JEE Physics - MCQ Practice Questions

For isothermal process: W = nRT ln(Vf/Vi) = 1 × 8.314 × 300 × ln($\frac{5}{2}$) = 2494.2 × 1.609 = 3.04 kJ

For adiabatic process: TVγ⁻¹ = constant. T₁V₁^(γ-1) = T₂V₂^(γ-1). 300 × 1^0.4 = T₂ × 2^0.4. T₂ = $\frac{300}{2}$^0.4 = $\frac{300}{1}$.6818 = 178.9 K

In Carnot cycle, isothermal processes involve heat exchange with reservoirs (Q₁ at hot reservoir, Q₂ at cold reservoir), while adiabatic processes have no heat exchange (Q=0)

For isobaric process: Q = nCₚΔT. For monoatomic gas, Cₚ = ($\frac{5}{2}$)R. Q = 5 × ($\frac{5}{2}$) × 8.314 × 100 = 62.355 kJ

The polytropic equation PVⁿ = constant covers isothermal (n=1), adiabatic (n=γ), isobaric (n=0), and isochoric (n=∞) processes as special cases

For isothermal expansion against constant pressure: W = PₑₓₜΔV = 1 × (10-1) = 9 L·atm = 9 × 101.325 = 910.125 J

For reversible cycle: Q_h/T_h = Q_c/T_c. $\frac{100}{400}$ = Q_c/300. Q_c = (100 × 300)/400 = 75 J

Van der Waals equation (P + a/V²)(V - b) = RT accounts for molecular volume (b term) and intermolecular attractive forces (a term)

Using first law: ΔU = Q - W = 300 - 500 = -200 J (internal energy decreases)

W = PΔV. Here W = 400 J, P = 2 × 101325 = 202650 Pa. So ΔV = W/P = $\frac{400}{202650}$ ≈ 0.00197 m³

In an isothermal process, temperature is constant. For an ideal gas, internal energy depends only on temperature, so ΔU = 0. Pressure and volume change according to PV = constant.

Internal energy U = nCvT. Without knowing the number of moles in each sample, the ratio cannot be determined. Same temperature doesn't mean same internal energy.

Efficiency η = (Qh - Qc)/Qh = (1000 - 600)/1000 = $\frac{400}{1000}$ = 0.40 = 40%

For reversible isothermal process: ΔS = Q/T = nR ln(Vf/Vi) since Q = nRT ln(Vf/Vi) for ideal gas isothermal expansion.

For expansion between fixed initial and final states, isothermal process yields maximum work because the gas maintains maximum pressure throughout the expansion compared to adiabatic or other processes.

For isobaric process: W = nRΔT = 2 × 8.314 × (600 - 300) = 2 × 8.314 × 300 = 4988.4 ≈ 4984 J

For diatomic gas: Cv = ($\frac{5}{2}$)R. Using Cp = Cv + R, we get Cp = ($\frac{5}{2}$)R + R = ($\frac{7}{2}$)R

COP = Tc/(Th - Tc) = 263/(303 - 263) = $\frac{263}{40}$ ≈ 6.575 ≈ 6.5

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.