Thermodynamic and Statistical Physics MCQ Quiz - Objective Question with Answer for Thermodynamic and Statistical Physics - Download Free PDF

Calculation:

Rewriting the energy expression:

E_n = ℎω (n₁ + 2n₂ + 3/2)

We need n₁ + 2n₂ + 3/2 < 4, which simplifies to n₁ + 2n₂ < 2.5.

The valid pairs (n₁, n₂) are: (0,0), (1,0), (2,0), and (0,1).

Next, we calculate the Boltzmann factors for these states using: α = exp(− ℎω / k_BT)

For (0,0): α^3/2

For (1,0): α^5/2

For (2,0): α^7/2

For (0,1): α^7/2

Summing these contributions gives the numerator:

α^3/2 + α^5/2 + 2α^7/2 = α^3/2(1 + α + 2α²)

The partition function Z is:

Z = α^3/2 ∑_n1=0^∞ αⁿ¹ ∑_n2=0^∞ α²ⁿ² = α^3/2 (1 / (1 − α)) (1 / (1 − α²))

Simplifying Z:

Z = α^3/2 / [(1 − α)(1 − α²)]

The probability P is the ratio of the numerator to the partition function:

P = [α^3/2(1 + α + 2α²)] / [α^3/2 / ((1 − α)(1 − α²))] = (1 + α + 2α²)(1 − α)(1 − α²)

Factoring 1 − α² as (1 − α)(1 + α):

P = (1 + α + 2α²)(1 − α)²(1 + α)

Calculation:

Energy of two Ising spins s = ±1/2 is given by:

E = s₁ s₂ + s₁ + s₂

List all configurations and their energies:

Using Boltzmann factors (β = 1 / k_B T):

P₊₊ ∝ e^−β(5/4), P_+- = P_-+ ∝ e^β/4, P_-- ∝ e^3β/4

Given:

P_-- = 16 ⋅ P₊₊ ⇒ (e^3β/4) / (e^−5β/4) = 16 ⇒ e^2β = 16 ⇒ β = 2 ln 2

Also,

(P_+- / P₊₊) = (e^β/4) / (e^−5β/4) = e^3ln2 = 8 ⇒ y = 8x

Let:

P₊₊ = x, P_-- = 16x, P_+- = P_-+ = y = 8x

Total probability:

x + 16x + 2 ⋅ 8x = 33x ⇒ x = 1/33, y = 8/33

So, the probability that the spins take opposite values is:

2y = 2 ⋅ (8/33) = 16/33

Calculation:

For a free Fermi gas in d-dimensions (ignoring spin):

The average energy per particle at T = 0 (i.e., ground state energy per particle) is:

= (d / (d + 2)) E_F

For 4D case:

Set d = 4,

So, = (4 / (4 + 2)) E_F = (4 / 6) E_F = (2 / 3) E_F

Calculation:

Two levels: E₁ = ω, E₂ = 3ω

At high temperature (T → ∞), the particles get equally distributed.

At thermal equilibrium:

The probability of being in state i is given by the Boltzmann distribution:

P_i = e^−βE_i / Z , Z = e^−βω + e^−β(3ω)

So average energy per particle:

ε_eq = [ω e^−βω + 3ω e^−β(3ω)] / [e^−βω + e^−β(3ω)]

Factor out e^−βω:

ε_eq = ω · [1 + 3 e^−2βω] / [1 + e^−2βω]

Let x = e^−2βω

Then, ε_eq = ω · (1 + 3x) / (1 + x)

Now take the high-temperature limit (β → 0 ⇒ x → 1):

lim_x→1 ε_eq = ω · (1 + 3) / (1 + 1) = ω · 4 / 2 = 2ω

Calculation:

For adiabatic process dQ = 0 

⇒ dU = -PdV

Thus for radiation, VT³ = constant.

The initial internal energy is U_i = aVT⁴ where a = 4σ/c.

U_i = (4σ/c)VT⁴ 

Using  VT3 = constant. we get

U_f = (4σ/c)(8V)(T/2)4 = (2σ/c)VT4 

The change in internal energy is Δ U = (4σ/c)VT4 - (2σ/c)VT4 = (2σ/c)VT4 

Concept:

The Carnot engine is a theoretical thermodynamic cycle proposed by Leonard Carnot. It estimates the maximum possible efficiency that a heat engine during the conversion process of heat into work and, conversely, working between two reservoirs can possess.

Calculation:

Q₁ = 100 J Q₂ = 40J

T₁ = ? T₂ = 300 K

\({Q_1\over Q_2} = {T_1 \over T_2}\)

⇒ \({100 \over 40} = {T_1 \over 300}\)

⇒ T₁ = \({(100 \times 300)\over 40}\)

= 750 K

The correct answer is option (3).

CONCEPT:

→In the isothermal process, when a gas container is compressed it is written as;

\(P_1V_1=P_2V_2\)

→For the adiabatic process it is written as;

\(P_1V_1 ^\gamma= P_2V_2^\gamma\)

Here P is the pressure and V is the volume.

EXPLANATION:

The volume of the container = \(V_1\)

When the is compressed to half of its volume = \(\frac{V_1}{2}\)

→By isothermal process when container A is compressed we have;

\(P_1V_1=P_2V_2\)

⇒ \(P_1V_1=P_2\frac{V_1}{2}\)

⇒ \(P_2=2P_1\) -----(1)

→By adiabatic process when B is compressed we have;

\(P_1V_1 ^\gamma= P_2'V_2'^\gamma\)

\(P_1V_1 ^\gamma= P_2'(\frac {V_1}{2})^\gamma\)

⇒ \(P_2' = 2^\gamma P_1\) ----(2)

On dividing equation (1) by (2) we have;

\(\frac{P_2}{P_2'}=2^{\gamma -1}\)

Hence, option 1) is the correct answer.

Concept:

Stefan-Boltzmann law- It states that the total radiant heat power emitted from a surface is proportional to the fourth power of its absolute temperature.

i.e., Power, P ∝ T4

where P and T is the power radiated and temperature respectively.

Calculation:

Radius of the spherical black body,  R₁ = 12 cm

Power radiated, P₁ = 450 watt 

Temperature,  t₁ = 500 K. 

New radius, R₂ = 6 cm

New temperature, t₂ = 2t₁

New power = P₂

Rate of power loss

P ∝  R²T⁴

\(\frac{P_1}{P_2} =\)\(\frac{R^2_1T^4_1}{R^2_2T^4_2}\)

= 4 × \(\frac{1}{16}\)

\(\frac{450}{r_2}=\frac{1}{4}\)

 P₂ = 1800 watt

We know,

λ_max T = constant (Wien's law)

So, \({λ _{{{\max }_1}}}\;{T_1} = {λ _{{{\max }_2}}}\;{T_2}\)

\( \Rightarrow {\lambda _0}T = \frac{{3{\lambda _0}}}{4}T'\)

\(\Rightarrow T' = \frac{4}{3}T\)

So, \(\frac{{{P_2}}}{{{P_1}}} = {\left( {\frac{{T'}}{T}} \right)^4} = {\left( {\frac{4}{3}} \right)^4} = \frac{{256}}{{81}}\)

Concept-  We will calculate Degeneracy and then co-relate the relation between spin and number of particles to find the ratio of fermi energy at respective spins.

Calculation-

Here, we have to find the ratio of E_f (S =  \(3\over2\) )   and E_f (S =  \(1\over2\) )   
Given, E_n = n² E₀

• E_n =  Energy of n^th state
• n = number of particles
• E0 = Energy of ground state

As we know that, 

• Degeneracy g_s = 2S+1

And, S is spin of fermions

• n ∝ \(1\over(2S+1)\)

Now, E_f = n²E₀

• E_f ∝ \(1\over(2S+1)^2\)  E₀

Put respective spin values in the formula to find the ratio between both energies.

• (E_f (S=​​\(3\over2\) ))  / (E_f (S=\(1\over2 \)) )   = \(1\over(2\times 3/2+1)^2 \) E₀  / \(1\over(2\times 1/2+1)^2\) E₀  = \(4\over16\)    = \(1\over4\)​

The correct answer is 3, 300, 100 and 240  
Key Points

• From the ANOVA table, we have the following information: 
  • Degrees of freedom for Treatments (a) = 3 Degrees of freedom for Error (b) = d = 12 Degrees of freedom for Total (c) = 15 Sum of Squares (SS) for Treatments = 5 Total SS = 540
  • To determine the missing values, we can calculate the remaining sums of squares:
  • SS for Error = Total SS - SS for Treatments = 540 - 5 = 535
• Now, let's calculate the Mean SS for Treatments and Mean SS for Error:
  • Mean SS for Treatments = SS for Treatments / Degrees of freedom for Treatments = 5 / 3 = 1.67
  • Mean SS for Error = SS for Error / Degrees of freedom for Error = 535 / 12 = 44.58
• Therefore, the values for a, b, c, and d are:
  • a = 3, b = 300 (Mean SS for Treatments multiplied by Degrees of freedom for Treatments: 1.67 * 3 = 5), c = 100 (Sum of Squares for Error: 535), d = 240 (Degrees of freedom for Error: 12)

Hence, the values of a, b, c and d are, respectively, 3, 300, 100, and 240.

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 \(P_{net}=\)\(\frac{11}{64}\)

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).

Calculation:

Heat radiated per unit time,

P = σAeT⁴

where, σ is Stefan's constant.

 For a black body, emissivity,

 e ~ 1

Heat energy radiated in time t by the black body,

E = σAtT⁴

where E = hν   

The correct answer is (4).

Concept:

Change in entropy of gas

Tds = CVdT + PdV

Calculation:

Change in entropy of gas

Tds = C_VdT + PdV

dV = 0

Tds = αT dT

Δ S_gas = α [T_f - T_i]

Change in entropy of reservoir 

T_fdS = αT dT 

dQ = αT dT

T_fΔ S_res = α \({T^2\over 2}\)

= \({\alpha \over 2}\)[\(T^2_f - T^2_i\)]

Δ S_Total = α(Tf - Ti) + \(\frac{{\rm{\alpha }}}{{{\rm{2}}{{\rm{T}}_{\rm{f}}}}}\left( {{\rm{T}}_{\rm{f}}^{\rm{2}}{\rm{ - T}}_{\rm{i}}^{\rm{2}}} \right)\)

The correct answer is option (4).