JEE Physics - MCQ Practice Questions

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

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

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

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%

By first law of thermodynamics: ΔU = Q − W = 500 − 200 = 300 J

In adiabatic compression (Q = 0), work is done on the gas (W < 0). By first law: ΔU = Q − W = 0 − W > 0. Since ΔU ∝ ΔT for ideal gas, temperature increases.

Carnot efficiency: η = 1 − (T_cold/T_hot) = 1 − ($\frac{300}{400}$) = 1 − 0.75 = 0.25 = 25%

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%

First law: ΔU = Q - W = 4000 - 1500 = 2500 J (where W is work done by system)

For adiabatic process: PV^γ = constant, TV^(γ-1) = constant, and T^γ P^(1-γ) = constant. All three relations are equivalent and correct.

For an ideal gas, internal energy U depends only on temperature (U = nC_vT). This is independent of pressure or volume, as shown by kinetic theory.

By first law of thermodynamics: ΔU = Q - W = 2400 - 600 = 1800 J (where W is work done by system)

First law: ΔU = Q - W, so W = Q - ΔU = 8000 - 5000 = 3000 J (work done by the system)