Thermodynamic coefficients#

In this section different related to first derivatives of thermodynamic state variables are introduced and discussed for different systems.

Thermodynamic coefficients of a single-component fluid#

For a 1-component fluid with constant composition, the first principle reads

\[de = T ds - P dv = T ds + \frac{P}{\rho^2} d \rho \ .\]

Heat capacity.

\[c_x := T \dfrac{\partial s}{\partial T}\Big|_x\]

For fluid systems, usually heat capacity at constant pressure or constant density are the most used.

Thermal expansion coefficients.

\[\alpha_x := \dfrac{1}{v} \dfrac{\partial v}{\partial T}\Big|_x = - \dfrac{1}{\rho} \dfrac{\partial \rho}{\partial T}\Big|_x\]

For fluid systems, usually thermal expansion coefficent at constant pressure is the most used.

Compressibility coefficients.

\[\beta_x := - \dfrac{1}{v} \dfrac{\partial v}{\partial P}\Big|_x = \dfrac{1}{\rho} \dfrac{\partial \rho}{\partial P}\Big|_x\]

For fluid systems, usually compressibility coefficent at constant temperature or entropy are the most used.

Relations between thermodynamic coefficients#

Relation between \(c_v\) and \(c_p\).

(2)#\[c_P - c_v = T \, v \, \dfrac{\alpha_P^2}{\beta_T} \ . \]

Relation between \(\beta_s\) and \(\beta_T\) - 1

(3)#\[\beta_s = \frac{c_v}{c_P} \beta_T \ .\]

Relation between \(\beta_s\) and \(\beta_T\) - 2

(4)#\[\beta_s - \beta_T = - \frac{v T}{c_P} \alpha_P^2 \ .\]

Relation between \(\beta_s\) and \(\beta_T\) - 3

(5)#\[\dfrac{1}{\beta_s} - \dfrac{1}{\beta_T} = \dfrac{T}{v c_v} \left( \partial_T P|_v \right)^2 \ .\]
Proof of the relation between heat capacities \(c_v\), \(c_P\)

Changing independent variables from \((T,v)\) to \((P,v)\) in the expression of entropy \(s(T,v) = s(T, P(v,T))\),

\[\begin{split}\begin{aligned} \dfrac{c_v}{T} := \left.\dfrac{\partial s}{\partial T}\right|_v & = \left.\dfrac{\partial s}{\partial T}\right|_P + \left.\dfrac{\partial s}{\partial P}\right|_T \left.\dfrac{\partial P}{\partial T}\right|_v = & \qquad \text{(Maxwell $\partial_P s|_T = - \partial_T v|_P$)} \\ & = \left.\dfrac{\partial s}{\partial T}\right|_P - \left.\dfrac{\partial v}{\partial T}\right|_P \left.\dfrac{\partial P}{\partial T}\right|_v = & \qquad \text{(relation below $\partial_T P|_v = \dots$)} \\ & = \partial_T s|_P + \dfrac{\left( \partial_P v |_T \right)^2}{\partial_P v |_T} = & \qquad \text{(def of TD coeffs)} \\ & = \dfrac{c_P}{T} - \dfrac{ v \alpha_P^2}{\beta_T} \end{aligned}\end{split}\]

having exploited the relation

\[\begin{aligned} \left.\dfrac{\partial P}{\partial T}\right|_v = \dfrac{\partial(P,v)}{\partial (T,v)} = \dfrac{\partial(P,v) / \partial(T,P)}{\partial (T,v) / \partial(T,P)} = - \dfrac{\partial_P v|_T}{\partial_P v|_T} \ . \end{aligned}\]
Proof of the relation between \(\beta_s\) and \(\beta_T\) - 1
\[\begin{split}\begin{aligned} v \beta_T & = \partial_P v |_T \\ v \beta_s & = \partial_P v |_s = \dfrac{\partial(v,s)}{\partial(P,s)} \dfrac{\partial(v,T)}{\partial(v,T)} \dfrac{\partial(P,T)}{\partial(P,T)} = \underbrace{\dfrac{\partial(P,T)}{\partial(P,s)}}_{= \dfrac{1}{\partial_T s|_P}} \underbrace{\dfrac{\partial(v,T)}{\partial(P,T)}}_{\partial_P v|_T} \underbrace{\dfrac{\partial(v,s)}{\partial(v,T)}}_{\partial_T s|_v} = \dfrac{c_v}{c_P} \left( v \beta_T \right) \end{aligned}\end{split}\]
Proof of the relation between \(\beta_s\) and \(\beta_T\) - 2
\[ \beta_s = \frac{c_v}{c_P} \beta_T = \dfrac{\beta_T}{c_P} \left[ c_P - T v \frac{\alpha^2_P}{\beta_T} \right] = \beta_T - T v \frac{\alpha^2_P}{c_P} \ . \]
Proof of the relation between \(\beta_s\) and \(\beta_T\) - 3
\[\begin{split}\begin{aligned} \dfrac{1}{\beta_s} = \frac{c_P}{c_v} \dfrac{1}{\beta_T} & = \dfrac{1}{c_v \, \beta_T} \left[ c_v + T v \frac{\alpha^2_P}{\beta_T} \right] = \\ & = \dfrac{1}{\beta_T} \, \dfrac{1}{c_v} \left[ c_v + T v \frac{\alpha^2_P}{\beta_T} \right] = \\ & = \dfrac{1}{\beta_T} + T v \frac{\frac{1}{v^2} (\partial_T v|_P)^2}{-\frac{1}{v} \partial_P v|_T} \frac{1}{-\frac{1}{v} \partial_P v|_T} = \\ & = \dfrac{1}{\beta_T} + T v \left( \frac{\partial_T v|_P}{\partial_P v|_T} \right)^2 = \dfrac{1}{\beta_T} + T v \left( \frac{\partial (v,P)}{\partial (T,P)} \frac{\partial (P,T)}{\partial (v,T)} \right)^2 = \dfrac{1}{\beta_T} + T v \left( \left.\dfrac{\partial P}{\partial T}\right|_v \right)^2 \end{aligned}\end{split}\]

Thermodynamic equilibrium#

Here thermodynamic equilibrium is discussed for a single-component fluid, for which the first principle reads

\[d e = T ds - P dv = T ds + \frac{P}{\rho^2} d \rho \ .\]

First conditions on energy, \(e(s, \rho)\), as a function of entropy and density then the equivalent conditions on entropy \(s(e, \rho)\), as a function of energy and density are discussed.

Conditions on energy#

Thermodynamic equilibrium implies conditions on the second order term in series expansion of the function \(e(s, \rho)\) (todo why? Spend few words. Use Landau as a reference if needed)

\[\begin{split}\begin{aligned} e(s+ds, \rho+d\rho) & = e(s, \rho) + \partial_s e|_{\rho} (s,\rho) d \rho + \partial_\rho e|_{s} (s, \rho) ds + \\ & + \dfrac{1}{2} \left[ \partial_{ss} e|_{\rho}(s,\rho) ds^2 + 2 \partial_{s\rho} e(s, \rho) ds d\rho + \partial_{\rho \rho} e(s, \rho) d \rho^2 \right] \ . \end{aligned}\end{split}\]

The equilibrium energy must be a minimum (todo check the function that must be a minimum, since \(\partial_s e|_{\rho} = T\),… and \(T\) not zero, and thus it can be a minimum!): the second order term must be positive for any increment of independent physical variables \(d \rho\), \(d s\), i.e.

\[\dfrac{1}{2} \left[ \partial_{ss} e|_{\rho} ds^2 + 2 \partial_{s\rho} e ds d\rho + \partial_{\rho \rho} e|_{s} d \rho^2 \right] \]

is a positive-definite quadratic form, i.e. the Hessian

\[\begin{split}\begin{bmatrix} \partial_{ss} e|_{\rho} & \partial_{s \rho} e \\ \partial_{s \rho } e & \partial_{\rho \rho} e|_{s} \end{bmatrix} \end{split}\]

is definite positive, and thus teh following conditions must hold

\[\begin{split}\begin{cases} \partial_{ss} e|_{\rho} > 0 \\ \partial_{ss} e|_{\rho} \partial_{\rho \rho} e|_{s} - \left( \partial_{s \rho} e \right)^2 < 0 \end{cases}\end{split}\]

Conditions on entropy#

Using \(e\), \(\rho\) as independent thermodynamic state variables, and the entropy \(s(e, \rho)\) it can be proved that the condition on \(e(s, \rho)\) for the thermodynamic equilibrium implies that \(s(e, \rho)\) is a maxiumum (todo check the meaning of maximum here), and thus

\[\dfrac{1}{2} \left[ \partial_{ee} s|_{\rho} de^2 + 2 \partial_{e\rho} s de d\rho + \partial_{\rho \rho} s|_{e} d \rho^2 \right] \]

is a positive-definite quadratic form, i.e. the Hessian

\[\begin{split}\begin{bmatrix} \partial_{ee} s|_{\rho} & \partial_{e \rho} s \\ \partial_{e \rho } s & \partial_{\rho \rho} s|_{e} \end{bmatrix} \end{split}\]

is definite positive, and thus teh following conditions must hold

\[\begin{split}\begin{cases} \partial_{ee} s|_{\rho} < 0 \\ \partial_{ee} s|_{\rho} \partial_{\rho \rho} s|_{e} - \left( \partial_{e \rho} s \right)^2 < 0 \end{cases}\end{split}\]
Relation between partial derivatives of \(e(s,\rho)\) and \(s(e,\rho)\)

The relation

\[e = e_{s \rho}(s_{e \rho}(e, \rho), \rho) \ ,\]

provides the link between the two representations, having written \(e_{s,\rho}()\), \(s_{e,\rho}()\) the functions with the arguent indicated as indices. This relation contains only \(e\), \(\rho\) as independent variables. All the required relations are evaluated computing partial derivatives of this relation.

First-order derivatives. It can be proved that

\[\begin{split}\begin{aligned} \partial_e s |_{\rho} & = \dfrac{1}{\partial_s e |_{\rho}} \\ \partial_{\rho} s |_e & = - \dfrac{\partial_{\rho} e |_{s}}{\partial_s e |_{\rho}} \\ \end{aligned}\end{split}\]

Second-order derivatives. It can be proved that

\[\begin{split}\begin{aligned} \partial_{ee} s|_{\rho} & = - \dfrac{1}{\left( \partial_s e|_{\rho} \right)^3} \partial_{ss} e |_{\rho} \\ \partial_{e\rho} s & = \dfrac{\partial_{\rho} e|_s}{\left( \partial_s e|_{\rho} \right)^3} \partial_{ss} e |_{\rho} - \dfrac{\partial_{s \rho} e}{(\partial_s e|_{\rho})^2} \\ \partial_{\rho\rho} s|_{e} & = - \dfrac{1}{\left( \partial_s e|_{\rho} \right)^3} \left[ \left( -\frac{\partial_\rho e_s}{\partial_s e|_\rho} \right)^2 \partial_{ss} e |_{\rho} + 2 \left(-\frac{\partial_\rho e_s}{\partial_s e|_\rho}\right) \partial_{s \rho} e + \partial_{\rho \rho} e|_s \right] \\ \end{aligned}\end{split}\]
Equivalence of conditions on \(e(s, \rho)\) and on \(s(e, \rho)\) for thermodynamic equilibrium

Exploiting first condition on partial derivatives of \(e(s,\rho)\), the first condition on the partial derivatives of \(s(e,\rho)\) is

\[\partial_{ee} s|_{\rho} = - \dfrac{1}{\left( \partial_s e|_{\rho} \right)^3} \partial_{ss} e |_{\rho} = - \dfrac{1}{T^3} \partial_{ss} e |_{\rho}< 0 \]

since \(T > 0\) and \(\partial_{ss} e|_{\rho} > 0\).

Second condition for derivatives of \(s(e, \rho)\) is

\[\begin{split}\begin{aligned} \partial_{\rho \rho} s|_{e} \partial_{ee} s |_{\rho} - \left( \partial_{\rho s} e \right)^2 & = \dots \\ & = \frac{1}{T^4} \left[ \partial_{ss} e|_{\rho} \partial_{\rho \rho} e|_{s} - \left( \partial_{s \rho} e \right)^2 \right] < 0 \ . \end{aligned} \end{split}\]