Updated By SaraNextGen

On March 11, 2024, 11:35 AM

Question 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?

Answer:

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 at the bottom, the mercury length that occupies the air space is: 100 – (76 + 15) = 9 cm

Hence, the total length of the air column = 15 + 9 = 24 cm

Let h cm of mercury flow out as a result of atmospheric pressure.

∴Length of the air column in the bore = 24 + h cm

And, length of the mercury column = 76 – h cm

Initial pressure, P₁ = 76 cm of mercury

Initial volume, 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.

∴P₁V₁ = P₂V₂

76 × 15 = h (24 + h)

h² + 24h – 1140 = 0

= 23.8 cm or –47.8 cm

Height cannot be negative. Hence, 23.8 cm of mercury will flow out from the bore and 52.2 cm of mercury will remain in it. The length of the air column will be 24 + 23.8 = 47.8 cm.

Question 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.]

Answer:

Rate of diffusion of hydrogen, R₁ = 28.7 cm³ s^–1

Rate of diffusion of another gas, R₂ = 7.2 cm³ s^–1

According to Graham’s Law of diffusion, we have:

Where,

M₁ is the molecular mass of hydrogen = 2.020 g

M₂ is the molecular mass of the unknown gas

= 32.09 g

32 g is the molecular mass of oxygen. Hence, the unknown gas is oxygen.

Question 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(ρ – P′) (h₂ –h₁)/ (ρRT)]

Where ρ is the density of the suspended particle, and ρ’ that of surrounding medium. [N_A is Avogadro’s number, and R the universal gas constant.] [Hint: Use Archimedes principle to find the apparent weight of the suspended particle.]

Answer:

According to the law of atmospheres, we have:

n₂ = n₁ exp [-mg (h₂ – h₁)/ kBT] … (i)

Where,

n₁ is thenumber density at height h₁, and n₂ is the number density at height h₂

mg is the weight of the particle suspended in the gas column

Density of the medium = ρ‘

Density of the suspended particle = ρ

Mass of one suspended particle = m‘

Mass of the medium displaced = m

Volume of a suspended particle = V

According to Archimedes’ principle for a particle suspended in a liquid column, the effective weight of the suspended particle is given as:

Weight of the medium displaced – Weight of the suspended particle

= mg – m‘g

Gas constant, R = kBN

 … (iii)

Substituting equation (ii) in place of mg in equation (i) and then using equation (iii), we get:

n₂ = n₁ exp [-mg (h₂ – h₁)/ kBT]

= n₁ exp [- (h₂ – h₁) ]

= n₁ exp [- (h₂ – h₁) ]

Question 13.14:

Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms:

[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].

Answer:

Density of the substance = ρAtomic mass of a substance = M

Avogadro’s number = N = 6.023 × 10²³

Volume of each atom 

Volume of N number of molecules  N … (i)

Volume of one mole of a substance =   … (ii)

N = 

For carbon:

M = 12.01 × 10^–3 kg,

ρ = 2.22 × 10³ kg m^–3

 = 1.29 Å

Hence, the radius of a carbon atom is 1.29 Å.

For gold:

M = 197.00 × 10^–3 kg

ρ = 19.32 × 10³ kg m^–3

 = 1.59 Å

Hence, the radius of a gold atom is 1.59 Å.

For liquid nitrogen:

M = 14.01 × 10^–3 kg

ρ = 1.00 × 10³ kg m^–3

 = 1.77 Å

Hence, the radius of a liquid nitrogen atom is 1.77 Å.

For lithium:

M = 6.94 × 10^–3 kg

ρ = 0.53 × 10³ kg m^–3

 = 1.73 Å

Hence, the radius of a lithium atom is 1.73 Å.

For liquid fluorine:

M = 19.00 × 10^–3 kg

ρ = 1.14 × 10³ kg m^–3

 = 1.88 Å

Hence, the radius of a liquid fluorine atom is 1.88 Å.