JEE Physics - MCQ Practice Questions

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

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

In adiabatic compression, no heat exchange occurs (Q=0). Work is done on the gas, so ΔU = W (positive). Since ΔU increases, temperature must increase.

Carnot efficiency: η = 1 - (Tc/Th) = 1 - ($\frac{300}{500}$) = 1 - 0.6 = 0.4 = 40%

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%