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201. The specific heat capacity at constant volume $C_v$ of an ideal gas can be calculated using:
ⓐ. $C_v = \frac{Q}{n\Delta T}$ (at constant volume)
ⓑ. $C_v = \frac{Q}{m}$
ⓒ. $C_v = \frac{P\Delta V}{\Delta T}$
ⓓ. $C_v = \frac{W}{nRT}$
Correct Answer: $C_v = \frac{Q}{n\Delta T}$ (at constant volume)
Explanation: At constant volume, no work is done. Hence, heat supplied goes into raising internal energy only. Therefore, $C_v = \frac{Q}{n\Delta T}$.
202. In a calorimeter experiment, 2 moles of a monatomic gas absorb 900 J of heat at constant volume. Find the rise in temperature.
ⓐ. 25 K
ⓑ. 30 K
ⓒ. 50 K
ⓓ. 75 K
Correct Answer: 30 K
Explanation: For a monatomic gas, $C_v = \tfrac{3}{2}R \approx 12.5 \,\text{J/mol·K}$. Total $C_v = 2 \times 12.5 = 25 \,\text{J/K}$. Then, $\Delta T = \tfrac{Q}{nC_v} = \tfrac{900}{25 \times 1} = 36 \,\text{K}$. Closest option is 30 K (approx).
203. For 1 mole of diatomic gas (without vibrations), the molar heat capacity at constant pressure is:
ⓐ. $\tfrac{5}{2}R$
ⓑ. $\tfrac{7}{2}R$
ⓒ. $3R$
ⓓ. $4R$
Correct Answer: $\tfrac{7}{2}R$
Explanation: Internal energy = $\tfrac{5}{2}RT$, so $C_v = \tfrac{5}{2}R$. Using Mayer’s relation $C_p = C_v + R = \tfrac{7}{2}R$.
204. Calculate the heat required to raise the temperature of 3 moles of a diatomic gas (ignoring vibrations) by 50 K at constant pressure.
ⓐ. $375R$
ⓑ. $525R$
ⓒ. $600R$
ⓓ. $750R$
Correct Answer: $525R$
Explanation: $Q = nC_p\Delta T$. For diatomic gas, $C_p = \tfrac{7}{2}R$. Substituting: $Q = 3 \times \tfrac{7}{2}R \times 50 = 525R$.
205. A gas has $C_v = 20 \,\text{J/mol·K}$. If 5 moles of this gas are heated at constant volume through 40 K, how much heat is absorbed?
ⓐ. 2,000 J
ⓑ. 4,000 J
ⓒ. 5,000 J
ⓓ. 6,000 J
Correct Answer: 4,000 J
Explanation: $Q = nC_v \Delta T = 5 \times 20 \times 40 = 4000 \, \text{J}$. This is the heat absorbed at constant volume.
206. The specific heat ratio $\gamma$ can be determined experimentally by:
ⓐ. Measuring pressure only
ⓑ. Measuring volume only
ⓒ. Using the speed of sound in gases
ⓓ. Using density of liquids
Correct Answer: Using the speed of sound in gases
Explanation: The velocity of sound in a gas is given by $v = \sqrt{\frac{\gamma RT}{M}}$. By measuring $v$, one can calculate $\gamma = C_p/C_v$, thus relating specific heats to measurable macroscopic properties.
207. The relation between molar heat capacity at constant pressure ($C_p$) and at constant volume ($C_v$) for ideal gases is:
ⓐ. $C_p = C_v – R$
ⓑ. $C_p = C_v + R$
ⓒ. $C_p = \tfrac{C_v}{R}$
ⓓ. $C_p = 2C_v$
Correct Answer: $C_p = C_v + R$
Explanation: Mayer’s relation gives $C_p – C_v = R$. Thus, $C_p$ is always greater than $C_v$ by $R$, due to extra work of expansion at constant pressure.
208. In an experiment, 1 mole of an ideal monatomic gas is heated at constant volume by 25 K. How much heat is required?
ⓐ. $25R$
ⓑ. $37.5R$
ⓒ. $50R$
ⓓ. $62.5R$
Correct Answer: $37.5R$
Explanation: For monatomic gas, $C_v = \tfrac{3}{2}R$. Heat $Q = nC_v\Delta T = 1 \times \tfrac{3}{2}R \times 25 = 37.5R$.
209. In calorimetry, why is a water equivalent of the calorimeter determined?
ⓐ. To calculate the density of gas
ⓑ. To account for heat absorbed by calorimeter walls
ⓒ. To reduce calculation errors by assuming zero heat loss
ⓓ. To measure pressure changes in gas
Correct Answer: To account for heat absorbed by calorimeter walls
Explanation: The calorimeter itself absorbs heat. To simplify calculations, its effect is represented as an equivalent mass of water, ensuring correct calculation of heat capacity and heat transfer.
210. The specific heat of helium at constant volume is closest to:
ⓐ. $3R$
ⓑ. $\tfrac{5}{2}R$
ⓒ. $\tfrac{3}{2}R$
ⓓ. $\tfrac{7}{2}R$
Correct Answer: $\tfrac{3}{2}R$
Explanation: Helium is a monatomic ideal gas, so it has only 3 translational degrees of freedom. Thus, $C_v = \tfrac{3}{2}R$. This is confirmed experimentally for noble gases.
211. Why does a monatomic ideal gas have $C_v = \tfrac{3}{2}R$?
ⓐ. It has both translational and rotational degrees of freedom.
ⓑ. It has only translational degrees of freedom (x, y, z motion).
ⓒ. It has only vibrational degrees of freedom.
ⓓ. It absorbs heat without changing energy.
Correct Answer: It has only translational degrees of freedom (x, y, z motion).
Explanation: Kinetic theory + equipartition theorem show that each quadratic degree of freedom contributes $\tfrac{1}{2}R$ per mole. A monatomic gas has 3 translational degrees, hence $C_v = 3 \times \tfrac{1}{2}R = \tfrac{3}{2}R$.
212. For a diatomic ideal gas, why is $C_v$ larger than for a monatomic gas?
ⓐ. Diatomic gases have more mass.
ⓑ. Diatomic molecules have extra rotational degrees of freedom.
ⓒ. Diatomic gases undergo chemical bonding.
ⓓ. Diatomic molecules have smaller volume.
Correct Answer: Diatomic molecules have extra rotational degrees of freedom.
Explanation: A diatomic molecule has 3 translational + 2 rotational degrees of freedom. Hence, $C_v = \tfrac{5}{2}R$. At higher temperatures, vibrational degrees also contribute, further increasing $C_v$.
213. At high temperatures, vibrational modes in molecules become active. How does this affect heat capacity?
ⓐ. Heat capacity decreases.
ⓑ. Heat capacity remains constant.
ⓒ. Heat capacity increases.
ⓓ. Heat capacity becomes zero.
Correct Answer: Heat capacity increases.
Explanation: Vibrational motion contributes 2 degrees of freedom (kinetic + potential) per mode. Each adds $R$ per mole to $C_v$. Thus, at high $T$, $C_v$ rises beyond values predicted at low $T$.
214. For a linear triatomic molecule at high temperature, the expected $C_v$ using kinetic theory is:
ⓐ. $\tfrac{3}{2}R$
ⓑ. $\tfrac{5}{2}R$
ⓒ. $3R$
ⓓ. $\tfrac{7}{2}R$
Correct Answer: $\tfrac{7}{2}R$
Explanation: Linear triatomic molecules (like $CO_2$) have 3 translational + 2 rotational + 2 vibrational quadratic modes = 7 DOF. Each adds $\tfrac{1}{2}R$. So $C_v = \tfrac{7}{2}R$.
215. The failure of classical predictions of specific heat at low temperatures is explained by:
ⓐ. Newton’s laws of motion
ⓑ. Kinetic theory with equipartition only
ⓒ. Quantum theory restricting excitation of energy modes
ⓓ. Conservation of mass
Correct Answer: Quantum theory restricting excitation of energy modes
Explanation: Classical equipartition predicts constant heat capacities, but experiments show $C_v$ decreases at low T. This is because quantum mechanics allows only discrete energy levels, so higher modes (rotational/vibrational) are “frozen out.”
216. Calculate the increase in internal energy when 4 moles of monatomic gas are heated by 50 K at constant volume.
ⓐ. $200R$
ⓑ. $250R$
ⓒ. $300R$
ⓓ. $400R$
Correct Answer: $300R$
Explanation: $\Delta U = n C_v \Delta T = 4 \times \tfrac{3}{2}R \times 50 = 300R$. Kinetic theory gives this directly from translational degrees of freedom.
217. For a diatomic gas (ignoring vibrations), calculate the heat required to raise 2 moles of the gas by 40 K at constant volume.
ⓐ. $100R$
ⓑ. $150R$
ⓒ. $200R$
ⓓ. $250R$
Correct Answer: $250R$
Explanation: $C_v = \tfrac{5}{2}R$. So $Q = nC_v\Delta T = 2 \times \tfrac{5}{2}R \times 40 = 250R$.
218. The experimental fact that $C_v$ of solids approaches zero at very low temperatures is explained by:
ⓐ. Newton’s third law
ⓑ. Equipartition theorem
ⓒ. Quantum theory (Debye model)
ⓓ. Avogadro’s law
Correct Answer: Quantum theory (Debye model)
Explanation: Classical kinetic theory predicts constant $C_v$ for solids, but the Debye model explains that only long-wavelength phonons are excited at low T, so heat capacity tends to zero, consistent with the third law of thermodynamics.
219. For 3 moles of a diatomic gas at 300 K (ignoring vibrations), the internal energy is:
ⓐ. $900R$
ⓑ. $1050R$
ⓒ. $1200R$
ⓓ. $2250R$
Correct Answer: $2250R$
Explanation: $U = n C_v T = 3 \times \tfrac{5}{2}R \times 300/300$. Careful: substituting directly, $U = 3 \times \tfrac{5}{2}R \times 300 = 2250R$.
220. Why do polyatomic gases generally have higher specific heat capacities than monatomic gases?
ⓐ. They contain more atoms per molecule.
ⓑ. They have more degrees of freedom (rotational and vibrational).
ⓒ. They are heavier than monatomic gases.
ⓓ. Their molecules are unstable.
Correct Answer: They have more degrees of freedom (rotational and vibrational).
Explanation: Equipartition theorem explains that each degree of freedom contributes $\tfrac{1}{2}R$ per mole. Polyatomic molecules have extra rotational and vibrational DOF, thus more ways to store energy, resulting in higher $C_v$ and $C_p$.
221. The mean free path of a gas molecule is defined as:
ⓐ. The distance between two gas containers
ⓑ. The average distance a molecule travels between successive collisions
ⓒ. The maximum distance a molecule can move in a second
ⓓ. The shortest path between two molecules
Correct Answer: The average distance a molecule travels between successive collisions
Explanation: From kinetic theory, the mean free path ($\lambda$) is the average distance covered by a molecule before colliding with another. It depends on number density of molecules and their effective collision cross-section.
222. The formula for mean free path is:
ⓐ. $\lambda = \frac{1}{\pi d^2 n}$
ⓑ. $\lambda = \frac{RT}{P}$
ⓒ. $\lambda = \frac{1}{n\sigma}$
ⓓ. $\lambda = \frac{k_B T}{\sqrt{2}\pi d^2 P}$
Correct Answer: $\lambda = \frac{k_B T}{\sqrt{2}\pi d^2 P}$
Explanation: Using kinetic theory, $\lambda = \frac{1}{\sqrt{2} n \pi d^2}$. Substituting $n = \tfrac{P}{k_BT}$, we get $\lambda = \frac{k_B T}{\sqrt{2}\pi d^2 P}$. Here $d$ is molecular diameter, $P$ pressure, $T$ temperature.
223. If the pressure of a gas is doubled at constant temperature, the mean free path will:
ⓐ. Double
ⓑ. Remain constant
ⓒ. Halve
ⓓ. Become zero
Correct Answer: Halve
Explanation: $\lambda \propto \tfrac{1}{P}$. At constant $T$, doubling $P$ doubles number density $n$, so molecules collide more often, reducing mean free path to half.
224. At constant pressure, increasing the temperature of a gas will cause mean free path to:
ⓐ. Decrease
ⓑ. Increase
ⓒ. Remain constant
ⓓ. Become zero
Correct Answer: Increase
Explanation: $\lambda \propto \tfrac{T}{P}$. At constant $P$, increasing $T$ increases $\lambda$. Molecules move faster, and density decreases, leading to longer average distances between collisions.
225. For a gas with molecular diameter $d = 3 \times 10^{-10} \,\text{m}$, at $P = 1 \,\text{atm} = 1.01 \times 10^5 \,\text{Pa}$, and $T = 300 \,\text{K}$, calculate the mean free path.
ⓐ. $1 \times 10^{-7} \,\text{m}$
ⓑ. $3 \times 10^{-8} \,\text{m}$
ⓒ. $6 \times 10^{-8} \,\text{m}$
ⓓ. $1 \times 10^{-9} \,\text{m}$
Correct Answer: $6 \times 10^{-8} \,\text{m}$
Explanation: Using $\lambda = \frac{k_B T}{\sqrt{2}\pi d^2 P}$. Substituting $k_B = 1.38 \times 10^{-23}, T = 300, d = 3 \times 10^{-10}, P = 1.01 \times 10^5$, we obtain $\lambda \approx 6 \times 10^{-8} \,m$.
226. The mean free path is inversely proportional to:
ⓐ. Temperature only
ⓑ. Pressure and square of molecular diameter
ⓒ. Molecular mass
ⓓ. Volume of container
Correct Answer: Pressure and square of molecular diameter
Explanation: From formula $\lambda = \tfrac{k_BT}{\sqrt{2}\pi d^2 P}$, we see $\lambda \propto \tfrac{1}{d^2 P}$. Larger molecules (higher $d$) and higher pressures reduce mean free path.
227. Why is the factor $\sqrt{2}$ included in the formula for mean free path?
ⓐ. To account for kinetic energy
ⓑ. To correct for relative motion of two colliding molecules
ⓒ. To correct for Avogadro’s number
ⓓ. To include intermolecular forces
Correct Answer: To correct for relative motion of two colliding molecules
Explanation: The probability of collisions depends not just on molecular speed but on relative velocity between two moving molecules. This adds a factor of $\sqrt{2}$ to the denominator in the mean free path expression.
228. If the diameter of molecules is halved while keeping pressure and temperature constant, the mean free path will:
ⓐ. Double
ⓑ. Increase four times
ⓒ. Decrease two times
ⓓ. Remain unchanged
Correct Answer: Increase four times
Explanation: $\lambda \propto \tfrac{1}{d^2}$. If $d$ is halved, $d^2$ decreases by 4, so $\lambda$ increases by 4.
229. In the upper atmosphere, mean free path of molecules is very large because:
ⓐ. Molecular diameter is very small
ⓑ. Pressure is extremely low
ⓒ. Temperature is very high
ⓓ. Molecules stop colliding
Correct Answer: Pressure is extremely low
Explanation: At high altitudes, atmospheric pressure decreases, reducing molecular density. Since $\lambda \propto \tfrac{1}{P}$, molecules travel long distances before colliding, sometimes several meters.
230. Which of the following quantities is NOT directly needed to calculate mean free path?
ⓐ. Molecular diameter
ⓑ. Gas pressure
ⓒ. Gas temperature
ⓓ. Molecular mass
Correct Answer: Molecular mass
Explanation: Mean free path depends on $d, P, T$ (through number density and collision cross-section). Molecular mass does not directly appear in the equation for $\lambda$, although it affects molecular speed distributions.
231. How does pressure affect the mean free path of a gas?
ⓐ. Mean free path increases with pressure.
ⓑ. Mean free path decreases with pressure.
ⓒ. Mean free path remains constant with pressure.
ⓓ. Mean free path first increases then decreases.
Correct Answer: Mean free path decreases with pressure.
Explanation: The mean free path is given by $\lambda = \tfrac{k_B T}{\sqrt{2}\pi d^2 P}$. At constant temperature, when pressure $P$ increases, the number density of molecules increases, leading to more frequent collisions. This reduces the average distance a molecule travels before colliding, hence the mean free path decreases inversely with pressure. For example, doubling pressure halves the mean free path.
232. If the pressure of a gas is reduced to half at constant temperature, what happens to mean free path?
ⓐ. It is reduced to half.
ⓑ. It doubles.
ⓒ. It remains constant.
ⓓ. It becomes zero.
Correct Answer: It doubles.
Explanation: Since $\lambda \propto \tfrac{1}{P}$, halving the pressure doubles the mean free path. At lower pressures, molecules are more spread out, so they travel longer distances before colliding. This is why gases in high-altitude or vacuum chambers have much larger mean free paths compared to gases at atmospheric pressure.
233. How does temperature affect the mean free path at constant pressure?
ⓐ. Mean free path decreases as temperature increases.
ⓑ. Mean free path increases as temperature increases.
ⓒ. Mean free path remains constant.
ⓓ. Mean free path becomes infinite.
Correct Answer: Mean free path increases as temperature increases.
Explanation: From $\lambda = \tfrac{k_B T}{\sqrt{2}\pi d^2 P}$, at constant pressure, $\lambda$ is directly proportional to $T$. Higher temperature increases molecular kinetic energy, and for a fixed pressure, this means the density of gas molecules becomes lower. Hence, molecules collide less often and travel farther between collisions, leading to a larger mean free path.
234. At very high altitudes, the mean free path of air molecules is extremely large because:
ⓐ. Temperature is very high.
ⓑ. Pressure is very low.
ⓒ. Molecular diameter decreases with height.
ⓓ. Mass of molecules decreases.
Correct Answer: Pressure is very low.
Explanation: At high altitudes, air pressure is much lower than at sea level. Since $\lambda \propto \tfrac{1}{P}$, lower pressure means molecules are less dense, so they travel long distances before colliding. This is why in the thermosphere and exosphere, mean free paths can reach several meters or even kilometers, compared to nanometers near sea level.
235. How does molecular diameter affect mean free path?
ⓐ. Larger diameter increases mean free path.
ⓑ. Larger diameter decreases mean free path.
ⓒ. Diameter has no effect.
ⓓ. Mean free path is inversely proportional to cube of diameter.
Correct Answer: Larger diameter decreases mean free path.
Explanation: The formula $\lambda = \tfrac{k_B T}{\sqrt{2}\pi d^2 P}$ shows that mean free path is inversely proportional to the square of molecular diameter. A larger diameter increases the effective collision cross-section, making collisions more frequent and reducing the distance molecules can travel without colliding. For example, helium atoms (small diameter) have larger mean free paths than oxygen molecules.
236. Why is the factor of $T/P$ important in determining mean free path?
ⓐ. Because it determines molecular size.
ⓑ. Because it reflects the balance between thermal energy and number density.
ⓒ. Because it shows mean free path is independent of temperature.
ⓓ. Because it cancels molecular mass from the equation.
Correct Answer: Because it reflects the balance between thermal energy and number density.
Explanation: Mean free path depends on how spread out molecules are (determined by pressure) and how much kinetic energy they have (determined by temperature). At higher $T$, molecules have more energy and are less dense at fixed pressure, increasing $\lambda$. At higher $P$, molecules are crowded, reducing $\lambda$. Thus, the ratio $T/P$ directly controls mean free path.
237. If the pressure of a gas is kept constant and the temperature is doubled, what happens to the mean free path?
ⓐ. It halves.
ⓑ. It doubles.
ⓒ. It remains the same.
ⓓ. It becomes zero.
Correct Answer: It doubles.
Explanation: Since $\lambda \propto T$ at constant $P$, doubling the temperature doubles the mean free path. Molecules move faster and are farther apart in terms of effective density, leading to less frequent collisions and longer average travel distance.
238. Which condition gives the smallest mean free path for a gas?
ⓐ. High pressure, high temperature
ⓑ. Low pressure, low temperature
ⓒ. High pressure, low temperature
ⓓ. Low pressure, high temperature
Correct Answer: High pressure, low temperature
Explanation: From the formula, $\lambda \propto \tfrac{T}{P}$. A high pressure means molecules are crowded, and low temperature means reduced kinetic energy, so collisions are very frequent. This combination results in the smallest possible mean free path.
239. At sea level (1 atm, 300 K), the mean free path of air molecules is approximately:
ⓐ. $10^{-10} \, m$
ⓑ. $10^{-7} \, m$
ⓒ. $10^{-5} \, m$
ⓓ. $10^{-2} \, m$
Correct Answer: $10^{-7} \, m$
Explanation: Using $\lambda = \tfrac{k_BT}{\sqrt{2}\pi d^2 P}$, with $ d \approx 3 \times 10^{-10} m$, $ T = 300K$, and $ P = 1.01 \times 10^5 Pa$, the mean free path for air molecules is about $6 \times 10^{-8} m$, i.e., $10^{-7} m$.
240. Which of the following statements best summarizes the effect of temperature and pressure on mean free path?
ⓐ. Higher temperature increases mean free path, while higher pressure decreases it.
ⓑ. Both higher temperature and higher pressure increase mean free path.
ⓒ. Both higher temperature and higher pressure decrease mean free path.
ⓓ. Mean free path is independent of both temperature and pressure.
Correct Answer: Higher temperature increases mean free path, while higher pressure decreases it.
Explanation: From the kinetic theory formula $\lambda = \tfrac{k_BT}{\sqrt{2}\pi d^2 P}$, temperature appears in numerator and pressure in denominator. Therefore, increasing $T$ increases $\lambda$, while increasing $P$ decreases $\lambda$. This is why gases in space (low $P$, high $T$) have very long mean free paths.
241. Which property of gases can be directly explained using the concept of mean free path?
ⓐ. Color of gases
ⓑ. Viscosity and diffusion
ⓒ. Atomic number
ⓓ. Melting point
Correct Answer: Viscosity and diffusion
Explanation: The mean free path determines how far molecules travel between collisions. This directly affects transport properties like viscosity (resistance to flow) and diffusion (spread of particles). Longer mean free paths mean molecules travel further before colliding, enhancing diffusion and lowering viscosity in dilute gases.
242. Graham’s law of diffusion is consistent with kinetic theory because:
ⓐ. Diffusion rate is proportional to the square of molecular mass.
ⓑ. Diffusion rate is inversely proportional to the square root of molar mass.
ⓒ. Diffusion is independent of molar mass.
ⓓ. Diffusion depends only on density, not mass.
Correct Answer: Diffusion rate is inversely proportional to the square root of molar mass.
Explanation: From kinetic theory, $v_{rms} = \sqrt{\tfrac{3RT}{M}}$. Lighter molecules have higher rms speeds, thus diffuse faster. Graham’s law: $\tfrac{r_1}{r_2} = \sqrt{\tfrac{M_2}{M_1}}$. This matches experimental diffusion behavior.
243. The rate of effusion of a gas through a small hole is proportional to:
ⓐ. Gas pressure only
ⓑ. Molecular size
ⓒ. Inverse square root of molar mass
ⓓ. Square of molecular velocity
Correct Answer: Inverse square root of molar mass
Explanation: Effusion is governed by kinetic theory. Lighter molecules (lower molar mass) move faster, thus escape more quickly through small holes. The relation is $r \propto \tfrac{1}{\sqrt{M}}$. This principle is applied in separating isotopes like $^{235}U$ and $^{238}U$.
244. Viscosity of a gas is related to mean free path by:
ⓐ. $\eta = \tfrac{1}{3} \rho \langle v \rangle \lambda$
ⓑ. $\eta = P \lambda$
ⓒ. $\eta = \rho \lambda^2$
ⓓ. $\eta = \tfrac{3}{2}RT \lambda$
Correct Answer: $\eta = \tfrac{1}{3} \rho \langle v \rangle \lambda$
Explanation: From kinetic theory, viscosity depends on gas density ($\rho$), mean molecular speed ($\langle v \rangle$), and mean free path ($\lambda$). Larger mean free paths increase viscosity, as molecules carry momentum farther before colliding.
245. Why does the viscosity of gases increase with temperature, unlike liquids?
ⓐ. Molecules stick together at high $T$.
ⓑ. Mean free path decreases at high $T$.
ⓒ. Molecular speeds increase at high $T$, enhancing momentum transfer.
ⓓ. Density of gas decreases at high $T$.
Correct Answer: Molecular speeds increase at high $T$, enhancing momentum transfer.
Explanation: In gases, viscosity depends on molecular motion and collisions. Higher $T$ increases molecular speed, so molecules transport momentum more efficiently across layers. Hence, viscosity increases, unlike in liquids where intermolecular attraction dominates.
246. Thermal conductivity of gases is explained by kinetic theory as:
ⓐ. Transfer of charge between ions
ⓑ. Transfer of energy through molecular collisions
ⓒ. Flow of photons in gases
ⓓ. Direct conduction by nuclei
Correct Answer: Transfer of energy through molecular collisions
Explanation: Gas molecules carry kinetic energy. When faster (hotter) molecules collide with slower (colder) ones, energy is transferred. The efficiency of this process depends on mean free path, molecular speed, and collision frequency.
247. In molecular dynamics simulations, why is mean free path important?
ⓐ. It sets the time step for collision calculations.
ⓑ. It determines atomic number.
ⓒ. It fixes the chemical bonding type.
ⓓ. It controls nuclear decay.
Correct Answer: It sets the time step for collision calculations.
Explanation: In simulations, mean free path and mean collision time decide how often molecules collide. This helps in setting time steps for algorithms. Too large a step misses collisions; too small makes simulations inefficient.
248. Which transport property of gases is NOT directly related to mean free path?
ⓐ. Diffusion
ⓑ. Effusion
ⓒ. Viscosity
ⓓ. Refractive index
Correct Answer: Refractive index
Explanation: Diffusion, effusion, and viscosity all involve molecular motion and collisions, hence depend on mean free path. Refractive index, however, is governed by electronic polarizability and not by collision dynamics.
249. Which equation relates diffusion coefficient $D$ to mean free path?
ⓐ. $D = \lambda$
ⓑ. $D = \tfrac{1}{3}\langle v \rangle \lambda$
ⓒ. $D = \tfrac{RT}{M}$
ⓓ. $D = \rho \lambda$
Correct Answer: $D = \tfrac{1}{3}\langle v \rangle \lambda$
Explanation: From kinetic theory, diffusion coefficient is proportional to average molecular speed and mean free path. Larger $\lambda$ and faster molecules increase mixing of gases, enhancing diffusion.
250. Why is separation of isotopes (e.g., $^{235}U$ and $^{238}U$) possible using gas effusion?
ⓐ. Isotopes have different charges.
ⓑ. Lighter isotopes effuse faster according to Graham’s law.
ⓒ. Heavy isotopes dissolve in liquids.
ⓓ. Isotopes chemically react differently.
Correct Answer: Lighter isotopes effuse faster according to Graham’s law.
Explanation: In uranium enrichment, $UF_6$ gas is used. Molecules with lighter isotope $^{235}U$ diffuse/effuse slightly faster than $^{238}U$ due to $r \propto 1/\sqrt{M}$. Repeated stages allow enrichment. This is a practical application of molecular dynamics and gas kinetics.
251. Brownian motion refers to:
ⓐ. The random motion of pollen grains in water observed under a microscope.
ⓑ. The uniform motion of molecules in a vacuum.
ⓒ. The systematic drift of particles in a liquid.
ⓓ. The rotation of Earth affecting fluids.
Correct Answer: The random motion of pollen grains in water observed under a microscope.
Explanation: In 1827, Robert Brown observed irregular, zig-zag motion of pollen grains suspended in water. This motion, called Brownian motion, is due to collisions with invisible water molecules. It provided experimental evidence for the molecular nature of matter.
252. Which scientist first explained the cause of Brownian motion theoretically?
ⓐ. Isaac Newton
ⓑ. Albert Einstein
ⓒ. Robert Brown
ⓓ. James Clerk Maxwell
Correct Answer: Albert Einstein
Explanation: While Robert Brown discovered the motion, Einstein (1905) provided the theoretical explanation, showing that it arises from random molecular collisions. His work quantitatively linked Brownian motion to molecular kinetic theory, later experimentally verified by Perrin.
253. Brownian motion provides strong evidence for:
ⓐ. Existence of ether
ⓑ. Discrete atomic and molecular structure of matter
ⓒ. Wave-particle duality
ⓓ. Conservation of angular momentum
Correct Answer: Discrete atomic and molecular structure of matter
Explanation: The erratic motion of suspended particles could only be explained if matter consists of tiny, constantly moving molecules. This confirmed the atomic theory of matter, which had been debated for centuries.
254. Why do smaller suspended particles show more vigorous Brownian motion compared to larger ones?
ⓐ. Smaller particles attract water molecules more strongly.
ⓑ. Smaller particles are lighter, so molecular collisions produce larger accelerations.
ⓒ. Larger particles have higher velocity.
ⓓ. Larger particles have higher density.
Correct Answer: Smaller particles are lighter, so molecular collisions produce larger accelerations.
Explanation: Collisions from surrounding molecules exert random impulses. Since force = mass × acceleration, lighter particles undergo greater acceleration from the same impulse, so they move more erratically. Larger particles appear more stable.
255. Brownian motion decreases in intensity when:
ⓐ. Temperature increases
ⓑ. Viscosity of medium decreases
ⓒ. Particle size decreases
ⓓ. Temperature decreases
Correct Answer: Temperature decreases
Explanation: Molecular motion depends on temperature. At lower $T$, average kinetic energy of molecules ($E_k = \tfrac{3}{2}k_BT$) decreases, so collisions are weaker. Hence, suspended particles move less vigorously.
256. Which equation derived by Einstein relates mean squared displacement in Brownian motion to time?
ⓐ. $\langle x^2 \rangle = 2Dt$
ⓑ. $\langle x^2 \rangle = v^2 t^2$
ⓒ. $\langle x^2 \rangle = \tfrac{RT}{P}$
ⓓ. $\langle x^2 \rangle = \tfrac{1}{2}at^2$
Correct Answer: $\langle x^2 \rangle = 2Dt$
Explanation: Einstein showed that the mean square displacement of a particle undergoing Brownian motion is proportional to time: $\langle x^2 \rangle = 2Dt$, where $D$ is diffusion coefficient. This formula allowed experimental determination of Avogadro’s number.
257. Which scientist experimentally verified Einstein’s theory of Brownian motion?
ⓐ. Robert Brown
ⓑ. Jean Perrin
ⓒ. James Clerk Maxwell
ⓓ. Niels Bohr
Correct Answer: Jean Perrin
Explanation: Jean Perrin (1908) verified Einstein’s theory experimentally. By analyzing Brownian motion, he determined Avogadro’s number, confirming the molecular-kinetic theory of heat. He was awarded the Nobel Prize in 1926 for this work.
258. Why does Brownian motion become less visible in very viscous liquids like honey?
ⓐ. Viscous liquids contain fewer molecules.
ⓑ. Viscosity dampens molecular motion, reducing displacement of suspended particles.
ⓒ. Larger molecules collide more strongly.
ⓓ. Suspended particles dissolve in viscous liquids.
Correct Answer: Viscosity dampens molecular motion, reducing displacement of suspended particles.
Explanation: In viscous liquids, molecular motion is hindered. Collisions with suspended particles produce weaker, smaller displacements, so random zig-zag motion is not easily observed.
259. Brownian motion cannot be observed in solids because:
ⓐ. Molecules in solids are stationary.
ⓑ. Collisions between molecules are perfectly elastic.
ⓒ. Particles are bound in fixed lattice positions.
ⓓ. Solids do not contain atoms.
Correct Answer: Particles are bound in fixed lattice positions.
Explanation: In solids, atoms and molecules vibrate about fixed lattice sites. They cannot impart large random motion to suspended particles, unlike in fluids where free molecular motion exists.
260. Which modern field uses Brownian motion as a fundamental concept?
ⓐ. Nuclear physics
ⓑ. Nanotechnology and particle tracking
ⓒ. Astronomy
ⓓ. Crystallography
Correct Answer: Nanotechnology and particle tracking
Explanation: Brownian motion principles are applied in nanotechnology, medical diagnostics, and colloid science. Tracking nanoparticles or biomolecules in solution relies on analyzing their Brownian motion, which gives insight into molecular interactions and diffusion properties.
261. According to Einstein’s 1905 explanation, the random motion of suspended particles in Brownian motion is caused by:
ⓐ. Earth’s gravitational pull
ⓑ. Continuous collisions with surrounding fluid molecules
ⓒ. Magnetic fields in the liquid
ⓓ. Radiation pressure of sunlight
Correct Answer: Continuous collisions with surrounding fluid molecules
Explanation: Einstein showed that suspended particles are constantly bombarded by uneven impacts of surrounding fluid molecules. These collisions are random in direction and magnitude, producing the irregular zig-zag motion observed under a microscope.
262. Einstein’s theory of Brownian motion provided a way to calculate:
ⓐ. Atomic number of elements
ⓑ. Avogadro’s number
ⓒ. Nuclear radius
ⓓ. Speed of light
Correct Answer: Avogadro’s number
Explanation: Einstein derived a relation between the mean square displacement of suspended particles and measurable quantities like temperature, viscosity, and particle radius. Jean Perrin verified this and determined Avogadro’s number experimentally, confirming the molecular theory of matter.
263. The mean square displacement in one dimension according to Einstein is given by:
ⓐ. $\langle x^2 \rangle = 2Dt$
ⓑ. $\langle x^2 \rangle = v_{rms}^2 t^2$
ⓒ. $\langle x^2 \rangle = \tfrac{RT}{P}$
ⓓ. $\langle x^2 \rangle = at^2$
Correct Answer: $\langle x^2 \rangle = 2Dt$
Explanation: Einstein derived that Brownian particles undergo diffusion-like motion. In 1D, their mean squared displacement is proportional to time, with proportionality constant equal to twice the diffusion coefficient, $D$.
264. Einstein’s expression for the diffusion coefficient $D$ of spherical particles of radius $r$ suspended in a fluid of viscosity $\eta$ is:
ⓐ. $D = \tfrac{k_BT}{\eta r}$
ⓑ. $D = \tfrac{k_BT}{6\pi \eta r}$
ⓒ. $D = \tfrac{RT}{\eta r}$
ⓓ. $D = \tfrac{RT}{6\pi \eta r}$
Correct Answer: $D = \tfrac{k_BT}{6\pi \eta r}$
Explanation: Einstein applied kinetic theory to suspended particles, showing their diffusion coefficient depends on Boltzmann’s constant, absolute temperature, fluid viscosity, and particle radius. This relation is crucial in molecular biophysics and colloid science.
265. The equation $\langle x^2 \rangle = \tfrac{2k_BT}{3\pi \eta r} t$ was used by Jean Perrin to:
ⓐ. Prove Newton’s second law
ⓑ. Verify the quantum theory of radiation
ⓒ. Experimentally confirm Einstein’s theory and calculate Avogadro’s number
ⓓ. Measure the density of water
Correct Answer: Experimentally confirm Einstein’s theory and calculate Avogadro’s number
Explanation: Perrin tracked suspended colloidal particles and measured their mean square displacements. His results matched Einstein’s predictions, allowing accurate calculation of Avogadro’s number and confirming atomic theory.
266. Why does Einstein’s theory link Brownian motion to the second law of thermodynamics?
ⓐ. It proves that heat flows from hot to cold.
ⓑ. It shows microscopic random fluctuations average out to macroscopic equilibrium.
ⓒ. It derives entropy directly.
ⓓ. It explains perpetual motion.
Correct Answer: It shows microscopic random fluctuations average out to macroscopic equilibrium.
Explanation: Brownian motion demonstrates that microscopic randomness produces macroscopic diffusion and thermal equilibrium, consistent with the second law. Energy is conserved at the micro-level, but disorder (entropy) increases overall.
267. What role did Einstein’s theory of Brownian motion play in physics history?
ⓐ. It confirmed the existence of photons.
ⓑ. It provided quantitative proof of atoms and molecules.
ⓒ. It led to Maxwell’s equations.
ⓓ. It explained nuclear reactions.
Correct Answer: It provided quantitative proof of atoms and molecules.
Explanation: Before Einstein’s work, atoms were still debated. His theory connected microscopic molecular motion to observable particle movement. Perrin’s verification convinced the scientific community of the atomic theory of matter.
268. Which physical constant can be determined from analyzing Brownian motion data?
ⓐ. Boltzmann constant ($k_B$)
ⓑ. Planck’s constant ($h$)
ⓒ. Universal gravitational constant ($G$)
ⓓ. Speed of light ($c$)
Correct Answer: Boltzmann constant ($k_B$)
Explanation: By measuring diffusion coefficient and mean squared displacement, $k_B$ can be experimentally determined. Once $k_B$ is known, Avogadro’s number can also be calculated using the gas constant relation $R = N_A k_B$.
269. Why does viscosity of the medium affect Brownian motion?
ⓐ. Viscosity reduces kinetic energy of suspended particles.
ⓑ. Viscosity resists molecular collisions, damping particle displacement.
ⓒ. Viscosity changes density of fluid.
ⓓ. Viscosity increases Avogadro’s number.
Correct Answer: Viscosity resists molecular collisions, damping particle displacement.
Explanation: Higher viscosity means greater resistance to motion of suspended particles. According to Einstein’s diffusion coefficient $D = \tfrac{k_BT}{6\pi \eta r}$, larger viscosity $\eta$ reduces $D$, hence the extent of Brownian motion.
270. Einstein’s explanation of Brownian motion is considered a bridge between:
ⓐ. Classical mechanics and relativity
ⓑ. Thermodynamics and statistical mechanics
ⓒ. Newton’s laws and electromagnetism
ⓓ. Quantum theory and optics
Correct Answer: Thermodynamics and statistical mechanics
Explanation: Einstein showed that Brownian motion is a macroscopic manifestation of microscopic molecular fluctuations. This connected thermodynamic laws with statistical descriptions, laying the foundation of statistical mechanics and modern molecular science.
271. In nanotechnology, Brownian motion is used to estimate the diffusion coefficient $D$ of nanoparticles. Which relation is applied?
ⓐ. $D = \frac{RT}{P}$
ⓑ. $D = \frac{k_B T}{6 \pi \eta r}$
ⓒ. $D = \frac{1}{2}at^2$
ⓓ. $D = \frac{v_{rms}^2}{3}$
Correct Answer: $D = \frac{k_B T}{6 \pi \eta r}$
Explanation: The Einstein–Stokes relation is used in nanoscience to compute diffusion coefficient of spherical particles of radius $r$ in a fluid of viscosity $\eta$. Smaller particles or lower viscosity lead to higher diffusion, important in drug delivery and colloid stability.
272. If a nanoparticle of radius $50 \, nm$ is suspended in water ($\eta = 1 \times 10^{-3} \, Pa \cdot s$) at room temperature ($T = 300 \, K$), calculate its diffusion coefficient.
ⓐ. $1.0 \times 10^{-11} \, m^2/s$
ⓑ. $4.4 \times 10^{-12} \, m^2/s$
ⓒ. $2.6 \times 10^{-13} \, m^2/s$
ⓓ. $5.0 \times 10^{-10} \, m^2/s$
Correct Answer: $4.4 \times 10^{-12} \, m^2/s$
Explanation: Using $D = \tfrac{k_B T}{6 \pi \eta r}$, with $k_B = 1.38 \times 10^{-23} \, J/K$, $T=300$, $ r=50 \times 10^{-9} \, m$, $\eta=10^{-3}$. Substitution gives $D \approx 4.4 \times 10^{-12} \, m^2/s$. This shows nanoscale particles diffuse slowly compared to molecules.
273. The root mean square displacement of a nanoparticle undergoing Brownian motion in time $t$ is given by:
ⓐ. $\sqrt{\langle x^2 \rangle} = \sqrt{2Dt}$
ⓑ. $\sqrt{\langle x^2 \rangle} = v_{rms}t$
ⓒ. $\sqrt{\langle x^2 \rangle} = \tfrac{1}{2}at^2$
ⓓ. $\sqrt{\langle x^2 \rangle} = \tfrac{RT}{P}$
Correct Answer: $\sqrt{\langle x^2 \rangle} = \sqrt{2Dt}$
Explanation: Einstein showed that Brownian motion resembles diffusion, where mean square displacement increases linearly with time. In 1D: $\langle x^2 \rangle = 2Dt$. Thus, longer observation time leads to larger particle displacement, used in nanoparticle tracking analysis.
274. A 100 nm gold nanoparticle in water has $D = 2.2 \times 10^{-12} \, m^2/s$. What is its root mean square displacement in 1 second?
ⓐ. $1.5 \times 10^{-6} \, m$
ⓑ. $2.1 \times 10^{-6} \, m$
ⓒ. $6.6 \times 10^{-7} \, m$
ⓓ. $4.7 \times 10^{-8} \, m$
Correct Answer: $6.6 \times 10^{-7} \, m$
Explanation: From $\sqrt{\langle x^2 \rangle} = \sqrt{2Dt} = \sqrt{2 \times 2.2 \times 10^{-12} \times 1} \approx 6.6 \times 10^{-7} \, m$. Nanoparticles move microscopically visible distances in fluids due to Brownian motion.
275. In particle physics, diffusion due to Brownian motion is important in gaseous detectors. The mean square displacement after time $t$ in 3D is:
ⓐ. $\langle r^2 \rangle = 2Dt$
ⓑ. $\langle r^2 \rangle = 4Dt$
ⓒ. $\langle r^2 \rangle = 6Dt$
ⓓ. $\langle r^2 \rangle = v_{rms}^2 t$
Correct Answer: $\langle r^2 \rangle = 6Dt$
Explanation: In three dimensions, mean squared displacement generalizes to $\langle r^2 \rangle = 6Dt$. This relation is used in analyzing charged particle drift and diffusion in cloud chambers and modern gaseous particle detectors.
276. Why is Brownian motion critical in nanomedicine for drug delivery systems?
ⓐ. It reduces drug solubility.
ⓑ. It enhances passive diffusion of nanoparticles through biological fluids.
ⓒ. It increases the binding energy of molecules.
ⓓ. It prevents nanoparticles from colliding.
Correct Answer: It enhances passive diffusion of nanoparticles through biological fluids.
Explanation: Nanoparticles carrying drugs rely on Brownian motion for transport in plasma and tissues. The random motion allows them to penetrate cellular structures without external energy, a key factor in targeted therapy design.
277. A nanoparticle of radius $r$ undergoes Brownian motion. If the viscosity of the medium is doubled, how does diffusion coefficient $D$ change?
ⓐ. $D$ becomes half.
ⓑ. $D$ doubles.
ⓒ. $D$ becomes one-fourth.
ⓓ. $D$ remains constant.
Correct Answer: $D$ becomes half.
Explanation: From Einstein–Stokes equation: $D = \tfrac{k_BT}{6 \pi \eta r}$. Doubling viscosity $\eta$ doubles the denominator, halving the diffusion coefficient. Thus, Brownian motion slows in thicker fluids, which is observed in nanoparticle suspensions.
278. In nuclear emulsions, Brownian-like motion of ions can be modeled using which formula?
ⓐ. $\langle x^2 \rangle = 2Dt$
ⓑ. $\langle x^2 \rangle = \tfrac{RT}{M}$
ⓒ. $\langle x^2 \rangle = \tfrac{1}{2}at^2$
ⓓ. $\langle x^2 \rangle = PVT$
Correct Answer: $\langle x^2 \rangle = 2Dt$
Explanation: Charged particles in emulsions experience random collisions with atoms, analogous to Brownian motion. Their diffusion-like spread can be described with Einstein’s displacement relation, useful in analyzing track broadening in detectors.
279. Why do nanoparticles exhibit stronger Brownian motion than microparticles?
ⓐ. They have larger mass.
ⓑ. They have smaller mass and radius, hence collisions impart greater acceleration.
ⓒ. They experience no fluid resistance.
ⓓ. They emit radiation energy.
Correct Answer: They have smaller mass and radius, hence collisions impart greater acceleration.
Explanation: Collisions from fluid molecules impart impulses independent of particle size. Smaller nanoparticles respond with greater acceleration, making their Brownian displacement much larger than bigger particles, which move sluggishly.
280. Which equation links diffusion coefficient of nanoparticles to temperature, radius, and viscosity, and is used in nanotechnology research?
ⓐ. $D = \tfrac{RT}{P}$
ⓑ. $D = \tfrac{k_BT}{6 \pi \eta r}$
ⓒ. $D = \tfrac{1}{2}mv^2$
ⓓ. $D = \tfrac{3RT}{2M}$
Correct Answer: $D = \tfrac{k_BT}{6 \pi \eta r}$
Explanation: The Einstein–Stokes relation is fundamental in nanoparticle research. It shows that diffusion coefficient depends on thermal energy ($k_BT$), particle radius ($r$), and medium viscosity ($\eta$). Smaller particles and lower viscosity enhance Brownian-driven diffusion.
281. Diffusion of gases is defined as:
ⓐ. Flow of gas molecules under gravity.
ⓑ. Random spreading of gas molecules into available space due to molecular motion.
ⓒ. Escape of gas molecules through a small hole without collisions.
ⓓ. Flow of gas molecules against a pressure gradient.
Correct Answer: Random spreading of gas molecules into available space due to molecular motion.
Explanation: Diffusion is the spontaneous mixing of gas molecules caused by their random motion. It is a slow process in gases compared to liquids because molecular collisions limit the net rate of mixing.
282. Effusion of gases is best described as:
ⓐ. Movement of gas molecules into another gas.
ⓑ. Escape of gas molecules through a small hole without collisions.
ⓒ. Compression of gases under high pressure.
ⓓ. Increase in gas volume due to temperature rise.
Correct Answer: Escape of gas molecules through a small hole without collisions.
Explanation: Effusion occurs when gas molecules pass through a tiny hole in a container into vacuum without colliding with each other or the walls. The rate depends on molecular speed and thus molar mass.
283. According to Graham’s law, the rate of diffusion or effusion is:
ⓐ. $r \propto M$
ⓑ. $r \propto \tfrac{1}{M}$
ⓒ. $r \propto \sqrt{M}$
ⓓ. $r \propto \tfrac{1}{\sqrt{M}}$
Correct Answer: $r \propto \tfrac{1}{\sqrt{M}}$
Explanation: Graham’s law states $\tfrac{r_1}{r_2} = \sqrt{\tfrac{M_2}{M_1}}$, where $M$ is molar mass. Lighter gases move faster, so they diffuse and effuse more quickly than heavier gases.
284. The ratio of rates of effusion of hydrogen and oxygen is:
ⓐ. 1:1
ⓑ. 2:1
ⓒ. 4:1
ⓓ. 8:1
Correct Answer: 8:1
Explanation: By Graham’s law, $\tfrac{r_H}{r_O} = \sqrt{\tfrac{M_O}{M_H}} = \sqrt{\tfrac{32}{2}} = \sqrt{16} = 4$. Careful: since $M_H = 2$ and $M_O = 32$, the ratio is $\sqrt{32/2} = \sqrt{16} = 4$. So the correct ratio is 4:1, not 8:1. Correct answer: C. 4:1.
285. Calculate the relative rate of diffusion of nitrogen ($M = 28$) to oxygen ($M = 32$).
ⓐ. 1.00
ⓑ. 1.07
ⓒ. 0.95
ⓓ. 1.20
Correct Answer: 1.07
Explanation: Using Graham’s law, $\tfrac{r_N}{r_O} = \sqrt{\tfrac{32}{28}} = \sqrt{1.14} \approx 1.07$. Thus nitrogen diffuses about 7% faster than oxygen under the same conditions.
286. A gas effuses 1.414 times faster than oxygen. What is its molar mass?
ⓐ. 8 g/mol
ⓑ. 16 g/mol
ⓒ. 22 g/mol
ⓓ. 32 g/mol
Correct Answer: 16 g/mol
Explanation: By Graham’s law: $\tfrac{r_{gas}}{r_O} = \sqrt{\tfrac{M_O}{M_{gas}}} = 1.414$. Squaring both sides: $\tfrac{32}{M_{gas}} = 2$. Hence, $M_{gas} = 16 \, g/mol$, which corresponds to methane.
287. Which of the following conditions increases the rate of diffusion of a gas?
ⓐ. Higher molar mass
ⓑ. Lower molar mass and higher temperature
ⓒ. Higher pressure
ⓓ. Strong intermolecular forces
Correct Answer: Lower molar mass and higher temperature
Explanation: Diffusion rate depends on molecular speed, which increases with temperature and decreases with molar mass ($v_{rms} = \sqrt{\tfrac{3RT}{M}}$). Lighter gases at higher temperatures diffuse faster.
288. If two gases $A$ and $B$ have molar masses $M_A$ and $M_B$, the ratio of their rates of diffusion is:
ⓐ. $\tfrac{r_A}{r_B} = \tfrac{M_A}{M_B}$
ⓑ. $\tfrac{r_A}{r_B} = \sqrt{\tfrac{M_B}{M_A}}$
ⓒ. $\tfrac{r_A}{r_B} = \sqrt{\tfrac{M_A}{M_B}}$
ⓓ. $\tfrac{r_A}{r_B} = \tfrac{M_B}{M_A}$
Correct Answer: $\tfrac{r_A}{r_B} = \sqrt{\tfrac{M_B}{M_A}}$
Explanation: Graham’s law states that the rate of diffusion/effusion is inversely proportional to the square root of molar mass. So lighter gases diffuse faster.
289. Why is effusion slower for heavy gases than for light gases?
ⓐ. Heavy gases have larger volume.
ⓑ. Heavy gases move with lower average speeds due to higher molar mass.
ⓒ. Heavy gases stick to the container walls.
ⓓ. Heavy gases are more viscous.
Correct Answer: Heavy gases move with lower average speeds due to higher molar mass.
Explanation: From kinetic theory, $v_{rms} = \sqrt{\tfrac{3RT}{M}}$. Larger $M$ means smaller speed, so fewer molecules strike the effusion hole per unit time. Thus, heavy gases effuse slower.
290. Which real-world process is based on gas effusion?
ⓐ. Electrolysis of water
ⓑ. Uranium isotope separation ($UF_6$)
ⓒ. Photosynthesis
ⓓ. Rocket propulsion
Correct Answer: Uranium isotope separation ($UF_6$)
Explanation: Effusion through porous membranes separates isotopes because lighter $^{235}UF_6$ molecules effuse slightly faster than heavier $^{238}UF_6$. Repeating many times enriches uranium, critical in nuclear technology.
291. Which kinetic theory expression relates pressure to molecular speed distribution?
ⓐ. $P = \tfrac{1}{3}\rho \langle v^2 \rangle$
ⓑ. $P = nRT$
ⓒ. $P = \rho g h$
ⓓ. $P = \tfrac{F}{A}$
Correct Answer: $P = \tfrac{1}{3}\rho \langle v^2 \rangle$
Explanation: Kinetic theory states that pressure of a gas arises from elastic collisions of molecules with the container walls. The formula $P = \tfrac{1}{3}\rho \langle v^2 \rangle$ connects macroscopic pressure with microscopic average of squared molecular speeds, where $\rho$ is mass density.
292. The Maxwell–Boltzmann distribution explains that:
ⓐ. All gas molecules move with the same speed at a given temperature.
ⓑ. Gas molecules have a spread of speeds, with most around a probable value.
ⓒ. Gas molecules stop moving at high temperatures.
ⓓ. Gas molecules move only with rms speed.
Correct Answer: Gas molecules have a spread of speeds, with most around a probable value.
Explanation: The distribution function shows that some molecules move slower, some faster, but the majority cluster around a most probable speed. This explains why macroscopic gas properties are averages over distributions.
293. Which of the following correctly represents root mean square (rms) speed?
ⓐ. $v_{rms} = \sqrt{\tfrac{2RT}{M}}$
ⓑ. $v_{rms} = \sqrt{\tfrac{3RT}{M}}$
ⓒ. $v_{rms} = \sqrt{\tfrac{8RT}{\pi M}}$
ⓓ. $v_{rms} = \sqrt{\tfrac{RT}{M}}$
Correct Answer: $v_{rms} = \sqrt{\tfrac{3RT}{M}}$
Explanation: From kinetic theory, mean square speed is $\langle v^2 \rangle = \tfrac{3RT}{M}$. Taking the square root gives rms speed. This is slightly larger than average and most probable speeds.
294. A sample of oxygen ($M = 32 \, g/mol$) is at 300 K. Calculate its rms speed.
ⓐ. $300 \, m/s$
ⓑ. $400 \, m/s$
ⓒ. $500 \, m/s$
ⓓ. $600 \, m/s$
Correct Answer: $500 \, m/s$
Explanation: Using $v_{rms} = \sqrt{\tfrac{3RT}{M}}$. Here $R = 8.314 \, J/molK$, $M = 0.032 \, kg/mol$. Substituting: $v_{rms} = \sqrt{\tfrac{3 \times 8.314 \times 300}{0.032}} \approx 483 \, m/s$, \~500 m/s.
295. The order of molecular speeds at a given temperature is:
ⓐ. $v_{avg} > v_{mp} > v_{rms}$
ⓑ. $v_{rms} > v_{avg} > v_{mp}$
ⓒ. $v_{mp} > v_{avg} > v_{rms}$
ⓓ. $v_{avg} = v_{rms} = v_{mp}$
Correct Answer: $v_{rms} > v_{avg} > v_{mp}$
Explanation: From equations:
$v_{mp} = \sqrt{\tfrac{2RT}{M}}$
$v_{avg} = \sqrt{\tfrac{8RT}{\pi M}}$
$v_{rms} = \sqrt{\tfrac{3RT}{M}}$.
Numerically, $v_{rms}$ is greatest, then $v_{avg}$, then $v_{mp}$.
296. The molecular speed distribution widens when:
ⓐ. Temperature decreases
ⓑ. Temperature increases
ⓒ. Molar mass increases
ⓓ. Pressure increases
Correct Answer: Temperature increases
Explanation: At higher $T$, molecules have more kinetic energy, so speeds spread over a broader range. The Maxwell–Boltzmann curve flattens and shifts right, showing more molecules with both very high and very low speeds.
297. A gas at 400 K has molecules with $v_{mp} = 400 \, m/s$. If temperature doubles, the new most probable speed is:
ⓐ. $400 \, m/s$
ⓑ. $565 \, m/s$
ⓒ. $800 \, m/s$
ⓓ. $1130 \, m/s$
Correct Answer: $800 \, m/s$
Explanation: Since $v_{mp} = \sqrt{\tfrac{2RT}{M}}$, it scales as $\sqrt{T}$. Doubling $T$ multiplies speed by $\sqrt{2}$. So new speed = $400 \times \sqrt{2} \approx 566 \, m/s$. Correcting: at 400 K, if $v_{mp}=400$, then at 800 K it is $400 \times \sqrt{2} \approx 566 \, m/s$. Correct answer: B. 565 m/s.
298. The kinetic theory connects temperature with:
ⓐ. rms momentum of molecules
ⓑ. Average translational kinetic energy per molecule
ⓒ. Potential energy of molecules
ⓓ. Rotational kinetic energy only
Correct Answer: Average translational kinetic energy per molecule
Explanation: Temperature is a direct measure of average translational kinetic energy: $\langle E_k \rangle = \tfrac{3}{2}k_BT$. Thus, macroscopic temperature scale is rooted in microscopic random motions.
299. Which phenomenon directly validates molecular speed distribution in gases?
ⓐ. Brownian motion
ⓑ. Diffusion and effusion
ⓒ. Photoelectric effect
ⓓ. Elasticity of solids
Correct Answer: Diffusion and effusion
Explanation: Graham’s law shows lighter gases diffuse/effuse faster, consistent with Maxwell–Boltzmann distribution prediction that molecular speed depends on molar mass. Thus, gas mixing experiments confirm molecular speed distribution.
300. Why does Maxwell–Boltzmann distribution predict that some molecules always exceed escape velocity from Earth?
ⓐ. Because all molecules have the same velocity.
ⓑ. Because distribution has no upper limit to speeds.
ⓒ. Because Earth’s gravity depends on molar mass.
ⓓ. Because molecular collisions stop at high speeds.
Correct Answer: Because distribution has no upper limit to speeds.
Explanation: The Maxwell–Boltzmann curve extends infinitely to high speeds, though with low probability. A tiny fraction of molecules always have speeds greater than Earth’s escape velocity. This explains gradual loss of light gases (like hydrogen, helium) from the atmosphere.
The Kinetic Theory is one of the most fundamental chapters in Class 11 Physics, explaining the connection between microscopic molecular motion and macroscopic properties of gases. Based on the NCERT/CBSE syllabus, this chapter discusses equipartition of energy, calculation of specific heats of gases, derivation of mean free path, and molecular speeds. It also highlights the significance of Boltzmann constant, Avogadro’s number, and gas constant, which are central to solving numerical problems in competitive exams. With a complete set of 498 MCQs and explanations, divided into 5 parts, this section (Part 3) presents another **100 important MCQs**, covering both conceptual and numerical problems for exam practice.
- 👉 Total MCQs in this chapter: 498.
- 👉 This page contains: Third set of 100 solved MCQs.
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