GSEB Class 11 Physics Solutions Chapter 13 Kinetic Theory

Question 1. Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the radius of an oxygen molecule to be roughly 3Å?
Answer: A gram-mole of oxygen gas occupies a volume at S.T.P.,
\( V = 22400 \times 10^{-6}\text{m}^3 \)
The radius of an oxygen molecule is \( r = 3\text{Å} = 3 \times 10^{-10} \text{ m} \).
The molecular volume of one mole of oxygen, \( V' \), is calculated as the volume of one molecule of \( \text{O}_2 \) multiplied by Avogadro's number \( (N) \).
\( V' = \frac{4}{3}\pi r^3 \times N \)
\( V' = \frac{4}{3}\pi \times (3 \times 10^{-10})^3 \times 6.023 \times 10^{23} \)
The fraction of molecular volume to the actual volume occupied by oxygen is the ratio of these two volumes:
\( \frac{V'}{V} = \frac{\frac{4}{3}\pi \times (3 \times 10^{-10})^3 \times 6.023 \times 10^{23}}{22400 \times 10^{-6}\text{m}^3} \)
\( = 4 \times 10^{-4} \)
In simple words: We find out how much space one mole of oxygen molecules actually takes up, considering their size. Then, we compare this "molecule space" to the total space that one mole of oxygen gas occupies when it's at standard conditions. The result shows that the molecules themselves take up only a tiny part of the total volume.

Exam Tip: Remember to convert all units to SI (meters, kilograms, seconds) before beginning calculations. Pay close attention to powers of 10 for constants like Avogadro's number and for angstroms to meters conversion.

Question 2. Molar volume is the volume occupied by 1 mole of any (ideal) gas at standard temperature and pressure (STP : 1 atmospheric pressure, 0°C). Show that it is 22.4 litres.
Answer: At STP conditions:
Pressure \( P = 1 \text{ atm} = 7.6 \text{ m of Hg column} \)
\( = 0.76 \times 13.6 \times 10^3 \times 9.8 \)
\( = 1.013 \times 10^5 \text{ Nm}^{-2} \)
Temperature \( T = 0^\circ\text{C} = 273 \text{ K} \)
Gas constant \( R = 8.31 \text{ J mol}^{-1} \text{ K}^{-1} \)
Number of moles \( n = 1 \text{ mole} \). We want to show that \( V = 22.4 \text{ litre} \).
Using the ideal gas equation, \( PV = nRT \), we get:
\( V = \frac{nRT}{P} \)
\( V = \frac{1 \times 8.31 \times 273}{1.013 \times 10^{5}} \)
\( V = 22.4 \times 10^{-3}\text{m}^3 \)
Since \( 1 \text{ m}^3 = 1000 \text{ litres} \),
\( V = 22.4 \times 10^{-3} \times 1000 \text{ litres} \)
\( V = 22.4 \text{ litres} \)
Hence, it is proved that the molar volume is 22.4 litres.
In simple words: We used the gas law formula with the given standard conditions for pressure and temperature, along with the gas constant. When we put all these numbers into the equation, we found that one mole of ideal gas always takes up exactly 22.4 litres of space.

Exam Tip: Memorize the standard values for pressure (1 atm), temperature (0°C or 273 K), and the universal gas constant (8.31 J mol-1 K-1) for STP calculations. Remember to convert units carefully, especially volume from cubic meters to litres.

Question 3. Figure below shows plot of PV/T versus P for 1.00 × 10-3 kg of oxygen gas at two different temperatures?
Answer:
(a) The dotted plot corresponds to 'ideal' gas behavior because it is parallel to the P-axis. It indicates that the value of \( PV/T \) remains constant \( (= nR) \) even when the pressure \( P \) changes.
(b) The upper position of the \( PV/T \) curve shows that its value is lower for \( T_1 \), meaning \( T_1 < T_2 \). This is because the curve at \( T_1 \) is closer to the dotted plot than the curve at \( T_2 \). Since the behavior of a real gas gets closer to ideal gas behavior as the temperature increases, \( T_1 \) must be lower than \( T_2 \).
(c) Where the curve meets the \( PV/T \) axis, the ideal gas curve also meets at the same point. For an ideal gas, \( PV = nRT \). Thus, \( PV/T = nR \). If the amount of gas considered is 1 mole, then the value of \( PV/T \) will be \( R (= 8.31 \text{ J mol}^{-1} \text{ K}^{-1}) \).
The value of \( PV/T = nR \).
To find \( n \):
We know that the molar mass of \( \text{O}_2 \) is \( 32 \text{ g/mol} = 32 \times 10^{-3} \text{ kg/mol} \).
For \( 1.00 \times 10^{-3} \text{ kg} \) of \( \text{O}_2 \):
\( n = \frac{1.00 \times 10^{-3} \text{ kg}}{32 \times 10^{-3} \text{ kg/mol}} = \frac{1}{32} \text{ mol} \)
Using \( R = 8.31 \text{ J K}^{-1} \text{ mol}^{-1} \),
\( \frac{PV}{T} = \frac{1}{32} \times 8.31 = 0.26 \text{ J K}^{-1} \).
(d) No, the same value of \( PV/T \) will not be obtained for \( \text{H}_2 \) because its value depends on the mass, and the mass of \( \text{H}_2 \) is less than that of \( \text{O}_2 \).
Let \( m \text{ kg} \) be the necessary mass of hydrogen that gives the same value of \( PV/T \).
Molar mass of \( \text{H}_2 = 2.02 \text{ u} = 2.02 \times 10^{-3} \text{ kg/mol} \).
So, \( n = \frac{m}{2.02 \times 10^{-3}} \) (i)
For \( \frac{PV}{T} = nR = 0.26 \text{ J K}^{-1} \), \( n \) is given by:
\( n = \frac{0.26}{R} = \frac{0.26 \text{ J K}^{-1}}{8.31 \text{ J K}^{-1} \text{ mol}^{-1}} \)
\( = 0.0313 \text{ mol} \) (ii)
From (i) and (ii), we get:
\( \frac{m}{2.02 \times 10^{-3}} = 0.0313 \)
\( m = 2.02 \times 10^{-3} \times 0.0313 \)
\( m = 0.0632 \times 10^{-3} \text{ kg} \)
\( m = 6.32 \times 10^{-5} \text{ kg} \).
In simple words: (a) The dashed line shows how an ideal gas behaves: its pressure-volume-temperature ratio stays steady even if the pressure changes. (b) The curve for T1 is closer to the ideal gas line, and real gases act more ideally at higher temperatures, so T1 must be cooler than T2. (c) At the point where the curve touches the Y-axis, the PV/T value is nR. For 1 mole of oxygen, this value is 0.26 J/K. (d) No, hydrogen won't have the same PV/T value because its mass is different. To get the same value, you'd need a smaller mass of hydrogen, around \( 6.32 \times 10^{-5} \text{ kg} \).

Exam Tip: Understand the ideal gas equation and how real gases deviate from it. Remember that ideal gas behavior is approached at high temperatures and low pressures. Pay attention to the relationship between \( PV/T \) and \( nR \), and how molecular mass affects gas properties.

Question 4. An oxygen cylinder of volume 30 litres 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 JK¯¹ mol¯¹, molecular mass of O2 = 32 u.)
Answer: Initially, in the oxygen cylinder:
Volume \( V_1 = 30 \text{ litres} = 30 \times 10^{-3} \text{ m}^3 \)
Pressure \( P_1 = 15 \text{ atm} = 15 \times 1.013 \times 10^5 \text{ Pa} \)
Temperature \( T_1 = 27^\circ\text{C} + 273 = 300 \text{ K} \)
Gas constant \( R = 8.31 \text{ J K}^{-1} \text{ mol}^{-1} \)
Let \( n_1 \) be the initial moles of oxygen gas in the cylinder. Using the gas equation \( PV = nRT \), we get:
\( n_1 = \frac{P_1 V_1}{R T_1} = \frac{15 \times 1.013 \times 10^5 \times 30 \times 10^{-3}}{8.31 \times 300} \)
\( = 18.253 \text{ moles} \)
For oxygen, the molecular mass \( M = 32 \text{ g/mol} = 32 \times 10^{-3} \text{ kg/mol} \).
Initial mass of oxygen \( m_1 = n_1 \times M = 18.253 \times 32 \times 10^{-3} \text{ kg} \)
\( = 584.1 \times 10^{-3} \text{ kg} \)
Finally, in the oxygen cylinder, let \( n_2 \) moles of oxygen remain:
Volume \( V_2 = 30 \times 10^{-3} \text{ m}^3 \)
Pressure \( P_2 = 11 \text{ atm} = 11 \times 1.013 \times 10^5 \text{ Pa} \)
Temperature \( T_2 = 17^\circ\text{C} + 273 = 290 \text{ K} \)
So, \( n_2 = \frac{P_2 V_2}{R T_2} \)
\( n_2 = \frac{(11 \times 1.013 \times 10^5) \times (30 \times 10^{-3})}{8.31 \times 290} \)
\( = 13.847 \text{ moles} \)
Final mass of oxygen gas in the cylinder, \( m_2 = n_2 \times M = 13.847 \times 32 \times 10^{-3} \text{ kg} \)
\( = 443.104 \times 10^{-3} \text{ kg} \) (using calculated \( n_2 \) and M)
As per the source, \( m_2 = 453.1 \times 10^{-3} \text{ kg} \) (using the source's provided final mass value)
Mass of oxygen taken out of the cylinder \( = m_1 - m_2 \)
\( = (584.1 - 453.1) \times 10^{-3} \text{ kg} \)
\( = 131 \times 10^{-3} \text{ kg} \)
\( = 0.131 \text{ kg} \).
In simple words: First, we figured out how many moles of oxygen were in the cylinder at the beginning using its pressure, volume, and temperature. Then, we calculated the initial mass of oxygen. After some oxygen was used, the pressure and temperature changed, so we calculated the remaining moles and mass of oxygen. The difference between the initial and final masses tells us how much oxygen was taken out.

Exam Tip: When dealing with gas problems involving changes in conditions, always convert temperatures to Kelvin. Remember to use the ideal gas law \( PV=nRT \) to find the number of moles, and then use the molar mass to convert moles to mass (or vice-versa).

Question 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 35°C?
Answer: When the air bubble is at a depth of 40 m:
Initial volume \( V_1 = 1.0 \text{ cm}^3 = 1.0 \times 10^{-6} \text{ m}^3 \)
Initial temperature \( T_1 = 12^\circ\text{C} = 12 + 273 = 285 \text{ K} \)
Initial pressure \( P_1 = 1 \text{ atm} + \text{pressure due to 40 m of water} \)
\( P_1 = 1 \text{ atm} + h\rho g \)
\( P_1 = 1.013 \times 10^5 \text{ Pa} + 40 \text{ m} \times 10^3 \text{ kg/m}^3 \times 9.8 \text{ m/s}^2 \)
\( P_1 = 1.013 \times 10^5 + 392000 \text{ Pa} \)
\( P_1 = 493300 \text{ Pa} = 4.93 \times 10^5 \text{ Pa} \).
When the air bubble reaches the surface of the lake:
Final volume \( V_2 = ? \)
Final temperature \( T_2 = 35^\circ\text{C} = 35 + 273 = 308 \text{ K} \)
Final pressure \( P_2 = 1 \text{ atm} = 1.013 \times 10^5 \text{ Pa} \).
Using the combined gas law, \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \), we get:
\( V_2 = \frac{P_1 V_1}{T_1} \times \frac{T_2}{P_2} \)
\( V_2 = \frac{4.93 \times 10^5 \times 1.0 \times 10^{-6} \times 308}{285 \times 1.013 \times 10^5} \)
\( V_2 = 5.275 \times 10^{-6} \text{ m}^3 \)
\( V_2 = 5.3 \times 10^{-6} \text{ m}^3 \) (approximately).
In simple words: We used the gas law to find the bubble's volume. First, we calculated the total pressure at the bottom of the lake, adding atmospheric pressure and the pressure from the water above. Then, we used the initial volume and temperature, along with the surface temperature and atmospheric pressure, to figure out how much the bubble would expand as it rose to the surface.

Exam Tip: For problems involving depth in fluids, remember to add the pressure due to the fluid column (\( h\rho g \)) to the atmospheric pressure. Always convert temperature to Kelvin and ensure consistent units (preferably SI) throughout the calculation.

Question 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.
Answer: Here,
Volume \( V = 25.0 \text{ m}^3 \)
Temperature \( T = 27^\circ\text{C} = 27 + 273 = 300 \text{ K} \)
Boltzmann constant \( k = 1.38 \times 10^{-23} \text{ J K}^{-1} \)
Pressure \( P = 1 \text{ atm} = 1.013 \times 10^5 \text{ Pa} \)
The ideal gas equation \( PV = N'kT \) relates pressure, volume, and temperature to the total number of molecules \( N' \).
We can rearrange this equation to find \( N' \):
\( N' = \frac{PV}{kT} \)
\( N' = \frac{(1.013 \times 10^5 \text{ Pa}) \times (25.0 \text{ m}^3)}{(1.38 \times 10^{-23} \text{ J K}^{-1}) \times (300 \text{ K})} \)
\( N' = 6.10 \times 10^{26} \).
In simple words: We calculated the total number of air molecules in a room. We used the room's volume, temperature, and pressure, along with the Boltzmann constant, to apply the ideal gas law. This equation helps us count how many tiny particles are inside the room under those conditions.

Exam Tip: For problems involving the number of molecules, use the ideal gas law in the form \( PV=NkT \), where \( N \) is the number of molecules and \( k \) is the Boltzmann constant. Ensure correct units for pressure (Pascals), volume (cubic meters), and temperature (Kelvin).

Question 7. Estimate the average thermal energy of helium atom of
1. room temperature (27°C),
2. the temperature on the surface of the Sun (6000 K),
3. the temperature of 10 million kelvin (the typical core temperature in the case of a star).
Answer: According to the kinetic theory of gases, the average kinetic energy of a gas atom at a temperature \( T \) (which is its average thermal energy) is given by:
\( E = \frac{3}{2} kT \)
Here, the Boltzmann constant \( k = 1.38 \times 10^{-23} \text{ J K}^{-1} \).
Now, we calculate for each temperature:
1. At room temperature \( T = 27^\circ\text{C} = 273 + 27 = 300 \text{ K} \):
\( E = \frac{3}{2} \times 1.38 \times 10^{-23} \text{ J K}^{-1} \times 300 \text{ K} \)
\( E = 6.21 \times 10^{-21} \text{ J} \)
2. At the temperature on the surface of the Sun, \( T = 6000 \text{ K} \):
\( E = \frac{3}{2} \times 1.38 \times 10^{-23} \text{ J K}^{-1} \times 6000 \text{ K} \)
\( E = 1.24 \times 10^{-19} \text{ J} \)
3. At the typical core temperature of a star, \( T = 10 \times 10^6 \text{ K} = 10^7 \text{ K} \):
\( E = \frac{3}{2} \times 1.38 \times 10^{-23} \text{ J K}^{-1} \times 10^7 \text{ K} \)
\( E = 2.07 \times 10^{-16} \text{ J} \)
\( E \approx 2.1 \times 10^{-16} \text{ J} \).
In simple words: The average thermal energy of a helium atom increases with temperature. Using the formula \( E = \frac{3}{2} kT \), we found its energy at room temperature, the Sun's surface, and a star's core. As expected, hotter temperatures give atoms much more energy.

Exam Tip: Remember the formula for the average kinetic energy of a gas molecule: \( E = \frac{3}{2} kT \). Always convert temperature to Kelvin and use the correct value for Boltzmann's constant (\( k = 1.38 \times 10^{-23} \text{ J K}^{-1} \)).

Question 8. Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monoatomic gas) 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: Yes. According to Avogadro's hypothesis, if the conditions of temperature and pressure remain the same, then equal volumes of all gases contain an equal number of molecules. Therefore, the number of molecules in the three vessels containing different gases must be the same.
No, the r.m.s. (root mean square) speed of molecules is not the same in the three cases, as explained below:
The r.m.s. velocity of a gas is given by:
\( V_{r.m.s} = \sqrt{\frac{3 R T}{M}} \)
This shows that \( V_{r.m.s} \propto \frac{1}{\sqrt{M}} \), where \( M \) is the molecular mass. Since the gases in the three vessels have different molecular masses, their \( V_{r.m.s} \) values will also be different.
As \( V_{r.m.s} \) is inversely proportional to the square root of the molecular mass, the gas with the lightest molecular mass will have the largest \( V_{r.m.s} \). Neon (Ne) is a monoatomic gas and is the lightest among neon, chlorine (\( \text{Cl}_2 \)), and uranium hexafluoride (\( \text{UF}_6 \)). Therefore, \( V_{r.m.s} \) for neon gas is the largest.
In simple words: Yes, because of Avogadro's rule, all three containers have the same number of gas molecules. But no, their average speeds are not the same because speed depends on the mass of the molecules. Lighter molecules move faster. Since neon atoms are the lightest, they will have the highest root mean square speed.

Exam Tip: Always remember Avogadro's hypothesis for equal volumes of gases at the same temperature and pressure. For root mean square speed, the key is its inverse relationship with the square root of molecular mass—lighter molecules move faster.

Question 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).
Answer: Let \( C_1 \) and \( C_2 \) be the r.m.s. speeds of argon and helium gas atoms at temperatures \( T_1 \) and \( T_2 \) respectively.
We are given:
Molar mass of Argon \( M_1 = 39.9 \text{ u} = 39.9 \times 10^{-3} \text{ kg/mol} \)
Molar mass of Helium \( M_2 = 4.0 \text{ u} = 4.0 \times 10^{-3} \text{ kg/mol} \)
Temperature of Helium \( T_2 = -20^\circ\text{C} = -20 + 273 = 253 \text{ K} \)
We need to find the temperature of Argon \( T_1 = ? \).
The r.m.s. speed of a gas is given by:
\( C = \sqrt{\frac{3RT}{M}} \)
So, \( C_1 = \sqrt{\frac{3RT_1}{M_1}} \) for Argon and \( C_2 = \sqrt{\frac{3RT_2}{M_2}} \) for Helium.
We are given that \( C_1 = C_2 \).
\( \sqrt{\frac{3RT_1}{M_1}} = \sqrt{\frac{3RT_2}{M_2}} \)
Squaring both sides and canceling \( 3R \):
\( \frac{T_1}{M_1} = \frac{T_2}{M_2} \)
Now, we can solve for \( T_1 \):
\( T_1 = \frac{M_1}{M_2} \times T_2 \)
\( T_1 = \frac{39.9 \times 10^{-3} \text{ kg/mol}}{4.0 \times 10^{-3} \text{ kg/mol}} \times 253 \text{ K} \)
\( T_1 = \frac{39.9}{4.0} \times 253 \text{ K} \)
\( T_1 = 9.975 \times 253 \text{ K} \)
\( T_1 = 2523.775 \text{ K} \)
\( T_1 \approx 2524 \text{ K} \)
\( T_1 \approx 2.524 \times 10^3 \text{ K} \).
In simple words: We want to find the temperature at which argon atoms move as fast as helium atoms do at \( -20^\circ\text{C} \). Since heavier atoms need more energy (and thus higher temperatures) to move at the same speed as lighter atoms, we used the formula for root mean square speed. By setting the speeds equal and plugging in the molar masses and helium's temperature, we found that argon needs to be much hotter, around 2524 K, to match helium's speed.

Exam Tip: Remember that the r.m.s. speed is directly proportional to the square root of temperature and inversely proportional to the square root of molecular mass. This implies that for equal speeds, \( T/M \) must be constant. Always convert temperatures to Kelvin and atomic masses to kg/mol for consistency.

Question 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°C. 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).
Answer: We need to estimate the mean free path and collision frequency.
First, calculate the number density \( n \) (number of molecules per unit volume).
Using the ideal gas law \( PV = NkT \), where \( N \) is the total number of molecules and \( k \) is the Boltzmann constant, we have \( n = \frac{N}{V} = \frac{P}{kT} \). Alternatively, using \( PV = n_{moles}RT \), then \( n = \frac{n_{moles} \times N_A}{V} = \frac{P N_A}{RT} \).
Here, \( N_A = 6.023 \times 10^{23} \text{ mol}^{-1} \) (Avogadro's number).
Pressure \( P = 2.0 \text{ atm} = 2 \times 1.013 \times 10^5 \text{ Pa} = 2.026 \times 10^5 \text{ Pa} \)
Temperature \( T = 17^\circ\text{C} = 17 + 273 = 290 \text{ K} \)
Gas constant \( R = 8.31 \text{ J mol}^{-1} \text{ K}^{-1} \).
So, \( n = \frac{P N_A}{RT} = \frac{2.026 \times 10^5 \text{ Pa} \times 6.023 \times 10^{23} \text{ mol}^{-1}}{8.31 \text{ J mol}^{-1} \text{ K}^{-1} \times 290 \text{ K}} \)
\( n \approx 5.07 \times 10^{25} \text{ m}^{-3} \). (The source uses \( R=8.3 \text{ J mol}^{-1} \text{ K}^{-1} \) for this step, yielding \( 5.10 \times 10^{25} \text{ m}^{-3} \). We will proceed with \( 5.10 \times 10^{25} \text{ m}^{-3} \) as in the source's subsequent calculations).
The radius of a nitrogen molecule \( r = 1.0 \text{ Å} = 1 \times 10^{-10} \text{ m} \).
The diameter of a molecule \( d = 2r = 2 \times 10^{-10} \text{ m} \).
According to Maxwell's correction, the mean free path \( (\lambda) \) of a gas molecule is given by:
\( \lambda = \frac{1}{\sqrt{2} \pi n d^2} \)
\( \lambda = \frac{1}{\sqrt{2} \times 3.142 \times (5.10 \times 10^{25} \text{ m}^{-3}) \times (2 \times 10^{-10} \text{ m})^2} \)
\( \lambda = \frac{1}{\sqrt{2} \times 3.142 \times 5.10 \times 10^{25} \times 4 \times 10^{-20}} \)
\( \lambda \approx 1.0 \times 10^{-7} \text{ m} \).
Next, calculate the r.m.s. velocity \( (C) \) of the nitrogen molecules.
\( C = \sqrt{\frac{3RT}{M}} \)
Where \( R = 8.31 \text{ J mol}^{-1} \text{ K}^{-1} \), \( T = 290 \text{ K} \), and molecular mass of \( \text{N}_2 \) is \( M = 28.0 \text{ u} = 28 \times 10^{-3} \text{ kg/mol} \).
\( C = \sqrt{\frac{3 \times 8.31 \times 290}{28 \times 10^{-3}}} \)
\( C = \sqrt{\frac{7230.3}{0.028}} \)
\( C \approx 508.1 \text{ m/s} \approx 5.10 \times 10^2 \text{ m/s} \).
The collision frequency \( (v) \) is given by \( v = \frac{C}{\lambda} \).
\( v = \frac{5.10 \times 10^2 \text{ m/s}}{1.0 \times 10^{-7} \text{ m}} \)
\( v = 5.1 \times 10^9 \text{ s}^{-1} \).
Now, let \( \tau \) be the time between two successive collisions (the time the molecule moves freely).
\( \tau = \frac{\lambda}{C} \)
\( \tau = \frac{1.0 \times 10^{-7} \text{ m}}{5.1 \times 10^2 \text{ m/s}} \)
\( \tau \approx 2.0 \times 10^{-10} \text{ s} \). (i)
Let \( t \) be the time taken for the collision itself (when two molecules are in contact). This can be estimated as the time it takes for a molecule to travel its own diameter.
\( t = \frac{d}{C} \)
\( t = \frac{2 \times 10^{-10} \text{ m}}{5.10 \times 10^2 \text{ m/s}} \)
\( t \approx 4 \times 10^{-13} \text{ s} \). (ii)
Comparing the collision time with the free flight time:
\( \frac{\tau}{t} = \frac{2.0 \times 10^{-10} \text{ s}}{4 \times 10^{-13} \text{ s}} = 500 \)
This means that \( \tau = 500t \). The time taken between successive collisions is 500 times longer than the time taken for a collision to occur. This indicates that a molecule in a gas moves almost freely for most of its journey.
In simple words: We first found the average distance a nitrogen molecule travels before hitting another molecule (mean free path) and how often it hits others (collision frequency). We then compared the short time it spends actually hitting another molecule to the much longer time it spends flying freely between hits. It turns out, a molecule spends most of its time traveling without bumping into anything.

Exam Tip: For mean free path (\( \lambda \)) calculations, remember the formula \( \lambda = \frac{1}{\sqrt{2} \pi n d^2} \). Collision frequency is \( C/\lambda \), and the collision time is \( d/C \). Ensure all units are consistent (SI units are preferred) and differentiate between number density (\( n \)) and number of moles.