Question

3. Consider the entropy as a function of temperature and pressure. Derive an expression for the change in entropy in terms of parameters such as Cp, a, K, T, and V.

Answer 1

- Entropy is a function of temperature and pressune Us=s(1,P) From 1st law of TO - du= aq faw - on, de a da to [ Assuming nevensable ich I process and P-u wool on du = Tds-pdv / i involved whene It ds = TGREU ids= I du tf dvL and Iw=_pdv on, da 'REÚ L T

Substituting equc into equ0 — ds= + [dw-pdv- vap] te du ods = dH - 2 dv. de te du -Po V-VdP T Now, we know H=H(Typ) O H= H Now, we know, (p = () So, dw = Cod Tt (1) de From egun lo and we can caite — ds = 141 cpd 17 (2), dp] - 7 de on, as - (pdt + + 1 (0) - 17 JP - Ald A t thermodynamic

Now from fundamental equinoy TD. - - / OT om )--- ) From @ and in ds = // CONT - Code - XV dp on, ds = op dT - avdp [ As we know Isotheromal expansion L Co-efficient ids = op dT - avde ( Change in entropy in teams of Cp, A, T, VI
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