NEET Physics - MCQ Practice Questions

Expansion in a rigid container is isochoric (constant volume), not isobaric. Phase changes and open systems typically occur at constant pressure.

For ideal gases, μ_JT = 0 because they have no intermolecular forces. Real gases show non-zero values depending on temperature and pressure conditions.

For gas mixtures: Cv(mix) = Σ(n_i × Cv_i)/Σn_i. The heat capacity is the weighted sum based on composition.

According to the second law of thermodynamics: ΔS_universe ≥ 0. For isolated systems, ΔS_system ≥ 0 (increases for irreversible, constant for reversible).

In a cyclic process, the system returns to initial state, so ΔU = 0 (state function). From first law: Q = W. Heat absorbed equals work done by the system.

For phase change at equilibrium: ΔS = ΔH/T = 6000 J / 300 K = 20 J/(mol·K). This applies at the melting point where both phases are in equilibrium.

For an adiabatic process: PV^γ = constant leads to TV^(γ-1) = constant when combined with ideal gas law.

At constant pressure, supplied heat does two things: increases internal energy and does expansion work. Hence Cp > Cv by amount R.

Second law of thermodynamics: For an isolated system, ΔS ≥ 0 (entropy increases for irreversible processes, constant for reversible).

For isothermal process: W = nRT ln(V_f/V_i) = nRT ln(P_i/P_f). Since ΔU = 0, q = W.

In adiabatic compression (q=0), work is done on gas, increasing both P and T without heat exchange. Other processes would have different P-T relationships.

COP = Q_cold/W. Therefore, Q_cold = COP × W = 5 × 100 = 500 J

For monatomic gas: Cv = ($\frac{3}{2}$)R, Cp = ($\frac{5}{2}$)R. Therefore γ = Cp/Cv = ($\frac{5}{2}$)/($\frac{3}{2}$) = $\frac{5}{3}$ ≈ 1.67

At equilibrium phase transition (melting point), ΔG = 0. This is the condition for phase equilibrium at constant T and P.

Free expansion is an irreversible process. For the isolated system, ΔS_universe > 0. The gas entropy increases as volume increases.

ΔG = ΔH - TΔS = -200 - (300)($\frac{50}{1000}$) = -200 - 15 = -185 kJ/mol

At same T and P, the chemical potential per molecule is identical. Internal energy and entropy depend on amount (volume), heat capacity depends on mass.

In adiabatic process, Q = 0. From first law: ΔU = Q - W = 0 - (-500) = 500 J (internal energy increases)

When n = γ (ratio of specific heats), the polytropic process is adiabatic. For isothermal n=1, isobaric n=0, isochoric n=∞.

At equilibrium: ΔG = 0, so ΔH = TΔS. T = ΔH/ΔS = (-150000)/(-100) = 1500 K