Question

1.

a)

A piston–cylinder assembly contains air, initially at 1.9 bar, 295 K, and a volume of 0.6 m³. The air undergoes a process to a state where the pressure is 1 bar, during which the pressure–volume relationship is pV = constant. Assuming ideal gas behavior for the air, determine the mass in kg.

b)

Argon contained in a closed, rigid tank, initially at 51.1°C, 2.1 bar, and a volume of 2.9 m³, is heated to a final pressure of 7.1 bar. Assuming the ideal gas model with k = 1.51 for the argon, determine the heat transfer, in kJ.

Can you plis solved and show me step by step this problem. Im having a hard time with this topic, thank you! If little notes can be provided or explanation can be given I would be really grateful, I really want to learn how to do it.

Answer 1

1. Page ① all formulas at the o end of I will provide the solution. - 0@ system Initial conditions libar = 100 kPa) Pi= 1.9 bar = 190 kPa Vi= 0.6 m3 ; 7,3295k Giren that air follows [PV= constant law Isothermal process CA Pra constant - to from Ideal gas equation PV=MRT - PIVI = MRT, 601 air R: 0.287 kJllcgk. S į. 190 x 0.6 = mx 0.287X (295) Saram= 1.346 kg of air CS scanned with CamScanner

Page ① Augon nos 1.51 Given Initial conditions Pi= 2.1 bara 210 kPa V = 2.9 m3 T1 = 5ijºc=51.1+273 T = 324.1 ķ final conditions Py = 71 bar = 710 lepa, here argon follows poly hopic process py!.5! constant Heat transter for to polyropic procen is given PINAMA LINIIIII ve here . n21.51 ; 821.4 Pulisi ac pullsi a Paulist 270x (2.9).51 = 110(w).5) I ve 21.294 m3 CS Scanned with CamScanner

Page ① i Heat transter 110 XL Q=1.4-151] [1210X27.9)-1 (1.4-1 J L 101-1 = 60.275) *(-607.333) Q 2167.01 les Heat ofequüved to the given process. CSScanned with CamScanner

EQ < : Scanned by CamScanner NOTE! - wolie done by the system = tve woolle done on the system a ve # cloned system codlk ( | Open System wolk (31) Non-Flow wolk (@) 1 steady flow wolk moving boundary wole (wmid) - [w="sp.ar Wo -, er dp o constant volume (0) Csometiccol@_ Isochronic (web) © constant pressure (o Psobaric r@ (81) Psopiestic iTwe-v[PE -Pe] Wz | uz= P(VF - Vo OOOOOOOOOOOOOOOOOOOOOOOOOO © en VA constant Temp. (01) Psothermal (OV Hyperbolic (pv=c) Wo= cenVF | Wo a C ln PE . Pf Wa Clo Wopen 28 Welosed (We = cln (P) Cz P p VD = PRVF = MRTI = MRTE ② Adiabatic process (Pv²g=c o PIVO-PA VF moltiTF) (8-1) poly hopic process (PVP=0) 1. Povg-PFVF - MR (T2-T6) 18-1 Wopen an. Welased. I |w2n-1) = [n-1) Scanned by CamScanner C NOTE! - nr=c(0-0) - AV pvac Area under por diagram gives "wolk AVEC (n21 P=C (n=0) Pacino) - pvec insi) pr^=c - prac يا Val (risoo) snenie cool-

EQ < : Ler * clockwise - wolk prodering device "Anti-clock wire =) wolk absorbing device Scanned by CamScanner procen! Heat - Heat is tom of Energy which crosses the boundary due to Temperatuve difterence. (a= mclot) NOTE! - Heat supplied to the syitem = tve . Heat rejected rom the system = ave. . * C= speútic heat (T/kg-K) Came = Heat capacity (T/k). for cloned system (8) Non- flow process - 0 v=C » Qu= mcr(07)] = du i » Pac - (Qp = molot) = dH T=C ») Q, z tv. = besc en wa) duzd H=0 ☺ pupec ) [=0 -> dua - dw] ® pupas - Q = ethan) (dwpoly OOOOOOOOOOOOOOOOOOOOOOOOOOOO T=C calculated NOTE! Heat transter 607 open systems by using first law of T-D NOTE!" fol Adiabatic process Pri=c) Scanned by CamScanner • @ similarities btw Heat and wölk! - 1) Both are boundary phenomena. ) Both are energies in Transit 3) Both are Transient w Roth 002 noth functions I wall Transient I Heats