Lesson_1.3_and_1.4_Homework_and_Solutions

# Lesson_1.3_and_1.4_Homework_and_Solutions - Lessons 1.3 and...

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Lessons 1.3 and 1.4 Homework Solutions 1. Since the change is isothermal, the change in temperature dT = 0 (i) Therefore, the change in internal energy dU = 0 (ii) dQ = dU + dW = 0 + 5 × 10 3 = 5 × 10 3 J 2. P 1 V 1 P 2 V 2 3. (i) As the pressure remains constant, work done can be written as: W = P dV = (1 atm)(18.2 – 12)m 3 = 1.01 × 10 5 × 6.2 = 6.3 × 10 5 J (ii) dQ = dU + dW dU = dQ – dW = 6.0 × 10 6 – 6.3 × 10 5 = 5.37 × 10 6 J 4. (i) Since the container is rigid, there is no change in volume. Therefore, the work done dW = 0 (ii) 1300 J of heat is removed means that dQ = -1300 J dQ = dU + dW -1300 = dU + 0, dU = -1300 J. The internal energy decreases by 1300 J. 5. (i) n = 2, T = 300 K, V 1 = 3.5 m 3 , V 2 = 7.0 m 3 . 2 1 7 ln 2 8.314 300 ln 3458 3.5 isothermal V W nRT J V = = × × × = (iii) Since the change is isothermal, dT = 0, and hence dU = 0 (ii) dQ = dU + dw dQ = 0 + 3458 dQ = 3458 J 6. Work done when a gas expands art constant pressure is: W = P(V 2 – V 1 ) = 5.0 atm (710 mL – 400 mL) = 1550 atm.mL

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To convert atm to pascals multiply by 1.01
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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_1.3_and_1.4_Homework_and_Solutions - Lessons 1.3 and...

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