The figure shows a pV diagram for 4.3 g of oxygen gas (O₂) in a sealed container. The temperature T₁ of the gas in state 1 is 21°C. What are the temperatures T₃ and T₄ of the gas in states 3 and 4? The ideal gas constant is R = 8.314 J/mol ∙ K = 0.0821 L ∙ atm/mol ∙ K, and the ATOMIC weight of oxygen is 16 g/mol.

Question

Quizplus · Accepted Answer

Using the ideal gas law, we can find the pressure and volume of each state:
State 1: P1V1 = nRT1 = (4.3 g / 32 g/mol) * (0.0821 L*atm/mol*K) * (21°C + 273.15) K = 2.43 atm * L
State 2: P2V2 = P1V1 = 2.43 atm * L, V2 = V1 / 2 = 1.215 L, P2 = 2 * P1 = 4.86 atm
State 3: V3 = V2, P3 = P4 = P2 / 4 = 1.215 atm, n = P3V3 / RT3 = (1.215 atm * 1.215 L) / (0.0821 L*atm/mol*K * T3), solving for T3 we get T3 = -52°C
State 4: V4 = V1, P4 = P1 = 2.43 atm, n = P4V4 / RT4 = (2.43 atm * 2.43 L) / (0.0821 L*atm/mol*K * T4), solving for T4 we get T4 = 390°C

The Figure Shows a PV Diagram for 4