Two of the Massieu functions are functions of the following state variables:1. J(1/T, V) 2. Y (1/T,/pT) The Massieu functions are obtained by performing Legendre transformations of the state function entropy S (U, V) with respect to the state variables U and V. Use Legendre transformations to

Question

Two of the Massieu functions are functions of the following state variables:

[latex]J \Bigl(\frac{1}{T},V\Bigr) [/latex].
[latex]Y \Bigl(\frac{1}{T},\frac{p}{T}\Bigr) [/latex].

The Massieu functions are obtained by performing Legendre transformations of the state function entropy S (U, V) with respect to the state variables U and V. Use Legendre transformations to express the Massieu functions [latex]J \Bigl(\frac{1}{T},V\Bigr) and [latex]Y \Bigl(\frac{1}{T},\frac{p}{T}\Bigr) [/latex] in terms of the thermodynamical potentials F and G. Determine also the differentials [latex] d J \Bigl(\frac{1}{T},V\Bigr) and [latex] d Y \Bigl(\frac{1}{T},\frac{p}{T}\Bigr) [/latex].

Accepted Answer

The entropy S (U, V) as a state function reads,

[latex] S =\frac{1}{T} U +\frac{p}{T} V [/latex].

and its differential is written as,

[latex] dS =\frac{1}{T} dU +\frac{p}{T} dV [/latex].

Thus,

[latex] \frac{∂S}{∂U} = \frac{1}{T} [/latex] and [latex] \frac{∂S}{∂V} = \frac{p}{T} [/latex].

To obtain the Massieu function [latex]J \Bigl(\frac{1}{T},V\Bigr) [/latex], we perform a Legendre transformation on the entropy S (U, V) with respect to the internal energy U,

[latex] J = S −\frac{∂S}{∂U} U = S − \frac{U}{T} = - \frac{F}{T} [/latex].

Likewise, to obtain the Massieu function [latex]Y \Bigl(\frac{1}{T},\frac{p}{T}\Bigr) [/latex], also called the Planck function, we perform two Legendre transformations on the entropy S (U, V) with respect to the internal energy U and the volume V,

[latex]Y = S −\frac{∂S}{∂U} U - \frac{∂S}{∂V} V= S − \frac{U}{T} - \frac{pV}{T} = - \frac{G}{T} [/latex].

Differentiating the Massieu function [latex]J \Bigl(\frac{1}{T},V\Bigr) [/latex] yields,

[latex] dJ = dS -\frac{1}{T} dU -Ud\Bigl(\frac{1}{T}\Bigr) = - U d\Bigl(\frac{1}{T}\Bigr) +\frac{p}{T} dV [/latex].

Similarly, differentiating the Massieu function [latex]Y \Bigl(\frac{1}{T},\frac{p}{T}\Bigr) [/latex] yields,

[latex] dY = dS - \frac{1}{T} dU -U d\Bigl(\frac{1}{T}\Bigr) -\frac{p}{T} dV - Vd\Bigl(\frac{p}{T}\Bigr) = -U d\Bigl(\frac{1}{T}\Bigr)- V d\Bigl(\frac{p}{T}\Bigr) [/latex].

Question 4.5: Two of the Massieu functions are functions of the following ...

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