Test02SOL-docAA

Since the process is isobaric heat is equal to a

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Unformatted text preview: versible adiabatic expansion under 5 constant external pressure from an initial state given by P1 = 1.0 x 10 Pa and T1 = 240.5 K until the final 4 equilibrium state where the pressure is P2 = 1.0 x 10 Pa. Part a. Show that the final temperature T2 is equal to 154 K (5 points). For an adiabatic expansion at constant external pressure, the work is equal to the change in energy, which for an ideal gas is dU = nCVm dT. Hence, - pext dV = nCVm dT This equation is integrated as: - pext (V2 – V1) = nCVm (T2 – T1) Expressing the volumes in terms of the temperature and pressure, we get: - pext (nRT2 / p2 – nRT1 / p1) = n(3/2)R (T2 – T1) which is simplified as: - pext (T2 / p2 – T1 / p1) = (3/2)(T2 – T1) and noting that pext = p2 and p1 = 10 p2 - T2 + T1 / 10 = (3/2)T2 – (3/2)T1 Hence, (5/2) T2 = (3/2)T1 + T1 / 10 and T2 = 153.92 K Part b. Calculate the change in entropy for the gas, for the surroundings and show that the process is spontaneous. The change in entropy must be calculated for a reversible path starting at P1, T1, V1 and ending at P2, V2, T2. 5 3 V1 = nRT1/P1 = 2.00 mol x 8.3145 J/mol.K x 240.5 K / 1.0x10 Pa = 0.039993 m 4 3 V2 = nRT2/P2 = 2.00 mol x 8.3145 J/mol.K x 153.92 K / 1.0x10 Pa = 0.255954 m The reversible path could consist of a reversible isochoric process from the initial state to the intermediate state A defined by VA = V1, PA = P2, TA and from an isob...
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This note was uploaded on 01/26/2014 for the course CHEM 3615 taught by Professor Aresker during the Spring '07 term at Virginia Tech.

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