Lesson_1.1_Homework_and_Solutions

Lesson_1.1_Homework_and_Solutions - Lesson 1.1 Homework...

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Lesson 1.1 Homework Solutions 1. (i) P 3 = 70 torr, t = 273 + 50 = 323 K 3 3 273.16 273.16 T P P P P T = = 82.8 273.16 323 70 323 70 273.16 273.16 ( ) 8 70 0 2 320 0 torr P ii K P T = × = = = × = 2. l o = 2.5 cm, l 100 = 22 cm 100 100 2.5 ( ) 35 100 22 2.5 9.3 16.7 2.5 ( ) 100 22 2.5 72.8 t o o o t t l l t l l l i l cm i C i t - = × - - = × - = - = × = - 3. P 1 = 2.5 atm, T 1 = 273 + 15 = 288 K, T 2 = 273 + 50 = 323 K, V 1 = 3.0 L (i) PV = nRT 2.5 3 0.32 0.08206 288 PV atm L n RT = = = × (ii) When volume is constant, 1 2 1 2 1 2 2 1 2.5 323 28 2 8 .8 P P T T PT at m m K t P K a T = × = = =
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4. V 1 = 61.5 L, T 1 = 273 + 18 = 291 K, P 1 = 2.45 atm V 2 = 48.8 L, T 2 = 273 + 50 = 323 K 1 1 2 2 1 2 1 1 2 2 1 2 2.45 61.5 323 291 48 3.4 .8 PV PV T T PVT P T atm V = × × = = = × 5. 1 atm = 101 kPa Absolute pressure = gauge pressure + atmospheric pressure P 1 = 240 + 101 = 341 kPa, V 1 = V 1 , T 1 = 273 + 22 = 295 K P 2 = 341 kPa, V 2 = V 2 , T 2 = 273 + 45 = 318 K 1 1 2 2 1 2 2 1 2 1 2 1 2 1
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This note was uploaded on 05/01/2011 for the course PHY 2049 taught by Professor George during the Spring '11 term at Edison State College.

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Lesson_1.1_Homework_and_Solutions - Lesson 1.1 Homework...

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