Unformatted text preview: d from ΔRU 298.15K using: qRXN = (0.0033754 mol RXN) x ( 3051.07 kJ/mol) qRXN =  10.2988 kJ qCAL is calculated using qCAL = CCAL ΔT where ΔT is the temperature rise in the calorimeter as a result of the exothermic combustion of 0.3177 g phenol. Hence, CCAL ΔT =  qRXN and ΔT =  qRXN / CCAL ΔT = (+10.299 kJ) / (8.30 kJ/ oC 1 ) = 1.2408 K With three significant figures, we obtain: Δ T = 1.24 K 4 Question 11: (10 points) 2.00 mol of a monoatomic ideal gas (CVm = 3/2 R) undergo an irreversible 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 = 4.0 x 10 Pa. Part a. Show that the final temperature T2 is equal to 183 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 = 2.5 p2  T2 + T1 / 2.5 = (3/2)T2 – (3/2)T1 Hence, (5/2) T2 = (3/2)T1 + T1 / 2.5 and T2 = 182.78 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 182.78 K / 4.0x10 Pa = 0.075986 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 isobaric process from A to the final state. Under these conditions, the change in entropy is equal to the change in entropy from “1” to “A” and from “A” to “2”. Note that TA is given by: PA VA / TA = P2 V1 / TA = nR 4
3
or TA = P2 V1 / nR = (4.0x10 Pa x 0.039993 m ) / (2.00 mol x 8.3145 J/mol.K) = 96.20 K ΔS(1à༎A) = nCVm ln (TA/TI) = 2.00 mol x (3/2) x 8.3145 J/mol.K x ln(96.20/240.5) Δ S(1à༎ A) =  22.85 J/K ΔS(Aà༎2) = nCPm ln (T2/TA) = 2.00 mol x (5/2) x 8.3145 J/mol.K x ln(182.78/96.20) Δ S(Aà༎ 2) = 26.68 J/K ΔS(1à༎2) =  22.85 + 26.68 J/K = 3.83 J/K Since the change in entropy for the surroundings is equal to 0 (adiabatic) process, then, the change in entropy for the universe is equal to 3.83 J/K, which is positive. Hence, the process is spontaneous. Question 12: (10 points) Using the appropriate entropy calculations, show that the following process is spontaneous: o
o
1.30 mol H2O(l) ( 50.0 C and 1.00 atm) à༎ 1.30 mol H2O(s) ( 50.0 C and 1.00 atm) o
To do so, you are given the value for the standard enthalpy...
View
Full Document
 Spring '07
 AREsker
 Physical chemistry, Thermodynamics, Enthalpy, pH

Click to edit the document details