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A closed rigid tank contains two compartments A and B, both filled with oxygen (ideal gas), separated by a fixed, perfectly heat-insulating partition. Compartment A: volume = 1 m³, pressure = 5 bar, temperature = 400 K. Compartment B: volume = 3 m³, pressure = 1 bar, temperature = 300 K. The fixed partition is replaced by a new partition that can slide freely and conduct heat but does not allow gas to leak between compartments. Find the volume of compartment A (in m³) after the system reaches equilibrium.

1. 1 m³
2. 2 m³
3. 3 m³
4. 4 m³

Correct answer: 2 m³

Solution

n_A = P_A*V_A/(R*T_A) = 5*1/(R*400); n_B = P_B*V_B/(R*T_B) = 1*3/(R*300). Using bar and m³ with consistent R: n_A/n_B = (5*1/400):(1*3/300) = (5/400):(3/300) = (1/80):(1/100) = 100:80 = 5:4. So n_A = 5k, n_B = 4k for some k. Total U = n_A*Cv*T_A + n_B*Cv*T_B = Cv*(5k*400 + 4k*300) = Cv*k*(2000+1200) = 3200*Cv*k. At equilibrium T_f: (n_A+n_B)*Cv*T_f = 3200*Cv*k → 9k*T_f = 3200k → T_f = 3200/9 K. Total volume = 4 m³ (rigid tank). At equilibrium P_f = (n_A+n_B)*R*T_f/(V_total) = 9k*R*(3200/9)/(4) = k*R*3200/4 = 800k*R. V_A = n_A*R*T_f/P_f = 5k*R*(3200/9)/(800k*R) = 5*(3200/9)/800 = 5*3200/(9*800) = 16000/7200 = 20/9 ≈ 2.22 m³. Hmm, not exactly 2. Let me use energy conservation more carefully with Cv. For diatomic O2: Cv = (5/2)R. Total internal energy: U = n_A*(5/2)R*T_A + n_B*(5/2)R*T_B. But n_A = 5/(400R)*bar*m³ and n_B = 3/(300R). Let's just use PV = nRT: n_A*R = 5*1/400 = 1/80 bar*m³/K; n_B*R = 1*3/300 = 1/100 bar*m³/K. U_total = Cv*(n_A*T_A + n_B*T_B) = Cv*((1/80)*400 + (1/100)*300)/... wait U = n*Cv*T and nRT = PV so nT = PV/R, thus n*Cv*T = Cv*PV/R = (Cv/R)*PV. For ideal diatomic: Cv/R = 5/2. U_total = (5/2)*(P_A*V_A + P_B*V_B) = (5/2)*(5*1 + 1*3) = (5/2)*8 = 20 bar*m³. At equilibrium: U_f = (5/2)*P_f*(V_A+V_B) = (5/2)*P_f*4 = 10*P_f. So P_f = 2 bar. n_total*R*T_f = P_f*V_total = 2*4 = 8 bar*m³. Also n_A*R = 1/80+... n_total*R = 1/80 + 1/100 = 5/400+4/400 = 9/400 bar*m³/K. T_f = 8/(9/400) = 8*400/9 = 3200/9 K. V_A = n_A*R*T_f/P_f = (1/80)*(3200/9)/2 = (3200/9)/(160) = 3200/1440 = 20/9 ≈ 2.22 m³. Closest option is 2 m³.

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