Statistical Thermodynamics Questions and Answers

Q: A particle can occupy either the ground state at E = 0 or an excited state at E \(\gt\) 0. At a temperature T, the probability of the particle being in the excited state is :

The probability is exp(−E/kBT) / (1 + exp(−E/kBT)).By the Boltzmann distribution, each state's population is proportional to exp(−E/kBT), divided by the partition function, the sum of these factors over all states. With levels at 0 and E,\[ \begin{aligned} q &= 1 + e^{-E/k_{B}T} \\ p_{\mathrm{exc}} &= \frac{e^{-E/k_{B}T}}{1 + e^{-E/k_{B}T}} \end{aligned} \]Option D is the probability of the ground state, option B is the partition function itself (which exceeds 1, so it cannot be a probability), and 0 holds only at absolute zero.

Q: According to Fermi-Dirac the statistics following assumption is made :

Fermi-Dirac statistics assumes the particles are indistinguishable and that only one particle may occupy a given state.Fermi-Dirac statistics applies to fermions, particles of half-integer spin such as electrons, protons and neutrons. Identical quantum particles cannot be labelled, so they are indistinguishable, and fermions obey the Pauli exclusion principle, which allows at most one particle in each quantum state. The resulting occupancy of a state of energy E never exceeds 1:\[ f(E) = \frac{1}{e^{(E-\mu)/kT} + 1} \]Distinguishable particles with any number per level is Maxwell-Boltzmann statistics. Indistinguishable particles with any number per state is Bose-Einstein statistics, which applies to bosons such as photons. Distinguishable particles with single occupancy belongs to none of the three standard statistics.

Q: In a canonical ensemble, a system X of fixed volume is in contact with a large reservoir Y, then :

In a canonical ensemble X can exchange only energy with Y.The canonical ensemble describes a system with fixed number of particles N, fixed volume V and fixed temperature T. The temperature is held constant by contact with a large heat reservoir, and the system and reservoir exchange energy as heat, so the system's energy fluctuates. The walls are closed to matter, so the number of particles cannot change.Exchange of neither energy nor particles defines the microcanonical ensemble (fixed N, V, E), an isolated system. Exchange of both energy and particles defines the grand canonical ensemble (fixed μ, V, T). An ensemble exchanging particles but not energy is not one of the standard three.

Q: For a system with non-degenerate zero energy state and doubly degenerate upper state (with energy E), the partition function is :

The partition function is 1 + 2e−E/kT.The molecular partition function sums the Boltzmann factor over all levels, each weighted by its degeneracy g:\[ \begin{aligned} q &= \sum_{i} g_{i}\,e^{-E_{i}/kT} \\ &= 1\cdot e^{0} + 2\,e^{-E/kT} = 1 + 2e^{-E/kT} \end{aligned} \]The ground level at zero energy is non-degenerate and contributes exactly 1. The doubly degenerate upper level contributes 2e−E/kT.2e−E/kT leaves out the ground state, 2 + e−E/kT puts the degeneracy on the wrong level, and 3e−E/kT treats all three states as if they had energy E.

Q: At low temperature, heat capacity of solids at constant volume is proportional to :

At low temperature CV is proportional to T3.In the Debye model of a solid, the heat capacity comes from lattice vibrations (phonons). At temperatures well below the Debye temperature θD, only low-frequency acoustic modes are excited, and their number grows as T3, giving the Debye T3 law:\[ \begin{aligned} C_{V} &= \frac{12\pi^{4}}{5}R \\ &\times \left(\frac{T}{\theta_{D}}\right)^{3} \end{aligned} \]The heat capacity falls to zero as T approaches 0, as the third law requires. (1/T)3 and (R/T)2 would make CV grow without limit as T falls, and (RT)2 does not match the Debye result. At high temperature CV instead approaches the constant 3R (Dulong and Petit).

Q: When in an ensemble the condition of constant temperature is replaced by the constant energy condition, each system behaves as :

Each system behaves as an isolated system.In the canonical ensemble each system has fixed N, V and T: it exchanges energy with a heat bath but not matter, so it is closed. If constant temperature is replaced by constant energy, the systems can exchange neither energy nor matter with their surroundings. That is the microcanonical (N, V, E) ensemble, and each member is an isolated system.'Closed' describes the canonical case being replaced. 'Open' systems exchange matter as well, as in the grand canonical ensemble (μ, V, T), and 'condensed' is a phase description, not a type of ensemble system.

Q: At what temperature, the population in v = 1 level of I2 would become half the population in ground energy level ? (Given : \(\dfrac{hc\bar{v}}{k}\) = 346.5 K)

The temperature is 500 K.For the vibrational levels (each non-degenerate) the Boltzmann ratio is set by the energy gap hcν̄ between v = 1 and v = 0:\[ \frac{N_{1}}{N_{0}} = e^{-hc\bar{\nu}/kT} = \frac{1}{2} \]\[ \begin{aligned} \frac{346.5}{T} &= \ln 2 = 0.693 \\ T &= \frac{346.5}{0.693} = 500\ \text{K} \end{aligned} \]773 K and 227 K are 500 K with 273 added or subtracted, a slip between kelvin and degrees Celsius, and 300 K is simply room temperature.

Q: The number of microstates of distributing five quanta of energy among four distinguishable particles is :

The number of microstates is 56.Distributing q identical quanta among N distinguishable particles is a stars-and-bars count: the number of ways is\[ \begin{aligned} W &= \frac{(q + N - 1)!}{q!\,(N - 1)!} \\ &= \frac{8!}{5!\,3!} = \frac{8 \times 7 \times 6}{6} = 56 \end{aligned} \]126 is what you get with N = 5 and q = 5 (9!/5!4!), mixing up the roles of particles and quanta, and 35 comes from 7!/4!3!, one too few in the numerator. 14 has no basis in the count.

11.

A particle can occupy either the ground state at E = 0 or an excited state at E \(\gt\) 0. At a temperature T, the probability of the particle being in the excited state is :

A.0
B.\(1 + \mathrm{exp} - E/K_{B}T\)
C.\(\mathrm{exp} - E/K_{B}T/(1 + \mathrm{exp} - E/K_{B}T)\)
D.\(1/(1 + \mathrm{exp} - E/K_{B}T)\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

The probability is exp(−E/k_BT) / (1 + exp(−E/k_BT)).

By the Boltzmann distribution, each state's population is proportional to exp(−E/k_BT), divided by the partition function, the sum of these factors over all states. With levels at 0 and E,

\[ \begin{aligned} q &= 1 + e^{-E/k_{B}T} \\ p_{\mathrm{exc}} &= \frac{e^{-E/k_{B}T}}{1 + e^{-E/k_{B}T}} \end{aligned} \]

Option D is the probability of the ground state, option B is the partition function itself (which exceeds 1, so it cannot be a probability), and 0 holds only at absolute zero.

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12.

According to Fermi-Dirac the statistics following assumption is made :

A.the particles are distinguishable and any number of particles may occupy the same energy level
B.the particles are indistinguishable and any number of particles may occupy a given energy level
C.the particles are indistinguishable but only one particle may occupy a given energy level
D.the particles are distinguishable and only one particle may occupy a given energy level

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: C

Fermi-Dirac statistics assumes the particles are indistinguishable and that only one particle may occupy a given state.

Fermi-Dirac statistics applies to fermions, particles of half-integer spin such as electrons, protons and neutrons. Identical quantum particles cannot be labelled, so they are indistinguishable, and fermions obey the Pauli exclusion principle, which allows at most one particle in each quantum state. The resulting occupancy of a state of energy E never exceeds 1:

\[ f(E) = \frac{1}{e^{(E-\mu)/kT} + 1} \]

Distinguishable particles with any number per level is Maxwell-Boltzmann statistics. Indistinguishable particles with any number per state is Bose-Einstein statistics, which applies to bosons such as photons. Distinguishable particles with single occupancy belongs to none of the three standard statistics.

Discuss in Forum

13.

In a canonical ensemble, a system X of fixed volume is in contact with a large reservoir Y, then :

A.X can exchange only energy with Y
B.X can exchange only particles with Y
C.X can exchange neither energy nor particles with Y
D.X can exchange both energy and particles with Y

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: A

In a canonical ensemble X can exchange only energy with Y.

The canonical ensemble describes a system with fixed number of particles N, fixed volume V and fixed temperature T. The temperature is held constant by contact with a large heat reservoir, and the system and reservoir exchange energy as heat, so the system's energy fluctuates. The walls are closed to matter, so the number of particles cannot change.

Exchange of neither energy nor particles defines the microcanonical ensemble (fixed N, V, E), an isolated system. Exchange of both energy and particles defines the grand canonical ensemble (fixed μ, V, T). An ensemble exchanging particles but not energy is not one of the standard three.

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14.

For a system with non-degenerate zero energy state and doubly degenerate upper state (with energy E), the partition function is :

A.\(2e^{-E/kT}\)
B.\(2 + e^{-E/kT}\)
C.\(3e^{-E/kT}\)
D.\(1 + 2e^{-E/kT}\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: D

The partition function is 1 + 2e^−E/kT.

The molecular partition function sums the Boltzmann factor over all levels, each weighted by its degeneracy g:

\[ \begin{aligned} q &= \sum_{i} g_{i}\,e^{-E_{i}/kT} \\ &= 1\cdot e^{0} + 2\,e^{-E/kT} = 1 + 2e^{-E/kT} \end{aligned} \]

The ground level at zero energy is non-degenerate and contributes exactly 1. The doubly degenerate upper level contributes 2e^−E/kT.

2e^−E/kT leaves out the ground state, 2 + e^−E/kT puts the degeneracy on the wrong level, and 3e^−E/kT treats all three states as if they had energy E.

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15.

At low temperature, heat capacity of solids at constant volume is proportional to :

A.\(T^{3}\)
B.\(\left(\frac{1}{T}\right)^{3}\)
C.\((RT)^{2}\)
D.\(\left(\frac{R}{T}\right)^{2}\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: A

At low temperature C_V is proportional to T³.

In the Debye model of a solid, the heat capacity comes from lattice vibrations (phonons). At temperatures well below the Debye temperature θ_D, only low-frequency acoustic modes are excited, and their number grows as T³, giving the Debye T³ law:

\[ \begin{aligned} C_{V} &= \frac{12\pi^{4}}{5}R \\ &\times \left(\frac{T}{\theta_{D}}\right)^{3} \end{aligned} \]

The heat capacity falls to zero as T approaches 0, as the third law requires. (1/T)³ and (R/T)² would make C_V grow without limit as T falls, and (RT)² does not match the Debye result. At high temperature C_V instead approaches the constant 3R (Dulong and Petit).

Discuss in Forum

16.

When in an ensemble the condition of constant temperature is replaced by the constant energy condition, each system behaves as :

A.Closed
B.Open
C.Isolated
D.Condensed

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

Each system behaves as an isolated system.

In the canonical ensemble each system has fixed N, V and T: it exchanges energy with a heat bath but not matter, so it is closed. If constant temperature is replaced by constant energy, the systems can exchange neither energy nor matter with their surroundings. That is the microcanonical (N, V, E) ensemble, and each member is an isolated system.

'Closed' describes the canonical case being replaced. 'Open' systems exchange matter as well, as in the grand canonical ensemble (μ, V, T), and 'condensed' is a phase description, not a type of ensemble system.

Discuss in Forum

17.

At what temperature, the population in v = 1 level of I₂ would become half the population in ground energy level ? (Given : \(\dfrac{hc\bar{v}}{k}\) = 346.5 K)

A.773 K
B.500 K
C.227 K
D.300 K

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: B

The temperature is 500 K.

For the vibrational levels (each non-degenerate) the Boltzmann ratio is set by the energy gap hcν̄ between v = 1 and v = 0:

\[ \frac{N_{1}}{N_{0}} = e^{-hc\bar{\nu}/kT} = \frac{1}{2} \]

\[ \begin{aligned} \frac{346.5}{T} &= \ln 2 = 0.693 \\ T &= \frac{346.5}{0.693} = 500\ \text{K} \end{aligned} \]

773 K and 227 K are 500 K with 273 added or subtracted, a slip between kelvin and degrees Celsius, and 300 K is simply room temperature.

Discuss in Forum

18.

The number of microstates of distributing five quanta of energy among four distinguishable particles is :

A.56
B.14
C.35
D.126

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

The number of microstates is 56.

Distributing q identical quanta among N distinguishable particles is a stars-and-bars count: the number of ways is

\[ \begin{aligned} W &= \frac{(q + N - 1)!}{q!\,(N - 1)!} \\ &= \frac{8!}{5!\,3!} = \frac{8 \times 7 \times 6}{6} = 56 \end{aligned} \]

126 is what you get with N = 5 and q = 5 (9!/5!4!), mixing up the roles of particles and quanta, and 35 comes from 7!/4!3!, one too few in the numerator. 14 has no basis in the count.

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