Lesson_1.3_and_1.4_Homework_and_Solutions

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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
To convert atm to pascals multiply by 1.01
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

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.

Page1 / 5

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online