Thermodynamic and Statistical Physics MCQ Quiz in मराठी - Objective Question with Answer for Thermodynamic and Statistical Physics - मोफत PDF डाउनलोड करा

Last updated on Apr 20, 2025

पाईये Thermodynamic and Statistical Physics उत्तरे आणि तपशीलवार उपायांसह एकाधिक निवड प्रश्न (MCQ क्विझ). हे मोफत डाउनलोड करा Thermodynamic and Statistical Physics एमसीक्यू क्विझ पीडीएफ आणि बँकिंग, एसएससी, रेल्वे, यूपीएससी, स्टेट पीएससी यासारख्या तुमच्या आगामी परीक्षांची तयारी करा.

Latest Thermodynamic and Statistical Physics MCQ Objective Questions

Top Thermodynamic and Statistical Physics MCQ Objective Questions

Thermodynamic and Statistical Physics Question 1:

A thermally isolated container, filled with an ideal gas at temperature T, is divided by a partition, which is clamped initially, as shown in the figure below.
F1 Vinanti Teaching 04.10.23 D7

The partition does not allow the gas in the two parts to mix. It is subsequently released and allowed to move freely with negligible friction. The final pressure at equilibrium is

  1. 5P/3
  2. 5P/4
  3. 3P/5
  4. 4P/5

Answer (Detailed Solution Below)

Option 1 : 5P/3

Thermodynamic and Statistical Physics Question 1 Detailed Solution

Explanation:

We have two chambers separated by a partition. The left chamber has double the initial pressure (2P₀) and the original volume (V). The right chamber has the original pressure (P₀) and double the original volume (2V).

After removing the partition, total volume of the system, V(sum)=V+2V=3V.

As the system is thermally isolated, energy before and after the partition is lifted should be the same, meaning the total pressure of the system should remain constant.

Expressing the pressure in terms of kinetic theory of gases: P=23×NV, where N is the number of molecules and is the mean kinetic energy.

As E=32KT  where K is Boltzmann constant and T is temperature, the expression becomes: P=NKTV

Initial pressure: P(sum)initial=NKTV+2NKTV=3NKTV

Final pressure: P(sum)final=PV(sum)=P×3V=3×NKT Because there's no heat exchange with the environment, T = constant, and hence NK = constant, so we can compare pressures directly: 

P(sum)initial=P(sum)final 

P=5P3

Thermodynamic and Statistical Physics Question 2:

A thermodynamic function G(T, P, N) = U - TS + PV is given in terms of the internal energy U, temperature T, entropy S, pressure P, volume V and the number of particles N. Which of the following relations is true? (In the following μ is the chemical potential.)

  1. S=GT|N,P
  2. S=GT|N,P
  3. V=GP|N,T
  4. μ=GN|P,T

Answer (Detailed Solution Below)

Option 1 : S=GT|N,P

Thermodynamic and Statistical Physics Question 2 Detailed Solution

G = U - TS + PV

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

dG = -SdT + VdP

(GT)N,P=S and GP|N,T=V

Thermodynamic and Statistical Physics Question 3:

when a gas expands adiabatically from volume V1 to V2 by a quasi-static reversible process, it cools from temperature T1 to T2. If now the same process is carried out adiabatically and irreversibly, and T2' is the temperature of the gas when it has equilibriated then?

  1. T2<T2
  2. T2>T2
  3. T2=T2
  4. T2=T2V1V2

Answer (Detailed Solution Below)

Option 2 : T2>T2

Thermodynamic and Statistical Physics Question 3 Detailed Solution

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.

T2T1>T2T1.

Hence, we get: T2>T2.

The correct option is option 2).

Thermodynamic and Statistical Physics Question 4:

Consider a system having three energy levels with energies 0, 2E, 3E with respective degeneracy 2,2,3. 4 bosons of spin 0 have to be accommodated in these levels such that energy of the system is 10E. The number of ways in which it can be done is:

  1. 15
  2. 18
  3. 9
  4. 36

Answer (Detailed Solution Below)

Option 2 : 18

Thermodynamic and Statistical Physics Question 4 Detailed Solution

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=(ni + gi  1)!ni!(gi  1)!.

Putting all the given values and solving we get:
W=(ni + gi  1)!ni!(gi  1)!=(2+21)!2!1!(2+31)!2!2!=3×4×32=18 .

The correct option is option (2).

Thermodynamic and Statistical Physics Question 5:

The angular frequency of oscillation of a quantum harmonic oscillator in two dimensions is ω. If it is in contact with an external heat bath at temperature T, its partition function is (in the following β = 1kBT)

  1. e2βhω(e2βhω1)2
  2. eβhω(eβhω1)2
  3. eβhωeβhω1
  4. e2βhωe2βhω1

Answer (Detailed Solution Below)

Option 2 : eβhω(eβhω1)2

Thermodynamic and Statistical Physics Question 5 Detailed Solution

Concept:

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

Calculation:

En = (n+1)hω 

The given quantum harmonic oscillator is two dimensional

∴ n = nx + ny

Partition Function of the system is

z = ∑ (n+1)exp-(n+1)hω 

where degeneracy = (n+1)

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

eβhω1eβhω+e2βhω(1e2βhω)2

eβhω(1eβhω)+e2βhω(1eβhω)2

eβhω(eβhω1)2

The correct answer is option (2).

Thermodynamic and Statistical Physics Question 6:

Falling drops of rain break up and coalesce with each other and finally achieve an approximately spherical shape in the steady state. The radius of such a drop scales with the surface tension σ as

  1. 1/√σ
  2. √σ
  3. σ
  4. σ2

Answer (Detailed Solution Below)

Option 1 : 1/√σ

Thermodynamic and Statistical Physics Question 6 Detailed Solution

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).

Thermodynamic and Statistical Physics Question 7:

An idealised atom has a non-degenerate ground state at zero energy and a g-fold degenerate excited state of energy E. In a non-interacting system of N such atoms, the population of the excited state may exceed that of the ground state above a temperature T > E2kBln2. The minimum value of g for which this is possible is

  1. 8
  2. 4
  3. 2
  4. 1

Answer (Detailed Solution Below)

Option 2 : 4

Thermodynamic and Statistical Physics Question 7 Detailed Solution

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 = (geϵ)/Z 

EXPLANATION:

F1 Vinanti Teaching 02.01.22 D12

The partition function Z = 1 + ge-βϵ 

(Where g = degeneracy factor, ϵ = energy)

The probability of the ground state 

⇒ P1 = (e×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 = E2kBln2

∴ N1 = N2 

⇒ N/Z =  N(ge-βE)/Z 

⇒ 1 = ge-E/kT = geEk(E2kBln2)      

⇒ g = exp (ln22 )

⇒ g = 4

Hence the correct answer is option 2.

Thermodynamic and Statistical Physics Question 8:

A polymer, made up of N monomers, is in thermal equilibrium at temperature T. Each monomer could be of length a or 2a. The first contributes zero energy, while the second one contributes ε. The average length (in units of Na) of the polymer at temperature T = ε / kB is

  1. 5+e4+e
  2. 4+e3+e
  3. 3+e2+e
  4. 2+e1+e

Answer (Detailed Solution Below)

Option 4 : 2+e1+e

Thermodynamic and Statistical Physics Question 8 Detailed Solution

CONCEPT:

A system in the canonical ensemble: We consider an ensemble of N identical systems (which may be labeled as 1, 2, 3,...N), sharing a total energy E. Let Er (r = 0, 1, 2, 3 ... ) denote the energy eigenvalues of the systems. If nr denotes the number of systems which, at any time t, have the energy value Er, then the set of numbers {nr} must satisfy the obvious conditions

rnr=NrnrEr=E=NU

Where U ( = E/N) denotes the average energy per system in the ensemble.

Some important statistical quantities of canonical ensemble,

Probability of State EiP(Ei)=eβEiieβEi

Mean value of any quantity A = = ∑P(Ei)Ai

where Ei and Ai are the energy and value of that quantity in ith state

EXPLANATION:

When the length of the monomer is "a", then the energy =  0

When the length of monomer is "2a", the the energy =  ϵ 

Now, the probability

P(ϵ)=eβEiieβEiP(ϵ=0)=eβ×0eβ×0+eβϵ=11+eβϵP(ϵ=ϵ)=eβ×ϵeβ×0+eβϵ=eβϵ1+eβϵmean length=L=P(0)×Na+P(ϵ)×2Na Letϵ=kT=1βϵβ=1L=Na[1+2eβϵ1+eβϵ]=Na[1+2e11+e1]L=Na[e+2e+1]

∴ The average length (in units of Na) is 2+e1+e

Hence the correct answer is option 4.

Thermodynamic and Statistical Physics Question 9:

A system has microstates with energies 0, 2kBT, 4kB T . What is the average energy E  of the system at temperature T?

  1. 2kBT 
  2. 0.29kBT
  3. 1.58kBT 
  4. 1.50kBT 

Answer (Detailed Solution Below)

Option 2 : 0.29kBT

Thermodynamic and Statistical Physics Question 9 Detailed Solution

Explanation:

The average energy is given by

E=iEieβEiieβEi 

For the given energy levels:

Z=e0+e2+e4=1+e2+e4E=0e0+2kBTe2+4kBTe41+e2+e4E=0+2kBTe2+4kBTe41+e2+e42kBT0.1353+4kBT0.01831+0.1353+0.0183E0.2706kBT+0.0732kBT1.1536E0.3438kBT1.15360.29kBT

Hence, the correct answer is option 2.

Thermodynamic and Statistical Physics Question 10:

Two energy levels, 0 (non-degenerate) and ϵ (doubly degenerate), are available to N non-interacting distinguishable particles. If U is the total energy of the system, for large values of N the entropy of the system is kB[NlnN(NUϵ)ln(NUϵ)+X] In this expression, X is

  1. UϵlnU2ϵ
  2. Uϵln2Uϵ
  3. 2Uϵln2Uϵ
  4. UϵlnUϵ

Answer (Detailed Solution Below)

Option 1 : UϵlnU2ϵ

Thermodynamic and Statistical Physics Question 10 Detailed Solution

Concept - We will find the Internal energy for both particles and then substitute the number of particles with the internal energy.

 

F1 Teaching Arbaz 23-10-23 D5

 

Entropy and probability relation can be related as : S=klnW

  • S is the entropy of the particle
  • W is probability

     

Calculation-          browse

  • Assume n1 and n2 are distinguishable particles with energy levels 0 and ϵ  with n1+n2=N.
  • N= non-interacting distinguishable particles
  • N=n1+n2

 

Now, U=n1×0+n2×ϵ    n2=Uϵ
n1=Nn2=N  Uϵ

  •  Probability of two particles n1 and n2  can written as W= N!n1!×n22!×n22! where W is probability
  • Entropy and the probability relation can be written as: S=klnW
  • S=kln(N!n1!×n22!×n22!)

Substitute values of n1 and n2, we get

  • S=kln[N! (NUϵ)!×(U2×ϵ)2!]
  • S=k[lnN!ln(NUϵ)!2ln(U2ϵ)!]
  • Now, use Stirling formula, lnN!=NlnNN

we get,

  • S=k[NlnNN[(NUϵ)ln(NUϵ)(NUϵ)]2[(U2ϵ)ln(U2ϵ)(Uϵ)]

  • S=k[NlnNN(NUϵ)ln(NUϵ)+NUϵ2(U2ϵ)ln(U2ϵ)+Uϵ]

 

By cancelling N and Uϵ we get, 

  • S=k[NlnN(NUϵ)ln(NUϵ)Uϵln(U2ϵ)]

Comparing this with the given equation, 

  • S=k[NlnN(NUϵ)ln(NUϵ)+X]

Here, value of X is = (Uϵ)ln(U2ϵ)

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