1. The differential form of the thermodynamic potentials for a fixed amount of a pure gas are: dU = TdS - pdV; dH = TdS + Vdp; dF = -SdT - pdV; dG = -SdT + Vdp The following properties are directly accessible through laboratory measurements: CV = (dU/dT)V; k = -1/V(dV/dP)T; a = 1/V(dV/dT)P Answer the following with the help of these expressions: a) Show that (dS/dV)T = (dp/dT)V = a/k b) Show that (dU/dV)p = Cv/Va + Ta/k - p
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First, let's recall the definitions of the heat capacities and compressibility: Cv = (∂U/∂T)_V (heat capacity at constant volume) K = -(1/V)(∂V/∂p)_T (isothermal compressibility) Now, let's differentiate the expressions for the thermodynamic potentials with Show more…
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