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HW10-key - Chemistry 111 Homework#10 Due(Wednesday 1 The...

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Chemistry 111 Homework #10 Due: November 16, 2005 (Wednesday) 1. The entropy of a substance _________ increases as it changes from a liquid to a gas. Type the word never , sometimes , always , often , or rarely in Blackboard. Always . 2. Entropy generally ___________ with increasing molecular structure. Type your answer in Blackboard. Increases . 3. When a pure liquid or solid dissolves in a solvent, the entropy of the substance generally… Once more, type the word in Blackboard. Increases . 4. The second law of thermodynamics states that… a) heat is energy. b) the enthalpy of the universe is increasing. c) Δ S of the universe is equal to zero. d) if the Δ G is negative, the reaction is spontaneous. e) the total entropy of the universe is increasing. 5. Which of the following statements summarizes the third law of thermodynamics?
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Chemistry 111 2 Homework #10 / Fall 2005 Answer Key 6. The change in entropy for the vaporization of CCl 4 is + 85.7 J/Kmol and the boiling temperature is 77°C. What is the heat of vaporization? In order to do this, we need to think about the vaporization process as a truly reversible process (it is!). Therefore, ! S = ! H T Of course, substitution yields… 85.7 J / Kmol = ! H 350 K Δ H = + 30.0 kJ/mol . 7. A 0.250 mole sample of O 2 gas (Cp = 7/2 R; assume ideality), is expanded adiabatically (no heat loss or exchange with the surroundings) from 300.0 K, 10.00 atm, and 0.616 L to a final state of 155.3 K, 1.00 atm, and 3.19 L. Determine Δ E for this transformation. What is the meaning of the term: “adiabatic”? This refers to a process in which there is no heat gained from or lost to the surroundings. Another name for this is "isolated system." For adiabatic processes, then, q is always equal to zero, and Δ E = W. Let’s analyze the information given at the start of the problem. P 1 = 10.00 atm P 2 = 1.00 atm V 1 = 0.616 L V 2 = 3.19 L T 1 = 300.0 K T 2 = 155.3 K 0.250 moles 0.250 moles adiabatic expansion q = 0 J Cp = 29.1 J/Kmol Unfortunately, although we said that path was adiabatic, we didn't talk at all about the external pressure. We need to know about the external pressure to be able to calculate work! On the other hand, we know the heat capacity, so we can calculate Δ E: Δ E = n Cv Δ T. (Don't forget that C P - C V = R Cv = 20.8 J/mol K).
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Chemistry 111 3 Homework #10 / Fall 2005 Answer Key Δ E = 0.250 mol (20.8 J/mol K) (155.3 K - 300.0 K) = - 752 J .
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