Question:

Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP : 1 atmospheric pressure, 0oC). Show that it is 22.4 litres.

Answer:

Step 1: Calculate the number of moles of a gas at standard temperature and pressure.

At STP, the number of moles of a gas is equal to 1 mole, since the temperature is 0oC and the pressure is 1 atmosphere.

Step 2: Calculate the volume of 1 mole of a gas at STP.

At STP, the volume of 1 mole of a gas is equal to 22.4 litres, since the ideal gas law states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

Step 3: Substitute the values into the ideal gas law equation.

PV = nRT

1 × V = 1 × 8.314 × 273.15

V = 22.4 litres

Therefore, the molar volume of a gas at STP is 22.4 litres.

Question:

An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27oC. After some oxygen is withdrawn from the cylinder the gauge pressure drops to 11 atm and its temperature drops to 17oC. Estimate the mass of oxygen taken out of the cylinder.(R=8.31Jmol^−1K^−1, molecular mass of O2​=32u).

Answer:

1. Calculate the number of moles of oxygen in the cylinder using the ideal gas law: n= PV/RT n = (15 atm x 30 L) / (8.31 J/molK x 300K) n = 0.86 mol

2. Calculate the mass of oxygen in the cylinder using the molar mass of oxygen: Mass = n x M Mass = 0.86 mol x 32 g/mol Mass = 27.52 g

3. Calculate the number of moles of oxygen after the pressure and temperature change: n = (11 atm x 30 L) / (8.31 J/molK x 290K) n = 0.78 mol

4. Calculate the mass of oxygen taken out of the cylinder: Mass taken out = (0.86 mol - 0.78 mol) x 32 g/mol Mass taken out = 2.56 g

Question:

Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m^3 at a temperatured of 27 ∘C and 1 atm pressure.

Answer:

1. Calculate the total number of moles of air molecules in the room: n = PV/RT n = (1 atm)(25.0 m^3)/(8.314 J/mol K)(300 K) n = 0.836 mol

2. Calculate the total number of air molecules in the room: N = nN_A N = (0.836 mol)(6.022 x 10^23 molecules/mol) N = 5.02 x 10^24 molecules

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:

1. The mercury thread will move down the tube, leaving the 15 cm column of air in the top part of the tube.

2. The mercury thread will come to rest at the bottom of the tube, leaving the 15 cm column of air trapped in the top part of the tube.

3. As the tube is held vertically, the air pressure in the tube increases due to the weight of the mercury thread.

4. This increased air pressure will push the 15 cm column of air out of the tube.

Question:

Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms : Substance Atomic Mass (u) Density (103 kg m^−3 Carbon (diamond) 12.01 2. 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:

Answer:

The size of the atoms can be estimated by using the known value of Avogadro’s number. Assuming the atoms to be tightly packed in a solid or liquid phase, the approximate size of the atoms can be calculated as follows:

Carbon (diamond): Size of the atom is approximately 0.3 A˚.

Gold: Size of the atom is approximately 0.3 A˚.

Nitrogen (liquid): Size of the atom is approximately 0.3 A˚.

Lithium: Size of the atom is approximately 0.2 A˚.

Fluorine (liquid): Size of the atom is approximately 0.3 A˚.

Question:

An air bubble of volume 1.0cm^3 rises from the bottom of a lake 40 m deep at a temperature of 12∘C. To what volume does it grow when it reaches the surface which is at a temperature of 35∘C ?

Answer:

1. Calculate the initial pressure of the air bubble at the bottom of the lake: P = density of water x gravity x depth P = 1000kg/m^3 x 9.81m/s^2 x 40m P = 392400 Pa

2. Calculate the final pressure of the air bubble at the surface: P = density of air x gravity x 0 P = 1.225kg/m^3 x 9.81m/s^2 x 0 P = 0 Pa

3. Calculate the initial temperature of the air bubble at the bottom of the lake: T1 = 12°C

4. Calculate the final temperature of the air bubble at the surface: T2 = 35°C

5. Calculate the ratio of the volume of the air bubble at the surface to the initial volume at the bottom of the lake: V2/V1 = (T2/T1) (P2/P1) V2/V1 = (35°C/12°C) (0/392400) V2/V1 = 0.0014

6. Calculate the volume of the air bubble at the surface: V2 = V1 x V2/V1 V2 = 1.0 cm^3 x 0.0014 V2 = 0.0014 cm^3

Question:

Estimate the average thermal energy of a helium atom at (i) room temperature (27 ∘C), (ii) the temperature on the surface of the Sun (6000 K), (iii) the temperature of 10 million kelvin (the typical core temperature in the case of a star).

Answer:

(i) Room temperature (27 ∘C): Average thermal energy of a helium atom at room temperature = 3/2 kT = (3/2) x (1.38 x 10^-23 J/K) x (300 K) = 2.07 x 10^-21 J

(ii) The temperature on the surface of the Sun (6000 K): Average thermal energy of a helium atom at the temperature on the surface of the Sun = 3/2 kT = (3/2) x (1.38 x 10^-23 J/K) x (6000 K) = 4.14 x 10^-20 J

(iii) The temperature of 10 million kelvin (the typical core temperature in the case of a star): Average thermal energy of a helium atom at the temperature of 10 million kelvin = 3/2 kT = (3/2) x (1.38 x 10^-23 J/K) x (10 x 10^6 K) = 6.9 x 10^-17 J

Question:

Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monoatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal number of respective molecules? Is the root mean square speed of molecules the same in the three cases? If not, in which case is vrms​ the largest?

Answer:

1. No, the vessels do not contain equal number of respective molecules. The first vessel contains only monoatomic molecules (neon), the second contains diatomic molecules (chlorine), and the third contains polyatomic molecules (uranium hexafluoride).

2. No, the root mean square speed of molecules is not the same in the three cases. The root mean square speed of molecules will be the largest in the vessel containing the monoatomic molecules (neon). This is because monoatomic molecules have the least number of atoms, and thus, the least amount of mass. Therefore, the molecules in the vessel containing neon will have the highest root mean square speed.

Question:

At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at −20oC?(Atomic mass of Ar=39u and He=400u). A 2.52×10^3K B 2.52×10^2K C 4.03×10^3K D 4.03×10^2K

Answer:

Step 1: Calculate the root mean square speed of an atom in an argon gas cylinder at -20oC.

Step 2: Calculate the root mean square speed of a helium gas atom at -20oC.

Step 3: Set the two root mean square speeds equal to each other and solve for the temperature.

Answer: C 4.03×10^3K

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:

Step 1: Calculate the ratio of the diffusion rates of the two gases: R1​/R2​= 28.7 cm^3 s^−1/7.2 cm^3 s^−1 = 4

Step 2: Calculate the ratio of the molecular masses of the two gases: M2​/M1 = (4)^2 = 16

Step 3: Use the periodic table to identify the gas with the molecular mass of 16. 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 so-called law of atmospheres n2​=n1​exp[−mg(h2​−h1​)/kB​T] 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[−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: Use Archimedes principle to find the apparent weight of the suspended particle.

The apparent weight of the suspended particle is equal to the weight of the displaced liquid, which is equal to the product of the volume of the particle, the density of the liquid, and the acceleration due to gravity.

Apparent weight = Volume of particle x Density of liquid x Acceleration due to gravity

Apparent weight = Vρg

Step 2: Use the law of atmospheres to derive the equation for sedimentation equilibrium of a suspension in a liquid column.

The law of atmospheres states that the density of a gas at a certain height is equal to the density at a lower height multiplied by the exponential of the product of the mass of the gas particle, the acceleration due to gravity, and the difference in heights.

n2 = n1exp[−mg(h2−h1)/kB​T]

Since the suspended particle is in a liquid column, we can replace the mass of the gas particle with the apparent weight of the suspended particle, and the density of the gas with the density of the suspended particle, and the density of the surrounding medium.

n2 = n1exp[−Vρg(h2−h1)/(ρRT)]

Therefore, the equation for sedimentation equilibrium of a suspension in a liquid column is:

n2​=n1​exp[−mgNA​(ρ−ρ′)(h2​−h1​)/(ρRT)]

Question:

Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 oC. Take the radius of a nitrogen molecule to be roughly 1.0 A. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N2​ = 28.0 u)

Answer:

1. Calculate the density of nitrogen in the cylinder: Density = Pressure/(Ideal Gas Constant x Temperature) Density = 2.0 atm/(0.082 atm L/mol K x 290 K) Density = 0.094 mol/L

2. Calculate the average distance between nitrogen molecules: Average distance = (1/density)^(1/3) Average distance = (1/0.094 mol/L)^(1/3) Average distance = 4.8 A

3. Calculate the mean free path of a nitrogen molecule: Mean free path = Average distance - 2 x Radius of nitrogen molecule Mean free path = 4.8 A - 2 x 1.0 A Mean free path = 2.8 A

4. Calculate the collision frequency of a nitrogen molecule: Collision frequency = 1/(Mean free path x density) Collision frequency = 1/(2.8 A x 0.094 mol/L) Collision frequency = 0.033 collisions/A

5. Calculate the time the molecule moves freely between two successive collisions: Time = Mean free path/Average Velocity Average Velocity = (3RT/M)^(1/2) Average Velocity = (3 x 0.082 atm L/mol K x 290 K/28.0 u)^(1/2) Average Velocity = 5.1 x 10^5 cm/s Time = 2.8 A/5.1 x 10^5 cm/s Time = 5.5 x 10^-7 s

Question:

Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3A˚.

Answer:

Step 1: Calculate the volume of an oxygen molecule. This can be done by using the formula for the volume of a sphere, V = 4/3πr3, where r is the radius of the oxygen molecule.

Step 2: Calculate the actual volume occupied by oxygen gas at STP. This can be done by using the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature.

Step 3: Divide the volume of the oxygen molecule by the actual volume occupied by oxygen gas at STP. This will give the fraction of molecular volume to the actual volume occupied by oxygen gas at STP.