Question

9. Derive the equation dH = TDS + VdP and show tha ASm = Cp.m In 2 - R In For the change of one mole of an ideal gas from Tu, P, to T2, P, assuming that Cp,m is independent of temperature. (Hint: Start from dU = Tds - pdV and consider what you need to add to both sides to get an expression for dH on the left side).

Answer 1

Solution: Using the fundamental equation given: dU = TdS - PdV, we will derive the required equation for dH. For this definition of Enthalpy will be used. Then, using the ideal gas equation, PV = nRT, the result obtained can be simplified. The integration of the simplified result would be done assuming that Cp,m is independent of the temperature in the given range of integration. Please see the attached image for detailed solution.

fundameatel eauahon: Stast hom Soluhon: the du Tds Pdv Ardd to both side s du+ dPv) Tds-Pdv + dlfv) Since Hlenthalpy) is depined as |H- U+PV, dM- dut diPv) Depinition dH LHS Tds- Pdv + PaV +Vaf Heur dM= Tds +vap proved Now Asanging eauahon abene The dH-vdp Tds dH VdP T 9n teams e molas quanhhes Vm dp dsm T idial gas for Pvm-RT DR Pv-nRT an Vm R detinihon H dH dut dPV) Aqgain fem Pv dnRT For an ideal gar, dn nRdT nydTnRdT LytR)dT du t 11 dHm

for an ideal gas du nCydT and CutR Heme C.m dT Ldm 1 Fem 3 and dsm pmdT R dp P Sntegaahing the equahen to state 1 we ger om Stale 2 2 Ce, m is T2 assumed CP,m dT 1 T fT ASm Rin /f2

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