Kinetic Theory Exercise 02
Question:
A 1 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:
 The mercury thread will move down towards the bottom of the tube.
 The trapped air column will move down with the mercury thread.
 The air column will eventually reach the bottom of the tube and be released.
 The mercury thread will take up the entire length of the tube, filling it completely.
Question:
Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms : Substance Atomic Mass (u) Density (10^3 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 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 A˚].
Answer:
Step 1: Calculate the number of atoms in 1 m3 of each substance by using the density and atomic mass given.
Carbon (diamond): 2.22 x 10^3 kg/m3 x (1/12.01 u) = 183.6 x 10^27 atoms/m3 Gold: 19.32 x 10^3 kg/m3 x (1/197.00 u) = 98.0 x 10^27 atoms/m3 Nitrogen (liquid): 1.00 x 10^3 kg/m3 x (1/14.01 u) = 71.4 x 10^27 atoms/m3 Lithium: 0.53 x 10^3 kg/m3 x (1/6.94 u) = 76.2 x 10^27 atoms/m3 Fluorine (liquid): 1.14 x 10^3 kg/m3 x (1/19.00 u) = 60.2 x 10^27 atoms/m3
Step 2: Calculate the volume of one atom.
Carbon (diamond): 1/183.6 x 10^27 m3/atom = 5.44 x 10^28 m3/atom Gold: 1/98.0 x 10^27 m3/atom = 1.02 x 10^26 m3/atom Nitrogen (liquid): 1/71.4 x 10^27 m3/atom = 1.40 x 10^26 m3/atom Lithium: 1/76.2 x 10^27 m3/atom = 1.31 x 10^26 m3/atom Fluorine (liquid): 1/60.2 x 10^27 m3/atom = 1.66 x 10^26 m3/atom
Step 3: Calculate the approximate size of the atom using Avogadro’s number.
Carbon (diamond): 5.44 x 10^28 m3/atom x (6.022 x 10^23 atoms/mol) = 3.27 x 10^5 m3/mol Gold: 1.02 x 10^26 m3/atom x (6.022 x 10^23 atoms/mol) = 6.12 x 10^4 m3/mol Nitrogen (liquid): 1.40 x 10^26 m3/atom x (6.022 x 10^23 atoms/mol) = 8.43 x 10^4 m3/mol Lithium: 1.31 x 10^26 m3/atom x (6.022 x 10^23 atoms/mol) = 7.88 x 10^4 m3/mol Fluorine (liquid): 1.66 x 10^26 m3/atom x (6.022 x 10^23 atoms/mol) = 9.99 x 10^4 m3/mol
Step 4: Convert the volume of one atom to approximate size in angstroms.
Carbon (diamond): 3.27 x 10^5 m3/mol x (10^10 A˚/m3) = 3.27 x 10^5 A˚ Gold: 6.12 x 10^4 m3/mol x (10^10 A˚/m3) = 6.12 x 10^6 A˚ Nitrogen (liquid): 8.43 x 10^4 m3/mol x (10^10 A˚/m3) = 8.43 x 10^6 A˚ Lithium: 7.88 x 10^4 m3/mol x (10^10 A˚/m3) = 7.88 x 10^6 A˚ Fluorine (liquid): 9.99 x 10^4 m3/mol x (10^10 A˚/m3) = 9.99 x 10^6 A˚
Therefore, the approximate size of the atoms of the given substances is in the range of a few A˚.
Question:
From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm^3 s^−1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm^3 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:

Use Graham’s law of diffusion to calculate the ratio of the diffusion rates: R1/R2 = 28.7 cm^3 s^−1/7.2 cm^3 s^−1 = 4.

Use the equation to calculate the ratio of the molecular masses: M2/M1 = (R1/R2)^2 = (4)^2 = 16.

Use the relative atomic mass of hydrogen (2.016 g/mol) to calculate the molecular mass of the other gas: M2 = M1 x 16 = 2.016 g/mol x 16 = 32.256 g/mol.

Use a periodic table to identify the gas with a molecular mass of 32.256 g/mol. The gas is Oxygen (O2).
Question:
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 socalled law of atmospheres n2=n1exp[−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=n1exp[−mgNA(ρ−ρ′)(h2−h1)/(ρRT)] where ρ is the density of the suspended particle and ρ′ 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.]
Answer:
Step 1: Understand the question.
Step 2: Use Archimedes principle to find the apparent weight of the suspended particle.
Step 3: Substitute the apparent weight of the particle into the law of atmospheres.
Step 4: Simplify the equation to get the equation for sedimentation equilibrium of a suspension in a liquid column.