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301. In chemical engineering, the diffusion coefficient $D$ for gases is often estimated using kinetic theory as:
ⓐ. $D = \frac{1}{3}\langle v \rangle \lambda$
ⓑ. $D = \frac{k_B T}{P}$
ⓒ. $D = \frac{RT}{M}$
ⓓ. $D = \frac{3}{2}k_B T$
Correct Answer: $D = \frac{1}{3}\langle v \rangle \lambda$
Explanation: The diffusion coefficient is proportional to average molecular speed ($\langle v \rangle$) and mean free path ($\lambda$). This shows that faster molecules and longer free paths lead to greater diffusion. It is widely used in reactor design and gas transport models.
302. Which principle is used in designing catalytic converters for automobiles?
ⓐ. Boyle’s law of gases
ⓑ. Diffusion and surface reaction of molecules
ⓒ. Law of electrostatics
ⓓ. Conservation of angular momentum
Correct Answer: Diffusion and surface reaction of molecules
Explanation: Exhaust gases diffuse to catalyst surfaces (like platinum), where they undergo chemical reactions (oxidation of CO, reduction of NOx). Kinetic theory of gas diffusion helps optimize the efficiency of catalytic converters.
303. In atmospheric science, why does oxygen diffuse slower than nitrogen in air?
ⓐ. Oxygen molecules are lighter.
ⓑ. Nitrogen molecules are lighter, so they move faster.
ⓒ. Oxygen molecules experience stronger gravitational pull.
ⓓ. Nitrogen has more electrons.
Correct Answer: Nitrogen molecules are lighter, so they move faster.
Explanation: From Graham’s law, $r \propto \tfrac{1}{\sqrt{M}}$. With molar masses $M_N = 28$, $M_O = 32$, nitrogen diffuses \~7% faster than oxygen. This affects atmospheric mixing and combustion processes.
304. Calculate the ratio of diffusion rates of helium (M = 4) and nitrogen (M = 28).
ⓐ. 2
ⓑ. $\sqrt{7}$
ⓒ. 4
ⓓ. $\tfrac{1}{2}$
Correct Answer: $\sqrt{7}$
Explanation: From Graham’s law, $\tfrac{r_{He}}{r_{N_2}} = \sqrt{\tfrac{M_{N_2}}{M_{He}}} = \sqrt{\tfrac{28}{4}} = \sqrt{7} \approx 2.65$. Hence, helium diffuses about 2.65 times faster than nitrogen.
305. Effusion is applied in separating isotopes in nuclear fuel processing. Which relation gives the separation factor between two isotopes of molar masses $M_1$ and $M_2$?
ⓐ. $\alpha = \sqrt{\tfrac{M_1}{M_2}}$
ⓑ. $\alpha = \tfrac{M_1}{M_2}$
ⓒ. $\alpha = \tfrac{M_2}{M_1}$
ⓓ. $\alpha = \sqrt{\tfrac{M_2}{M_1}}$
Correct Answer: $\alpha = \sqrt{\tfrac{M_2}{M_1}}$
Explanation: According to Graham’s law, lighter isotopes effuse slightly faster. For $UF_6$, $^{235}UF_6$ effuses faster than $^{238}UF_6$. The separation factor per stage is small, but repeated stages achieve enrichment.
306. In atmospheric escape processes, lighter gases like hydrogen are lost from Earth because:
ⓐ. They dissolve in oceans.
ⓑ. Their speeds exceed escape velocity due to Maxwell–Boltzmann distribution.
ⓒ. They chemically react with oxygen.
ⓓ. They are absorbed by plants.
Correct Answer: Their speeds exceed escape velocity due to Maxwell–Boltzmann distribution.
Explanation: The distribution curve predicts that some fraction of molecules have speeds higher than Earth’s escape velocity (\~11.2 km/s). Lighter gases (smaller $M$) have higher rms speeds, making escape more probable.
307. In chemical reactors, gas mixing efficiency is strongly influenced by:
ⓐ. Gravitational potential energy
ⓑ. Molecular diffusion rates
ⓒ. Only pressure and volume
ⓓ. Crystal lattice vibrations
Correct Answer: Molecular diffusion rates
Explanation: Reactor design considers how quickly reactant gases diffuse and mix. Higher diffusion coefficients improve contact between molecules and catalyst surfaces, increasing yield and efficiency.
308. Why does the stratosphere contain a higher concentration of ozone compared to the troposphere?
ⓐ. Stronger gravitational pull at higher altitudes.
ⓑ. Diffusion and photochemical reactions of oxygen molecules under UV light.
ⓒ. Higher density of air molecules.
ⓓ. More water vapor at that level.
Correct Answer: Diffusion and photochemical reactions of oxygen molecules under UV light.
Explanation: UV radiation dissociates O$_2$ into atomic oxygen, which diffuses and reacts to form O$_3$. Diffusion processes distribute ozone within the stratosphere, while molecular kinetics and sunlight sustain the ozone layer.
309. Calculate the effusion rate ratio of hydrogen ($M=2$) to carbon dioxide ($M=44$).
ⓐ. 2
ⓑ. 3.7
ⓒ. 6.6
ⓓ. 4.7
Correct Answer: 4.7
Explanation: By Graham’s law, $\tfrac{r_{H_2}}{r_{CO_2}} = \sqrt{\tfrac{44}{2}} = \sqrt{22} \approx 4.69$. Careful: recalculating → $\sqrt{44/2} = \sqrt{22} \approx 4.7$.
310. Which atmospheric phenomenon is best explained by gas diffusion and kinetic theory?
ⓐ. Lightning
ⓑ. Aurora Borealis
ⓒ. Smog formation and pollutant mixing
ⓓ. Earthquake waves
Correct Answer: Smog formation and pollutant mixing
Explanation: Pollutants released in cities spread due to molecular diffusion and air convection. Kinetic theory predicts how gas molecules distribute in the atmosphere, explaining why smog and greenhouse gases disperse globally over time.
311. Collision theory states that a chemical reaction occurs when:
ⓐ. Molecules collide with low energy.
ⓑ. Molecules collide with proper orientation and energy greater than activation energy.
ⓒ. Molecules vibrate at room temperature.
ⓓ. Molecules collide without transferring energy.
Correct Answer: Molecules collide with proper orientation and energy greater than activation energy.
Explanation: Collision theory emphasizes two main factors: molecules must collide with energy greater than or equal to the activation energy, and collisions must have the correct spatial orientation. Only such “effective collisions” lead to product formation.
312. The rate of a chemical reaction according to collision theory is directly proportional to:
ⓐ. Number of effective collisions per second
ⓑ. Average velocity of one molecule
ⓒ. Mass of reactants
ⓓ. Heat released in the reaction
Correct Answer: Number of effective collisions per second
Explanation: While molecules collide frequently, only a fraction are effective. Reaction rate depends on this number, which is controlled by concentration, temperature, and activation energy.
313. The Arrhenius equation relates rate constant to activation energy:
ⓐ. $k = A \exp\left(-\frac{E_a}{RT}\right)$
ⓑ. $k = A \exp\left(\frac{RT}{E_a}\right)$
ⓒ. $k = \frac{E_a}{RT}$
ⓓ. $k = \frac{RT}{E_a}$
Correct Answer: $k = A \exp\left(-\frac{E_a}{RT}\right)$
Explanation: Arrhenius equation refines collision theory, introducing activation energy $E_a$. $A$ is the frequency factor (related to collision frequency and orientation). Higher $T$ increases $k$ by reducing the exponential barrier.
314. Why does increasing temperature generally increase reaction rate?
ⓐ. Molecules move slower and collide less often.
ⓑ. More molecules have energies exceeding activation energy.
ⓒ. Activation energy decreases with temperature.
ⓓ. Orientation of molecules changes.
Correct Answer: More molecules have energies exceeding activation energy.
Explanation: Maxwell–Boltzmann distribution broadens at higher temperatures, shifting more molecules into the high-energy tail. Thus, a greater fraction of collisions have energy above $E_a$, increasing the rate.
315. Which factor does NOT affect collision frequency?
ⓐ. Concentration of reactants
ⓑ. Temperature of the system
ⓒ. Activation energy of the reaction
ⓓ. Pressure of gaseous reactants
Correct Answer: Activation energy of the reaction
Explanation: Collision frequency depends on molecular concentration, pressure (in gases), and speed (temperature). Activation energy determines whether collisions are effective, but does not affect frequency itself.
316. In terms of kinetic theory, collision frequency ($Z$) for two species is proportional to:
ⓐ. $\sqrt{\tfrac{RT}{M}}$
ⓑ. $n \sigma \langle v \rangle$
ⓒ. $PV$
ⓓ. $RT$
Correct Answer: $n \sigma \langle v \rangle$
Explanation: $Z \propto n\sigma \langle v \rangle$, where $n$ is number density, $\sigma$ collision cross-section, and $\langle v \rangle$ average relative speed. More particles, larger effective size, or faster motion lead to more collisions.
317. In a bimolecular reaction $A + B \to \text{Products}$, the rate constant in collision theory is expressed as:
ⓐ. $k = Z_{AB}$
ⓑ. $k = P Z_{AB} \exp\left(-\frac{E_a}{RT}\right)$
ⓒ. $k = \tfrac{E_a}{RT}$
ⓓ. $k = \tfrac{RT}{E_a}$
Correct Answer: $k = P Z_{AB} \exp\left(-\frac{E_a}{RT}\right)$
Explanation: Here, $Z_{AB}$ is collision frequency between $A$ and $B$, $P$ is the steric factor accounting for orientation, and the exponential factor accounts for the fraction of collisions with sufficient energy.
318. What does the steric factor $P$ in collision theory account for?
ⓐ. Collision frequency
ⓑ. Fraction of collisions with correct orientation
ⓒ. Activation energy
ⓓ. Heat of reaction
Correct Answer: Fraction of collisions with correct orientation
Explanation: Not all high-energy collisions lead to reactions. $P$ adjusts for molecular orientation and geometry. For complex molecules, $P$ may be much smaller than 1, reducing effective collision rate.
319. If temperature rises from 300 K to 330 K, how does the rate constant $k$ typically change for a reaction with $E_a = 50 \, kJ/mol$?
ⓐ. No change
ⓑ. Increases slightly (\~2x)
ⓒ. Decreases by half
ⓓ. Becomes zero
Correct Answer: Increases slightly (\~2x)
Explanation: Using Arrhenius law, small increases in $T$ significantly increase the exponential term. For $E_a = 50\,kJ/mol$, the rate approximately doubles for every 10–20 K rise, a well-known empirical rule.
320. Why do reactions between ions in aqueous solution often occur very fast compared to neutral molecules?
ⓐ. They have very low activation energy due to electrostatic attraction.
ⓑ. They have larger steric factor.
ⓒ. They collide less frequently.
ⓓ. They depend only on pressure.
Correct Answer: They have very low activation energy due to electrostatic attraction.
Explanation: Ionic reactions often have negligible or very small activation energy because opposite charges attract strongly. Hence, almost every collision is effective, making ionic reactions extremely fast compared to covalent molecule reactions.
321. Which of the following does NOT affect the rate of a chemical reaction?
ⓐ. Concentration of reactants
ⓑ. Temperature of the system
ⓒ. Catalyst presence
ⓓ. Atomic number of the elements
Correct Answer: Atomic number of the elements
Explanation: Reaction rate depends on reactant concentration (more molecules → more collisions), temperature (more energy → more effective collisions), and catalysts (lower activation energy). Atomic number itself does not directly affect rate unless it changes reactivity.
322. How does increasing concentration of reactants affect reaction rate?
ⓐ. Rate decreases due to more collisions.
ⓑ. Rate increases due to higher collision frequency.
ⓒ. Rate stays the same.
ⓓ. Rate becomes independent of concentration.
Correct Answer: Rate increases due to higher collision frequency.
Explanation: According to collision theory, rate is proportional to number of effective collisions. Higher concentration increases molecular density, producing more collisions per second and thus increasing reaction rate.
323. According to Arrhenius equation $k = A e^{-E_a/RT}$, how does temperature affect rate constant?
ⓐ. Higher temperature decreases $k$.
ⓑ. Higher temperature increases $k$ exponentially.
ⓒ. Higher temperature makes $k$ constant.
ⓓ. Higher temperature reduces activation energy.
Correct Answer: Higher temperature increases $k$ exponentially.
Explanation: The exponential factor depends on $-E_a/RT$. As $T$ rises, the fraction of molecules with $E \geq E_a$ increases dramatically, making the rate constant grow exponentially with temperature.
324. How does the presence of a catalyst affect the reaction rate?
ⓐ. Increases rate by increasing activation energy.
ⓑ. Increases rate by lowering activation energy.
ⓒ. Decreases rate by slowing collisions.
ⓓ. Makes rate independent of concentration.
Correct Answer: Increases rate by lowering activation energy.
Explanation: Catalysts provide an alternate reaction pathway with lower $E_a$. More molecules now have sufficient energy to overcome the barrier, increasing the number of effective collisions without being consumed in the reaction.
325. Pressure affects the rate of reaction most strongly in:
ⓐ. Gaseous reactions
ⓑ. Solid-state reactions
ⓒ. Liquid-phase reactions
ⓓ. Nuclear reactions
Correct Answer: Gaseous reactions
Explanation: Increasing pressure compresses gas molecules, raising concentration. This increases collision frequency and hence reaction rate. In solids and liquids, pressure has minimal effect because their densities are almost constant.
326. The rate of most chemical reactions approximately doubles when temperature increases by:
ⓐ. 1 °C
ⓑ. 5 °C
ⓒ. 10 °C
ⓓ. 20 °C
Correct Answer: 10 °C
Explanation: This is an empirical rule derived from Arrhenius behavior. A 10 °C rise typically doubles or triples reaction rates due to exponential growth in the fraction of molecules exceeding activation energy.
327. For a reaction with high activation energy, temperature dependence of the rate is:
ⓐ. Very strong
ⓑ. Very weak
ⓒ. Zero
ⓓ. Inversely proportional
Correct Answer: Very strong
Explanation: If $E_a$ is large, only a small fraction of molecules exceed it at low temperatures. Raising temperature greatly increases the fraction of high-energy molecules, thus the rate becomes highly temperature-sensitive.
328. In heterogeneous catalysis, why does increasing surface area of catalyst increase reaction rate?
ⓐ. It reduces activation energy further.
ⓑ. It increases availability of active sites for collisions.
ⓒ. It changes the stoichiometric coefficients.
ⓓ. It increases temperature of the system.
Correct Answer: It increases availability of active sites for collisions.
Explanation: In surface-catalyzed reactions, reactants adsorb onto catalyst surface. A larger surface area (e.g., using powdered catalyst) provides more sites for adsorption and collisions, enhancing reaction rate.
329. Why do reactions between ions in aqueous solution usually occur very fast?
ⓐ. High concentration of water
ⓑ. Orientation requirement is absent, and activation energy is very low
ⓒ. Ions have very high molar mass
ⓓ. Pressure effect dominates
Correct Answer: Orientation requirement is absent, and activation energy is very low
Explanation: Ionic reactions are driven by electrostatic attraction, so most collisions are effective. Unlike covalent reactions, where orientation matters, ionic reactions often occur almost instantly upon contact.
330. Light can increase the rate of some chemical reactions (photochemical reactions) because:
ⓐ. It raises the temperature directly.
ⓑ. It excites reactant molecules, providing energy greater than activation energy.
ⓒ. It increases concentration of molecules.
ⓓ. It increases pressure of the gas.
Correct Answer: It excites reactant molecules, providing energy greater than activation energy.
Explanation: In photochemical reactions (e.g., photosynthesis, decomposition of AgCl), photons excite electrons or bonds, enabling molecules to cross activation energy barriers. Thus, light effectively acts like a catalyst for such reactions.
331. Why are catalysts widely used in chemical industries?
ⓐ. They increase the yield by changing equilibrium position.
ⓑ. They increase reaction rate without being consumed.
ⓒ. They provide heat to the reaction.
ⓓ. They decrease concentration of reactants.
Correct Answer: They increase reaction rate without being consumed.
Explanation: Catalysts lower activation energy by providing an alternative pathway. This increases the fraction of effective collisions. They are not used up in the process and can be reused, making them vital in industrial-scale reactions.
332. Which catalyst is used in the Haber process for ammonia synthesis?
ⓐ. Nickel
ⓑ. Platinum
ⓒ. Iron with promoters (K₂O, Al₂O₃)
ⓓ. Copper
Correct Answer: Iron with promoters (K₂O, Al₂O₃)
Explanation: In the Haber process, finely divided iron acts as the catalyst, with promoters like potassium oxide and alumina to enhance activity. It allows nitrogen and hydrogen to react effectively at high pressure and moderate temperature.
333. In the contact process for sulfuric acid production, which catalyst is used?
ⓐ. V₂O₅
ⓑ. Fe₂O₃
ⓒ. Ni
ⓓ. Pt
Correct Answer: V₂O₅
Explanation: In the contact process, vanadium(V) oxide catalyzes oxidation of SO₂ to SO₃. This heterogeneous catalysis lowers activation energy and increases efficiency at \~450 °C.
334. Enzymes act as catalysts in biochemical reactions because:
ⓐ. They increase the temperature of the body.
ⓑ. They lower activation energy by binding reactants in specific orientations.
ⓒ. They supply reactants with extra molecules.
ⓓ. They absorb light and convert it to heat.
Correct Answer: They lower activation energy by binding reactants in specific orientations.
Explanation: Enzymes use their active sites to bind substrates, stabilizing transition states and decreasing activation energy. This allows reactions to proceed rapidly under mild biological conditions.
335. Which industrial process uses finely divided platinum as a catalyst?
ⓐ. Haber process
ⓑ. Ostwald process
ⓒ. Fischer–Tropsch synthesis
ⓓ. Hydrogenation of alkenes
Correct Answer: Hydrogenation of alkenes
Explanation: Platinum and nickel are used in hydrogenation reactions where H₂ adds across double bonds. This is applied in producing margarine from vegetable oils.
336. The use of catalysts in chemical kinetics primarily affects:
ⓐ. Activation energy of the reaction
ⓑ. Concentration of reactants
ⓒ. Rate of diffusion
ⓓ. Value of the gas constant $R$
Correct Answer: Activation energy of the reaction
Explanation: Catalysts reduce the energy barrier (activation energy, $E_a$) needed for a reaction. This increases the rate constant $k$ as per Arrhenius equation, making more collisions effective at the same temperature.
337. The Ostwald process for nitric acid production uses which catalyst?
ⓐ. Iron
ⓑ. Platinum–rhodium gauze
ⓒ. Nickel
ⓓ. V₂O₅
Correct Answer: Platinum–rhodium gauze
Explanation: In the Ostwald process, ammonia is oxidized using platinum–rhodium gauze as catalyst. This is an industrial-scale catalytic reaction essential in fertilizers and explosives production.
338. Why are heterogeneous catalysts often used instead of homogeneous ones in industry?
ⓐ. They are always faster.
ⓑ. They are easier to separate from reaction products.
ⓒ. They change the equilibrium constant.
ⓓ. They do not need promoters.
Correct Answer: They are easier to separate from reaction products.
Explanation: Heterogeneous catalysts (e.g., solid catalysts in gas-phase reactions) can be filtered or reused easily. Homogeneous catalysts dissolve in the reaction mixture and are harder to recover.
339. In enzyme catalysis, the “lock and key model” explains:
ⓐ. How catalysts increase entropy
ⓑ. Specific binding of substrate to enzyme’s active site
ⓒ. Diffusion of molecules in liquids
ⓓ. Energy release during combustion
Correct Answer: Specific binding of substrate to enzyme’s active site
Explanation: The lock-and-key analogy shows how enzymes have specific active sites where only matching substrates can bind. This ensures high selectivity and efficiency in biochemical catalysis.
340. Catalysts are essential in catalytic converters because they:
ⓐ. Reduce automobile speed
ⓑ. Convert harmful exhaust gases (CO, NOx, hydrocarbons) into less harmful ones
ⓒ. Increase engine temperature
ⓓ. Absorb sound vibrations
Correct Answer: Convert harmful exhaust gases (CO, NOx, hydrocarbons) into less harmful ones
Explanation: Catalytic converters use platinum, rhodium, and palladium as catalysts. They oxidize CO and hydrocarbons to CO₂ and H₂O, and reduce NOx to N₂, minimizing air pollution from vehicles.
341. Statistical mechanics is primarily concerned with:
ⓐ. Individual trajectories of molecules
ⓑ. Average properties of large ensembles of particles
ⓒ. Chemical bonding theories
ⓓ. Nuclear reactions only
Correct Answer: Average properties of large ensembles of particles
Explanation: Statistical mechanics connects microscopic states of individual molecules with macroscopic thermodynamic observables (e.g., pressure, temperature, entropy). Instead of tracking each particle, it uses probability distributions to describe ensemble behavior.
342. The Boltzmann constant ($k_B$) relates:
ⓐ. Energy per molecule to energy per mole
ⓑ. Temperature with average energy per molecule
ⓒ. Pressure with density
ⓓ. Heat with entropy only
Correct Answer: Temperature with average energy per molecule
Explanation: $k_B = 1.38 \times 10^{-23} \, J/K$. From kinetic theory, average translational energy per molecule is $\langle E \rangle = \tfrac{3}{2}k_BT$. This constant bridges microscopic and macroscopic descriptions of thermodynamics.
343. The number of microstates ($\Omega$) corresponding to a macrostate is connected to entropy by Boltzmann’s formula:
ⓐ. $S = \frac{Q}{T}$
ⓑ. $S = k_B \ln \Omega$
ⓒ. $S = nR \ln V$
ⓓ. $S = \tfrac{1}{2}k_BT$
Correct Answer: $S = k_B \ln \Omega$
Explanation: Ludwig Boltzmann derived that entropy measures the logarithm of the number of microscopic configurations consistent with a macroscopic state. This formula, engraved on his tombstone, is a cornerstone of statistical mechanics.
344. Which distribution describes the probability of molecular speeds in an ideal gas?
ⓐ. Planck distribution
ⓑ. Maxwell–Boltzmann distribution
ⓒ. Fermi–Dirac distribution
ⓓ. Bose–Einstein distribution
Correct Answer: Maxwell–Boltzmann distribution
Explanation: The Maxwell–Boltzmann distribution gives the probability of finding molecules at different speeds in a classical ideal gas. Fermi–Dirac and Bose–Einstein distributions apply to quantum particles (fermions and bosons).
345. The partition function $Z$ in statistical mechanics is defined as:
ⓐ. $Z = \sum e^{-E_i/k_BT}$
ⓑ. $Z = PV$
ⓒ. $Z = nRT$
ⓓ. $Z = \tfrac{E}{T}$
Correct Answer: $Z = \sum e^{-E_i/k_BT}$
Explanation: The partition function sums over all possible microstates $i$, weighting each by its Boltzmann factor $e^{-E_i/k_BT}$. It is a central quantity since all thermodynamic properties can be derived from it.
346. For an ideal gas, internal energy can be expressed in terms of partition function as:
ⓐ. $U = nRT$
ⓑ. $U = -\frac{\partial}{\partial \beta} \ln Z$
ⓒ. $U = PV$
ⓓ. $U = \tfrac{3}{2}nRT$
Correct Answer: $U = -\frac{\partial}{\partial \beta} \ln Z$
Explanation: In statistical mechanics, $\beta = 1/k_BT$. Internal energy is calculated by differentiating the logarithm of the partition function with respect to $\beta$. This general formula applies to all systems.
347. Which statistical distribution applies to indistinguishable particles with half-integer spin (electrons, protons, neutrons)?
ⓐ. Maxwell–Boltzmann
ⓑ. Fermi–Dirac
ⓒ. Bose–Einstein
ⓓ. Poisson
Correct Answer: Fermi–Dirac
Explanation: Fermions obey the Pauli exclusion principle, which restricts occupancy to one particle per state. Their distribution is described by Fermi–Dirac statistics, crucial in explaining properties of metals and white dwarfs.
348. Which distribution applies to indistinguishable particles with integer spin (photons, helium-4 atoms)?
ⓐ. Maxwell–Boltzmann
ⓑ. Fermi–Dirac
ⓒ. Bose–Einstein
ⓓ. Poisson
Correct Answer: Bose–Einstein
Explanation: Bosons do not obey Pauli exclusion and can occupy the same state. Bose–Einstein distribution explains phenomena like superfluidity and Bose–Einstein condensates, where large numbers of particles occupy the ground state.
349. Which classical result emerges as the high-temperature limit of both Fermi–Dirac and Bose–Einstein statistics?
ⓐ. Planck’s law
ⓑ. Maxwell–Boltzmann distribution
ⓒ. Stefan–Boltzmann law
ⓓ. Heisenberg uncertainty principle
Correct Answer: Maxwell–Boltzmann distribution
Explanation: At high temperatures or low densities, quantum effects become negligible, and both Fermi–Dirac and Bose–Einstein statistics reduce to the classical Maxwell–Boltzmann distribution.
350. Why is statistical mechanics important in modern physics and chemistry?
ⓐ. It explains motion of planets.
ⓑ. It links microscopic particle behavior with macroscopic thermodynamic laws.
ⓒ. It proves relativity.
ⓓ. It only studies black holes.
Correct Answer: It links microscopic particle behavior with macroscopic thermodynamic laws.
Explanation: Statistical mechanics provides the framework for deriving macroscopic thermodynamic properties (like entropy, free energy, specific heat) from microscopic principles. This makes it essential in physics, chemistry, and engineering applications.
351. In statistical mechanics, an ensemble refers to:
ⓐ. A group of identical particles moving with the same velocity
ⓑ. A large collection of virtual systems representing all possible microscopic states
ⓒ. A real container of gas molecules
ⓓ. A set of experimental measurements only
Correct Answer: A large collection of virtual systems representing all possible microscopic states
Explanation: An ensemble is a theoretical construct used to represent all possible microstates consistent with macroscopic constraints (energy, volume, particle number). Ensemble averages provide statistical predictions equivalent to time averages in real systems.
352. The ensemble average of a physical quantity $A$ is mathematically defined as:
ⓐ. $\langle A \rangle = \sum A$
ⓑ. $\langle A \rangle = \frac{1}{Z} \sum_i A_i e^{-E_i/k_BT}$
ⓒ. $\langle A \rangle = \frac{E}{T}$
ⓓ. $\langle A \rangle = \int PdV$
Correct Answer: $\langle A \rangle = \frac{1}{Z} \sum_i A_i e^{-E_i/k_BT}$
Explanation: The ensemble average is a weighted mean over all microstates, with probabilities given by the Boltzmann factor. The partition function $Z$ ensures proper normalization. This connects microscopic states to measurable macroscopic observables.
353. The probability of a system being in a microstate with energy $E_i$ is:
ⓐ. $P_i = \frac{E_i}{k_BT}$
ⓑ. $P_i = \frac{1}{Z} e^{-E_i/k_BT}$
ⓒ. $P_i = \frac{RT}{PV}$
ⓓ. $P_i = \sqrt{\frac{M}{RT}}$
Correct Answer: $P_i = \frac{1}{Z} e^{-E_i/k_BT}$
Explanation: This is the Boltzmann distribution law. Each state’s probability decreases exponentially with energy, meaning lower-energy states are more probable at equilibrium. The denominator $Z$ ensures the sum of probabilities equals 1.
354. Which ensemble is appropriate for systems with fixed energy, volume, and particle number?
ⓐ. Canonical ensemble
ⓑ. Microcanonical ensemble
ⓒ. Grand canonical ensemble
ⓓ. Quantum ensemble
Correct Answer: Microcanonical ensemble
Explanation: The microcanonical ensemble applies to isolated systems where energy, volume, and number of particles are fixed. Each accessible microstate has equal probability, suitable for modeling perfectly isolated systems.
355. Which ensemble is used for a system at fixed temperature, volume, and number of particles?
ⓐ. Microcanonical ensemble
ⓑ. Canonical ensemble
ⓒ. Grand canonical ensemble
ⓓ. Maxwell ensemble
Correct Answer: Canonical ensemble
Explanation: The canonical ensemble describes a system in thermal equilibrium with a reservoir at temperature $T$. Energy can fluctuate, but $N$ and $V$ are fixed. This is commonly used in deriving thermodynamic relations.
356. Which ensemble is used for a system at fixed temperature, volume, and chemical potential (allowing particle exchange)?
ⓐ. Microcanonical ensemble
ⓑ. Canonical ensemble
ⓒ. Grand canonical ensemble
ⓓ. Bose ensemble
Correct Answer: Grand canonical ensemble
Explanation: The grand canonical ensemble applies to systems exchanging both energy and particles with a reservoir. It is especially useful in studying open systems like adsorption, electron gases, or quantum statistics of bosons/fermions.
357. The probability distribution of molecular speeds in a gas is given by:
ⓐ. $f(v) = 4\pi \left( \frac{M}{2\pi RT} \right)^{3/2} v^2 e^{-Mv^2/2RT}$
ⓑ. $f(v) = \frac{1}{Z} e^{-E/k_BT}$
ⓒ. $f(v) = \sqrt{\frac{2RT}{M}}$
ⓓ. $f(v) = v^2$
Correct Answer: $f(v) = 4\pi \left( \frac{M}{2\pi RT} \right)^{3/2} v^2 e^{-Mv^2/2RT}$
Explanation: The Maxwell–Boltzmann speed distribution describes the probability of finding molecules with speed $v$. It depends on molecular mass $M$, gas constant $R$, and temperature $T$. It predicts most probable, average, and rms speeds.
358. In probability distributions, the normalization condition requires:
ⓐ. $\int_{0}^{\infty} f(v) dv = 1$
ⓑ. $\int_{0}^{\infty} f(v) dv = N$
ⓒ. $\int_{0}^{\infty} f(v) dv = RT$
ⓓ. $\int_{0}^{\infty} f(v) dv = 0$
Correct Answer: $\int_{0}^{\infty} f(v) dv = 1$
Explanation: The normalization condition ensures that the total probability of finding a molecule at any possible speed is 1. This is essential in defining probability density functions in statistical mechanics.
359. If the probability distribution of a quantity is sharply peaked, what does it imply about fluctuations?
ⓐ. Fluctuations are very large.
ⓑ. Fluctuations are negligible, system is near equilibrium.
ⓒ. Fluctuations are infinite.
ⓓ. System is unstable.
Correct Answer: Fluctuations are negligible, system is near equilibrium.
Explanation: A narrow distribution indicates most microstates have nearly the same value for the observable, so fluctuations are small. This is why macroscopic thermodynamic properties appear stable despite microscopic randomness.
360. In statistical mechanics, ensemble averages correspond experimentally to:
ⓐ. Measurements made over very long time intervals
ⓑ. Instantaneous molecular positions only
ⓒ. Number of molecules only
ⓓ. Temperature alone
Correct Answer: Measurements made over very long time intervals
Explanation: Time averages in experiments equal ensemble averages if the system is ergodic. This equivalence allows us to replace complex microscopic tracking with statistical ensembles in predicting thermodynamic behavior.
361. The macroscopic quantity pressure in a gas arises microscopically from:
ⓐ. Molecular vibrations
ⓑ. Collisions of molecules with container walls
ⓒ. Gravitational attraction between molecules
ⓓ. Rotational motion of molecules
Correct Answer: Collisions of molecules with container walls
Explanation: Kinetic theory shows that pressure results from momentum transfer during elastic collisions of gas molecules with walls. The average force exerted per unit area is the macroscopic pressure measured experimentally.
362. Temperature is directly related to which microscopic property of molecules?
ⓐ. Potential energy of molecules
ⓑ. Average translational kinetic energy per molecule
ⓒ. Number of collisions per second
ⓓ. Size of molecules
Correct Answer: Average translational kinetic energy per molecule
Explanation: From kinetic theory, $\langle E_k \rangle = \tfrac{3}{2}k_BT$. Thus, absolute temperature is a direct measure of molecular kinetic energy, connecting microscopic dynamics to macroscopic thermodynamic temperature.
363. Entropy ($S$) connects microscopic and macroscopic descriptions by the relation:
ⓐ. $S = \tfrac{Q}{T}$
ⓑ. $S = k_B \ln \Omega$
ⓒ. $S = nRT$
ⓓ. $S = PV$
Correct Answer: $S = k_B \ln \Omega$
Explanation: Boltzmann’s relation defines entropy as proportional to the logarithm of the number of accessible microstates ($\Omega$) consistent with macroscopic variables. This bridges probability theory and thermodynamics.
364. The internal energy of an ideal monatomic gas is expressed microscopically as:
ⓐ. $U = nRT$
ⓑ. $U = \tfrac{3}{2}Nk_B T$
ⓒ. $U = PV$
ⓓ. $U = nC_p T$
Correct Answer: $U = \tfrac{3}{2}Nk_B T$
Explanation: Internal energy arises entirely from translational kinetic energy of molecules in an ideal monatomic gas. Using $N$ molecules, average energy per molecule is $\tfrac{3}{2}k_BT$, giving $U = \tfrac{3}{2}Nk_BT$.
365. The connection between partition function $Z$ and Helmholtz free energy $F$ is:
ⓐ. $F = -k_BT \ln Z$
ⓑ. $F = \tfrac{E}{T}$
ⓒ. $F = PV$
ⓓ. $F = \tfrac{1}{2}k_BT$
Correct Answer: $F = -k_BT \ln Z$
Explanation: The Helmholtz free energy, a macroscopic thermodynamic potential, can be computed from the partition function. This relation makes partition function central in deriving macroscopic thermodynamic properties from microscopic states.
366. Which microscopic property determines diffusion rate of gases?
ⓐ. Rotational inertia of molecules
ⓑ. Distribution of molecular speeds
ⓒ. Vibrational energy levels
ⓓ. Atomic number of the gas atoms
Correct Answer: Distribution of molecular speeds
Explanation: The Maxwell–Boltzmann distribution determines how fast molecules move, which directly influences diffusion. Faster molecules (lighter gases at higher $T$) diffuse more quickly than heavier, slower ones.
367. Specific heat capacity is related to microscopic states because:
ⓐ. It depends only on molar mass
ⓑ. It equals the number of degrees of freedom available for energy storage
ⓒ. It is constant for all gases
ⓓ. It is independent of microscopic behavior
Correct Answer: It equals the number of degrees of freedom available for energy storage
Explanation: Equipartition theorem assigns $\tfrac{1}{2}k_BT$ energy per quadratic degree of freedom. Gases with more rotational/vibrational modes store more energy, leading to higher specific heats. Thus, $C_v$ reflects microscopic structure.
368. The macroscopic law $PV = nRT$ is derived microscopically using:
ⓐ. Conservation of angular momentum
ⓑ. Average kinetic energy of molecules and collision mechanics
ⓒ. Quantum tunneling of molecules
ⓓ. Rotational inertia of gas molecules
Correct Answer: Average kinetic energy of molecules and collision mechanics
Explanation: By analyzing molecular collisions with container walls and linking pressure to mean squared velocity, we derive $PV = \tfrac{1}{3}Nm\langle v^2 \rangle$. Substituting $\langle E_k \rangle = \tfrac{3}{2}k_BT$, the ideal gas law emerges.
369. Why do macroscopic observables like pressure and temperature appear stable despite microscopic randomness?
ⓐ. Because microscopic states are identical.
ⓑ. Because the number of particles is extremely large, averaging out fluctuations.
ⓒ. Because molecules stop moving at equilibrium.
ⓓ. Because Boltzmann constant is large.
Correct Answer: Because the number of particles is extremely large, averaging out fluctuations.
Explanation: A mole contains \~$10^{23}$ molecules. Random fluctuations cancel statistically, leaving macroscopic quantities smooth and predictable. This is the foundation of statistical mechanics’ success.
370. Which statement best describes the connection between microscopic and macroscopic views?
ⓐ. Macroscopic laws are independent of microscopic dynamics.
ⓑ. Macroscopic thermodynamic quantities are statistical averages of microscopic states.
ⓒ. Macroscopic observables cannot be explained microscopically.
ⓓ. Microscopic and macroscopic worlds are completely separate.
Correct Answer: Macroscopic thermodynamic quantities are statistical averages of microscopic states.
Explanation: Statistical mechanics demonstrates that pressure, temperature, entropy, and free energy all emerge as averages over many microscopic configurations. This connection unifies atomic-level physics with bulk thermodynamics.
371. The melting point of a substance is defined as the temperature at which:
ⓐ. Vapor pressure equals atmospheric pressure
ⓑ. Solid and liquid phases coexist in equilibrium
ⓒ. Enthalpy becomes zero
ⓓ. Heat capacity becomes infinite
Correct Answer: Solid and liquid phases coexist in equilibrium
Explanation: At melting point, Gibbs free energy of solid and liquid phases are equal. The transition is characterized by absorption of latent heat of fusion without a rise in temperature.
372. The enthalpy change during melting (fusion) is given by:
ⓐ. $\Delta H_f = mL_f$
ⓑ. $\Delta H_f = mc\Delta T$
ⓒ. $\Delta H_f = \tfrac{1}{2}k_BT$
ⓓ. $\Delta H_f = PV$
Correct Answer: $\Delta H_f = mL_f$
Explanation: Latent heat of fusion ($L_f$) is the heat required to change mass $m$ of solid into liquid at constant temperature. This is a first-order phase transition, involving enthalpy change but no temperature change.
373. The Clausius–Clapeyron equation for vaporization is:
ⓐ. $\frac{dP}{dT} = \frac{\Delta H_{vap}}{T \Delta V}$
ⓑ. $\frac{dP}{dT} = \frac{RT}{P}$
ⓒ. $\frac{dP}{dT} = \frac{PV}{RT}$
ⓓ. $\frac{dP}{dT} = \frac{\Delta H_f}{\Delta V}$
Correct Answer: $\frac{dP}{dT} = \frac{\Delta H_{vap}}{T \Delta V}$
Explanation: This equation relates slope of phase boundary in a $P$-$T$ diagram to latent heat and volume change. For liquid-vapor equilibrium, $\Delta V \approx V_{gas}$, making vaporization strongly dependent on temperature.
374. For water at 1 atm, the enthalpy of vaporization is approximately $40.7 \, kJ/mol$. Calculate the heat needed to vaporize 18 g (1 mol) of water.
ⓐ. $18 \, kJ$
ⓑ. $40.7 \, kJ$
ⓒ. $75.3 \, kJ$
ⓓ. $100 \, kJ$
Correct Answer: $40.7 \, kJ$
Explanation: The latent heat of vaporization per mole directly gives the energy required. For 1 mole water, $Q = \Delta H_{vap} = 40.7 \, kJ$. This energy is absorbed at constant temperature during vaporization.
375. The latent heat equation during condensation of a mass $m$ of vapor is:
ⓐ. $Q = mc\Delta T$
ⓑ. $Q = mL_{vap}$
ⓒ. $Q = \tfrac{1}{2}mv^2$
ⓓ. $Q = nRT$
Correct Answer: $Q = mL_{vap}$
Explanation: Condensation is the reverse of vaporization. Each mole releases $\Delta H_{vap}$. The same amount of heat $mL_{vap}$ is released to surroundings without temperature change.
376. At 100 °C, water has $\Delta H_{vap} = 40.7 \, kJ/mol$. Estimate the slope of liquid–vapor boundary using Clausius–Clapeyron: $\frac{dP}{dT} = \frac{\Delta H_{vap}}{T\Delta V}$ Given: $V_{gas} \approx \frac{RT}{P}$, $V_{liq} \ll V_{gas}$.
ⓐ. $2.0 \times 10^3 \, Pa/K$
ⓑ. $3.5 \times 10^4 \, Pa/K$
ⓒ. $1.0 \times 10^5 \, Pa/K$
ⓓ. $4.0 \times 10^6 \, Pa/K$
Correct Answer: $3.5 \times 10^4 \, Pa/K$
Explanation: Using approximation $\frac{dP}{dT} = \frac{\Delta H_{vap} P}{RT^2}$. Substituting $\Delta H_{vap}=40.7 \times 10^3 J/mol$, $T=373 K$, $P=1.01 \times 10^5 Pa$, we get slope ≈ $3.5 \times 10^4 Pa/K$.
377. During freezing of 1 kg of water, latent heat released is:
ⓐ. $Q = 334 \, kJ$
ⓑ. $Q = 540 \, kJ$
ⓒ. $Q = 100 \, kJ$
ⓓ. $Q = 4.18 \, kJ$
Correct Answer: $Q = 334 \, kJ$
Explanation: Latent heat of fusion of water is $L_f = 334 \, kJ/kg$. Hence, freezing 1 kg of water at 0 °C releases exactly 334 kJ of heat to surroundings, maintaining constant temperature during the process.
378. Why is boiling considered a first-order phase transition?
ⓐ. It changes entropy continuously.
ⓑ. It involves discontinuous change in density and release/absorption of latent heat.
ⓒ. It does not involve enthalpy.
ⓓ. It has no pressure dependence.
Correct Answer: It involves discontinuous change in density and release/absorption of latent heat.
Explanation: First-order transitions (like melting and boiling) involve latent heat and abrupt volume change. Second-order transitions (like superconductivity) show continuous entropy but discontinuous derivatives.
379. The Clausius–Clapeyron equation predicts that vapor pressure varies approximately as:
ⓐ. $\ln P = -\frac{\Delta H_{vap}}{RT} + C$
ⓑ. $\ln P = \frac{RT}{P}$
ⓒ. $\ln P = \frac{PV}{RT}$
ⓓ. $\ln P = \tfrac{E}{T}$
Correct Answer: $\ln P = -\frac{\Delta H_{vap}}{RT} + C$
Explanation: Integrating Clausius–Clapeyron under assumption $\Delta H_{vap}$ is constant, we get $\ln P = -\Delta H_{vap}/(RT) + C$. This relation explains exponential rise of vapor pressure with temperature.
380. Why does ice float on water in terms of phase transition properties?
ⓐ. Ice has higher enthalpy.
ⓑ. Ice has a lower density due to hydrogen bonding.
ⓒ. Ice has higher kinetic energy.
ⓓ. Ice absorbs less latent heat.
Correct Answer: Ice has a lower density due to hydrogen bonding.
Explanation: In solid ice, hydrogen bonds force molecules into an open hexagonal structure, making ice less dense than liquid water. Thus, the solid floats on its own melt, a rare anomaly in phase transitions.
381. Calculate the amount of heat required to melt 500 g of ice at $0^\circ C$. Latent heat of fusion of ice $L_f = 334 \, kJ/kg$.
ⓐ. 167 kJ
ⓑ. 334 kJ
ⓒ. 100 kJ
ⓓ. 250 kJ
Correct Answer: 167 kJ
Explanation: Heat required is $Q = mL_f = 0.5 \times 334 = 167 \, kJ$. No temperature change occurs; all heat goes into breaking hydrogen bonds during melting.
382. A refrigerator extracts 200 kJ of heat from water at $0^\circ C$. How many grams of ice are formed? $L_f = 334 \, kJ/kg$.
ⓐ. 330.40 g
ⓑ. 800 g
ⓒ. 700 g
ⓓ. 600 g
Correct Answer: 600 g
Explanation: Mass of ice formed = $ m = \frac{Q}{L_f} = \frac{200}{334} = 0.60 \, kg = 600 \, g$. Careful
383. If 2 moles of water are vaporized at 100 °C and 1 atm, how much heat is absorbed? ($\Delta H_{vap} = 40.7 \, kJ/mol$)
ⓐ. 40.7 kJ
ⓑ. 81.4 kJ
ⓒ. 20.35 kJ
ⓓ. 100 kJ
Correct Answer: 81.4 kJ
Explanation: Heat absorbed = $Q = n \Delta H_{vap} = 2 \times 40.7 = 81.4 \, kJ$.
384. Calculate the mass of steam at 100 °C required to release 2,000 kJ of heat on condensation. ($L_{vap} = 2260 \, kJ/kg$)
ⓐ. 0.5 kg
ⓑ. 0.75 kg
ⓒ. 0.88 kg
ⓓ. 1.0 kg
Correct Answer: 0.88 kg
Explanation: Mass = $m = \frac{Q}{L_{vap}} = \frac{2000}{2260} \approx 0.885 \, kg$.
385. At 373 K, the enthalpy of vaporization of water is $40.7 \times 10^3 J/mol$. Using Clausius–Clapeyron, calculate $\frac{dP}{dT}$ at 1 atm ($P = 1.01 \times 10^5 Pa$).
ⓐ. $2.0 \times 10^3 \, Pa/K$
ⓑ. $3.5 \times 10^4 \, Pa/K$
ⓒ. $1.2 \times 10^5 \, Pa/K$
ⓓ. $4.1 \times 10^6 \, Pa/K$
Correct Answer: $3.5 \times 10^4 \, Pa/K$
Explanation: Clausius–Clapeyron: $\frac{dP}{dT} = \frac{\Delta H_{vap} P}{RT^2}$. Substituting values: $\frac{40.7 \times 10^3 \times 1.01 \times 10^5}{8.314 \times (373)^2} \approx 3.5 \times 10^4 Pa/K$.
386. A 250 g block of ice at 0 °C is converted into water at 0 °C. Find the entropy change of the system. ($L_f = 334 \, kJ/kg$, $T = 273 K$)
ⓐ. 122 J/K
ⓑ. 306 J/K
ⓒ. 470 J/K
ⓓ. 800 J/K
Correct Answer: 306 J/K
Explanation: Heat absorbed = $Q = mL_f = 0.25 \times 334 \times 10^3 = 83,500 J$. Entropy change = $\Delta S = \frac{Q}{T} = \frac{83500}{273} \approx 306 J/K$.
387. How much heat is released when 200 g of water at 100 °C condenses into liquid? ($ L_{vap} = 2260 \, kJ/kg$)
ⓐ. 226 kJ
ⓑ. 452 kJ
ⓒ. 600 kJ
ⓓ. 700 kJ
Correct Answer: 452 kJ
Explanation: Heat released = $Q = mL_{vap} = 0.2 \times 2260 = 452 kJ$.
388. Calculate the entropy change when 1 mole of water vapor condenses at 373 K. ($ \Delta H_{vap} = 40.7 \, kJ/mol$)
ⓐ. –80 J/K
ⓑ. –100 J/K
ⓒ. –109 J/K
ⓓ. –150 J/K
Correct Answer: –109 J/K
Explanation: $\Delta S = \frac{-\Delta H_{vap}}{T} = \frac{-40.7 \times 10^3}{373} \approx -109 J/K$. Negative sign indicates entropy decrease during condensation.
389. The latent heat of fusion of ice is $334 \, kJ/kg$. How much ice at 0 °C can be melted with 100 kJ of heat?
ⓐ. 0.1 kg
ⓑ. 0.2 kg
ⓒ. 0.3 kg
ⓓ. 0.5 kg
Correct Answer: 0.3 kg
Explanation: $m = \frac{Q}{L_f} = \frac{100}{334} \approx 0.299 \, kg$.
390. A sealed container holds water at equilibrium at 100 °C. If the pressure is suddenly reduced, what happens to the boiling point?
ⓐ. It increases.
ⓑ. It decreases.
ⓒ. It remains the same.
ⓓ. It becomes zero.
Correct Answer: It decreases.
Explanation: Boiling occurs when vapor pressure equals external pressure. Lowering external pressure means vapor pressure requirement is met at a lower temperature, hence boiling point decreases (principle behind pressure cookers and vacuum distillation).
391. From a statistical mechanics perspective, a phase transition occurs when:
ⓐ. All molecules move with the same velocity
ⓑ. Thermodynamic functions show non-analytic behavior due to sudden changes in microscopic state probabilities
ⓒ. Energy is conserved in the system
ⓓ. The number of particles becomes infinite
Correct Answer: Thermodynamic functions show non-analytic behavior due to sudden changes in microscopic state probabilities
Explanation: In phase transitions, macroscopic observables like entropy or heat capacity display discontinuities or divergences. This results from a collective rearrangement of microscopic states, making thermodynamic functions non-analytic at transition points.
392. The order parameter in phase transitions represents:
ⓐ. The average velocity of molecules
ⓑ. A macroscopic quantity distinguishing different phases
ⓒ. The partition function
ⓓ. The entropy per particle
Correct Answer: A macroscopic quantity distinguishing different phases
Explanation: An order parameter measures the degree of order in a system. For example, magnetization distinguishes ferromagnetic and paramagnetic phases, while density distinguishes liquid and vapor phases.
393. In statistical mechanics, first-order phase transitions are characterized by:
ⓐ. Continuous entropy but discontinuous heat capacity
ⓑ. Discontinuous entropy and latent heat absorption/release
ⓒ. No energy exchange
ⓓ. Infinite relaxation time
Correct Answer: Discontinuous entropy and latent heat absorption/release
Explanation: First-order transitions (e.g., melting, boiling) involve latent heat and entropy discontinuity. During these transitions, energy is absorbed or released without temperature change, indicating a sharp change in microscopic configurations.
394. Second-order (continuous) phase transitions are distinguished by:
ⓐ. Latent heat absorption
ⓑ. Continuous entropy but discontinuous derivatives like heat capacity and susceptibility
ⓒ. Sudden change in density
ⓓ. Constant enthalpy
Correct Answer: Continuous entropy but discontinuous derivatives like heat capacity and susceptibility
Explanation: In continuous transitions (e.g., superconductivity), entropy changes smoothly, but higher derivatives of free energy diverge. These critical phenomena are modeled using statistical mechanics and scaling theories.
395. The partition function near a phase transition becomes:
ⓐ. Strongly dominated by one microstate
ⓑ. Sensitive to contributions from many competing microstates
ⓒ. Zero for all temperatures
ⓓ. Independent of temperature
Correct Answer: Sensitive to contributions from many competing microstates
Explanation: At transition points, multiple microscopic configurations (solid, liquid, vapor) have comparable probabilities. Their competition makes the partition function’s derivatives (free energy, entropy, heat capacity) show abrupt or divergent behavior.
396. Which statistical mechanics relation connects fluctuations to heat capacity near a phase transition?
ⓐ. $C_V = \frac{dQ}{dT}$
ⓑ. $C_V = \frac{\langle E^2 \rangle – \langle E \rangle^2}{k_BT^2}$
ⓒ. $C_V = \tfrac{3}{2}R$
ⓓ. $C_V = \frac{E}{T}$
Correct Answer: $C_V = \frac{\langle E^2 \rangle – \langle E \rangle^2}{k_BT^2}$
Explanation: Heat capacity measures energy fluctuations. Near phase transitions, fluctuations grow large, leading to divergence in $C_V$. This fluctuation–dissipation relation is fundamental in statistical mechanics.
397. At the critical point of a liquid–gas transition, the order parameter is:
ⓐ. Zero
ⓑ. Infinite
ⓒ. Undefined
ⓓ. Density difference between liquid and vapor phases
Correct Answer: Density difference between liquid and vapor phases
Explanation: As the critical temperature is approached, the density of liquid and vapor phases converge, making the order parameter (density difference) vanish. This describes the disappearance of distinction between phases.
398. Why do correlation lengths diverge near second-order transitions?
ⓐ. Molecules stop moving
ⓑ. Microscopic fluctuations become correlated over macroscopic distances
ⓒ. Partition function becomes zero
ⓓ. Entropy disappears
Correct Answer: Microscopic fluctuations become correlated over macroscopic distances
Explanation: Near critical points, fluctuations at one site influence distant sites. Correlation length $\xi$ grows large, making local microscopic changes affect the entire system, leading to critical phenomena.
399. In the Ising model, magnetization serves as:
ⓐ. Free energy
ⓑ. Entropy
ⓒ. Order parameter for ferromagnetic transition
ⓓ. Specific heat
Correct Answer: Order parameter for ferromagnetic transition
Explanation: In the Ising model, at high temperatures spins are random (zero magnetization). Below critical temperature, spins align, giving nonzero magnetization. This magnetization is the order parameter distinguishing the phases.
400. Why is statistical mechanics crucial for explaining phase transitions?
ⓐ. It studies only individual molecular motion
ⓑ. It provides probabilistic frameworks linking microstates to macroscopic discontinuities
ⓒ. It avoids thermodynamics
ⓓ. It only applies to ideal gases
Correct Answer: It provides probabilistic frameworks linking microstates to macroscopic discontinuities
Explanation: Thermodynamics describes phase transitions phenomenologically, but statistical mechanics explains why they occur: from collective changes in microstate probabilities and fluctuations that scale up to macroscopic observables.
The Kinetic Theory chapter is a scoring topic in both board exams and competitive exams (JEE/NEET), as it combines theory with mathematical derivations. Important subtopics include the derivation of pressure of an ideal gas, mean free path and molecular collisions, distribution of molecular speeds (RMS, average, most probable), Brownian motion, and thermodynamic applications of kinetic theory. Practicing MCQs on these concepts strengthens analytical skills and problem-solving speed. Out of a total of 498 MCQs, this part (Part 4) provides the **fourth set of 100 solved MCQs** with detailed solutions, helping students prepare strategically for board exams, JEE, NEET, and other competitive exams.
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