Two thermally insulated vessels are connected by a narrow tube fitted with a val
ID: 2972373 • Letter: T
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 = 17.4 L, contains oxygen at a temperature of T1 = 310 K and a pressure of P1 = 1.78 atm. The other vessel of volume V2 = 22.0 L contains oxygen at a temperature of T2 = 460 K and a pressure of P2 = 2.75 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
The single most important point in solving this problem is that the total number of moles must remain constant and the total internal energy must remain constant before and after opening the valve. So, first find the number of moles in each vessel, and then find the internal energy of each vessel, then the total number of moles and energy, and consider the total volume: V(total) = 15.2 L + 22.4 L = 37.6 L In the first vessel, the number of moles: n = pV/RT = 1.75(15.2) / 0.08206(300) = 1.08 mol And in the second vessel, the number of moles: n = pV/RT = 1.95(22.4) / 0.08206(450) = 1.183 mol Therefore, the total # of moles: n(total) = 1.08+1.183 = 2.263 mol For the first vessel, the internal energy: U = (5/2)nRT = (5/2)(1.08)(8.314)(300) = 6734 J and for the second vessel, the internal energy: U = (5/2)nRT = (5/2)(1.183)(8.314)450 = 11065.9 J So that, the total energy: U = 6734 + 11065.9 = 17800 J Then the final temperature can be obtained by use of the eqn for the total energy: U(total) = (5/2)nRT = 17800. T = (2/5)U/nR = 0.4(17800) / (2.263)(8.314) = 378 K, and the final pressure: P = nRT/ V = (2.263)(0.08207)(378) / 37.6 = 1.869 atm
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