Thermodynamic and Statistical Physics MCQ Quiz in தமிழ் - Objective Question with Answer for Thermodynamic and Statistical Physics - இலவச PDF ஐப் பதிவிறக்கவும்

Last updated on Apr 6, 2025

பெறு Thermodynamic and Statistical Physics பதில்கள் மற்றும் விரிவான தீர்வுகளுடன் கூடிய பல தேர்வு கேள்விகள் (MCQ வினாடிவினா). இவற்றை இலவசமாகப் பதிவிறக்கவும் Thermodynamic and Statistical Physics MCQ வினாடி வினா Pdf மற்றும் வங்கி, SSC, ரயில்வே, UPSC, மாநில PSC போன்ற உங்களின் வரவிருக்கும் தேர்வுகளுக்குத் தயாராகுங்கள்.

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.

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, .

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: , where N is the number of molecules and is the mean kinetic energy.

As   where K is Boltzmann constant and T is temperature, the expression becomes: 

Initial pressure: 

Final pressure: Because there's no heat exchange with the environment, T = constant, and hence NK = constant, so we can compare pressures directly: 

 

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.)

Answer (Detailed Solution Below)

Option 1 :

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

 and 

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. T_2 \)

Answer (Detailed Solution Below)

Option 2 : T_2 \)

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)

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 \).

Hence, we get:  T_2 \).

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:

.

Putting all the given values and solving we get:
 .

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 = )

Answer (Detailed Solution Below)

Option 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)+...

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 > . 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:

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 = 

∴ N1 = N2 

⇒ N/Z =  N(ge-βE)/Z 

⇒ 1 = ge-E/kT =      

⇒ g = exp (ln22 )

⇒ g = 4

Hence the correct answer is option 2.

Thermodynamic and Statistical Physics Question 8:

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

Answer (Detailed Solution Below)

Option 2 :

Thermodynamic and Statistical Physics Question 8 Detailed Solution

Explanation:

The average energy is given by

For the given energy levels:

Hence, the correct answer is option 2.

Thermodynamic and Statistical Physics Question 9:

The heat capacity CV at constant volume of a metal, as a function of temperature, is αT + βT3, where α and β are constants. The temperature dependence of the entropy at constant volume is

Answer (Detailed Solution Below)

Option 1 :

Thermodynamic and Statistical Physics Question 9 Detailed Solution

Explanation:

We know that:

Cv = αT + βT3

From the first law of thermodynamics, we know that:

.

By integrating the differential equation, we have:

The correct option is option (1).

Thermodynamic and Statistical Physics Question 10:

The average energy U of a 1-D quantum oscillator of frequency  and in contact with a heat bath at temperature T is given by:

Answer (Detailed Solution Below)

Option 1 :

Thermodynamic and Statistical Physics Question 10 Detailed Solution

Explanation:

The energy for the harmonic oscillator is given as:

.

The partition function can be written as:

.

Now simplifying it a bit we get:

.

The average energy is given by:

.

Therefore .

The correct option is option (1).

Hot Links: teen patti joy 51 bonus teen patti cash master teen patti teen patti palace