Isothermal and Adiabatic Processes — Practice Quiz

1.67

Step 1: For a monoatomic ideal gas (like He, Ar, Ne), each molecule has 3 degrees of freedom (only translational). Step 2: Using the equipartition theorem, Cv = (3/2)R and Cp = Cv + R = (5/2)R. Step 3: Therefore, γ = Cp/Cv = (5/2)R / (3/2)R = 5/3 ≈ 1.67. Step 4: For diatomic gases (like H₂, O₂, N₂), γ = 7/5 = 1.40. For triatomic gases, γ ≈ 1.33. γ can never be 1 because Cp is always greater than Cv.

W = μRT × ln(V₂/V₁)

Step 1: For an isothermal process, PV = μRT = constant, so P = μRT/V. Step 2: Work done = ∫P dV (from V₁ to V₂) = ∫(μRT/V) dV = μRT ∫dV/V. Step 3: Integrating, W = μRT [ln V] from V₁ to V₂ = μRT × ln(V₂/V₁). Step 4: Option (a) is the formula for adiabatic work. Option (c) is for isobaric (constant pressure) process. Option (d) relates to change in internal energy.

1727 J

Step 1: Given: μ = 1 mol, T = 300 K, V₂/V₁ = 2, R = 8.31 J mol⁻¹ K⁻¹, ln 2 = 0.693. Step 2: Formula: W = μRT ln(V₂/V₁). Step 3: W = 1 × 8.31 × 300 × 0.693. Step 4: W = 8.31 × 300 × 0.693 = 8.31 × 207.9 = 1727 J. Step 5: Option (a) uses wrong temperature. Option (c) uses ln 4 instead of ln 2. Option (d) halves the answer incorrectly. The correct answer is 1727 J.

W = μR(T₁ - T₂) / (γ - 1)

Step 1: For an adiabatic process, PV^γ = K (constant), and integrating gives W = (P₁V₁ - P₂V₂)/(γ - 1). Step 2: Using ideal gas law, P₁V₁ = μRT₁ and P₂V₂ = μRT₂. Step 3: Substituting, W = (μRT₁ - μRT₂)/(γ - 1) = μR(T₁ - T₂)/(γ - 1). Step 4: If T₂ < T₁ (expansion), W is positive (work done by gas). If T₂ > T₁ (compression), W is negative (work done on gas). Option (a) is for isothermal. Option (c) incorrectly uses Cp. Option (d) is for isobaric.