H14- Thermo 3-solutions

H14- Thermo 3-solutions - phan(dtp377 – H14 Thermo 3 –...

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Unformatted text preview: phan (dtp377) – H14: Thermo 3 – mccord – (51600) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This is your LAST homework on Quest for CH301. 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 II 2. V 3. I and IV 4. II and III correct 5. III and 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 =- . 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( ℓ ) IV) 2 H 2 O( ℓ ) + O 2 (g) → 2 H 2 O 2 ( ℓ ) phan (dtp377) – H14: Thermo 3 – mccord – (51600) 2 which one(s) is/are likely to exhibit a positive Δ S ? 1. III and IV only 2. I and II only 3. I, III and IV only 4. All have a positive Δ S ....
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This note was uploaded on 03/26/2012 for the course CHEM 301 taught by Professor Wandelt during the Spring '08 term at University of Texas.

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H14- Thermo 3-solutions - phan(dtp377 – H14 Thermo 3 –...

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