Chapter 13 Kinetic Theory Solutions
Question - 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?
Answer - 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, la= 15 cm
Since the bore is held vertically in air with the open end atthe bottom, the mercury length that occupies the air space is:
= 100 – (76 + 15)
= 9 cm
Therefore,
The total length of the air column = 15 + 9 = 24 cm
Let h cm of mercury flow out as a result of atmospheric pressure
So,
Length of the air column in the bore = 24 + h cm
And,
Length of the mercury column = 76 – h cm
Initial pressure, V1 = 15 cm3
Final pressure, P2 =76 – (76 – h)
= h cm of mercury
Final volume, V2 =(24 + h) cm3
Temperature remains constant throughout the process
Therefore,
P1V1 = P2V2
On substituting, we get,
76 x 15 = h (24 + h)
h2 + 24h – 11410= 0
On solving further, we get,
= 23.8 cm or -47.8 cm
Since height cannot be negative. Hence, 23.8 cm of mercury willflow out from the bore
Length of the air column = 24 + 23.8 = 47.8 cm
Question - 12 : - From a certainapparatus, 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 anaverage rate of 7.2 cm3 s-1. Identify the gas.
[Hint: Use Graham’s law ofdiffusion: R1/R2 = ( M2 /M1 )1/2, where R1 , R2 arediffusion rates of gases 1 and 2, and M1 and M2 their respective molecular masses. The law is asimple consequence of kinetic theory.]
Answer - 12 : -
Given
Rate of diffusion of hydrogen, R1 = 28.7 cm3s-1
Rate of diffusion of another gas, R2 = 7.2 cm3s-1
According to Graham’s Law of diffusion,
We have,
R1 /R2 = √M2 / M1
Where,
M1 isthe molecular mass of hydrogen = 2.020g
M2 isthe molecular mass of the unknown gas
Hence,
M2 =M1 (R1 / R2)2
= 2.02 (28.7 / 7.2)2
We get,
= 32.09 g
32 g is the molecular mass of oxygen.
Therefore, the unknown gas is oxygen.
Question - 13 : - A gas inequilibrium has uniform density and pressure throughout its volume. This isstrictly true only if there are no external influences. A gas column undergravity, for example, does not have a uniform density (and pressure). Asyou might expect, its density decreases with height. The precisedependence is given by the so-called law of atmospheres
n2= n1 exp[ -mg (h2 – h1)/ kBT]
where n2, n1 referto number density at heights h2 andh1
respectively. Use this relationto derive the equation for sedimentation equilibrium of a suspension in a
liquid column:
n2 = n1 exp[ -mg NA (ρ – ρ′) (h2–h1)/ (ρ RT)]
where ρ is the density of thesuspended particle, and ρ′ ,that of surrounding medium. [NA is Avogadro’snumber, and R the universal gas constant.] [Hint : Use Archimedes principleto find the apparent weight of the suspended particle.]
Answer - 13 : -
Law of atmosphere
n2= n1 exp [ -mg (h2 – h1)/ kBT] ———(1)
The suspended particle experiences an apparent weight because ofthe liquid displaced
According to Archimedes principle
Apparent weight = Weight of the water displaced – weight of thesuspended particle
= mg – m’g
= mg – Vρ’g = mg – (m/ρ) ρ’g
= mg (1 – (ρ’/ρ)) ——-(2)
ρ′= Density of the water
ρ = Density of the suspended particle
m′ = Mass of the suspended particle
m = Mass of the water displaced
V = Volume of a suspended particle
Boltzmann’s constant (K) = R/NA ——(3)
Substituting equation (2) and equation (3) in equation (1)
n2= n1 exp [ -mg (h2 – h1)/ kBT]
n2 =n1 exp [ -mg (1– ρ′/ ρ ) (h2–h1)NA / (RT)]
n2 =n1 exp [ -mg NA (ρ – ρ′ ) (h2–h1)/ (ρ RT)]
Question - 14 : - Given below aredensities of some solids and liquids. Give rough estimates of the size of theiratoms :
| Substance | Atomic Mass (u) | Density (103 Kg m-3) |
| 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 ofAvogadro’s number. You should, however, not take the actual numbers you obtainfor various atomic sizes too literally. Because of the crudeness of the tightpacking approximation, the results only indicate that atomic sizes are in therange of a few Å].
Answer - 14 : -
If r is the radius of the atom then the volume of each atom =(4/3)πr3
Volume of all the substance = (4/3)πr3 xN = M/ρ
M is the atomic mass of the substance
ρ is the density of the substance
One mole of the substance has 6.023 x 1023 atoms
r = (3M/4πρ x 6.023 x 1023)1/3
For carbon, M = 12. 01 x 10-3 kgand ρ = 2.22 x 103 kg m-3
R = (3 x 12. 01 x 10-3/4x 3.14 x 2.22 x 103 x 6.023 x 1023)1/3
= (36.03 x 10-3 /167.94x 1026)1/3
1.29 x 10 -10 m= 1.29 Å
For gold, M = 197 x 10-3 kgand ρ = 19. 32 x 103 kg m-3
R = (3 x 197 x 10-3/4x 3.14 x 19.32 x 103 x 6.023 x 1023)1/3
= 1.59 x 10 -10 m= 1.59 Å
For lithium, M = 6.94 x 10-3 kg and ρ = 0.53 x 103 kg/m3
R = (3 x 6.94 x 10-3/4x 3.14 x 0.53 x 103 x 6.023 x 1023)1/3
= 1.73 x 10 -10 m= 1.73 Å
For nitrogen (liquid), M = 14.01 x 10-3 kg and ρ = 1.00 x 103 kg/m3
R = (3 x 14.01 x 10-3/4x 3.14 x 1.00 x 103 x 6.023 x 1023)1/3
= 1.77 x 10 -10 m= 1.77 Å
For fluorine (liquid), M = 19.00 x 10-3 kg and ρ = 1.14 x 103 kg/m3
R = (3 x 19 x 10-3/4x 3.14 x 1.14 x 103 x 6.023 x 1023)1/3
= 1.88 x 10 -10 m= 1.88 Å