Govt. Exams
Entrance Exams
# Solution: Heating Ideal Diatomic Gas in Rigid Container
In a rigid container, the volume remains constant, so we use Gay-Lussac's Law to find the final temperature, and the first law of thermodynamics to calculate heat supplied.
Step 1: Find Final Temperature Using Gay-Lussac's Law
For a constant volume process with an ideal gas, pressure is directly proportional to absolute temperature.
Substituting the given values (P₁ = 1 atm, T₁ = 300 K, P₂ = 2 atm):
Step 2: Calculate Heat Supplied Using First Law of Thermodynamics
For a constant volume process, no work is done (W = 0), so all heat goes into changing internal energy.
Where ΔT = T₂ - T₁ = 600 - 300 = 300 K, and Cᵥ = 5R/2
Using R ≈ 8.31 J/(mol·K): Q = 3 × 2.5 × 8.31 × 300 = 3742.5 J ≈ 3750 J
**Final Answer: (A) T_f =
Efficiency η = 1 - T_c/T_h = 1 - 300/600 = 0.5. Work done W = η × Q_h = 0.5 × 1200 = 600 J. Heat rejected Q_c = Q_h - W = 1200 - 600 = 600 J. Verification: Q_c/Q_h = T_c/T_h → 600/1200 = 300/600 ✓
For adiabatic process: TV^(γ-1) = constant. T₁V₁^(γ-1) = T₂V₂^(γ-1). 400 × V^0.4 = T₂ × (2V)^0.4. T₂ = 400/(2^0.4) = 400/1.3195 ≈ 252.1 K
Isobaric process: V₁/T₁ = V₂/T₂. If V₂ = 2V₁, then T₂ = 2T₁ = 2 × 300 = 600 K
Efficiency η = W/Qh = 0.30, so W = 0.30 × 5000 = 1500 J. Heat rejected: Qc = Qh - W = 5000 - 1500 = 3500 J
For diatomic gas at constant pressure: Cp = (7/2)R. Q = nCpΔT = 3 × (7/2) × 8.314 × 300 = 37,413 J
At equilibrium: ΔG = 0, so ΔH = TΔS. T = ΔH/ΔS = (-150000)/(-100) = 1500 K
When n = γ (ratio of specific heats), the polytropic process is adiabatic. For isothermal n=1, isobaric n=0, isochoric n=∞.
For Carnot refrigerator: COP = Qc/W = Tc/(Th - Tc) = 250/100 = 2.5, so W = 500/2.5 = 200 J. Wait, recalculating: W = Qc × (Th - Tc)/Tc = 500 × 100/250 = 200 J. Hmm, let me verify: W = Qc(Th/Tc - 1) = 500(350/250 - 1) = 500 × 0.4 = 200 J.
In adiabatic process, Q = 0. From first law: ΔU = Q - W = 0 - (-500) = 500 J (internal energy increases)
