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H13 Thermo 3

# H13 Thermo 3 - saldana(avs387 H13 Thermo 3 Mccord(52385...

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saldana (avs387) – H13: Thermo 3 – Mccord – (52385) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Consider the following processes. (Treat all gases as ideal.) I) The pressure of one mole of oxygen gas is allowed to double isothermally. II) Carbon dioxide is allowed to expand isothermally to 10 times its original vol- ume. III) The temperature of one mole of helium is increased 25 C at constant pressure. IV) Nitrogen gas is compressed isothermally to one half its original volume. V) A glass of water loses 100 J of energy reversibly at 30 C. Which of these processes leads to an increase in entropy? 1. I and IV 2. III and V 3. II and III correct 4. I and II 5. V Explanation: R = 8 . 314 J · mol 1 · K 1 Assume 1 mol in each case. Entropy de- creases if Δ S is negative. For the oxygen gas pressure doubling isothermally, P 2 = 2 P 1 and Δ S = nR ln parenleftbigg P 1 P 2 parenrightbigg = (1 mol)(8 . 314 J · mol 1 · K 1 ) ln parenleftbigg 1 2 parenrightbigg = - 5 . 76 J · K 1 We expect a negative answer since pressure increased. For the CO 2 gas volume expanding 10 × isothermally, V 2 = 10 V 1 and Δ S = nR ln parenleftbigg V 2 V 1 parenrightbigg = (1 . 00 mol)(8 . 314 J · mol 1 · K 1 ) × ln(10) = +38 . 29 J · K 1 We expect a positive answer since volume increased. For the nitrogen gas compressed to 1 2 origi- nal volume isothermally, V 1 = 2 V 2 and Δ S = nR ln parenleftbigg V 2 V 1 parenrightbigg = (1 mol)(8 . 314 J · mol 1 · K 1 ) ln parenleftbigg 1 2 parenrightbigg = - 5 . 76 J · K 1 We expect a negative answer since volume decreased. For the cooling glass of water, T = 30 C + 273 . 15 = 303 . 15 K Δ S = q T = - 200 J 303 . 15 K = - 0 . 6597 J · K 1 The last situation (heating the 1 mol of He) does not give enough data to calculate an answer but from the formula Δ S = nC p , m ln parenleftbigg T 2 T 1 parenrightbigg n = 1 mol and for a monoatomic ideal gas C p , m = 2 . 5 R . Finally if the temperature increases this means T 2 > T 1 so ln parenleftbigg T 2 T 1 parenrightbigg will be positive. We expect a positive answer since tempera- ture increased. 002 10.0 points For the four chemical reactions I) 3 O 2 (g) 2 O 3 (g) II) 2 H 2 O(g) 2 H 2 (g) + O 2 (g) III) H 2 O(g) H 2 O( ) IV) 2 H 2 O( ) + O 2 (g) 2 H 2 O 2 ( ) which one(s) is/are likely to exhibit a positive Δ S ?

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saldana (avs387) – H13: Thermo 3 – Mccord – (52385) 2 1. II only correct 2. III and IV only 3. I and II only 4. I, III and IV only 5. All have a positive Δ S . Explanation: The Third Law of Thermodynamics states that the entropy of a perfect pure crystal at 0 K is 0. As disorder, randomness, and de- grees of freedom increase, so does S . Entropy can increase by changing phase from solid to liquid to gas, and by increasing temperature, volume, or number of particles.
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