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Test02SOL-docAA

# Hence the heat is transferred out of the surroundings

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Unformatted text preview: aric 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 = (1.0x10 Pa x 0.039993 m ) / (2.00 mol x 8.3145 J/mol.K) = 24.05 K ΔS(1à༎A) = nCVm ln (TA/TI) = 2.00 mol x (3/2) x 8.3145 J/mol.K x ln(24.05/240.5) ΔS(1à༎A) = - 57.43 J/K ΔS(Aà༎2) = nCPm ln (T2/TA) = 2.00 mol x (5/2) x 8.3145 J/mol.K x ln(153.92/24.05) ΔS(Aà༎2) = 77.17 J/K ΔS(1à༎2) = - 57.43+77.17 J/K = 19.74 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 19.74 J/K, which is positive. Hence, the process is spontaneous. 4 Question 12: (10 points) Using the appropriate entropy calculations, show that the following process is spontaneous: o o 1.30 mol H2O(l) (- 30.0 C and 1.00 atm) à༎ 1.30 mol H2O(s) (- 30.0 C and 1.00 atm) o To do so, you are given the value for the standard enth...
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