Question

PART B(1) A gas mixture expands with a known pressure-volume relation, PVConst. The gas mixture behaves as an ideal gas. The process is polytropic with n=1.3. Kinetic and Potential energy effects are or (plv1)in(v2/v1), dU=mCvdT, 1 bar-10 N/sq.m, 1 kJ-10 N.m, Universal gas constant(R)- 8.314 kJ/(kmol.K)= MR, Please refer to the given data. enegligible (Hints: PV-mRT, Work-(p2v2-p1v1)/(1-n) , 0.3m P-2 bars Va.72 laFind the mass (m) of gas in kg T s00x AU-m di- 2.5 Kkg (o.s3/R 300 K) ma.6 kg 2.5 K3/ (b)Find the final pressure(P2) in bars (2 hars(0.3 P bar ( c)Find the final temperature(T2) in K 0.010.5KK)(300K) 2.5rkg 612X 5 K 1 4Find the work done for n-1.3 in kJ (2. in) (0.3.1) KJ 6.37 e) Find the heat transfer(Q) in kJ w1o.503 KJ

P1= 5 bars, P2 = ?, V1 = .3 m^3 , V2 = .7 m^3, Cv = .5 KJ/kgK, dv = 2.5 KJ/kg

*du = 2.5 KJ/ Kg

Answer 1

P=2 bar V = 0.3 m² Ti=book Vg = 0.7m² CA222222222 P, v, 13 - Constant P2 V = P, U13 | Ca = , vº (( .3 . [12 = 0.66 bar] ③ we have pu=mRT P. V Pa - Ta= T Peve PV, 0.3 0.66x105 0.7 = 300 X 0.3 2x05 x To = 231k/ du=madT = mc [T-Tg ) 2.5 = mx 0.5 *[300-2313 m = 0. 093 4 57 @ work done W- Pa Vg - PV, 1-n W - 0.66+105x0.7 - 2x0.34105 1-1.3 Wr. - 46k 1 l

@ Heat transfer a= U+W A= 2.5-46 To = - 43.5 kJ) ro bar V = 0.3 m² T=3ook ugo.am Cr=0.5 kJ/kg k dU= 2.5 kJ/kg @ . [P2 = 1.66 bar o Tg = Ti Pove you - 300 x1.56x105 x ot 28 Joº X 03 [T2=581 kl du = mlrdt 25= mx 0.5x (581 - 300] Im=0.0178 kg

& W = Peve Piv, len W Jouls 1.66x105x0.7 - 54105 x 0.3 1-1.3 W = – 112.67 KI 8- U+W - 2.5 - 112.67 [Q=-110.16 kJ]