Kinetic Theory is a model that explains the macroscopic properties of matter, such as pressure, temperature, and volume, in terms of the microscopic behavior of its constituent particles (atoms or molecules).
Key Concepts:
Molecular Motion: The molecules of a gas are in constant, random motion.
Pressure: The pressure exerted by a gas is due to the collisions of its molecules with the walls of the container.
Temperature: The temperature of a gas is a measure of the average kinetic energy of its molecules.
Ideal Gas Equation: This equation relates the pressure, volume, temperature, and number of moles of an ideal gas: PV = nRT.
Specific Heat Capacity: This is the amount of heat energy required to raise the temperature of a unit mass of a substance by one degree Celsius.
Mean Free Path: The average distance traveled by a molecule between two successive collisions.
Law of Equipartition of Energy: This law states that the average kinetic energy associated with each degree of freedom of a molecule is (1/2)kT, where k is the Boltzmann constant and T is the absolute temperature.
By understanding these concepts, we can explain various phenomena like gas laws, thermal expansion, and heat conduction
1. 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 3 Å.
Ans : Understanding the Problem:
We are given the diameter of an oxygen molecule (d = 3 Å = 3 x 10^-10 m).
We need to calculate the ratio of the actual volume occupied by oxygen molecules to the molar volume of oxygen gas at STP.
Calculations:
1. Actual Volume of Oxygen Molecules:
Assuming oxygen molecules to be spherical, the volume of one molecule (V_molecule) is:
V_molecule = (4/3)πr³ = (4/3)π(d/2)³
The total volume of one mole of oxygen molecules (V_actual) is the product of the volume of one molecule and the Avogadro number (N_A):
V_actual = N_A × V_molecule = N_A × (4/3)π(d/2)³
Substituting the values:
V_actual = 6.023 × 10^23 × (4/3) × 3.14 × (1.5 × 10^-10 m)³
Calculating the value:
V_actual = 8.51 × 10^-6 m³ = 8.51 × 10^-3 liters
2. Molar Volume of Oxygen Gas at STP:
One mole of any gas occupies a volume of 22.4 liters at Standard Temperature and Pressure (STP).
3. Ratio of Actual to Molar Volume:
The ratio of the actual volume to the molar volume is:
Ratio = V_actual / V_molar = (8.51 × 10^-3 liters) / 22.4 liters
≈ 3.8 × 10^-4
Conclusion:
The ratio of the actual volume occupied by oxygen molecules to the molar volume of oxygen gas at STP is approximately 3.8 × 10^-4. This indicates that the actual volume occupied by the molecules is very small compared to the total volume of the gas
2. Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP : 1 atmospheric pressure, 0 °C). Show that it is 22.4 litres.
Ans : *Understanding the Problem:**
We need to show that 1 mole of an ideal gas occupies 22.4 liters at STP.
*Solution:
We can use the ideal gas equation to derive this:
PV = nRT
Where:
* P denote pressure (1 atm = 101325 Pa)
* V denote volume (in m³)
* n denote number of moles (1 mol)
* R denote universal gas constant (8.314 J/mol K)
* T denote temperature in Kelvin (0°C = 273.15 K)
Rearranging the equation to solve for V:
V = nRT / P
Substituting the values:
V = (1 mol)(8.314 J/mol K)(273.15 K) / (101325 Pa)
Calculating the volume:
V ≈ 0.0224 m³
Converting cubic meters to liters:
V ≈ 0.0224 m³ × (1000 L/m³)
= 22.4 L
Therefore, 1 mole of an ideal gas occupies 22.4 liters at STP.
3. Figure 12.8 shows plot of PV/T versus P for 1.00×10–3 kg of oxygen gas at two different temperatures. Fig. 12.8
(a) What does the dotted plot signify?
(b) Which is true: T1 > T2 or T1 < T2 ?
(c) What is the value of PV/T where the curves meet on the y-axis?
(d) If we obtained similar plots for 1.00×10–3 kg of hydrogen, would we get the same value of PV/T at the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of PV/T (for low pressure high temperature region of the plot) ? (Molecular mass of H2 = 2.02 u, of O2 = 32.0 u, R = 8.31 J mo1–1 K–1.)
Ans : Analyzing the PV/T vs P Plot for Oxygen Gas
Understanding the Plot:
The given plot shows the variation of PV/T with pressure (P) for a fixed mass of oxygen gas at two different temperatures, T₁ and T₂.
Answering the Questions:
(a)The dotted plot represents the ideal gas behavior, where PV/T is a constant. This is the ideal gas equation: PV = nRT. For a fixed mass of gas, nR is constant, so PV/T is constant.
(b)From the plot, we can see that the curve corresponding to T₁ is above the curve for T₂. This implies that at a given pressure, the value of PV/T is higher for T₁. Since PV/T is directly proportional to temperature, we can conclude that T₁ > T₂.
(c)At the y-axis, the pressure P is zero. So, the value of PV/T at this point is independent of the temperature and is solely determined by the mass of the gas and the gas constant R.
To calculate this value, we can use the ideal gas equation:
PV/T = nR
For 1.00 × 10⁻³ kg of oxygen gas:
n = mass / molar mass = (1.00 × 10⁻³ kg) / (32 × 10⁻³ kg/mol) = 3.125 × 10⁻⁵ mol
Therefore, PV/T = nR = (3.125 × 10⁻⁵ mol)(8.31 J/mol K) ≈ 2.6 × 10⁻⁴ J/K
(d) No, we would not get the same value of PV/T for hydrogen at the same point on the y-axis. The value of PV/T depends on the number of moles of the gas.
To find the mass of hydrogen that would yield the same value of PV/T, we can equate the expressions for PV/T for oxygen and hydrogen:
(n_O₂)(R) = (n_H₂)(R)
Where n_O₂ and n_H₂ are the number of moles of oxygen and hydrogen, respectively.
Since the R values are the same, we can equate the number of moles:
n_H₂ = n_O₂
Determine, mass of hydrogen:
mass_H₂ = n_H₂ × molar mass of H₂ = (3.125 × 10⁻⁵ mol) × (2.02 g/mol)
≈ 6.3 × 10⁻⁵ kg
Therefore, 6.3 × 10⁻⁵ kg of hydrogen would yield the same value of PV/T at the point where the curves meet on the y-axis..
4 An oxygen cylinder of volume 30 litre has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J mol–1 K–1, molecular mass of O2 = 32 u).
Ans : Understanding the Problem:
We’re given a cylinder of gas with an initial volume of 30 liters, an initial pressure of 15 atm, and an initial temperature of 27°C. The gas is allowed to cool and the pressure drops to 11 atm at a temperature of 17°C. We need to determine the mass of gas removed from the cylinder.
Calculations:
Initial State:
Initial volume (V₁): 30 liters = 30 x 10^-3 m³
Initial pressure (P₁): 15 atm = 15 x 1.013 x 10^5 Pa
Initial temperature (T₁): 27°C = 300 K
Initial number of moles (n₁):
n₁ = (P₁V₁) / (RT₁) = (15 x 1.013 x 10^5 x 30 x 10^-3) / (8.31 x 300) ≈ 18.3 moles
Final State:
Final pressure (P₂): 11 atm = 11 x 1.013 x 10^5 Pa
Final volume (V₂): 30 liters = 30 x 10^-3 m³
Final temperature (T₂): 17°C = 290 K
Final number of moles (n₂):
n₂ = (P₂V₂) / (RT₂)
= (11 x 1.013 x 10^5 x 30 x 10^-3) / (8.31 x 290) ≈ 13.9 moles
Number of Moles Removed:
Number of moles removed = n₁ – n₂ = 18.3 – 13.9 = 4.4 moles
Mass of Gas Removed:
Assuming the gas is oxygen (molar mass of 32 g/mol):
Mass of gas removed = 4.4 moles × 32 g/mol = 140.8 g
≈ 0.141 kg
Therefore, approximately 0.141 kg of oxygen gas was removed from the cylinder.
5. An air bubble of volume 1.0 cm3 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 ?
Ans : Understanding the Problem:
We have a bubble of air trapped inside a lake at a depth of 40 m.
We need to find the volume of the bubble when it reaches the surface of the lake.
Given Information:
Initial volume of the bubble (V₁) = 1 cm³ = 1 x 10⁻⁶ m³
Initial pressure (P₁) = Pressure of water + Atmospheric pressure
P₁ = ρgh + P₀ = 1000 kg/m³ × 9.8 m/s² × 40 m + 1.01 x 10⁵ Pa
≈ 4.93 x 10⁵ Pa
Initial temperature (T₁) = 12°C = 285 K
Final pressure (P₂) = Atmospheric pressure = 1.01 x 10⁵ Pa
Final temperature (T₂)
= 35°C
= 308 K
Using the Ideal Gas Law:
We can use the ideal gas law to relate the initial and final states of the bubble:
(P₁V₁) / T₁ = (P₂V₂) / T₂
To find final volume (V₂); Rearrange equation as
V₂ = (P₁V₁T₂) / (P₂T₁)
Substituting the given values:
V₂ = (4.93 x 10⁵ Pa × 1 x 10⁻⁶ m³ × 308 K) / (1.01 x 10⁵ Pa × 285 K)
≈ 5.3 x 10⁻⁶ m³
Therefore, the volume of the bubble when it reaches the surface of the lake is approximately 5.3 x 10⁻⁶ m³.
6. Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m3 at a temperature of 27 °C and 1 atm pressure.
Ans : Understanding the Problem:**
We’re asked to estimate the total number of air molecules in a room of a given volume at a specific temperature and pressure.
*Solution:
1. Use the Ideal Gas Law:
The ideal gas law is given by:
PV = nRT
Where:
* P denote pressure (1 atm = 101325 Pa)
* V denote volume (25 m³)
* n denote number of moles
* R denote gas constant (8.314 J/mol K)
* T denote temperature in Kelvin (27°C = 300 K)
2. Calculate the Number of Moles:
To find n, Rearrange ideal gas law as
n = PV / RT
Substituting the values:
n = (101325 Pa × 25 m³) / (8.314 J/mol K × 300 K) ≈ 1010 moles
3. Calculate the Number of Molecules:
One mole of any substance contains Avogadro’s number (6.022 × 10²³) of particles. Therefore, the total number of molecules in the room is:
Number of molecules = n × Avogadro’s number ≈ 1010 mol × 6.022 × 10²³ molecules/mol
≈ 6.07 × 10²⁶ molecules
Therefore, there are approximately 6.07 × 10²⁶ air molecules in the room.
7. 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).
Ans : The average thermal energy of a particle in a gas is given by the equipartition theorem:
Average kinetic energy per molecule = (3/2)kT
Where:
k denote Boltzmann constant (1.38 × 10⁻²³ J/K)
T denote temperature in Kelvin
Calculations:**
(i) Room Temperature (27°C = 300 K):
Average kinetic energy = (3/2)(1.38 × 10⁻²³ J/K)(300 K) ≈ 6.21 × 10⁻²¹ J
(ii) Surface of the Sun (6000 K):
Average kinetic energy = (3/2)(1.38 × 10⁻²³ J/K)(6000 K) ≈ 1.24 × 10⁻¹⁹ J
(iii) Core Temperature of a Star (10 million K):
Average kinetic energy = (3/2)(1.38 × 10⁻²³ J/K)(10⁷ K) ≈ 2.07 × 10⁻¹⁶ J
Therefore, the average thermal energy of a helium atom at room temperature, the surface of the Sun, and the core of a star are approximately 6.21 × 10⁻²¹ J, 1.24 × 10⁻¹⁹ J, and 2.07 × 10⁻¹⁶ J, respectively.
8. Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), 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 ?
Ans : According to Avogadro’s hypothesis, equal volumes of gases under the same conditions of temperature and pressure contain the same number of molecules. Therefore, the number of molecules in each case is the same.
Given,
root mean square (rms) velocity of gas molecules by following formula
v_rms = √(3kT/m)
where:
v_rms is the rms velocity
k is the Boltzmann constant
T is the temperature in Kelvin
m is the mass of one molecule
From the formula, we can see that the rms velocity is inversely proportional to the square root of the molecular mass. Since neon has the lowest atomic mass among the given gases, it will have the highest rms velocity.
9 .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 – 20 °C ? (atomic mass of Ar = 39.9 u, of He = 4.0 u).
Ans : Understanding the Problem:**
We’re given the temperature of helium gas and its rms speed. We need to find the temperature of argon gas at which its rms speed is equal to that of the helium gas.
Solution:
Givenm
root-mean-square speed of a gas molecule
vrms = √(3RT/M)
Where:
* R denote gas constant
* T = temperature in Kelvin
*(M)molar mass of the gas molecul
For the rms speeds of argon (Ar) and helium (He) to be equal, their vrms² values must be equal:
(3RT_Ar / M_Ar) = (3RT_He / M_He)
The R and 3 terms cancel out, leaving:
(T_Ar / M_Ar) = (T_He / M_He)
We can now substitute the given values and solve for T_Ar:
(T_Ar / 39.9) = ((-20 + 273.15) K / 4)
Solving for T_Ar:
T_Ar ≈ 2530 K
Therefore, the temperature of the argon gas should be approximately 2530 K for its atoms to have the same rms speed as helium atoms at -20°C.
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 0C. 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 N2 = 28.0 u).
Ans : Understanding the Problem:
We have a gas at a pressure of 2 atm and a temperature of 17°C.
The gas molecules have a radius of 1 Å.
We need to calculate the mean free path, root mean square velocity, and collision frequency of the gas molecules.
Calculations:
1. Volume per Molecule:
Using the ideal gas equation, we can calculate the volume occupied by one mole of the gas:
PV = nRT
Rearranging for volume (V):
V = (RT) / P
Substituting the given values:
V = (8.31 J/mol K × 290 K) / (2.026 × 10^5 Pa) ≈ 1.189 × 10^-2 m³
The molecular number density (n) is:
n = N_A / V = 6.023 × 10^23 / 1.189 × 10^-2 ≈ 5.06 × 10^25 m^-3
2. Mean Free Path:
The average distance a molecule travels between collisions is called the mean free path (λ).
λ = 1 / (√2πnd²)
Substituting the values:
λ = 1 / (√2π × 5.06 × 10^25 m^-3 × (2 × 1 × 10^-10 m)²) ≈ 1.1 × 10^-7 m
3. Root Mean Square Velocity:
given,
root-mean-square (rms) speed of gas molecules
v_rms = √(3RT / M)
Substituting the values:
v_rms = √(3 × 8.31 J/mol K × 290 K / (28 × 1.66 × 10^-27 kg)) ≈ 5.08 × 10^2 m/s
4. Collision Frequency:
The collision frequency (ν) is the number of collisions a molecule undergoes per second:
ν = v_rms / λ
Substituting the values:
ν = 5.08 × 10^2 m/s / 1.1 × 10^-7 m ≈ 4.62 × 10^9 s^-1
5. Time Between Collisions:
The time between successive collisions (τ) is the reciprocal of the collision frequency:
τ = 1 / ν = 1 / 4.62 × 10^9 s^-1 ≈ 2.17 × 10^-10 s
6. Collision Time:
The collision time (δt) is the time taken for a collision:
δt = d / v_rms = (2 × 1 × 10^-10 m) / (5.08 × 10^2 m/s) ≈ 3.92 × 10^-13 s
Conclusion:
The mean free path of the gas molecules is 1.1 × 10^-7 m.
The root mean square velocity is 5.08 × 10^2 m/s.
The collision frequency is 4.62 × 10^9 collisions per second.
The time between successive collisions is 2.17 × 10^-10 seconds.
The collision time is 3.92 × 10^-13 seconds.
