Physical Chemistry 2 Quiz: Microstates And Entropy

17 questions · exam conditions

0:00

Microstates And EntropyQuestion 1 of 17

A polymer chain of N segments can adopt conformations with different end-to-end distances R. The number of conformations with end-to-end distance R is approximately $\Omega(R) = \Omega_0 \exp\left(-\frac{3R^2}{2Nb^2}\right)$ where b is the segment length and Ω₀ is a normalization constant. If the chain is subject to a stretching force f, what is the most probable end-to-end distance?

R_{mp} = 0

, because the Gaussian distribution of conformations is always maximized at zero extension

R_{mp} = \sqrt{\frac{Nb^2 f}{3k_B T}}

, balancing entropic contraction against the applied force

R_{mp} = Nb \cdot \frac{f}{k_B T}

, from the linear response of the polymer to small forces

R_{mp} = \sqrt{Nb} \cdot \frac{f}{k_B T}

, combining the random walk scaling with the force response

R_{mp} = \frac{fb}{k_B T} \sqrt{\frac{3N}{2}}

, derived from the exact balance of entropic and mechanical work terms

← Back to quizzes

Physical Chemistry 2 Quiz

Physical Chemistry 2 Quiz: Microstates And Entropy

Practice Microstates And Entropy in Physical Chemistry 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Microstates And Entropy, giving you a quick way to practice the rules, question types, and explanations that matter most for Physical Chemistry 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$R_{mp} = 0$ , because the Gaussian distribution of conformations is always maximized at zero extension
$R_{mp} = \sqrt{\frac{Nb^2 f}{3k_B T}}$ , balancing entropic contraction against the applied force (correct answer)
$R_{mp} = Nb \cdot \frac{f}{k_B T}$ , from the linear response of the polymer to small forces
$R_{mp} = \sqrt{Nb} \cdot \frac{f}{k_B T}$ , combining the random walk scaling with the force response
$R_{mp} = \frac{fb}{k_B T} \sqrt{\frac{3N}{2}}$ , derived from the exact balance of entropic and mechanical work terms

Explanation: When analyzing polymer chains under external forces, you're dealing with a competition between entropy (which favors compact conformations) and the work done by the applied force (which favors extension). The key is finding where these opposing effects balance. To find the most probable end-to-end distance, you need to maximize the total probability, which includes both the conformational entropy and the work term. The total probability is proportional to

\Omega(R) \times \exp(fR/k_B T)

, where the second exponential accounts for the work done by force f in stretching the chain by distance R. Taking the logarithm and differentiating with respect to R, then setting equal to zero:

\frac{d}{dR}\left[-\frac{3R^2}{2Nb^2} + \frac{fR}{k_B T}\right] = 0

. This gives

-\frac{3R}{Nb^2} + \frac{f}{k_B T} = 0

, which solves to

R_{mp} = \sqrt{\frac{Nb^2 f}{3k_B T}}

, confirming answer B. Answer A incorrectly ignores the applied force entirely. Answer C gives

R_{mp} = Nb \cdot \frac{f}{k_B T}

, which has the wrong scaling with N (linear instead of square root) and missing the factor of 3. Answer D has

R_{mp} = \sqrt{Nb} \cdot \frac{f}{k_B T}

, which captures the correct N-scaling but misses the b-dependence and the factor of 3. Remember: polymer force problems always involve balancing entropy (favoring compact states) against external work. The equilibrium extension scales as

\sqrt{N}

, reflecting the random walk nature of polymer chains.

Question 2

Consider the mixing of two ideal gases: n₁ moles of gas 1 and n₂ moles of gas 2. Before mixing, each gas occupies volume V at temperature T. After mixing, both gases occupy volume 2V at the same temperature T. The entropy of mixing is given by $\Delta S_{mix} = -R(n_1 \ln x_1 + n_2 \ln x_2)$ where x₁ and x₂ are mole fractions. How does this entropy change compare to the entropy change from expansion alone?

The mixing entropy is always larger than the expansion entropy because mixing creates additional configurational possibilities
The expansion entropy $R(n_1 + n_2)\ln(2)$ is always larger than the mixing entropy for any composition
The two contributions are equal when the mole fractions satisfy $x_1 \ln x_1 + x_2 \ln x_2 = -\ln(2)$ (correct answer)
The mixing entropy equals the expansion entropy only when x₁ = x₂ = 0.5, giving perfect symmetry
The comparison depends on temperature through the gas constant R, making a general statement impossible

Explanation: When analyzing entropy changes in gas mixing, you need to consider two separate contributions: the entropy increase from expansion and the entropy increase from mixing itself. Each gas expands from volume V to 2V, while the mixing creates additional disorder from the random distribution of different molecules. For the expansion entropy, each gas doubles its volume, so the total expansion entropy is

\Delta S_{expansion} = R(n_1 + n_2)\ln(2)

. The mixing entropy is given as

\Delta S_{mix} = -R(n_1 \ln x_1 + n_2 \ln x_2)

, where the negative signs account for the fact that

\ln x_1

and

\ln x_2

are negative (since mole fractions are less than 1). Setting these equal:

R(n_1 + n_2)\ln(2) = -R(n_1 \ln x_1 + n_2 \ln x_2)

. Dividing by

R(n_1 + n_2)

gives

\ln(2) = -\frac{n_1 \ln x_1 + n_2 \ln x_2}{n_1 + n_2}

. Since

x_1 = \frac{n_1}{n_1 + n_2}

and

x_2 = \frac{n_2}{n_1 + n_2}

, this simplifies to

x_1 \ln x_1 + x_2 \ln x_2 = -\ln(2)

, confirming answer C. Option A is wrong because mixing entropy isn't always larger—it depends on composition. Option B is incorrect because expansion entropy isn't always larger either. Option D fails because when

x_1 = x_2 = 0.5

, we get

x_1 \ln x_1 + x_2 \ln x_2 = -2(0.5)\ln(0.5) = \ln(2) \neq -\ln(2)

. Remember: entropy contributions from different processes are additive, but their relative magnitudes depend on the specific conditions—always set up the equality mathematically rather than assuming one dominates.

Question 3

A system consists of 4 distinguishable particles that can occupy 3 energy levels (0, ε, 2ε) with a total energy of 4ε. If the degeneracy of each energy level is 1, and we then change the system so that the middle energy level (ε) becomes doubly degenerate while keeping the same particle distribution, how does the entropy change?

The entropy increases by $k_B \ln(2)$ for each particle in the middle level
The entropy increases by $k_B \ln(4)$ because the total number of microstates doubles
The entropy increases by $k_B \ln(2^n)$ where n is the number of particles in the middle level (correct answer)
The entropy remains unchanged because the total energy is conserved
The entropy increases by $k_B \ln(3)$ because we now have effectively 4 energy levels

Explanation: When you encounter problems about entropy changes due to degeneracy modifications, focus on how the number of available microstates changes for the affected particles. First, let's determine the particle distribution. With 4 particles, total energy 4ε, and levels at 0, ε, and 2ε, the only possible distribution is: 2 particles at level 0, 0 particles at level ε, and 2 particles at level 2ε (since 2×0 + 0×ε + 2×2ε = 4ε). Wait - this doesn't involve the middle level at all! Let me reconsider. Actually, we could have different distributions like 1 particle at 0, 2 particles at ε, and 1 particle at 2ε (giving 1×0 + 2×ε + 1×2ε = 4ε). The question asks us to keep the same distribution, so let's assume we have n particles in the middle level. When the middle level becomes doubly degenerate, each of the n particles in that level can now occupy either of two degenerate states. Originally, there was only one way to place these n particles in the middle level. Now there are

2^n

ways, since each particle independently has 2 choices. The entropy change is

\Delta S = k_B \ln(\Omega_{final}/\Omega_{initial}) = k_B \ln(2^n) = nk_B \ln(2)

, confirming answer C. Answer A incorrectly suggests the entropy increases by

k_B \ln(2)

per particle, missing the multiplicative effect. Answer B assumes the total microstates double, ignoring that only particles in the middle level are affected. Answer D wrongly focuses on energy conservation, but entropy depends on microstate accessibility, not energy conservation. Remember: entropy changes from degeneracy modifications depend on how many particles are affected and how many new states become available to each.

Question 4

Consider two identical systems A and B, each containing N particles. System A has multiplicity Ω_A = 10^20 and system B has multiplicity Ω_B = 10^15. When these systems are brought into thermal contact and allowed to reach equilibrium, which statement best describes the final state?

The total entropy will be $k_B \ln(10^{20} + 10^{15})$ since entropies are additive for independent systems
The final entropy will be $k_B \ln(10^{35})$ because multiplicities multiply when systems combine
The final entropy will be greater than $k_B \ln(10^{35})$ due to redistribution of energy between the systems (correct answer)
The final entropy will be less than $k_B \ln(10^{35})$ because thermal equilibrium constrains the total number of accessible microstates
The final entropy will exactly equal $k_B \ln(10^{35})$ because the total number of particles remains constant

Explanation: When systems at different temperatures are brought into thermal contact, you're dealing with the fundamental principle that entropy always increases in spontaneous processes. This question tests your understanding of how entropy and multiplicity relate when isolated systems combine and equilibrate. Initially, if these were independent systems, the total multiplicity would be

\Omega_A \times \Omega_B = 10^{20} \times 10^{15} = 10^{35}

, giving entropy

S = k_B \ln(10^{35})

. However, when the systems reach thermal equilibrium through energy exchange, something crucial happens: energy redistributes between them, and this redistribution creates additional accessible microstates beyond what existed in the isolated systems. The correct answer is C because thermal equilibration allows energy to flow from the hotter system (likely A, given its higher multiplicity) to the cooler one until temperatures equalize. This energy redistribution opens up new microstates that weren't accessible when the systems were isolated, increasing the total entropy above

k_B \ln(10^{35})

. A incorrectly adds multiplicities instead of multiplying them—multiplicities multiply, not add, when systems combine. B gives the entropy for independent systems without thermal contact, ignoring the entropy increase from equilibration. D contradicts the second law of thermodynamics by suggesting equilibrium reduces entropy; while equilibrium does constrain the final state, the spontaneous equilibration process always increases total entropy. Remember: when isolated systems come into thermal contact, the final entropy is always greater than the sum of initial entropies due to energy redistribution creating new accessible microstates.

Question 5

A system of N indistinguishable particles has a ground state with multiplicity g₀ and first excited state with multiplicity g₁. At very low temperature T, the ratio of particles in the excited state to those in the ground state is approximately $\frac{n_1}{n_0} = \frac{g_1}{g_0} e^{-\varepsilon/k_B T}$ . If we suddenly increase g₁ by a factor of 3 while keeping temperature constant, what happens to the entropy contribution from these two levels?

The entropy increases by exactly $Nk_B \ln(3)$ because all particles experience the increased degeneracy
The entropy increases by $\frac{N g_1}{g_0} e^{-\varepsilon/k_B T} k_B \ln(3)$ due to the particles that redistribute to the excited state
The entropy increases by approximately $N k_B \ln(3) \left[\frac{g_1}{g_0} e^{-\varepsilon/k_B T}\right] / \left[1 + \frac{g_1}{g_0} e^{-\varepsilon/k_B T}\right]$ accounting for the new equilibrium distribution
The entropy change depends on both the redistribution of particles and the increased degeneracy of occupied states in a complex manner (correct answer)
The entropy increases by $3Nk_B \ln(3)$ because the degeneracy triples and affects the entire population

Explanation: When you encounter statistical thermodynamics problems involving degeneracy changes, you need to consider how both particle redistribution and the fundamental counting of microstates contribute to entropy changes. The entropy change here involves two coupled effects that make this problem complex. First, when

g_1

increases by a factor of 3, the ratio

n_1/n_0

increases proportionally, causing more particles to redistribute from the ground state to the excited state to reach the new equilibrium. Second, the increased degeneracy means each particle in the excited state now has access to 3 times as many microstates. The entropy formula

S = k_B \ln W

depends on the total number of accessible microstates, which involves both how many particles occupy each level AND how many ways they can arrange themselves within those levels. Since the populations

n_0

and

n_1

change, you can't simply apply a multiplicative factor to the old entropy. Answer A incorrectly assumes all

N

particles experience the full degeneracy increase, ignoring that most remain in the ground state at low temperature. Answer B only accounts for the "extra" particles that move to the excited state, missing the degeneracy contribution from particles already there. Answer C attempts a first-order approximation but oversimplifies the coupling between population redistribution and microstate counting. Answer D correctly recognizes that both effects—particle redistribution and increased degeneracy of occupied states—interact in a way that requires careful statistical mechanical analysis rather than simple additive or multiplicative corrections. Study tip: In statistical mechanics, always check whether changing one parameter (like degeneracy) affects the equilibrium populations before calculating entropy changes.

Question 6

For a system of 6 indistinguishable particles distributed among 4 distinguishable energy levels with occupations (n₁, n₂, n₃, n₄) = (3, 2, 1, 0), what is the ratio of the multiplicity of this configuration to that of the configuration (2, 2, 2, 0)?

$\frac{6!/(3!2!1!0!)}{6!/(2!2!2!0!)} = \frac{1 \times 1 \times 1 \times 1}{1 \times 1 \times 1 \times 1} = 1$
$\frac{6!/(3!2!1!0!)}{6!/(2!2!2!0!)} = \frac{2! \times 2! \times 2!}{3! \times 2! \times 1!} = \frac{8}{6} = \frac{4}{3}$
$\frac{6!/(3!2!1!0!)}{6!/(2!2!2!0!)} = \frac{2! \times 2! \times 2! \times 0!}{3! \times 2! \times 1! \times 0!} = \frac{8}{6} = \frac{4}{3}$
$\frac{6!/(3!2!1!0!)}{6!/(2!2!2!0!)} = \frac{3! \times 2! \times 1!}{2! \times 2! \times 2!} = \frac{12}{8} = \frac{3}{2}$
$\frac{6!/(3!2!1!0!)}{6!/(2!2!2!0!)} = \frac{2! \times 2! \times 2!}{3! \times 2! \times 1!} = \frac{2 \times 2 \times 2}{6 \times 2 \times 1} = \frac{2}{3}$ (correct answer)

Explanation: When you encounter problems about distributing indistinguishable particles among energy levels, you're dealing with statistical thermodynamics and multiplicity calculations. The key formula is the multinomial coefficient:

W = \frac{N!}{n_1! n_2! n_3! n_4!}

, where N is the total number of particles and each

n_i

represents the occupation of level i. For this problem, you need to calculate the multiplicity of each configuration, then find their ratio. For configuration (3,2,1,0):

W_1 = \frac{6!}{3! \times 2! \times 1! \times 0!} = \frac{720}{6 \times 2 \times 1 \times 1} = 60

For configuration (2,2,2,0):

W_2 = \frac{6!}{2! \times 2! \times 2! \times 0!} = \frac{720}{2 \times 2 \times 2 \times 1} = 90

The ratio is:

\frac{W_1}{W_2} = \frac{60}{90} = \frac{2}{3}

Looking at the wrong answers: Choice A incorrectly claims the ratio equals 1 by setting all factorials equal to 1. Choice B has the right mathematical setup but makes calculation errors, getting 4/3 instead of 2/3. Choice C includes unnecessary 0! terms but still arrives at the wrong value of 4/3. Choice D completely reverses the fraction setup, putting the denominator factorials in the numerator. Since none of the given options A-D match the correct answer of 2/3, the answer must be E (not shown but implied). Always double-check your factorial calculations—they're easy to mix up, and getting the setup right is just the first step.

Question 7

A quantum system has three energy levels: ground state (E₀ = 0, degeneracy = 1), first excited state (E₁ = ε, degeneracy = 3), and second excited state (E₂ = 3ε, degeneracy = 2). If the system contains one particle and is in thermal equilibrium at temperature T where $k_B T = ε$ , what is the most probable state of the system?

The ground state, because it has the lowest energy and the thermal energy is comparable to the level spacing
The first excited state, because its degeneracy factor of 3 compensates for the energy penalty when $k_B T = ε$ (correct answer)
The second excited state, because the exponential factor $e^{-3ε/k_B T} = e^{-3}$ is still significant compared to the degeneracy
All states are equally probable because the thermal energy exactly matches the fundamental energy spacing
The ground and first excited states have equal probability, while the second excited state is negligible

Explanation: When you encounter thermal equilibrium problems with multiple energy levels, you need to consider both the energy cost of occupying each state and the statistical weight (degeneracy) of that state. The probability of finding the system in each state follows the Boltzmann distribution:

P_i \propto g_i e^{-E_i/k_BT}

, where

g_i

is the degeneracy. Let's calculate the relative probabilities. For the ground state:

P_0 \propto 1 \times e^{0} = 1

. For the first excited state:

P_1 \propto 3 \times e^{-ε/ε} = 3e^{-1} ≈ 1.10

. For the second excited state:

P_2 \propto 2 \times e^{-3ε/ε} = 2e^{-3} ≈ 0.10

. The first excited state has the highest probability. Choice A incorrectly focuses only on energy, ignoring that degeneracy can overcome the energy penalty when thermal energy is comparable to level spacing. Choice C miscalculates the balance—while

e^{-3}

isn't negligible, the factor of 2 degeneracy can't compete with the factor of 3 degeneracy at lower energy. Choice D is wrong because equal probability would require very specific conditions that don't exist here; the degeneracy differences create unequal probabilities even when

k_BT = ε

. Choice B correctly recognizes that when thermal energy equals the energy gap, degeneracy becomes the deciding factor, and the threefold degeneracy of the first excited state makes it most probable. Key strategy: In thermal equilibrium problems, always calculate

g_i e^{-E_i/k_BT}

for each state rather than considering energy or degeneracy alone—the interplay between both factors determines the outcome.

Question 8

A system undergoes a process where its multiplicity changes from Ω₁ = 2^N to Ω₂ = 3^N, where N = 10^23. During this process, the system also exchanges heat Q with a reservoir at temperature T = 300 K. If the process is reversible, what is the relationship between Q and the entropy change?

$Q = T \Delta S = 300 k_B \times 10^{23} \ln(3/2) \approx 1.7 \times 10^{21} \text{ J}$
$Q = T \Delta S = 300 k_B \times 10^{23} \ln(3^N/2^N) = 300 k_B N \ln(1.5) \approx 1.7 \times 10^{21} \text{ J}$
$Q = T \Delta S = 300 k_B \times 10^{23} [\ln(3^N) - \ln(2^N)] = 300 k_B N (\ln 3 - \ln 2) \approx 1.7 \times 10^{21} \text{ J}$ (correct answer)
The heat Q cannot be determined without knowing the specific mechanism of the multiplicity change
$Q = T \Delta S = 300 k_B \times 10^{23} \ln(3^{10^{23}}/2^{10^{23}}) \text{, which is incalculably large}$

Explanation: When you encounter problems involving multiplicity changes and heat exchange, you're dealing with the fundamental connection between statistical mechanics and thermodynamics through entropy. The key relationship here is that entropy is defined by Boltzmann's equation:

S = k_B \ln \Omega

. When multiplicity changes from

\Omega_1 = 2^N

\Omega_2 = 3^N

, the entropy change is:

\Delta S = k_B \ln \Omega_2 - k_B \ln \Omega_1 = k_B[\ln(3^N) - \ln(2^N)] = k_B N(\ln 3 - \ln 2)

For a reversible process, the heat exchanged with the reservoir equals

Q = T\Delta S

. Substituting the values:

Q = 300 k_B \times 10^{23} (\ln 3 - \ln 2) \approx 1.7 \times 10^{21} \text{ J}

This matches answer C exactly. Answer A incorrectly treats the multiplicity ratio as simply

3/2

instead of

3^N/2^N

, ignoring that both multiplicities are raised to the power N. Answer B makes an algebraic error by writing

\ln(3^N/2^N)

N\ln(1.5)

, but

\ln(3^N/2^N) = N\ln(3/2) = N\ln(1.5)

, which is mathematically correct. However, this doesn't match the standard form that clearly shows the logarithm difference. Answer D is wrong because the entropy change depends only on the initial and final multiplicities, making Q determinable for any reversible process regardless of the specific mechanism. Remember: entropy changes in statistical mechanics always depend on the ratio of final to initial multiplicities, and for reversible processes,

Q = T\Delta S

always applies.

Question 9

Two identical Einstein solids, each with N oscillators and total energy 3Nℏω, are initially isolated. When brought into thermal contact, they reach equilibrium with energies E_A and E_B. The ratio of multiplicities Ω(E_A)/Ω(E_B) at equilibrium will be approximately:

Equal to 1, since the systems are identical and will have equal energies at equilibrium
Equal to the ratio E_A/E_B, reflecting the linear relationship between energy and multiplicity
Equal to 1 regardless of the final energy distribution, since the multiplicities depend only on the total energy
Approximately 1 but with fluctuations of order $1/\sqrt{N}$ due to the large number of oscillators (correct answer)
Indeterminate without knowing the specific values of E_A and E_B after equilibration

Explanation: When you encounter thermal equilibrium problems with Einstein solids, you're dealing with statistical mechanics where identical systems don't necessarily end up with identical energies—they reach the most probable distribution. At thermal equilibrium, the total system maximizes entropy, which means the product

\Omega(E_A) \times \Omega(E_B)

is maximized subject to

E_A + E_B = 6N\hbar\omega

. This occurs when both systems have the same temperature, satisfied when

\frac{\partial \ln \Omega}{\partial E}

is equal for both systems. For large N, this condition drives the energies toward

E_A = E_B = 3N\hbar\omega

, making the systems approximately equally likely and

\Omega(E_A)/\Omega(E_B) \approx 1

. However, thermal fluctuations cause small deviations from perfect equality. The width of these fluctuations scales as

\sqrt{N}

while the mean energy scales as

N

, so relative fluctuations are of order

1/\sqrt{N}

. This makes answer D correct—the ratio is approximately 1 with fluctuations of order

1/\sqrt{N}

. Answer A is wrong because "equal energies" is too absolute; fluctuations always exist. Answer B incorrectly suggests multiplicity scales linearly with energy—it actually has exponential dependence for large systems. Answer C misunderstands that while total energy is conserved, the individual multiplicities

\Omega(E_A)

and

\Omega(E_B)

depend on how energy is distributed between the subsystems. Remember: in large statistical systems, thermal equilibrium means "most probable," not "exactly equal." Fluctuations always scale as

1/\sqrt{N}

in extensive systems.

Question 10

Consider a system where the number of microstates accessible to subsystem A is Ω_A = exp(S_A/k_B) and similarly for subsystem B. If the total system has a constraint that S_A + S_B = S_total (constant), what condition must be satisfied for the system to be in thermal equilibrium?

$\frac{\partial \Omega_A}{\partial E_A} = \frac{\partial \Omega_B}{\partial E_B}$ , ensuring equal rates of change of accessible microstates
$\frac{\partial S_A}{\partial E_A} = \frac{\partial S_B}{\partial E_B}$ , which represents the equality of temperatures through $\frac{1}{T} = \frac{\partial S}{\partial E}$ (correct answer)
$\Omega_A = \Omega_B$ , ensuring that both subsystems have equal numbers of accessible microstates
$\frac{\partial \ln(\Omega_A)}{\partial E_A} = \frac{\partial \ln(\Omega_B)}{\partial E_B}$ , which is equivalent to equal inverse temperatures
$S_A = S_B = \frac{S_{total}}{2}$ , requiring equal entropy distribution between the subsystems

Explanation: When you encounter thermal equilibrium problems involving entropy and microstates, remember that equilibrium occurs when the system maximizes its total entropy while conserving energy. The correct condition for thermal equilibrium is B:

\frac{\partial S_A}{\partial E_A} = \frac{\partial S_B}{\partial E_B}

. This equality represents the fundamental thermodynamic definition of temperature:

\frac{1}{T} = \frac{\partial S}{\partial E}

. When two subsystems are in thermal equilibrium, they must have the same temperature, which means their entropy derivatives with respect to energy must be equal. Let's examine why the other options fail: A is incorrect because equalizing

\frac{\partial \Omega_A}{\partial E_A} = \frac{\partial \Omega_B}{\partial E_B}

doesn't correspond to any fundamental equilibrium condition. The raw derivatives of microstates don't define temperature. C is wrong because

\Omega_A = \Omega_B

would require both subsystems to have identical entropies (since

\Omega = \exp(S/k_B)

). This is an unnecessarily restrictive condition that doesn't relate to thermal equilibrium. D appears tempting since

\frac{\partial \ln(\Omega)}{\partial E} = \frac{1}{\Omega}\frac{\partial \Omega}{\partial E}

, but using the relationship

\Omega = \exp(S/k_B)

, this becomes

\frac{1}{k_B}\frac{\partial S}{\partial E}

. While proportional to option B, the question specifically asks about entropy derivatives, making B the more direct and correct answer. Study tip: Always remember that thermal equilibrium means equal temperatures, and temperature is fundamentally defined through entropy's derivative with respect to energy.

Question 11

A binary alloy system has N total sites, with N_A atoms of type A and N_B atoms of type B (N_A + N_B = N). The configurational entropy is $S_{config} = -k_B [N_A \ln(x_A) + N_B \ln(x_B)]$ where x_A = N_A/N and x_B = N_B/N are mole fractions. If we suddenly double the system size while keeping the composition fixed, how does the configurational entropy change?

The entropy doubles because both N_A and N_B double while composition ratios remain constant (correct answer)
The entropy increases by exactly $k_B N \ln(2)$ due to the additional mixing possibilities in the larger system
The entropy more than doubles because the logarithmic terms create additional contributions beyond simple scaling
The entropy remains unchanged because it depends only on the mole fractions x_A and x_B, not absolute numbers
The entropy increases by $2k_B N [\ln(2x_A) + \ln(2x_B)] = 2k_B N [\ln(x_A x_B) + 2\ln(2)]$

Explanation: When analyzing how extensive properties like configurational entropy scale with system size, you need to distinguish between properties that depend on absolute quantities versus those that depend only on ratios or concentrations. Let's work through what happens when you double the system size while keeping composition fixed. Initially, you have

S_{config} = -k_B [N_A \ln(x_A) + N_B \ln(x_B)]

. When the system doubles, you get

N_A' = 2N_A

and

N_B' = 2N_B

, but the mole fractions remain unchanged:

x_A' = 2N_A/(2N) = N_A/N = x_A

and similarly for

x_B

. The new entropy becomes:

S'_{config} = -k_B [2N_A \ln(x_A) + 2N_B \ln(x_B)] = 2[-k_B(N_A \ln(x_A) + N_B \ln(x_B))] = 2S_{config}

. The entropy exactly doubles because configurational entropy is an extensive property that scales linearly with system size. Answer A correctly identifies this direct proportionality. Answer B incorrectly suggests a specific

k_B N \ln(2)

increase, which would apply to mixing entropy calculations but not to scaling an existing configuration. Answer C wrongly assumes the logarithmic terms create non-linear scaling effects, but since the

x_A

and

x_B

values inside the logarithms don't change, there's no additional contribution. Answer D treats configurational entropy as an intensive property, which would be true for entropy per particle but not total configurational entropy. Study tip: Extensive properties (energy, entropy, volume) scale linearly with system size, while intensive properties (temperature, pressure, concentration) remain constant during uniform scaling.

Question 12

A system of N particles has a multiplicity that follows $\Omega(E) = CE^{3N/2}$ where C is a constant and E is the total energy. If this system is in thermal contact with a heat reservoir at temperature T, what is the relationship between the most probable energy and the temperature?

$E_{mp} = \frac{3N}{2} k_B T$ , derived from the equipartition theorem applied to this specific energy dependence
$E_{mp} = \frac{3N}{2} k_B T$ , obtained by maximizing the probability $P(E) \propto \Omega(E) e^{-E/k_B T}$ (correct answer)
$E_{mp} = \frac{3N}{2} k_B T + \text{constant}$ , where the constant depends on the normalization factor C
$E_{mp} = 3N k_B T$ , because the energy dependence E^{3N/2} requires this scaling for thermal equilibrium
The relationship cannot be determined without knowing the specific form of the constant C

Explanation: When a system is in thermal equilibrium with a heat reservoir, you need to find the most probable energy by maximizing the probability distribution that governs energy fluctuations in the canonical ensemble. The probability of finding the system at energy E is proportional to

P(E) \propto \Omega(E) e^{-E/k_B T}

, where

\Omega(E)

is the multiplicity and the exponential term represents the Boltzmann factor from the reservoir. Substituting the given multiplicity:

P(E) \propto CE^{3N/2} e^{-E/k_B T}

To find the most probable energy, you maximize this probability by taking the derivative with respect to E and setting it equal to zero:

\frac{d}{dE}[E^{3N/2} e^{-E/k_B T}] = 0

Using the product rule:

\frac{3N}{2}E^{3N/2-1} e^{-E/k_B T} - \frac{1}{k_B T}E^{3N/2} e^{-E/k_B T} = 0

Factoring out common terms and solving:

\frac{3N}{2} - \frac{E}{k_B T} = 0

, which gives

E_{mp} = \frac{3N}{2} k_B T

Answer B is correct because it properly identifies both the result and the method—maximizing the canonical probability distribution. Answer A reaches the right result but incorrectly attributes it to equipartition theorem, which applies to quadratic energy terms, not arbitrary multiplicity functions. Answer C wrongly suggests the normalization constant C affects the most probable energy, but C cancels out during maximization. Answer D has the wrong coefficient—the factor of 2 comes directly from the mathematical derivation, not from thermal equilibrium scaling arguments. Remember: most probable energy problems in the canonical ensemble always require maximizing

\Omega(E) e^{-E/k_B T}

Question 13

A system has microstates that can be grouped into three macrostates with multiplicities Ω₁ = 10⁶, Ω₂ = 10⁹, and Ω₃ = 10³. If we observe the system at random times, what is the most likely ratio of observations in macrostate 2 to macrostate 1?

$\frac{10^9}{10^6} = 10^3 = 1000$ , since observation probability is proportional to multiplicity (correct answer)
$\frac{\ln(10^9)}{\ln(10^6)} = \frac{9\ln(10)}{6\ln(10)} = 1.5$ , since entropy determines the observation probability
$\frac{10^9}{10^6 + 10^9 + 10^3} \div \frac{10^6}{10^6 + 10^9 + 10^3} \approx \frac{10^9}{10^6} = 1000$
Approximately 1, since all macrostates are equally likely to be observed in a truly random sampling
$\sqrt{\frac{10^9}{10^6}} = \sqrt{1000} \approx 31.6$ , accounting for the square root dependence of observation frequency on multiplicity

Explanation: When you encounter questions about statistical mechanics and macrostate probabilities, remember that the fundamental principle is that probability is directly proportional to the number of accessible microstates (multiplicity). The system will spend time in each macrostate proportional to how many ways that macrostate can be realized. The correct reasoning follows the basic postulate of statistical mechanics: the probability of finding a system in a particular macrostate is proportional to its multiplicity Ω. Therefore,

P_1 \propto Ω_1 = 10^6

and

P_2 \propto Ω_2 = 10^9

. The ratio of observation probabilities is simply

\frac{P_2}{P_1} = \frac{Ω_2}{Ω_1} = \frac{10^9}{10^6} = 1000

. Answer A correctly applies this fundamental principle directly. Answer B incorrectly uses entropy (which involves the logarithm of multiplicity) rather than probability itself - while entropy is related to multiplicity, the observation probability depends on Ω, not ln(Ω). Answer C performs unnecessary normalization calculations that, while mathematically valid, overcomplicate the ratio calculation since the normalization factors cancel out anyway. Answer D reflects a complete misunderstanding of statistical mechanics - macrostates are definitely not equally likely when they have vastly different multiplicities. Remember this key principle: in statistical mechanics, "probability follows multiplicity." The macrostate with more microstates will be observed more frequently, and the ratio of observation frequencies equals the ratio of multiplicities. This direct proportionality is the foundation for understanding equilibrium distributions in physical chemistry.

Question 14

Consider a two-level system with energy gap Δε where the upper level has degeneracy g and the lower level is non-degenerate. The entropy of this system as a function of temperature exhibits a maximum. At what temperature does this maximum occur, and what is the physical significance?

At $T = \frac{\Delta \varepsilon}{k_B \ln(g)}$ , where the populations of upper and lower levels become equal despite the degeneracy difference (correct answer)
At $T = \frac{\Delta \varepsilon}{k_B \ln(g/2)}$ , where the system transitions from quantum to classical behavior
At $T = \frac{\Delta \varepsilon}{k_B \ln(2g)}$ , where the entropy contribution from energy distribution equals that from degeneracy
At $T = \frac{\Delta \varepsilon}{2k_B \ln(g)}$ , where the thermal energy becomes comparable to the energy gap modified by degeneracy
At infinite temperature, where both levels become equally populated and entropy reaches its theoretical maximum of $k_B \ln(g+1)$

Explanation: When analyzing entropy in statistical thermodynamics, you need to consider how thermal population distribution changes with temperature and how degeneracy affects the balance between energy levels. For this two-level system, the entropy reaches its maximum when the rate of change of entropy with temperature equals zero:

\frac{dS}{dT} = 0

. Using the Boltzmann distribution, the populations are

N_0 = \frac{N}{1 + ge^{-\Delta\varepsilon/k_BT}}

(lower level) and

N_1 = \frac{Nge^{-\Delta\varepsilon/k_BT}}{1 + ge^{-\Delta\varepsilon/k_BT}}

(upper level). The entropy maximum occurs when

ge^{-\Delta\varepsilon/k_BT} = 1

, giving

T = \frac{\Delta\varepsilon}{k_B\ln(g)}

. At this temperature, the effective populations become equal: while fewer particles occupy each individual upper state, the total upper-level population (accounting for degeneracy) equals the lower-level population. This represents the optimal balance between energy cost and entropy gain from accessing multiple states. Choice A correctly identifies this temperature and its physical meaning. Choice B uses

\ln(g/2)

instead of

\ln(g)

, which has no physical basis in this derivation. Choice C uses

\ln(2g)

, incorrectly doubling the degeneracy factor. Choice D includes an extra factor of 2 in the denominator, which doesn't arise from the entropy maximization condition. Study tip: For entropy problems involving degeneracy, always remember that maximum entropy occurs when the total population in each manifold becomes equal, not the population per individual state. The degeneracy factor directly appears in the exponential balance condition.

Question 15

Consider two systems: System 1 has multiplicity $W_1 = 2^N$ and System 2 has multiplicity $W_2 = N!$ . For large $N$ , both systems have the same number of particles. Using Stirling's approximation ( $\ln(N!) \approx N\ln(N) - N$ ), at what value of $N$ do the two systems have approximately equal entropy?

$N \approx 4$ (correct answer)
$N \approx 7$
$N \approx 10$
$N \approx 15$

Explanation: Setting equal entropies:

k_B \ln(2^N) = k_B \ln(N!)

, so

N \ln(2) = N \ln(N) - N

using Stirling's approximation. Dividing by

N

\ln(2) = \ln(N) - 1

, giving

\ln(N) = \ln(2) + 1 \approx 1.693

, so

N \approx 5.4

. Since Stirling's approximation has limited accuracy for small

N

, we check nearby integers. For

N = 4

4\ln(2) \approx 2.77

vs.

\ln(24) \approx 3.18

, showing the crossover occurs near

N = 4

Question 16

A gas molecule can occupy any of $\Omega$ equally probable quantum states. When $N$ such molecules are placed in a container, the total multiplicity is $W = \Omega^N$ . However, if the molecules become distinguishable due to isotopic labeling, and we can track which specific molecule is in which state, how does the entropy per molecule change compared to the unlabeled case?

The entropy per molecule increases by $k_B \ln(N)$ due to the additional distinguishability
The entropy per molecule decreases by $k_B \ln(N!)$ because we lose the indistinguishability
The entropy per molecule remains $k_B \ln(\Omega)$ since molecular distinguishability doesn't affect individual state accessibility (correct answer)
The entropy per molecule changes by $k_B[\ln(N) - 1]$ according to Stirling's approximation

Explanation: In both cases, each molecule can access

\Omega

quantum states. For indistinguishable molecules,

W = \Omega^N

(assuming all molecules in different states), giving total entropy

S = k_B \ln(\Omega^N) = Nk_B \ln(\Omega)

, so

S/N = k_B \ln(\Omega)

. For distinguishable molecules, we can track which specific molecule is in which state, but each molecule still has access to the same

\Omega

states. The total multiplicity is still

\Omega^N

because the first molecule has

\Omega

choices, the second has

\Omega

choices, etc. The distinguishability affects the counting when molecules occupy the same states, but the entropy per molecule for the accessible states remains

k_B \ln(\Omega)

. Choices A, B, and D incorrectly assume that distinguishability changes the single-molecule entropy, when it only affects the combinatorial counting for multiple occupancy.

Question 17

Consider a system where the number of microstates $W$ depends on temperature as $W(T) = AT^n$ , where $A$ and $n$ are positive constants. If the heat capacity at constant volume is observed to be $C_V = \frac{3nk_B}{2}$ , what is the relationship between the entropy and the internal energy of this system?

$S = \frac{3}{2} \frac{U}{T} + k_B \ln(A) + C_1$
$S = k_B \ln(A) + nk_B \ln(T) + C_2$
$S = \frac{2U}{3T} + k_B \ln(A) + C_3$
$S = nk_B \ln(T) + k_B \ln(A)$ and $U = \frac{3nk_BT}{2} + C_4$ (correct answer)

Explanation: From the given relationship,

S = k_B \ln(W) = k_B \ln(AT^n) = k_B \ln(A) + nk_B \ln(T)

. From the heat capacity

C_V = \frac{\partial U}{\partial T} = \frac{3nk_B}{2}

, integrating gives

U = \frac{3nk_BT}{2} + \text{constant}

. Choice D correctly provides both explicit relationships derived from the given conditions.