Question

A piston/cylinder arrangement with an cross sectional area of 1 mạ contains carbon dioxide at 200 kPa and 290 K. The cylinder has a thick piston that initially held by a pin as shown in Figure 1. The metal piston has a density of 8000 kg/mº, standard acceleration of gravity, &c is 9.81 m/s and the atmospheric pressure is 101 kPa. The pin is now removed, allowing the piston to move and after a while the gas returns to 290 K. SO mm 100 mm 100 mm Figure 1: Piston/cylinder arrangement a. b. Find the external pressure and determine is the piston moves up or stays at the initial position once the pin removed. (3 Marks) Find the location (the height) of the piston once the pin removed and gas temperature returns to 290 K. (2 Marks) Sketch the P.V diagram of the process. (2 Marks) Calculate the work done for the process. (3 Marks) c. d.

Answer 1

3. Somu room 3 Are peet premure solution given T = 290k Pi= 200k pa - 200x10?pq Ni = AnexL 100X10x1 Patu 101 kPa Viz olm? density of material episton Pp 3 mons f = dooo kg/m3 loouh = 100x10 yt 4,m? Odm3 a external prentun pressure due to atmospheric prot produe to piston meight = 101x10² + mg = 101x10²t fuge A 101x103 8000 x 100x10 Ft X 9101 108.848 kPa - pr. inside -(prodre to erton, of pre due to piston) (290 - 108.8481kpa 101.152 kPa "pet pressures external prettune, than pilton mows up. b) According to ideal gas law PV = nRTO similarly, while temp, returns to Pasul A = 1m2 Ti=290k 1. Pz V₂ = nRT, -mg Pinside loom la =183.74mm P, VI = P2 V2 V₂ = XX Area = x Arena F XX l I am PIVI 200x103 xo V2 when + Piston will stop than a Pinsiche - Patrat Ppiston cylinder 1 Ppiston low P2= &

200x1080.1 - 101x10²+ Jooox 100x10 & 1x4.01 20 108. ous 90 108.848 0.1837um x = 183.74mm height llocation of piston will be 183.74mm, P-v diagram of the proces V₂ = x x A sea V2 = 183.74x108 x 1 To=290k 3 a Pal = 0.18374 m² Vi colm Isothermal curte Pi = 200k pa P2 To =290k P2 = PIVI 200x1000X1 vo 183.74X103 P2 = 108. & 49 kepa. P2 = 108.849kpa V = 0.1 m3 12=0.18374 MB V d work done for prous (I so thermal prouss) W- - Pav V2 W =-SMRTI N = nRT S dr. VI W = -ART lu 2 iP2 from ideal gas equation, PivianRT, W =-Pi y en for PI W = 200x103 x Pojxlu ( 100849) W = -12.167 Kb lovel sign shows that a done by gas on piston, r -- HRT, lu fi - Pivlu Per Ar work