Mean Free Path & Molecular Collisions — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Mean Free Path & Molecular Collisions MCQs with step-by-step solutions (30 questions). Part of Kinetic Theory of Gases & Radiation. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Mean Free Path & Molecular Collisions · medium · theory
The mean free path $\lambda$ of gas molecules of diameter $d$ and number density $n$ is given by:
A. $\dfrac{1}{\sqrt{2}\pi n d^2}$ ✓ Correct
B. $\dfrac{1}{2\pi n d^2}$
C. $\dfrac{1}{\sqrt{2}\pi n^2 d}$
D. $\dfrac{\sqrt{2}}{\pi n d^2}$
Solution: The standard kinetic-theory result is $\lambda = \dfrac{1}{\sqrt{2}\pi n d^2}$, the $\sqrt{2}$ arising from the relative motion of the colliding molecules.
Q2 — Mean Free Path & Molecular Collisions · easy · theory
The mean free path of gas molecules is defined as:
A. The average distance between two neighbouring molecules
B. The distance a molecule travels before reaching the container wall
C. The average distance travelled by a molecule between two successive collisions ✓ Correct
D. The total distance travelled by a molecule in one second
Solution: It is the mean of the free path lengths between successive molecular collisions, and it sets the scale for transport properties like viscosity and conduction.
Q3 — Mean Free Path & Molecular Collisions · easy · theory
The mean free path of gas molecules varies with the number density $n$ as:
A. $\lambda \propto \dfrac{1}{n^2}$
B. $\lambda \propto \dfrac{1}{n}$ ✓ Correct
C. $\lambda \propto n^2$
D. $\lambda \propto n$
Solution: From $\lambda = \dfrac{1}{\sqrt{2}\pi n d^2}$, a denser gas offers more collision partners and therefore a shorter free path.
Q4 — Mean Free Path & Molecular Collisions · medium · theory
The mean free path of gas molecules depends on the molecular diameter $d$ as:
A. $\lambda \propto \dfrac{1}{d^2}$ ✓ Correct
B. $\lambda \propto d^2$
C. $\lambda \propto d$
D. $\lambda \propto \dfrac{1}{d}$
Solution: The collision cross-section is proportional to $d^2$, so a larger molecule is struck more often and the free path falls as $\dfrac{1}{d^2}$.
Q5 — Mean Free Path & Molecular Collisions · medium · numerical
If the molecular diameter of a gas were doubled without changing anything else, the mean free path would become:
A. $4$ times its original value
B. $\dfrac{1}{4}$ of its original value ✓ Correct
C. $2$ times its original value
D. $\dfrac{1}{2}$ of its original value
Solution: $\lambda \propto \dfrac{1}{d^2}$, so doubling $d$ reduces the mean free path by a factor of $4$.
Q6 — Mean Free Path & Molecular Collisions · hard · theory
In terms of pressure $P$ and absolute temperature $T$, the mean free path of gas molecules is:
A. $\lambda = \dfrac{\sqrt{2}\pi d^2 P}{k_B T}$
B. $\lambda = \dfrac{k_B P}{\sqrt{2}\pi d^2 T}$
C. $\lambda = \dfrac{k_B T}{\sqrt{2}\pi d^2 P}$ ✓ Correct
D. $\lambda = \dfrac{\sqrt{2}\pi d^2 T}{k_B P}$
Solution: Substituting $n = \dfrac{P}{k_B T}$ into $\lambda = \dfrac{1}{\sqrt{2}\pi n d^2}$ gives $\lambda = \dfrac{k_B T}{\sqrt{2}\pi d^2 P}$.
Q7 — Mean Free Path & Molecular Collisions · medium · numerical
At constant temperature, if the pressure of a gas is doubled, its mean free path:
A. Becomes four times
B. Remains unchanged
C. Halves ✓ Correct
D. Doubles
Solution: At fixed $T$, $\lambda \propto \dfrac{1}{P}$. Doubling the pressure packs the molecules twice as densely, halving the free path.
Q8 — Mean Free Path & Molecular Collisions · medium · numerical
At constant pressure, if the absolute temperature of a gas is doubled, its mean free path:
A. Doubles ✓ Correct
B. Halves
C. Remains unchanged
D. Becomes four times
Solution: At fixed $P$, $\lambda \propto T$. Heating at constant pressure thins the gas out, so the molecules travel further between collisions.
Q9 — Mean Free Path & Molecular Collisions · medium · theory
The number density $n$ of molecules of an ideal gas at pressure $P$ and absolute temperature $T$ is:
A. $\dfrac{P}{R T}$
B. $\dfrac{k_B T}{P}$
C. $\dfrac{P T}{k_B}$
D. $\dfrac{P}{k_B T}$ ✓ Correct
Solution: Rearranging $P = n k_B T$ gives $n = \dfrac{P}{k_B T}$ molecules per unit volume.
Q10 — Mean Free Path & Molecular Collisions · medium · theory
The collision frequency of a gas molecule with mean speed $\bar{v}$ and mean free path $\lambda$ is:
A. $\dfrac{\lambda}{\bar{v}}$
B. $\bar{v}\lambda$
C. $\dfrac{\bar{v}}{\lambda}$ ✓ Correct
D. $\dfrac{1}{\bar{v}\lambda}$
Solution: A molecule covers $\bar{v}$ metres each second and suffers one collision every $\lambda$ metres, so it undergoes $\dfrac{\bar{v}}{\lambda}$ collisions per second.
Q11 — Mean Free Path & Molecular Collisions · medium · theory
The average time between two successive molecular collisions, in terms of mean free path $\lambda$ and mean speed $\bar{v}$, is:
A. $\dfrac{1}{\lambda\bar{v}}$
B. $\dfrac{\bar{v}}{\lambda}$
C. $\dfrac{\lambda}{\bar{v}}$ ✓ Correct
D. $\lambda\bar{v}$
Solution: Time equals distance over speed, so the mean time of relaxation is $\tau = \dfrac{\lambda}{\bar{v}}$ — the reciprocal of the collision frequency.
Q12 — Mean Free Path & Molecular Collisions · medium · theory
The mean free path of air molecules at a high altitude compared with that at sea level is:
A. Larger, because the pressure and hence the number density is lower ✓ Correct
B. Zero, since there are no collisions
C. Smaller, because the temperature is lower
D. The same, since it depends only on molecular diameter
Solution: The atmosphere thins with height, so $n$ falls sharply. Since $\lambda \propto \dfrac{1}{n}$, the free path grows — it becomes enormous in the upper atmosphere.
Q13 — Mean Free Path & Molecular Collisions · easy · theory
The dimensional formula of the mean free path is:
A. $[M^1L^1T^{-1}]$
B. $[M^0L^0T^1]$
C. $[M^0L^2T^0]$
D. $[M^0L^1T^0]$ ✓ Correct
Solution: The mean free path is an average distance, so it has the dimension of length only.
Q14 — Mean Free Path & Molecular Collisions · medium · theory
The collision frequency of gas molecules is directly proportional to:
A. The mean free path
B. The square of the mean free path
C. The number density of molecules ✓ Correct
D. The molar mass of the gas
Solution: Collision frequency is $\dfrac{\bar{v}}{\lambda}$, and since $\lambda \propto \dfrac{1}{n}$, the frequency rises in direct proportion to $n$.
Q15 — Mean Free Path & Molecular Collisions · hard · theory
The typical order of magnitude of the mean free path of air molecules at room temperature and atmospheric pressure is:
A. $10^{-12}\text{ m}$
B. $10^{2}\text{ m}$
C. $10^{-7}\text{ m}$ ✓ Correct
D. $10^{-2}\text{ m}$
Solution: At NTP the mean free path of air molecules is roughly $10^{-7}\text{ m}$, which is some hundreds of times larger than the molecular diameter itself.
Q16 — Mean Free Path & Molecular Collisions · easy · theory
In kinetic theory, collisions between gas molecules are assumed to be:
A. Perfectly inelastic and of long duration
B. Perfectly elastic and of negligible duration ✓ Correct
C. Absent altogether
D. Partially elastic with a fixed energy loss
Solution: Elastic collisions of negligible duration conserve both momentum and kinetic energy, which is what allows the gas to stay in equilibrium at a fixed temperature indefinitely.
Q17 — Mean Free Path & Molecular Collisions · hard · numerical
A gas has $10^{25}$ molecules per cubic metre, each of diameter $2 \times 10^{-10}\text{ m}$. Its mean free path is approximately:
A. $1.8 \times 10^{-6}\text{ m}$
B. $5.6 \times 10^{-7}\text{ m}$ ✓ Correct
C. $5.6 \times 10^{-5}\text{ m}$
D. $2.8 \times 10^{-7}\text{ m}$
Solution: $\lambda = \dfrac{1}{\sqrt{2}\pi n d^2} = \dfrac{1}{4.443 \times 10^{25} \times 4 \times 10^{-20}} = \dfrac{1}{1.78 \times 10^6} \approx 5.6 \times 10^{-7}\text{ m}$.
Q18 — Mean Free Path & Molecular Collisions · medium · numerical
At constant temperature, the pressure of a gas is tripled. Its mean free path becomes:
A. $\dfrac{1}{9}$ of its original value
B. $3$ times its original value
C. $\dfrac{1}{3}$ of its original value ✓ Correct
D. Unchanged
Solution: At fixed $T$, $\lambda \propto \dfrac{1}{P}$, so tripling the pressure reduces the mean free path to one-third.
Q19 — Mean Free Path & Molecular Collisions · hard · numerical
Both the absolute temperature and the pressure of a gas are doubled. Its mean free path:
A. Halves
B. Becomes four times
C. Doubles
D. Remains unchanged ✓ Correct
Solution: $\lambda = \dfrac{k_BT}{\sqrt{2}\pi d^2 P} \propto \dfrac{T}{P}$. Doubling both leaves the ratio, and hence $\lambda$, unaltered.
Q20 — Mean Free Path & Molecular Collisions · medium · numerical
Gas molecules with a mean speed of $500\text{ m/s}$ have a mean free path of $10^{-7}\text{ m}$. Their collision frequency is:
A. $2 \times 10^{-10}\text{ s}^{-1}$
B. $5 \times 10^{11}\text{ s}^{-1}$
C. $5 \times 10^9\text{ s}^{-1}$ ✓ Correct
D. $5 \times 10^7\text{ s}^{-1}$
Solution: Collision frequency $= \dfrac{\bar{v}}{\lambda} = \dfrac{500}{10^{-7}} = 5 \times 10^9$ collisions per second.
Q21 — Mean Free Path & Molecular Collisions · hard · numerical
A gas has $2 \times 10^{25}$ molecules per cubic metre, each of diameter $3 \times 10^{-10}\text{ m}$. Its mean free path is approximately:
A. $2.50 \times 10^{-7}\text{ m}$
B. $5.63 \times 10^{-7}\text{ m}$
C. $1.25 \times 10^{-5}\text{ m}$
D. $1.25 \times 10^{-7}\text{ m}$ ✓ Correct
Solution: $\lambda = \dfrac{1}{\sqrt{2}\pi nd^2} = \dfrac{1}{4.443 \times 2 \times 10^{25} \times 9 \times 10^{-20}} = \dfrac{1}{8.0 \times 10^6} \approx 1.25 \times 10^{-7}\text{ m}$.
Q22 — Mean Free Path & Molecular Collisions · easy · numerical
At constant temperature, the pressure of a gas is halved. Its mean free path:
A. Halves
B. Remains unchanged
C. Doubles ✓ Correct
D. Becomes four times
Solution: At fixed $T$, $\lambda \propto \dfrac{1}{P}$, so halving the pressure doubles the mean free path.
Q23 — Mean Free Path & Molecular Collisions · medium · numerical
At constant pressure, the absolute temperature of a gas is tripled. Its mean free path:
A. Triples ✓ Correct
B. Remains unchanged
C. Becomes one-third
D. Becomes nine times
Solution: At fixed $P$, $\lambda \propto T$, so tripling the temperature triples the mean free path.
Q24 — Mean Free Path & Molecular Collisions · medium · numerical
If the molecular diameter of a gas were tripled with everything else unchanged, the mean free path would become:
A. $3$ times its original value
B. $\dfrac{1}{3}$ of its original value
C. $\dfrac{1}{9}$ of its original value ✓ Correct
D. $9$ times its original value
Solution: $\lambda \propto \dfrac{1}{d^2}$, so tripling the diameter divides the mean free path by $9$.
Q25 — Mean Free Path & Molecular Collisions · medium · numerical
Gas molecules with a mean speed of $400\text{ m/s}$ have a mean free path of $2 \times 10^{-7}\text{ m}$. Their collision frequency is:
A. $2 \times 10^9\text{ s}^{-1}$ ✓ Correct
B. $5 \times 10^8\text{ s}^{-1}$
C. $2 \times 10^7\text{ s}^{-1}$
D. $8 \times 10^{-5}\text{ s}^{-1}$
Solution: Collision frequency $= \dfrac{\bar{v}}{\lambda} = \dfrac{400}{2 \times 10^{-7}} = 2 \times 10^9\text{ s}^{-1}$.
Q26 — Mean Free Path & Molecular Collisions · medium · numerical
Gas molecules of mean speed $500\text{ m/s}$ have a mean free path of $5 \times 10^{-7}\text{ m}$. The average time between successive collisions is:
A. $2 \times 10^{-9}\text{ s}$
B. $10^{-9}\text{ s}$ ✓ Correct
C. $10^{-10}\text{ s}$
D. $10^{-7}\text{ s}$
Solution: $\tau = \dfrac{\lambda}{\bar{v}} = \dfrac{5 \times 10^{-7}}{500} = 10^{-9}\text{ s}$.
Q27 — Mean Free Path & Molecular Collisions · hard · numerical
The number density of molecules of an ideal gas at $10^5\text{ Pa}$ and $273\text{ K}$ is approximately ($k_B = 1.38 \times 10^{-23}$):
A. $2.65 \times 10^{25}\text{ m}^{-3}$ ✓ Correct
B. $6.02 \times 10^{23}\text{ m}^{-3}$
C. $2.42 \times 10^{25}\text{ m}^{-3}$
D. $3.77 \times 10^{21}\text{ m}^{-3}$
Solution: $n = \dfrac{P}{k_BT} = \dfrac{10^5}{1.38 \times 10^{-23} \times 273} = \dfrac{10^5}{3.767 \times 10^{-21}} \approx 2.65 \times 10^{25}\text{ m}^{-3}$.
Q28 — Mean Free Path & Molecular Collisions · hard · numerical
The absolute temperature of a gas is doubled while its pressure is tripled. Its mean free path becomes:
A. Unchanged
B. $\dfrac{2}{3}$ of its original value ✓ Correct
C. $6$ times its original value
D. $\dfrac{3}{2}$ of its original value
Solution: $\lambda \propto \dfrac{T}{P}$, so the new value is $\dfrac{2}{3}$ of the original.
Q29 — Mean Free Path & Molecular Collisions · easy · numerical
The number density of molecules in a gas is doubled at constant temperature. Its mean free path:
A. Doubles
B. Becomes one-fourth
C. Remains unchanged
D. Halves ✓ Correct
Solution: $\lambda \propto \dfrac{1}{n}$, so doubling the number density halves the mean free path.
Q30 — Mean Free Path & Molecular Collisions · hard · numerical
The mean free path of molecules of diameter $2 \times 10^{-10}\text{ m}$ at $300\text{ K}$ and $10^5\text{ Pa}$ is approximately ($k_B = 1.38 \times 10^{-23}$):
A. $2.33 \times 10^{-7}\text{ m}$ ✓ Correct
B. $5.63 \times 10^{-7}\text{ m}$
C. $1.25 \times 10^{-7}\text{ m}$
D. $2.33 \times 10^{-5}\text{ m}$
Solution: $\lambda = \dfrac{k_BT}{\sqrt{2}\pi d^2P} = \dfrac{4.14 \times 10^{-21}}{4.443 \times 4 \times 10^{-20} \times 10^5} = \dfrac{4.14 \times 10^{-21}}{1.777 \times 10^{-14}} \approx 2.33 \times 10^{-7}\text{ m}$.