Andrews' experiment revealed the existence of a critical temperature above which gases cannot be liquefied regardless of applied pressure. At this point, the distinction between liquid and gas disappears and the substance becomes a supercritical fluid.
At the critical point, the isotherm has an inflection point defined by:
\[ \left( \frac{\partial P}{\partial V} \right)_T = 0, \quad \left( \frac{\partial^2 P}{\partial V^2} \right)_T = 0 \]
Correct Answer: (C) Distinction between gas and liquid vanishes
The flat region on the isotherms in Andrews' experiment represents the phase transition from gas to liquid, i.e., condensation. This region corresponds to a constant pressure at which both liquid and gas phases coexist in equilibrium.
Correct Answer: (B) Condensation of gas into liquid
The van der Waals equation for one mole of a real gas is:
\[ \left( P + \frac{a}{V^2} \right)(V - b) = RT \]
At the critical point, the isotherm has a point of inflection:
\[ \left( \frac{\partial P}{\partial V} \right)_T = 0, \quad \left( \frac{\partial^2 P}{\partial V^2} \right)_T = 0 \]
Rewriting the van der Waals equation in terms of pressure:
\[ P = \frac{RT}{V - b} - \frac{a}{V^2} \]
First derivative with respect to \( V \):
\[ \frac{\partial P}{\partial V} = -\frac{RT}{(V - b)^2} + \frac{2a}{V^3} \]
Set it to zero at the critical point:
\[ -\frac{RT_c}{(V_c - b)^2} + \frac{2a}{V_c^3} = 0 \tag{1} \]
Second derivative:
\[ \frac{\partial^2 P}{\partial V^2} = \frac{2RT}{(V - b)^3} - \frac{6a}{V^4} \]
Set it to zero at the critical point:
\[ \frac{2RT_c}{(V_c - b)^3} - \frac{6a}{V_c^4} = 0 \tag{2} \]
From (1):
\[ RT_c = \frac{2a (V_c - b)^2}{V_c^3} \]
From (2):
\[ RT_c = \frac{3a (V_c - b)^3}{V_c^4} \]
Equating both expressions for \( RT_c \):
\[ \frac{2a (V_c - b)^2}{V_c^3} = \frac{3a (V_c - b)^3}{V_c^4} \]
Canceling \( a \) and \( (V_c - b)^2 \):
\[ \frac{2}{V_c^3} = \frac{3(V_c - b)}{V_c^4} \]
Multiplying both sides by \( V_c^4 \):
\[ 2V_c = 3(V_c - b) \]
Simplifying:
\[ 2V_c = 3V_c - 3b \Rightarrow V_c = 3b \]
\[ \boxed{V_c = 3b} \]
Also, we can now find the critical temperature and pressure:
\[ T_c = \frac{8a}{27Rb}, \quad P_c = \frac{a}{27b^2} \]
Correct Answer: (C)
To compute the work done during an isothermal compression of a real gas following the van der Waals equation, we begin with the general formula for work:
\[ W = \int_{V_A}^{V_B} P \, dV \]
For 1 mole of a van der Waals gas, the pressure is given by:
\[ P = \frac{RT}{V - b} - \frac{a}{V^2} \]
Substituting into the work integral:
\[ W = \int_{V_A}^{V_B} \left( \frac{RT}{V - b} - \frac{a}{V^2} \right) dV \]
Now evaluate each term separately:
So the total work is:
\[ W = RT \ln\left( \frac{V_B - b}{V_A - b} \right) - a \left( \frac{1}{V_B} - \frac{1}{V_A} \right) \]
This is the expression for the isothermal work done on a van der Waals gas.
\[ \boxed{ W = RT \ln\left( \frac{V_B - b}{V_A - b} \right) - a \left( \frac{1}{V_B} - \frac{1}{V_A} \right) } \]
Correct Answer: (B)
Correct Option: (B)
The work done during isothermal expansion is given by:
\[ W = \int_{V_1}^{V_2} P \, dV \]
In isothermal expansion, the gas expands from \( V_1 \) to \( V_2 \), with \( V_2 > V_1 \). Since pressure \( P \) is always positive and \( dV > 0 \), this integral is positive:
\[ W > 0 \quad \text{if} \quad V_2 > V_1 \]
Hence, the condition for positive work is that the final volume is greater than the initial volume.
Correct Option: (B)
The van der Waals equation for 1 mole of a gas is:
\[ \left( P + \frac{a}{V^2} \right)(V - b) = RT \]
To find the critical point, apply the inflection conditions:
\[ \left( \frac{\partial P}{\partial V} \right)_T = 0, \quad \left( \frac{\partial^2 P}{\partial V^2} \right)_T = 0 \]
From solving these conditions, we obtain:
\[ V_c = 3b, \quad T_c = \frac{8a}{27Rb}, \quad P_c = \frac{a}{27b^2} \]
Therefore:
\[ \boxed{P_c = \frac{a}{27b^2}} \]
Correct Option: (C)
At temperatures below the critical point (\( T < T_c \)), van der Waals isotherms show a region with positive slope:
\[ \left( \frac{\partial P}{\partial V} \right)_T > 0 \]
This indicates instability and is physically not realizable. In reality, a phase transition occurs at constant pressure.
To correct this, Maxwell’s construction replaces the unstable part of the curve with a horizontal segment representing liquid-gas coexistence. The resulting curve includes:
So, the isotherm has a curve with a region of negative slope and a flat portion.
Correct Option: (C)
The compressibility factor at the critical point is defined as:
\[ Z_c = \frac{P_c V_c}{R T_c} \]
Using the van der Waals critical constants:
\[ V_c = 3b, \quad T_c = \frac{8a}{27Rb}, \quad P_c = \frac{a}{27b^2} \]
Substitute into the expression for \( Z_c \):
\[ Z_c = \frac{ \left( \frac{a}{27b^2} \right)(3b) }{ R \cdot \left( \frac{8a}{27Rb} \right) } \]
Simplifying:
\[ Z_c = \frac{3a}{27b} \cdot \frac{27Rb}{8a} = \frac{8}{27} \]
So the critical compressibility factor is:
\[ \boxed{Z_c = \frac{8}{27}} \]
Correct Option: (C)
At the critical point, the van der Waals isotherm just touches the horizontal line that defines phase coexistence — this point is an inflection point.
This is defined mathematically by:
\[ \left( \frac{\partial P}{\partial V} \right)_T = 0, \quad \left( \frac{\partial^2 P}{\partial V^2} \right)_T = 0 \]
The isotherm has no local maximum or minimum, but it does have a change in curvature (i.e., from concave up to concave down or vice versa). This is the defining feature of an inflection point.
\[ \boxed{\text{Inflection point at the critical point}} \]
Correct Option: (C)
Below the critical temperature \( T_c \), the van der Waals equation predicts a region in the isotherm where:
\[ \left( \frac{\partial P}{\partial V} \right)_T > 0 \]
This implies negative compressibility, which is unphysical. In reality, the system undergoes a phase transition between liquid and gas at a constant pressure \( P_{\text{eq}} \).
To correct this, Maxwell proposed that the pressure during the phase transition be determined by ensuring the areas above and below the flat portion of the isotherm (between volumes \( V_l \) and \( V_g \)) are equal:
\[ \int_{V_l}^{V_g} \left( P(V) - P_{\text{eq}} \right) \, dV = 0 \]
Where:
This rule replaces the unstable part of the curve with a horizontal line, ensuring mechanical and thermodynamic stability.
The area under the flat segment of the isotherm from \( V_l \) to \( V_g \) represents the phase transition at constant pressure and corresponds to the coexistence of the gas and liquid phases. This is the essence of the phase transition as corrected by Maxwell’s construction.
The flat segment represents the pressure of phase coexistence determined using Maxwell's equal area rule.
Correct Answer: (C)
Correct Option: (B)
The Zeroth Law of Thermodynamics forms the foundation for the concept of temperature. It states:
If two systems A and B are in thermal equilibrium with a third system C, then A and B are in thermal equilibrium with each other.
This implies that temperature is a transitive property. Mathematically, if:
\[ T_A = T_B \quad \text{and} \quad T_B = T_C \quad \Rightarrow \quad T_A = T_C \]
Therefore, system A must be in thermal equilibrium with system C.
Correct answer: (B)
Correct Option: (C)
A system is in thermodynamic equilibrium if all three types of equilibrium exist simultaneously:
These must hold throughout the system for equilibrium to persist. Any imbalance leads to a net change (heat flow, volume change, or mass transfer).
Correct answer: (C)
Correct Option: (B)
Isothermal compressibility is defined as:
\[ \kappa_T = -\frac{1}{V} \left( \frac{\partial V}{\partial P} \right)_T \]
Start from:
\[ PV = nRT \Rightarrow V = \frac{nRT}{P} \]
Now, take the partial derivative at constant temperature:
\[ \left( \frac{\partial V}{\partial P} \right)_T = -\frac{nRT}{P^2} = -\frac{V}{P} \]
Substitute into the definition:
\[ \kappa_T = -\frac{1}{V} \cdot \left( -\frac{V}{P} \right) = \frac{1}{P} \]
So although the correct numerical value for an ideal gas is \( \kappa_T = \frac{1}{P} \), the definitional form is:
\[ \boxed{ \kappa_T = -\frac{1}{V} \left( \frac{\partial V}{\partial P} \right)_T } \]
Correct answer: (B)
Correct Option: (B)
Derive the expression for the work done by an ideal gas during a reversible isothermal expansion from volume \( V_1 \) to \( V_2 \).
In an isothermal process, the temperature remains constant throughout the transformation. For an ideal gas, internal energy \( U \) depends only on temperature. Thus:
\[ \Delta U = 0 \quad \Rightarrow \quad Q = W \]
According to the first law of thermodynamics, any heat added to the system is fully converted to work.
\[ W = \int_{V_1}^{V_2} P \, dV \]
For an ideal gas undergoing an isothermal process, pressure as a function of volume is: \[ P = \frac{nRT}{V} \] where \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is the constant temperature.
\[ W = \int_{V_1}^{V_2} \frac{nRT}{V} \, dV \] Since \( nRT \) is constant (as \( T \) is constant), we factor it out: \[ W = nRT \int_{V_1}^{V_2} \frac{1}{V} \, dV \] \[ W = nRT \left[ \ln V \right]_{V_1}^{V_2} \] \[ W = nRT \ln\left( \frac{V_2}{V_1} \right) \]
If \( V_2 > V_1 \) (expansion), then \( \ln(V_2/V_1) > 0 \Rightarrow W > 0 \): work is done by the gas.
\[ \boxed{W = nRT \ln\left( \frac{V_2}{V_1} \right)} \]
Correct option: (B)
Correct Option: (B)
In an isothermal process, temperature remains constant. From the first law: \[ \Delta U = 0 \Rightarrow Q = W \]
Work done by the gas: \[ W = \int_{V_1}^{V_2} P \, dV \] Substitute the ideal gas law \( P = \frac{nRT}{V} \): \[ W = nRT \int_{V_1}^{V_2} \frac{1}{V} dV = nRT \ln\left( \frac{V_2}{V_1} \right) \]
Final Answer: \[ \boxed{W = nRT \ln\left( \frac{V_2}{V_1} \right)} \]
Correct Option: (D)
For adiabatic processes: \( Q = 0 \)
The relation: \[ PV^\gamma = \text{constant} \] Work done: \[ W = \int_{V_1}^{V_2} P \, dV \] Use: \[ P = \frac{P_1V_1^\gamma}{V^\gamma} \Rightarrow W = \frac{P_1V_1^\gamma}{1 - \gamma} \left( V_2^{1-\gamma} - V_1^{1-\gamma} \right) \] Other valid forms:
Final Answer: All are valid. \[ \boxed{W = \frac{nR(T_2 - T_1)}{\gamma - 1}} \quad \text{or similar} \]
Correct Option: (B)
Start from definitions: \[ C_V = \left( \frac{\partial U}{\partial T} \right)_V, \quad C_P = \left( \frac{\partial H}{\partial T} \right)_P \] where \( H = U + PV \). For an ideal gas: \[ H = U + nRT \Rightarrow dH = dU + nR dT \] So: \[ C_P = C_V + nR \Rightarrow \boxed{C_P - C_V = nR} \]
Correct Option: (B)
Using first law: \[ \Delta U = Q - W = 300\,\text{J} - 100\,\text{J} = \boxed{200\,\text{J}} \]
Correct Option: (C)
\[ \eta = 1 - \frac{T_C}{T_H} \] This arises from: \[ \frac{Q_C}{Q_H} = \frac{T_C}{T_H} \Rightarrow \eta = \frac{W}{Q_H} = \frac{Q_H - Q_C}{Q_H} = 1 - \frac{Q_C}{Q_H} \]
Final Answer: \[ \boxed{\eta = 1 - \frac{T_C}{T_H}} \]
Correct Option: (B)
\[ W = Q_H - Q_C = 800\,\text{J} - 500\,\text{J} = \boxed{300\,\text{J}} \]
Correct Option: (C)
It is impossible to transfer heat from a colder body to a hotter one without external work.
This prohibits spontaneous heat flow against gradient.
Answer: No process is possible in which heat is transferred from a colder to a hotter body without work.
Correct Option: (B)
\[ \frac{Q_C}{Q_H} = \frac{T_C}{T_H} \Rightarrow T = \text{proportional to heat exchanged reversibly} \]
Answer: \[ \boxed{ \frac{Q_C}{Q_H} = \frac{T_C}{T_H} } \]
Correct Option: (B)
\[ \text{COP} = \frac{Q_C}{W} = \frac{T_C}{T_H - T_C} \]
Answer: \[ \boxed{ \text{COP} = \frac{T_C}{T_H - T_C} } \]
Correct Option: (B)
For a first-order phase transition (e.g., liquid to vapor), the slope of the coexistence curve is: \[ \frac{dP}{dT} = \frac{L}{T(V_v - V_l)} \]
Derived from equality of Gibbs free energies and Maxwell relations.
Answer: \[ \boxed{ \frac{dP}{dT} = \frac{L}{T(V_v - V_l)} } \]
Correct Option: (C)
\[ \text{COP}_{\text{heat pump}} = \frac{Q_H}{W} = \frac{1500}{500} = \boxed{3} \]
Correct Option: (C)
From: \[ dS = \frac{dQ_{\text{rev}}}{T} \] At constant volume: \[ dQ = nC_V dT \Rightarrow dS = \frac{nC_V dT}{T} \Rightarrow \Delta S = \int_{T_1}^{T_2} \frac{nC_V dT}{T} = nC_V \ln\left( \frac{T_2}{T_1} \right) \]
Final Answer: \[ \boxed{ \Delta S = nC_V \ln\left( \frac{T_2}{T_1} \right) } \]
Correct Option: (D)
The second law of thermodynamics has two equivalent forms:
These statements imply that:
Both Clausius and Kelvin-Planck statements reinforce the idea that complete conversion of heat to work is impossible without compensation.
Correct Answer: (D) Both B and C
Correct Option: (B)
\[ \eta = 1 - \frac{T_C}{T_H} = 1 - \frac{300}{600} = 1 - 0.5 = 0.5 \]
\[ W = \eta Q_H = 0.5 \times 900 = \boxed{450\,\mathrm{J}} \]
Answer: (B)
Correct Option: (A)
A heat pump extracts heat \( Q_C \) from a cold body and delivers \( Q_H \) to a hot body using work \( W \). From energy balance: \[ Q_H = Q_C + W \]
The coefficient of performance (COP) is: \[ \text{COP} = \frac{Q_H}{W} \Rightarrow \frac{Q_C + W}{W} = \frac{Q_C}{W} + 1 \Rightarrow \text{COP} = \frac{T_H}{T_H - T_C} \] (Using Carnot relations: \( \frac{Q_C}{Q_H} = \frac{T_C}{T_H} \))
Final Answer: \[ \boxed{ \text{COP}_{\text{heat pump}} = \frac{T_H}{T_H - T_C} } \]
Correct Option: (B)
At phase equilibrium (e.g., liquid-vapour), Gibbs free energy per mole is equal:
\[ dG = V dP - S dT = 0 \Rightarrow \frac{dP}{dT} = \frac{\Delta S}{\Delta V} \]
But \( \Delta S = \frac{L}{T} \), where \( L \) is latent heat per mole:
\[ \frac{dP}{dT} = \frac{L}{T(V_v - V_l)} \]
This is the Clausius-Clapeyron equation, and it gives the slope of the phase boundary in a P–T diagram.
Final Answer: \[ \boxed{ \frac{dP}{dT} = \frac{L}{T(V_v - V_l)} } \]
Correct Option: (A)
\[ \text{COP}_{\text{refrigerator}} = \frac{T_C}{T_H - T_C} = \frac{277}{300 - 277} = \frac{277}{23} \approx \boxed{12.04} \]
Rounding to one decimal place: \( \boxed{12.0} \)
Answer: (A)
Correct Option: (B)
Entropy is a thermodynamic state function that quantifies the degree of disorder or the number of microstates associated with a macroscopic state.
\[ dS = \frac{dQ_{\text{rev}}}{T} \]
This relation is valid only for reversible processes, where the system is always in thermodynamic equilibrium with its surroundings.
Correct Answer: (B)
Correct Option: (C)
For an isolated system, the second law states:
If heat flows from a hot reservoir at \( T_1 \) to a cold reservoir at \( T_2 \), with \( T_1 > T_2 \), then:
\[ \Delta S = \frac{-Q}{T_1} + \frac{Q}{T_2} = Q\left( \frac{1}{T_2} - \frac{1}{T_1} \right) > 0 \]
Therefore, entropy of the universe always increases for irreversible processes.
Correct Answer: (C)
Correct Option: (B)
Let heat \( Q \) transfer from a body at temperature \( T_1 \) to another at \( T_2 \), where \( T_1 > T_2 \).
Entropy loss of hot body: \( \Delta S_1 = -\frac{Q}{T_1} \)
Entropy gain of cold body: \( \Delta S_2 = \frac{Q}{T_2} \)
Total entropy change: \[ \Delta S = \Delta S_1 + \Delta S_2 = Q \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]
This is always positive for \( T_1 > T_2 \), confirming irreversibility.
Answer: \[ \boxed{ \Delta S = Q \left( \frac{1}{T_2} - \frac{1}{T_1} \right) } \]
Correct Option: (C)
For a reversible process at constant volume:
\[ dQ = nC_V dT \Rightarrow dS = \frac{dQ}{T} = \frac{nC_V dT}{T} \Rightarrow \Delta S = nC_V \int_{T_1}^{T_2} \frac{dT}{T} = nC_V \ln\left( \frac{T_2}{T_1} \right) \]
Final Answer: \[ \boxed{ \Delta S = nC_V \ln\left( \frac{T_2}{T_1} \right) } \]
Correct Option: (B)
In a TS diagram for a Carnot cycle:
Work done per cycle: \[ W = Q_H - Q_C = T_H \Delta S - T_C \Delta S = \Delta S (T_H - T_C) \]
Efficiency: \[ \eta = \frac{W}{Q_H} = \frac{T_H - T_C}{T_H} \Rightarrow \boxed{ \eta = 1 - \frac{T_C}{T_H} } \]
Correct Answer: (B)
Correct Option: (A)
Let heat \( Q \) flow in steady state through a slab of thickness \( dx \) and area \( A \) across a temperature difference \( dT \).
\[ q = \frac{dQ}{A \, dt} \]
The rate of heat flow per unit area (heat flux) is proportional to the negative temperature gradient:
\[ q = -k \frac{dT}{dx} \]
where:
Answer: \( \boxed{q = -k \frac{dT}{dx}} \)
Correct Option: (B)
A bad conductor is placed between a heated metallic disc and a cold one. After steady state, the top disc is removed, and its cooling rate is measured.
Let:
\[ \text{Rate of heat loss} = ms \frac{dT}{dt} \]
In steady state, this is equal to the heat flow through the poor conductor of thickness \( x \), area \( A \), and temperature difference \( T_1 - T_2 \):
\[ \frac{Q}{t} = k A \frac{T_1 - T_2}{x} \Rightarrow k = \frac{ms \frac{dT}{dt} \cdot x}{A(T_1 - T_2)} \]
Answer: \( \boxed{k = \frac{ms \cdot \frac{dT}{dt} \cdot x}{A(T_1 - T_2)} } \)
Correct Option: (A)
Thermal resistance is defined as the resistance to heat flow through a material.
\[ R_{th} = \frac{\Delta T}{Q/t} \Rightarrow R_{th} = \frac{x}{kA} \]
Answer: \( \boxed{R_{th} = \frac{x}{kA}} \)
Correct Option: (B)
Total energy radiated per second per unit area by a black body: \[ Q = \sigma A T^4 \]
Where:
Answer: \( \boxed{Q = \sigma A T^4} \)
Correct Option: (D)
\[ Q \propto T^4 \Rightarrow Q_2 = Q_1 \left( \frac{T_2}{T_1} \right)^4 \]
\[ Q_2 = Q_1 \cdot 2^4 = 16 Q_1 \Rightarrow \boxed{\text{Power increases 16 times}} \]
Answer: (D)
Correct Option: (B)
\[ S = k \ln(\Omega) \]
This defines entropy statistically.
Correct Option: (B)
Each particle has 3 position + 3 momentum coordinates → 6 coordinates.
For \( N \) particles: \( 6N \)-dimensional phase space
Hence, the system is represented as one point in \( \Gamma \)-space, but \( N \) points in \( \mu \)-space.
Correct Option: (B)
If a system is isolated and in equilibrium, each accessible microstate is equally probable.
In the microcanonical ensemble: \[ p_i = \frac{1}{\Omega}, \quad \text{for all states within energy shell} \]
This allows us to define entropy as: \[ S = -k \sum p_i \ln p_i = k \ln \Omega \]
Correct Option: (D)
Time average of a quantity = Ensemble average:
\[ \langle A \rangle_{\text{time}} = \langle A \rangle_{\text{ensemble}} \]This means the system explores all accessible microstates over time.
Correct Option: (A)
The number of accessible states with energy in \( [E, E + \delta E] \) is:
\[ \Omega(E) = \frac{1}{h^{3N} N!} \int \delta(E - H(p,q)) \, d^{3N}p \, d^{3N}q \]
This integral is over the constant energy hypersurface in phase space.
Correct Option: (C)
Describes systems at fixed \( N, V, T \). Probability of a microstate with energy \( E_i \):
\[ p_i = \frac{e^{-\beta E_i}}{Z}, \quad \beta = \frac{1}{kT} \]
Discrete case: \[ Z = \sum_i e^{-\beta E_i} \]
Continuous case: \[ Z = \int e^{-\beta H(p,q)} \, d\Gamma \]
Correct Option: (A)
\[ \langle E \rangle = \sum_i p_i E_i = \frac{1}{Z} \sum_i E_i e^{-\beta E_i} \]
This can be written as: \[ \langle E \rangle = -\frac{\partial \ln Z}{\partial \beta} \]
Correct Option: (A)
Each quadratic degree of freedom contributes: \[ \langle E \rangle = \frac{1}{2}kT \]
For energy of form \( E = \frac{1}{2} a x^2 \), the probability: \[ P(x) \propto e^{-\beta \frac{1}{2} a x^2} \Rightarrow \langle x^2 \rangle = \frac{kT}{a} \Rightarrow \langle E \rangle = \frac{1}{2}kT \]
Correct Option: (C)
\[ \mathcal{Z} = \sum_N \sum_i e^{\beta(\mu N - E_i)} \]
\[ \langle N \rangle = \frac{\partial \ln \mathcal{Z}}{\partial (\beta \mu)} = \frac{1}{\beta} \frac{\partial \ln \mathcal{Z}}{\partial \mu} \]
Correct Option: (D)
Answer: \( \boxed{\text{All of the above}} \)
Correct Option: (C)
For distinguishable, non-interacting particles, the number of ways to arrange \( n_i \) particles in energy level \( \epsilon_i \) is:
\[ W = \frac{N!}{\prod_i n_i!} \]
Subject to constraints: \[ \sum_i n_i = N, \quad \sum_i n_i \epsilon_i = E \] Using Stirling's approximation and Lagrange multipliers, we get: \[ n_i = A e^{-\epsilon_i / kT} \Rightarrow f(\epsilon) = A e^{-\epsilon / kT} \]
Correct Option: (A)
For indistinguishable fermions (obey Pauli exclusion), the number of ways to arrange \( n_i = 0 \text{ or } 1 \) particles is: \[ f(\epsilon) = \frac{1}{e^{(\epsilon - \mu)/kT} + 1} \]
\[ f(\epsilon) = \begin{cases} 1, & \epsilon < \mu = \epsilon_F \\ 0, & \epsilon > \epsilon_F \end{cases} \]
Correct Option: (A)
For bosons: \[ f(\epsilon) = \frac{1}{e^{(\epsilon - \mu)/kT} - 1} \]
Bose-Einstein condensation occurs when the chemical potential \( \mu \to 0 \), and the ground state is macroscopically occupied at \( T < T_c \).
Correct Option: (B)
\[ \langle E \rangle = \int_0^\infty \epsilon f(\epsilon) g(\epsilon) d\epsilon \] For ideal gas in 3D: \[ g(\epsilon) \propto \epsilon^{1/2}, \quad f(\epsilon) \propto e^{-\epsilon/kT} \Rightarrow \langle E \rangle = \frac{3}{2}kT \]
Correct Option: (A)
At \( T = 0 \), all states up to \( \epsilon_F \) are filled. Using density of states: \[ N = \int_0^{\epsilon_F} g(\epsilon) d\epsilon \Rightarrow \epsilon_F = \frac{\hbar^2}{2m} \left( 3\pi^2 n \right)^{2/3} \]
Correct Option: (B)
\[ n(\epsilon) = g(\epsilon) f(\epsilon) \] For classical gas in 3D: \[ g(\epsilon) \propto \epsilon^{1/2}, \quad f(\epsilon) \propto e^{-\epsilon/kT} \Rightarrow n(\epsilon) = A \epsilon^{1/2} e^{-\epsilon/kT} \] When converted to momentum space: \[ n(\epsilon) = A \epsilon^2 e^{-\epsilon/kT} \, d\epsilon \]
Correct Option: (C)
For \( \epsilon - \mu \gg kT \):
Conclusion: All reduce to Maxwell-Boltzmann form.
Correct Option: (A)
\[ n = \frac{1}{\lambda_T^3} \zeta(3/2) \Rightarrow T_c = \frac{2\pi \hbar^2}{mk} \left( \frac{n}{\zeta(3/2)} \right)^{2/3} \]
Correct Option: (B)
\[ E = \int_0^{\epsilon_F} \epsilon g(\epsilon) d\epsilon \Rightarrow E = \frac{3}{5} N \epsilon_F \Rightarrow \langle E \rangle = \frac{E}{N} = \frac{3}{5} \epsilon_F \]
Correct Option: (A)
From thermodynamics: \[ P = -\left( \frac{\partial E}{\partial V} \right)_N \Rightarrow P = \frac{2}{5} n \epsilon_F \]
This comes from evaluating energy density and differentiating w.r.t volume.