Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 17 - Work, Heat, and the First Law of Thermodynamics - Conceptual Questions - Page 496: 11

Answer

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Work Step by Step

a) We are cooling down the system (by contacting the ice to the cylinder) which makes the volume compresses since the locking is plugged out and the piston now is moving freely. This means that the work done on the system is positive since the final volume is less than the initial volume. $$\boxed{W\gt 0}$$ Also, due to the ice contact, the temperature decreases which means that $$\boxed{\Delta T\lt 0}$$ This also means that the heat escapes from the system which means that $$\boxed{Q\lt 0}$$ --- b) From the given figure, and since the ice remains in touch with the cylinder so the final temperature of the gas after a long time when it reaches a new equilibrium point is zero Celsius. $$T_f=0^\circ\rm C$$ This means that the temperature decreases while the volume also decreases. And in this case, $$\dfrac{ P_1 V_1}{T_1}=\dfrac{ P_2 (V_2\downarrow)}{(T_2\downarrow)}$$ And since the mass did not change above the position, the pressure is constant during this process. It is an isobaric process. See the graph below.
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