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b) When the gas goes from A to B, how much work WAB is done by it on its environment? • Hint: The area of an ellipse with semimajor axis v and semiminor axis w is given by puw. c) How much heat QA.B flows into the gas during the transformation A-B?

b) When the gas goes from A to B, how much work WAB is done by it on its environment? • Hint: The area of an ellipse with semimajor axis v and semiminor axis w is given by puw. c) How much heat QA.B flows into the gas during the transformation A-B?

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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answer a b and c pls
A monatomic ideal gas undergoes a cyclic transformation as shown. a) When the gas goes from A to B, what is the change in its internal energy AEA.B? b) When the gas goes from A to B, how much work WA.B is done by it on its environment? 3p. • Hint: The area of an ellipse with semimajor axis v and semiminor axis w is given by 2po B puw. c) How much heat QA„B flows into the gas during the transformation A→B? Po d) Answer the same questions for the return transformation B→A. V V. 3V. 5V. e) What is the net work done by the gas on its environment over the cycle? **
Transcribed Image Text:A monatomic ideal gas undergoes a cyclic transformation as shown. a) When the gas goes from A to B, what is the change in its internal energy AEA.B? b) When the gas goes from A to B, how much work WA.B is done by it on its environment? 3p. • Hint: The area of an ellipse with semimajor axis v and semiminor axis w is given by 2po B puw. c) How much heat QA„B flows into the gas during the transformation A→B? Po d) Answer the same questions for the return transformation B→A. V V. 3V. 5V. e) What is the net work done by the gas on its environment over the cycle? **
Expert Solution
Step 1

Area of the ellipse is  = (5V0-V0 ) × (3P0 - P0 )×π

                                      =8P0V0π 

Work done for A to B is = 8P0V0 + 4P0V0π 

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