solution_13pdf - yoon(jy4326 HW13 markert(58840 This...

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yoon (jy4326) – HW13 – markert – (58840) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A copper telephone wire has essentially no sag between poles 28 m apart on a winter day when the temperature is 15 C. How much longer is the wire on a summer day when the temperature is 33 C? As- sume that the thermal coefficient of copper is constant throughout this range at its room temperature value of 1 . 7 × 10 5 ( C) 1 . Correct answer: 2 . 2848 cm. Explanation: Given : T 0 = 15 C , L 0 = 28 m , T 1 = 33 C , and α = 1 . 7 × 10 5 ( C) 1 . The relation between length and tempera- ture is Δ L = L 1 L 0 = L 0 α ( T 1 T 0 ) , = (28 m) bracketleftbig 1 . 7 × 10 5 ( C) 1 bracketrightbig × [33 C ( 15 C)] parenleftbigg 100 cm 1 m parenrightbigg = 2 . 2848 cm 002 10.0 points The absolute temperature of a sample of monatomic ideal gas is doubled at constant volume. What effect, if any, does this have on the pressure and density of the sample of gas? Pressure Density 1. Multiplied by 4 Doubles 2. Doubles Multiplied by 4 3. Remains the same Doubles 4. Remains the same Remains the same 5. Doubles Remains the same correct Explanation: Since the volume remains the same, the density remains the same: P V = n R T . Thus doubling the absolute temperature of an ideal gas at constant volume will double the pressure. 003 (part 1 of 2) 10.0 points An air bubble originating from an under water diver has a radius of 7 . 2 mm at some depth h . When the bubble reaches the surface of the water, it has a radius of 5 mm. Assuming the temperature of the air in the bubble remains constant, determine the abso- lute pressure at this depth h . The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 3 . 01584 × 10 5 Pa. Explanation: Let : ρ = 1000 kg / m 3 , g = 9 . 8 m / s 2 , P atm = 1 . 01 × 10 5 Pa , R 1 = 5 mm = 0 . 005 m , and R 2 = 7 . 2 mm = 0 . 0072 m . From the ideal gas law P V = n R T = constant , so P 1 V 1 = P atm V atm P 1 P atm = V atm V 1 P 1 P atm = 4 3 π R 3 2 4 3 π R 3 1 P 1 = P atm R 3 2 R 3 1
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yoon (jy4326) – HW13 – markert – (58840) 2 = (1 . 01 × 10 5 Pa) (0 . 0072 m) 3 (0 . 005 m) 3 = 3 . 01584 × 10 5 Pa . 004 (part 2 of 2) 10.0 points Determine the depth of the diver. Correct answer: 20 . 4678 m. Explanation: The absolute pressure is P 1 = P atm + ρ g h h = P 1 P atm ρ g = 3 . 01584 × 10 5 Pa 1 . 01 × 10 5 Pa (1000 kg / m 3 ) (9 . 8 m / s 2 ) = 20 . 4678 m . 005 10.0 points An ideal gas is held in a container at constant volume. Initially, its temperature is 8 C and its pressure is 1 . 3 atm.
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