In class Professor Carter showed that the entropy of an ideal gas is a function of state as it is a perfect di erential. C nR v dS = dT dV T V From this an alternate de nition of the heat capacity at constant volume and molar volume are: ! ! @S C @S R v = and = @T T @V V V T He also introduced the Gibbs Free energy G = U TS + PV . Using the di erential form of the Gibbs Free energy write alternate expressions for the entropy and volume. Solution 6.2 First lets write the Gibbs Free energy in it's di erential form for an ideal gas. G = U TS + PV dG = dU TdS SdT + PdV + VdP dG =(TdS PdV ) TdS SdT + PdV + VdP dG = VdP SdT @G @G Now on examining the rst derivatives we can see: S = and V = . There @T @P P T are many other consequences of the form of the Gibbs Free energy that will be examined later in the course. Example Problem 6.3 A kilo of liquid water transforms to ice in a giant freezer at atmospheric pressure and the melting temperature. Given the latent heat of fusion, L = 334kJ=kg, calculate the f molar entropy of fusion for water, S . f Solution 6.3 The latent heat of fusion is the heat given o by a liquid when it transforms to a solid at L f .This is a reversible the melting temp erature and pressure. For the freezer: S = freezer T transformation. So S = 0 and therefore S = S . The latententropy universe w ater freezer of fusion is the entropy change of the solid up on melting. 334kJ=kg 0:018kg J S = S =S = =22 f w ater f r eez er 273:15K 1mole mole K It should be noted that the latent heat of fusion and entropy of fusion are rep orted as p ositive numb ers. So the interpretation could be that each is the change in that prop erty of the system on melting or the change in that prop erty of the surroundings on freezing..