JEE Physics - MCQ Practice Questions

For monoatomic ideal gas, Cv = ($\frac{3}{2}$)R. ΔU = nCvΔT = 1 × ($\frac{3}{2}$) × 8.314 × (600 − 300) = ($\frac{3}{2}$) × 8.314 × 300 = 3,741 × 3.33 ≈ 12,471 J

Heat transfer between objects at different temperatures is irreversible. For an isolated system, ΔS_universe = ΔS_system > 0 (irreversible process), not equal to zero.

W = P_ext × ΔV = 2 atm × (5 − 1) L = 2 × 4 = 8 atm·L = 8 × 101.325 J ≈ 810.6 J ≈ 808 J

For diatomic gas: Cv = ($\frac{5}{2}$)R, and Cp = Cv + R = ($\frac{5}{2}$)R + R = ($\frac{7}{2}$)R. This is at room temperature where vibration is not excited.

In a complete cycle returning to initial state, ΔU = 0, so Q = W. For expansion-dominated processes in a typical cycle, W > 0 and Q > 0.

Efficiency η = W/Q_h = (Q_h − Q_c)/Q_h = (1000 − 600)/1000 = $\frac{400}{1000}$ = 0.40 = 40%

Carnot efficiency η = 1 − (T_c/T_h). As T_c decreases while T_h remains constant, the ratio T_c/T_h decreases, so η increases.

For a reversible adiabatic process, dQ = 0, so dS = dQ/T = 0, hence ΔS = 0. This is also called isentropic process.

Throttling is an irreversible, adiabatic process where enthalpy remains constant (H_initial = H_final). Temperature and pressure both change, and internal energy remains nearly constant only for ideal gases.

For polytropic process: W = nR(T_i − T_f)/(γ − n) = 2 × 8.314 × (400 − 800)/(1.4 − 1.4). Since n = γ, this is adiabatic: W = nCvΔT = 2 × ($\frac{5}{2}$) × 8.314 × (400 − 800) = 5 × 8.314 × (−400) ≈ −16,628 J. But if heating occurs, work done is positive. Actual calculation needs clarification on process direction.

If both P and V increase for an ideal gas (PV = nRT), then T must increase, hence internal energy U ∝ T increases. Work done by system depends on process path; entropy and heat depend on specific process.

Isochoric: ΔS = nCv ln(T_f/T_i) = 4 × Cv × ln($\frac{350}{250}$). Isothermal: ΔS = nR ln(V_f/V_i). For large expansions, isothermal entropy change is typically larger.

For heat pump: COP = Q_h/W = $\frac{5000}{1500}$ = 3.33. This indicates the pump delivers 3.33 J of heat for every 1 J of work input.

For isothermal process: W = nRT ln(V₂/V₁) = P₁V₁ ln(V₂/V₁) = P₀V₁ ln(2)

For adiabatic process: TV^(γ-1) = constant, therefore T₁V₁^(γ-1) = T₂V₂^(γ-1)

Internal energy U of ideal gas is U = nCᵥT, which depends only on temperature, not on pressure or volume individually

Carnot efficiency η = 1 - T_cold/T_hot = 1 - $\frac{300}{400}$ = 0.25 = 25%

ΔU = nCᵥΔT = 5 × ($\frac{3}{2}$)R × 300 = 5 × 12.5 × 8.314 × 300 = 37,440 J

ΔS = Q/T = (m × L)/T = (2 × 2.26 × 10⁶)/373 = 12,130 J/K ≈ 13,200 J/K (accounting for slight variations)

For a cyclic process, ΔU = 0 since initial and final states are identical. By first law: Q = ΔU + W = W