Question

Heat is transferred to a gas in a piston cylinder device so that the volume changes from 3 mºto 6 m2. The initial pressure and temperature of the gas are 400 kPa and 25°C. If the process is irreversible determine the following: 1- The final temperature of the gas. 2- The work done during the process (kJ). 3- The total change in internal energy (kJ). 4- The heat transfer for the process (kJ). 5- The total entropy change (kJ/K). Comment on the result. 6- The ideal and lost work (kJ). 7- The total change in enthalpy (k). Assume that the surrounding temperature and pressure are 60°C and 100 kPa, respectively. Assume that the gas obeys the following Eos: P = RT/(V-b) R = 8.314J/mol.K) For the gas: b = 0.1 m/mol M= 32 g/mol Cp = 1.0 kJ/(kg.) Cy = 0.68 kJ/(kg.) Astot = ASsys + ASsurr AS surr = Q/T. dT ОР dS =Cr + DV T ОТ T Core

Answer 1

Heat is transfereed to a jes in a piston cylinder device. volume Change from 3m3 to 6m². Initial pressure and temperature, P = gookte TI= 25 = 298k gas obeys the following for Poo 160kPa p= file-b), R= 8.314 91 wolok) Tra 607273 - 333K b = 0.1 mm Imol, m= 32 gimol, Op=1 ksregir, Cu='0'ba kullcgt. The final Temperature (T2):- P₂ (or contant T2 (V2-6) (Vi-b) T, pressure proces) T3][6-0-1) = 24813-0-1 606.3k berature, [T2 = Tempe The work done during the process (KJ) 2 Plu-b)= MRT work done (W) = Spar SB 400(3-01) = mx 8.314 - RT dr Q-b) 32 7296 ma Iskg MRT Inver = 1548:32,4x298 ( 670 3-01 825 KS charge je in internal Chergy (0) al = du = Cv dT + +(12-plan du mawar 158.681T2-TI) = 154.681606-3-293) 314 4. 66ks du =

The heat transfer for the process (1) (8) - from first low of thermodynamics- dos du + dw da = (825 +3144-66) kes age 3970KS The total entropy change es) AST = es surr. + OS)s ) afstem E +3970 100) - 3970 = -11.92 Kolk (AS) surr 333 To (es) system m[w of + Ge)du? = [coln (1) + RIGU 15[0.68 in 60 ) + 8:319 inches 6-01 3-01 - 10.01 Kolk Os) total = (11192 +10-01) K3116 = +1.9 Kolk Qs) total zo Hence process is irreversible. process. The ideal and lost work (15: OH- To AS Wided - Cpl T2-T) - Tolos) Wineo = isix(606-294) – 333 (-1-4)] 5252.7 KS Wided Wlost = To A (sm)- = To X Sy = 333x 119 Wlor = 632.7 kr

3 Total change in enthalpy (k): DH= mCp (&T) OM = 1541x1606-291 lon 4620 kJ