Question

15. For one mole of a real gas, consider the change of enthalpy as the reduced pressure is increased from zero pressure (P, >0) to the given reduced pressure, Pr, while the temperature is held constant. The equation of state of a real gas can be expressed in terms of the compression factor (Z) in the following equation. PV RT Under this process, show that the change of enthalpy is as follows: () -RT; ("667), dr

Answer 1

cu IG Tn, pp. Te Coro Tp, Pn 20 tea (Divided by To , and Pe ) (9) HP АН 12 4 - Cordon (more) I dpn (Proved) T2Tp IP Where TH - HIG) 7.2=0H Peve - Ze amd RTe Pe= Critican Priessine Ve= Critican volume Te = criticen temperatione Zcà Critican Compression factor. o loub

state (T₂, P2) (T,,P=0) (T2, P₂) Solution : To Compute the change in enthalpy in going from the Stade (ti, Pg to the We start from. To, P2 AH (2,1)=H (Tz , P2) -H (T, P,) = J dh (Enthalpy departure function Tomap, tip, : H (Ta, Pz.) - H (TU, PI] = H264-T2, P2) - H 56 (F1P,) + X-TC+)70p + [v-Tep] dp (T, ) - ( H - u 16 ) ,, + u*CE.) - HP (T, Ps) – (*-*-*)0, Enthalpy deportare (vesi decor) (9 - 12" ).P="" [v-Teplot. The Compression factor (2) in the following equation Tz, P=0 T, P TP-0 With respent tot 7 a = RT Z (To, Pro). u V-TC)p=-- RI? )[differentiation both side Perling the value in equation : we get- (H-4*%) 7, p = 272 ( 2 ) p dp TP. T, p=0 Tr. Pr Toz laro lodpr. Topr = - RTe l o Tm, Pn=0
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