Thermodynamic and Statistical Physics MCQ Quiz in తెలుగు - Objective Question with Answer for Thermodynamic and Statistical Physics - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

G = U - TS + PV

dG = dU - Tds - sdT + PdV + VdP = TdS - PdV - TdS - SdT + PdV + VdP

dG = -SdT + VdP

\(\left(\frac{\partial G}{\partial T}\right)_{N, P}=-S \) and \(\left.\frac{\partial G}{\partial P}\right|_{N, T}=V\)

Explanation:

Here both process undergoes adiabatic expansion hence,

dQ = 0, which means dU = - dW.

As dV is positive as there is an expansion hence, we have dU to be negative.

That is U(final) < U(initial), therefore T(final) < T(initial).

Hence there is a cooling.

Again, for the same level of expansion, the work done in an irreversible process is larger than that of a reversible process.

As in an irreversible process, it takes lots of effort to expand the same way keeping in mind that

there will be additional friction restricting it from doing the same.

Hence dW (Irr) > dW(rev).

Therefore, we can conclude that as dW is proportional to dU which is again proportional to dT hence,

dT (Irr) > dT(rev).

Now putting the value of temperature given in the question that 

 it cools from temperature T1 to T₂ in the reversible case and T2' is the temperature of the gas for the irreversible case.

\(T_2'-T_1 > T_2 - T_1 \).

Hence, we get: \(T_2' > T_2 \).

The correct option is option 2).

Explanation:

There are N= 4 bosons.

They have to be arranged in such a way that the total energy would be = 10E.

where the three energy levels have energies 0, 2E, and 3E with respective degeneracy 2,2,3.

The possible way for such a configuration is 0 particles in 0 E, 2 in 2E, and 2 in 3E energies.

The total number of possible ways can be further calculated using the relation:

\(W = \frac{(n_i~ +~ g_i~ -~1)!}{n_i! (g_i~-~1)!} \).

Putting all the given values and solving we get:
\(W = \frac{(n_i~ +~ g_i~ -~1)!}{n_i! (g_i~-~1)!} = \frac{(2+2-1)!}{2! 1!} \frac{(2+3-1)!}{2! 2!} = \frac{3\times 4 \times 3}{2} = 18 \) .

The correct option is option (2).

Concept:

Partition functions are functions of the thermodynamic state variables, such as the temperature and volume.

Calculation:

E_n = (n+1)hω 

The given quantum harmonic oscillator is two dimensional

∴ n = n_x + n_y

Partition Function of the system is

z = ∑ (n+1)exp-(n+1)hω 

where degeneracy = (n+1)

z = exp(-hω)+2exp(-2hω)+3exp(-3hω)+...

= \({e^{-\beta h\omega}\over1- e^{-\beta h\omega}} + {e^{-2\beta h\omega}\over (1- e^{-2\beta h\omega})^2}\)

= \({e^{-\beta h\omega} (1-e^{-\beta h\omega})+ e^{-2\beta h\omega}\over(1-e^{-\beta h\omega})^2}\)

= \({e^{\beta h\omega}\over (e^{\beta h\omega}-1)^2}\)

The correct answer is option (2).

Concept:

A falling drop of rainwater acquires a spherical shape due to Surface tension. This phenomenon can be observed in the nearly spherical shape of tiny drops of liquids and of soap bubbles.

Calculation:

Work done due to a combination of raindrops is

W = σ × change in area 

= σ × (4πR²-n4πr²)

Taking small r negligible:

W = σ × 4πR²

∴ R ∝ 1/√σ

The correct answer is option (1).

CONCEPT:

If an atom has n excited states with energy E₁ E₂ E₃ ...

and number of particles N₁, N₂, N₃ ...

then the partition function 

⇒ Z = 1 + ge-βϵ 

(Where g = degeneracy factor)

If an energy value E corresponds to more than one energy level with the same energy E then the energy level is said to be degenerate and it is represented by a degeneracy factor. 

If the total number of particles is N then 

Number of particles in the particular state '1'

⇒ N1 = NP1

Where, P₁ is the probability of the state = P1 = (ge-β^ϵ)/Z 

EXPLANATION:

The partition function Z = 1 + ge^-βϵ 

(Where g = degeneracy factor, ϵ = energy)

The probability of the ground state 

⇒ P₁ = (e-β^×0)/Z =1/Z

Number of particles in the ground state 

N₁ = NP₁ = N/Z

Similarly, The probability of the excited state 

⇒ P₂ = (ge-βE)/Z 

Number of particles in the excited state 

N2 = NP2 = N(ge-βE)/Z

For the minimum value of g, both the population should be equal at T = \(\rm\frac{E}{2 k_B \ln 2}\)

∴ N₁ = N₂ 

⇒ N/Z =  N(ge-βE)/Z 

⇒ 1 = ge-E/kT = \( ge^{-\frac{E}{k(\rm\frac{E}{2 k_B \ln 2})}}\)      

⇒ g = exp (ln2² )

⇒ g = 4

Hence the correct answer is option 2.

Explanation:

The average energy is given by

\(\langle E \rangle = \frac{\sum_i E_i e^{-\beta E_i}}{\sum_i e^{-\beta E_i}}\ \)

For the given energy levels:

\( Z = e^{-0} + e^{-2} + e^{-4} = 1 + e^{-2} + e^{-4} \\ \langle E \rangle = \frac{0 \cdot e^{-0} + 2k_B T \cdot e^{-2} + 4k_B T \cdot e^{-4}}{1 + e^{-2} + e^{-4}} \\ \langle E \rangle = \frac{0 + 2 k_B T \cdot e^{-2} + 4 k_B T \cdot e^{-4}}{1 + e^{-2} + e^{-4}} \approx \frac{2 k_B T \cdot 0.1353 + 4 k_B T \cdot 0.0183}{1 + 0.1353 + 0.0183}\\ \langle E \rangle \approx \frac{0.2706 k_B T + 0.0732 k_B T}{1.1536} \\ \langle E \rangle \approx \frac{0.3438 k_B T}{1.1536} \approx 0.29 k_B T \)

Hence, the correct answer is option 2.

Concept:

We will use the Probability formula for a random walker in one-dimensional lattice which is given by 

• \(\frac{N!}{n_1! n_2!}.p^{n_1} q^{n_2}\)

Explanation:

Here, \(p=\frac{1}{2}\) and \(q=\frac{1}{2}\)

Case-1- A and B meet when A takes four right steps & zero left steps and B takes three right steps and one left step. So, probability becomes

• \(\) \(P_{12}=\frac{4!}{0! 4!}.(\frac{1}{2})^{4} (\frac{1}{2})^{0}\times \frac{4!}{3! 1!}.(\frac{1}{2})^{3} (\frac{1}{2})^{1}=\frac{1}{16}\times\frac{4}{16} \ \)

Case-2- A and B meet when A takes two right steps & two left steps and B takes two right steps and two left step. So, probability becomes

• \(\) \(P_{33}=\frac{4!}{2! 2!}.(\frac{1}{2})^{2} (\frac{1}{2})^{2}\times \frac{4!}{2! 2!}.(\frac{1}{2})^{2} (\frac{1}{2})^{2} \ \)\(=\frac{6}{16}\times\frac{6}{16}\)

• Net Probability is\(P_{net}=\)\(\frac{1}{16}\times\frac{4}{16}+\frac{6}{16}\times\frac{6}{16}+\frac{1}{16}\times\frac{4}{16}\)
• \(P_{net}=\)\(( \frac{44}{16\times 16})=\frac{11}{64}\ \ \)

So, the correct answer is \(P_{net}=\)\(\frac{11}{64}\)

Explanation:

We know that:

C_v = αT + βT3

From the first law of thermodynamics, we know that:

\( d S=\frac{d Q}{T}=\frac{C_V d T}{T}\).

By integrating the differential equation, we have:

\(\int d S=\int\left(\alpha+\beta T^2\right) d T\)

\(S=\alpha T+\frac{1}{3} \beta T^3\)

The correct option is option (1).

Explanation:

The energy for the harmonic oscillator is given as:

\(E_n = (n+ \frac{1}{2})\hbar \omega \).

The partition function can be written as:

\(Z = \sum_n e^{-\beta E_n} =e^{-\beta \hbar \omega /2} \sum_n e^{-n\beta \hbar \omega} \).

Now simplifying it a bit we get:

\(Z = \frac{e^{-\beta \hbar \omega /2}}{ 1- e^{-\beta \hbar \omega}} = \frac{1}{2 \sinh (\frac{\beta \hbar \omega}{2}) } \).

The average energy is given by:

\(U = - \frac{\partial ln Z}{\partial \beta} = - \frac{\partial ln (\frac{1}{2 \sinh (\frac{\beta \hbar \omega}{2}) } )}{\partial \beta}\).

Therefore \(U = \frac{\hbar \omega }{2} \coth (\beta \hbar \omega/2) \).

The correct option is option (1).