h2 – h1)/ kBT] ….(i)
where is the number of density at height and is the number of density at height
mg is the weight of the particle suspended in the gas column
Density of the medium =
Density of the suspended particle
13.12 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.
[Hint : Use Graham’s law of diffusion: R1/R2 = ( M2 /M1 )1/2, where R1, R2 are diffusion rates of gases 1 and 2, and M1 and M2 their respective molecular masses. The law is a simple consequence of kinetic theory.]
13.12 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.
[Hint : Use Graham’s law of diffusion: R1/R2 = ( M2 /M1 )1/2, where R1, R2 are diffusion rates of gases 1 and 2, and M1 and M2 their respective molecular masses. The law is a simple consequence of kinetic theory.]
13.12 Rate of diffusion of hydrogen, = 28.7 cm3 s–1
Rate of diffusion of another gas, = 7.2 cm3 s–1
According to Graham's law of diffusion, we have:
= , where = molecular mass of hydrogen = 2.02 g and is the molecular mass of the unknown gas
= 2.02= 32.09 = Molecular mass of Oxygen
Henc
13.11 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?
13.11 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?
13.11 Length of the narrow bore, L = 1 m = 100 cm
Length of the mercury thread, l = 76 cm
Length of the air column between mercury and the closed end, = 15 cm
Since the bore is held vertically in air with the open end at the bottom, the mercury length that occupies the air space is 100 – (7
13.10 Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 . 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 = 28.0 u).
13.10 Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 . 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 = 28.0 u).
13.10 Pressure inside the cylinder containing nitrogen, P = 2.0 atm = 2 Pa
Temperature inside the cylinder, T = 17
Radius of nitrogen molecule, r = 1.0 Å = 1 m
Diameter of nitrogen molecule, d = 2 m
Molecular mass of nitrogen molecule, M = 28 u = 28 g (assume)