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

Carnot efficiency: η = 1 - (Tc/Th) = 1 - ($\frac{300}{500}$) = 1 - 0.60 = 0.40 or 40%

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

The first law states: ΔU = Q + W, where ΔU is change in internal energy, Q is heat absorbed, and W is work done on the system

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 an ideal gas, internal energy U depends only on temperature (U = nCᵥT), not on pressure or volume independently

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

Mayer's relation: Cₚ - Cᵥ = R for one mole of ideal gas, hence Cₚ = Cᵥ + R

For isochoric process (constant volume): PV = nRT with V constant gives P/T = nR/V = constant (Gay-Lussac's Law)

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

In series arrangement with identical rods, temperature difference is equally distributed. ΔT_total = 100°C, so ΔT per rod = $\frac{100}{3}$ = 33.3°C. Second junction = 100 - 2(33.3) = 33.3°C

Clausius statement: Heat cannot spontaneously transfer from a colder body to a hotter body without external work being done on the system

If W_net < 0 (work done on gas), then from first law: ΔU_cycle = 0 = Q - W, so Q = W < 0, meaning net heat flows out

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)

For diatomic gas: Cᵥ = ($\frac{5}{2}$)R and Cₚ = ($\frac{7}{2}$)R. γ = Cₚ/Cᵥ = $\frac{7}{5}$ = 1.40

Heat lost by body 1: Q = C(T₁ - T_f) = 1000(400 - T_f). Heat gained by body 2: Q = 1000(T_f - 300). ΔS_univ = C ln(T_f/T₁) + C ln(T_f/T₂) = 1000[ln(T_f/400) + ln(T_f/300)] = 0.575. Solving: T_f = 350 K

By first law: ΔU = Q - W. Here ΔU = 150 J, W = 100 J (work done by system). So Q = ΔU + W = 150 + 100 = 250 J