Two thermally insulated vessels are connected by a narrow tube fitted with a val

Question

Two thermally insulated vessels are connected by a narrow tube fitted with a valve that is initially closed as shown in the figure below. One vessel of volume V1 = 16.6 L, contains oxygen at a temperature of T1 = 305 K and a pressure of P1 = 1.79 atm. The other vessel of volume V2 = 22.5 L contains oxygen at a temperature of T2 = 470 K and a pressure of P2 = 2.35 atm. When the valve is opened, the gases in the two vessels mix and the temperature and pressure become uniform throughout. (a) What is the final temperature? (b) What is the final pressure?

Explanation / Answer

V(total) = 16.6 L + 22.5 L = 39.1 L In the first vessel, the number of moles: n = pV/RT = 1.79(16.6) / 0.08206(305) = 1.18 mol And in the second vessel, the number of moles: n = pV/RT = 2.35(22.5) / 0.08206(470) = 1.37 mol Therefore, the total # of moles: n(total) = 1.18+1.37 = 2.55 mol For the first vessel, the internal energy: U = (5/2)nRT = (5/2)(1.18)(8.314)(305) = 7480 J and for the second vessel, the internal energy: U = (5/2)nRT = (5/2)(1.37)(8.314)470 = 13383 J So that, the total energy: U = 7480 + 13383 = 20863 J Then the final temperature can be obtained by use of the eqn for the total energy: U(total) = (5/2)nRT = 20863. T = (2/5)U/nR = 0.4(20863) / (2.55)(8.314) = 393 K, and the final pressure: P = nRT/ V = (2.55)(0.08207)(393) / 39.1 = 2.10 atm

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