Statistical Thermodynamics MCQ Quiz in मल्याळम - Objective Question with Answer for Statistical Thermodynamics - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

The correct answer is X can exchange only energy with Y

Concept:-

Thermal Equilibrium and Boltzmann Distribution: 

• One of the fundamental concepts associated with the canonical ensemble is the idea of thermal equilibrium with a reservoir. Systems in a canonical ensemble follow the Boltzmann distribution, which dictates the probability of the system being in a specific energy state, with higher energy states being exponentially less probable as energy increases.
• The temperature of the system, which is determined by the reservoir, plays a crucial role in this distribution.

Partition Function (Z):

• The partition function, denoted as (Z), is a central concept in the study of canonical ensembles. It is a summation (or integral, in the continuous case) over all possible states of the system, weighted by their Boltzmann factor, (\(e^{-E_i/k_BT}\)), where (Ei) is the energy of state (i), (k_B) is the Boltzmann constant, and (T) is the temperature of the ensemble.
• The partition function plays a key role in determining macroscopic quantities such as free energy, entropy, and specific heat.

Explanation:-

Canonical Ensemble: A canonical ensemble is a theoretical construct in statistical mechanics where a system is considered to be in thermal equilibrium with a heat reservoir. The defining characteristics of such a system include:

• The volume ((V)) of the system is fixed.
• The number of particles ((N)) in the system is fixed.
• The system is allowed to exchange energy ((E)) with the heat reservoir, leading to variations in the internal energy of the system.

The concept of a canonical ensemble is crucial in understanding systems at constant temperature, where fluctuations in energy are permissible as long as the overall temperature remains constant.

In a canonical ensemble, the system is allowed to exchange energy with its surroundings or a reservoir but not particles. The volume of the system is kept constant, and the system is in thermal equilibrium with the reservoir. This exchange of energy allows the system to explore different microstates while maintaining a constant number of particles and volume.

Additional Information 
- Option 2 (X can exchange only particles with Y): This description fits the grand canonical ensemble, not the canonical ensemble. In a grand canonical ensemble, the system can exchange both particles and energy with the reservoir, and both the chemical potential and temperature are equilibrated between the system and the reservoir.

- Option 3 (X can exchange neither energy nor particles with Y): This situation describes an isolated system, where there is no exchange of energy or particles with the surroundings. This is better represented by the microcanonical ensemble, where the energy, volume, and number of particles in the system are all fixed.

- Option 4 (X can exchange both energy and particles with Y): This is incorrect for a canonical ensemble as it suggests a scenario more akin to the grand canonical ensemble, where both energy and particles can be exchanged between the system and the reservoir. In a canonical ensemble, only energy exchange is permitted.

Conclusion:-

So, In a canonical ensemble, a system X of fixed volume is in contact with a large reservoir Y, then X can exchange only energy with Y

The correct answer is \(\frac{RT_c}{8p_c}\)

Concept:-

1. Van der Waals Equation
For a non-ideal gas, the Van der Waals equation provides corrections for molecular volume and intermolecular forces, represented as: 
\([ \left( P + \frac{n^2 a}{V^2} \right) (V - nb) = nRT ]\)

(a) and (b) are Van der Waals constants pertaining to intermolecular forces and the finite size of gas molecules, respectively.
2. Critical Properties
At the critical point:

(Tc) = Critical temperature
(Pc) = Critical pressure
(Vc) = Critical volume
The critical constants (a), (b), (Tc), (Pc), and (Vc) are related to the substance's individual properties.

Explanation:-

We know that,

\(T_c=\frac{8 a}{27 R b}\)

So,

\(27 b =\frac{8 a}{T_c \cdot R}\)                               …(i)

\(p_c =\frac{a}{27 b^2}\)                                 …(ii) 

From Eq. (i) and (ii)

\(p_c =\frac{a \times T_c \times R}{b \times 8 a}\)

\(b =\frac{T_c \cdot R}{8 \cdot p_c}\)

Conclusion:-

So, The volume correction factor for a non-ideal gas in terms of critical pressure (pc), critical molar volume (Vc), critical temperature (Tc) and gas constant (R) is \(\frac{RT_c}{8p_c}\)

The correct answer is hv/(In 2 kB)

Concept:-

Quantum Mechanics: The foundation for this question lies in quantum mechanics. Quantum mechanics is a key principle in physics that describes nature at the smallest level - including atomic and subatomic particles. One takeaway from quantum mechanics is the concept of energy quantization, which means that a system can only occupy certain discrete energy levels.

Vibrational Modes: In the context of molecules, vibrational modes describe the different ways a molecule can vibrate assuming it behaves like a simple harmonic oscillator. Each vibrational mode is associated with a unique energy level. The vibrational energy levels of a diatomic molecule are quantized and defined by (v+1/2)hν, where v is a quantum number, h is Planck's constant, and ν is the frequency.

Boltzmann Distribution: This is a statistical law in thermodynamics that predicts the probability of a particle being in a certain state as a function of that state's energy and the system's temperature. According to this distribution, at thermal equilibrium, the ratio of the number of atoms in two states (N2/N1) is equal to the exponential of the negative ratio between energy difference and the product of Boltzmann's constant and temperature (exp(-ΔE/kT)).

Thermal Equilibrium: This is a state in which all parts of a system are at a uniform temperature and no energy transfer is taking place. In thermal equilibrium conditions, Boltzmann Distribution applies to describe the population of different energy states.

Energy Levels and Temperature: At higher temperatures, more molecules will be able to overcome the activation energy to reach the excited energy levels. This question is essentially asking about the temperature at which the population of the first excited state will be half that of the ground state based on Boltzmann Distribution.

Explanation:-

According to Boltzmann distribution, the ratio of the population of two energy states under thermal equilibrium at temperature T is defined by:

\(N2/N1 = exp^{(-ΔE/kT)}\)

where:

N2 and N1 are the numbers of molecules in the higher and lower energy states, respectively
ΔE is the energy difference between the two states
k is Boltzmann constant
T is the absolute temperature
In the case of vibrational modes of molecules, the energy levels are quantized and given by (v+1/2)hν for v = 0, 1, 2, 3, ...

Given that the population of the first excited state (v = 1) is half that of the ground state (v = 0), the Boltzmann distribution simplifies to:

\(1/2 = e^{[-(2hν/2)/kT]}\)

Taking natural logarithm on both sides:

\(ln(1/2) = -(hν/kT)\)

Solving for T gives you the temperature at which the population of the first excited state is half that of the ground state:

\(T = -hν/[k\times ln(1/2)]\)

Since ln(1/2) is -ln2,

\(T = hν/[k \times ln2]\)

which corresponds to the option (b), hv/(ln 2 kB).

It means the temperature at which the population of the first excited state is half that of the ground state is given by hv/(ln 2 kB).

Remember that h is Planck's constant, ν is the frequency of vibration (v in the question), k is the Boltzmann constant, and T is the absolute temperature.

Conclusion:-

 The temperature at which the population of the first excited state will be half that of the ground state is given by hv/(ln 2 kB).

The correct answer is 320K

Concept:-

• Kinetic Theory of Gases: According to this theory, gas particles are in constant random motion and collisions between these particles are perfectly elastic.
• Average Molecular Speed: Different molecules in a gas have different speeds at any point in time due to their continuous, random motion. However, an 'average' speed can be calculated which gives an indication of how quickly the molecules are moving overall.
• The effect of Mass on Gas Speeds: Lighter gas molecules will travel at higher average speeds than heavier gas molecules at the same temperature.
• Root Mean Square Speed: This is a measure of the average speed of particles in a gas. It's directly proportional to the square root of the absolute temperature and inversely proportional to the square root of the molar mass.
• The Temperature-Speed Relationship: Increasing the temperature of a gas increases the average speed of its molecules. This is because temperature is a measure of the average kinetic energy of the particles in a substance, so raising the temperature increases their kinetic energy, and therefore their speed.

Explanation:-

we have the average velocity of a gas molecule as  \(\sqrt{\frac{3RT}{M}}\)

Such that the average velocity \(\sqrt{\frac{T}{M}}\)

\(\frac{V_{O_2}}{V_{H_2}}=\sqrt{\frac{T_{O_2 }. M_{H_2}}{T_{H_2 }. M_{O_2}}}\)

M_H2 is 2 and M_O2 is 32 also T_H2 is 20K

Also, the velocity of hydrogen and the velocity of oxygen is equal.

\(1=\sqrt{\frac{T_{O_2 } \times 2}{20 \times 32}}\)

squaring both sides.

\(T_{O_2}= 320K\)

Conclusion:-

So,  the temperature at which the average velocity of oxygen equals that of hydrogen at 20 K. is 320K

Explanation:-

• The rotational partition function (q) is given by,

\(q=\frac{k_BT}{\sigma hcB}\)

where, \(B=\frac{h}{8\pi^2 Ic}\)

or, \(q=\frac{k_BT}{\sigma hc\frac{h}{8\pi^2 Ic}}\)

or, \(q=\frac{8\pi^2 IkT}{\sigma h^2}\)

or, \(q\propto \frac{1}{\sigma}\)

Hence, two factors should be considered,

(1) \(I \propto \mu\)

(2) \(\sigma (symmetry \;element)\)

• In case of \(H-C\equiv C-H\), it has the least reduced mass and also for it \(\sigma =2\)

Conclusion:-

• Hence, the molecule with the smallest rotation partition function at any temperature among the following is

\(H-C\equiv C-H\)

Concept:-

• For a system with i component the partition function (Z) is given by,

\(Z = \sum_{i}^{}g_{i}e^{-\frac{E_{i}}{K_{B}T}}\)

\(\frac{n_j}{n_i}=\frac{g_j}{g_i}e^{-\frac{\Delta \varepsilon }{k_B T}}\)

n_i = number of particles in i^th state,

gi = degeneracy of the ith state,

Ei = energy of the ith state,

kB = Boltzmann constant,

and T is the absolute temperature of the system.

Explanation:-

• The equilibrium population ratio \(\left ( \frac{n_j}{n_i} \right )\) of a doubly-degenerate energy level (Ej) lying at energy 2 units higher than a lower non-degenerate energy level (Ej) and Assuming kBT = 1 unit, we got

\(\frac{n_j}{n_i}=\frac{g_j}{g_i}e^{-\frac{\Delta \varepsilon }{k_B T}}\)

\(\frac{n_j}{n_i}=\frac{2}{1}e^{-\frac{2 }{1}}\) (g_j=2, gi=1, \(\Delta \varepsilon =2\))

= 2e-2

Conclusion:-

• Hence, the equilibrium population ratio will be 2e-2

Concept:-

• Statistical Thermodynamics is a branch of Statistical Mechanics that deals with equilibrium (time-independent) processes.
• Statistical mechanics enables us to understand thermodynamics from a molecular perspective through the concept of ensembles.
• Statistical mechanics allows us to know the absolute magnitude of thermodynamic quantities like S, U, H, A, G…in terms of microscopic parameters of the system, and not just their change, dS, dU, dH, dA, dG (which as we know from CT only depend on a few macroscopic parameters).
• Theories for both classical and statistical thermodynamics are governed by the second law of thermodynamics through the medium of entropy.
• Entropy in classical thermodynamics is determined empirically.
• Entropy in statistical mechanics is a function of the distribution of microstates of the system.
• An example of different possible microstates for a particle is shown below:

• Entropy (S) is a state function and extensive property of a system since the magnitude of dq for a given change in temperature depends upon the mass.
• Now, for a system with w number of orientations / arrangements / microstates is given by, 

S = kb ln w

• For Distinguishable particles with occupation number {a, b, c, d} in M number of energy states, the number of microstates (w) will be

\(w =\frac{N!}{\Pi_i !\times n_i !}\)

where, N =  number of particles,

\(\Pi _i\) = number of particles in that energy state,

and n_i = occupation number

Explanation:-

• The number of microstates (w) for 6 identical particles with their occupation number {0, 1, 2, 3} in four states is

\(w =\frac{6!}{0!1!2!3!}\)

\(=\frac{6\times5\times4\times3\times2\times1}{1 \times1\times 2\times1\times3\times2\times1}\)

= 60

• Thus, the correct entropy for 6 identical particles with their occupation number {0, 1, 2, 3} in four states is

S = KB ln w

S = KB ln 60 (as w = 60)

Conclusion:-

• Hence, the correct entropy for 6 identical particles with their occupation number {0, 1, 2, 3} in four states is KBln60

Calculations:
To calculate the work done by an ideal gas as it expands against a constant pressure, you can use the formula:\(W = -P \cdot Δ V\)

Where, (W) is the work done, (P) is the constant pressure,ΔV is the change in volume.
Given: 

\(P = 1 \times 10^5 , \text{N/m}^2 \)
\(Δ V = (1 \times 10^{-2} , \text{m}^3) - (1 \times 10^{-3} , \text{m}^3) = 9 \times 10^{-3} , \text{m}^3\)
Now, plug in the values and calculate the work done:

\(W = - (1 \times 10^5 , \text{N/m}^2) \cdot (9 \times 10^{-3 } , \text{m}^3) = -9 \times 10^2 , \text{N} \cdot \text{m} = -900 , \text{J}\)

So, the work done by the ideal gas is -900 J.

The correct answer is option 1) -900 J.

Concept:-

• Statistical Thermodynamics is a branch of Statistical Mechanics that deals with equilibrium (time-independent) processes.
• Statistical mechanics enables us to understand thermodynamics from a molecular perspective through the concept of ensembles.
• Statistical mechanics allows us to know the absolute magnitude of thermodynamic quantities like S, U, H, A, G…in terms of microscopic parameters of the system, and not just their change, dS, dU, dH, dA, dG (which as we know from CT only depend on a few macroscopic parameters).
• Theories for both classical and statistical thermodynamics are governed by the second law of thermodynamics through the medium of entropy.
• Entropy in classical thermodynamics is determined empirically.
• Entropy in statistical mechanics is a function of the distribution of microstates of the system.
• An example of different possible microstates for a particle is shown below:

Explanation:-

• The possible microstates having total energy E = 3a is shown below

Conclusion:-

Hence, the number of microstates having total energy E = 3a is 2.

Explanation:-

To determine the correct ordering of entropy for the three cases A, B, and C, we need to consider the different ways in which the particles can occupy the available energy levels.

Case A: Spin-1/2 Fermions

For spin-1/2 fermions, the Pauli exclusion principle applies, meaning that no two identical fermions can occupy the same quantum state simultaneously. Let's analyze the possibilities:

• All three particles occupy the energy level ε: There is only one way to arrange this configuration, as all particles are indistinguishable fermions. The total energy for this configuration is 4ε.
• One particle occupies each of the energy levels 0, ε, and 2ε: Each particle can be assigned to one of the three energy levels, resulting in

3! = 6 possible arrangements for this configuration.

• The total number of microstates for case A is 1  + 6 = 7. Since entropy is related to the logarithm of the number of microstates, the entropy for case A is larger than the other cases.

Case B: Spin-0 Bosons

For Spin-0 bosons, such as photons, do not obey the Pauli exclusion principle and can occupy the same quantum state simultaneously. Let's analyze the possibilities:

• All three particles occupy the energy level ε: Again, there is only one way to arrange this configuration, as all particles are indistinguishable bosons. The total energy for this configuration is 3ε.
• One particle occupies each of the energy levels 0, ε, and 2ε: Again, there are 2 possible arrangements for this configuration.
• The total number of microstates for case B is 1  + 6 = 7, which is the same as case A. Therefore, the entropy for case B is also the same as case A.

Case C: Classically Distinguishable Particles

In this case, the particles are distinguishable from each other, and there are no restrictions on their occupancy of energy levels. Let's analyze the possibilities:

• All three particles occupy the energy level ε: Each particle can be assigned to the energy level ε independently, resulting in

3 × 3 × 3 = 27 possible arrangements.

• Two particles occupy the energy level ε and one particle occupies the energy level 2ε: Each particle can be assigned to one of the three energy levels independently, resulting in 3 × 3 × 3

= 27 possible arrangements.

• One particle occupies each of the energy levels 0, ε, and 2ε: Each particle can be assigned to one of the three energy levels independently, resulting in 3! = 6 possible arrangements.
• The total number of microstates for case C is 27 + 27 + 6 = 60. Since entropy is related to the logarithm of the number of microstates, the entropy for case C is larger than for cases A and B.

In summary, the correct ordering of entropy for the three cases is:

∵ \((\text { no. of arrangements })_{\text {classical }}^C\) > \((\text { no. of arrangements })_{\text {fermions }}^A\) > \((\text { no. of arrangements })_{\text {bosons }}^B\)

⇒ \(S_C>S_A=S_B\)

Conclusion:-

• Hence, the correct option is (d)