3) Consider 2 kg of air has a decrease of internal energy of 25kJ while its temperature is reduced to one-third of the initialtemperature during a reversible non flow isobaric process.Determine a) Initial temperature b) final temperature c) Heat(Q) d) Work and e) Change in Entropy

Answered: Image /qna-images/answer/4822e02d-6b4f-4434-a0d6-d98c5d57525b.jpg

Answered: 3) Consider 2 kg of air has a decrease…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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One-quarter Ibmol of oxygen gas (O2) undergoes a process from p1 = 20 lb;/in?, T1 = 500°R to p2 = 150 Ib/in?. For the process W = -500 Btu and Q = -152.5 Btu. Assume the oxygen behaves as an ideal gas. Determine T2, in °R, and the change in entropy, in Btu/PR.
m= 1 kg Process path (straight line) R= 0.50 2 C, = 0.5 kJ/kg K P V= 1 m3 P= 300 kPa V= 0.5 m3 P= 600 kPa An ideal gas in a piston undergoes a process from state 1 to state 2 above. d) The work to move the piston is: a) performed ON the system (Win) b) performed BY the system (Wout) e) If no other work is done on or by the system during the process, circle the term that must be non-zero (i.e. what else occurs during the process) to satisfy the Conservation of Energy Equation for a closed system having no change in KE or PE. W - W, + Qin - Qout = AU in out
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Air initially at 15 psia and 60 deg. F is brought to a final temperature of 200 deg. F by a reversible non-flow process in each case and find the work required in Btu/lb if a) n = 0; b) n = infinity; c) n = 1.2; d) n = k

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3) Consider 2 kg of air has a decrease of internal energy of 25 kJ while its temperature is reduced to one-third of the initial temperature during a reversible non flow isobaric process. Determine a) Initial temperature b) final temperature c) Heat (Q) d) Work and e) Change in Entropy