Question

A. Please provide your assumptions, a diagram (with system boundary), and a major formula at each question B. Please provide the table number if you need to use tables at Appendix. 1. As shown in the figure, an insulated box is initially divided into halves by a frictionless, massless, thermally conducting piston. On one side of the piston is 1.5 m of air at 400 K at 400 kPa. On the other side is 1.5 m of air at 400 K at 200 kPa. The piston is now removed and equilibrium is attained. Employing the ideal gas model for the air, (a) Find the final temperature in K. (10 pt). (b) Determine the final pressure, in kPa (10 pt). (c) Determine the total amount of entropy produced, in kJ/K. (20 pt). Do NOT assume a constant specific heat. Insulation Massless piston V = 1.5 m P = 400 kPa T= 400K V = 1.5 m P = 200 kPa T = 400 K

Thermodynamics problem.

please show detail work including which table you used, thanks.

Answer 1

I datar P = hookla v = 1.sm? P2 = 2ook Pa Aiz V₂ = 1.5m3 1 - Yook v-15 m3 (p=mookh v=1.5m3 P=2ook hall T-yook Assumption 2 ③ ( Als is behave as an Ideal gas. No heat transfer btw system is surroundings mass is conserve Free expansion process work done zero -(a) From conservation of mass m3= mitma (from Ideal gas RT, PV + P2 V₂ equation Pu=MRT P3 V3 From Pest law of thermodynamics so - dots duro 03 - 0; = 0 L of the system where vis Internal energy up-ui zl. Tz- matty & might Mz RT3 - MaRT &m, RT, PV = Pave + pv

P22 Put Pave V3 - Pa = P. v. A P2 V2 e Vitv₂ From eq 0 a T3 = P. V + Bolz PV + P2 V 2 TL 12 T3 = 4ooxl.st 200x1.5 Yooxis 200x1.5 yoo T T3 = look (3) Final favore pressure it Yoox is t 2008 1.3 from equation her istis P2 Book.pl

nopy of system Somies bonivens - AS system + 4 sagouding adlabata A Sgypten - Dot AS Do = m (pla Ig - m, Blue from Ideal gas equation s P, Via MRT, mit 5,226 kg m2 = 2.613 64 - 5.276 0907 400 A F 5.226 x 1.005 en 400 AS, = 0.43)41 Roy Asa - Me Cplen Iz m2 Blue Tor – 0.3040 RJ/x As system 0.4314+ (- 03040) As seen 0.1274) The total weant of entropy produs