Reduced temperature

Equation

>Top, >Model


The reduced temperature is a normalized scale calculated using two reference temperatures in order to obtain values between 0 and 1.

Therefore, using $T_0$ as a base temperature, $T_r$ as the temperature range, and $T$ as the temperature in question, the reduced temperature can be defined as:

$ \tau = \displaystyle\frac{ T - T_0 }{ T_r }$

$T$
Temperatura
$K$
9343
$T_0$
Temperatura base
$K$
9345
$T_r$
Temperatura de referencia
$K$
9344
$\tau$
Temperatura reducida
$-$
9346

ID:(12350, 0)



Reduced pressure

Equation

>Top, >Model


The reduced pressure is a normalized scale calculated using two reference pressures in order to obtain values between 0 and 1.

Therefore, with $p_0$ as a base pressure, $p_r$ as the pressure range, and $p$ as the pressure, the reduced pressure can be defined as:

$ \pi = \displaystyle\frac{ p - p_0 }{ p_r }$

$p$
Presión
$Pa$
9347
$p_0$
Presión de base
$Pa$
9349
$p_r$
Presión de referencia
$Pa$
9350
$\pi$
Presión reducida
$-$
9348

ID:(12351, 0)



Reduced salinity

Equation

>Top, >Model


The reduced salinity is a normalized scale calculated using a reference salinity in order to obtain values around 1.

Therefore, using $i_r$ as the salinity range and $i$ as the salinity in question, the reduced salinity can be defined as:

$ \xi = \sqrt{ \displaystyle\frac{ i }{ i_r } }$

$i$
Salinidad
$-$
9352
$i_r$
Salinidad de referencia
$-$
9353
$\xi$
Salinidad reducida
$-$
9351

ID:(12352, 0)



Definition of the first derivative of the Gibbs potential

Equation

>Top, >Model


In order to calculate the various parameters, it is necessary to be able to differentiate the Gibbs potential, which corresponds to the slopes of this function with respect to pressure or temperature.

In general, the Gibbs potential factors, denoted as $g_x$, are defined with $x$ representing the variable and $g$ representing the molar Gibbs free energy, as follows:

$ g_x =\displaystyle\frac{\partial g }{\partial x }$

$g$
Energía libre de Gibbs molar
$J/mol$
9356
$g_x$
Primera derivada de la energía libre de Gibbs molar
$J/mol$
9357
$x$
Primera variable termodinámica
$-$
9354

ID:(12356, 0)



Second derivative of the Gibbs potential

Equation

>Top, >Model


For the calculation of various parameters, it is necessary to be able to take second-order derivatives of the Gibbs potential, which corresponds to the curvatures of this function with respect to pressure and/or temperature.

In general, the factors of the Gibbs potential are defined as follows:

$ g_{xy} =\displaystyle\frac{\partial^2 g }{\partial x \partial y }$

$g$
Energía libre de Gibbs molar
$J/mol$
9356
$x$
Primera variable termodinámica
$-$
9354
$g_{xy}$
Segunda derivada de la energía libre de Gibbs molar
$J/mol$
9358
$y$
Segunda variable termodinámica
$-$
9355

ID:(12357, 0)



Gibbs free energy density for water and steam

Equation

>Top, >Model


The density of Gibbs free energy, or specifically Gibbs free energy, as a function of temperature and pressure $g(T,p)$, can be expressed as a polynomial in reduced temperature $\tau$ and reduced pressure $\pi$, which is written as follows:

$\displaystyle\frac{ g^w(T,p) }{ g_r } = \displaystyle\sum_{j=0}^7\displaystyle\sum_{k=0}^6 g_{jk} \tau ^ j \pi ^ k $

$g_r$
Energía libre de Gibbs molar de referencia
$J/mol$
9379
$g_w$
Energía libre de Gibbs molar del agua
$J/mol$
9341
$\pi$
Presión reducida
$-$
9348
$\tau$
Temperatura reducida
$-$
9346

The equation for the specific Gibbs energy as a function of temperature and pressure, $g(T,p)$, can be expressed as a polynomial in the reduced temperature $\tau$ and reduced pressure $\pi$:

$ \tau = \displaystyle\frac{ T - T_0 }{ T_r }$



$ \pi = \displaystyle\frac{ p - p_0 }{ p_r }$



which can be written as:

$\displaystyle\frac{ g^w(T,p) }{ g_r } = \displaystyle\sum_{j=0}^7\displaystyle\sum_{k=0}^6 g_{jk} \tau ^ j \pi ^ k $

where $g_r$ is the value of the specific Gibbs free energy for the reference temperature and pressure.

ID:(12349, 0)



Gibbs free energy density for the salt component

Equation

>Top, >Model


To account for the specific Gibbs free energy of oceanic water, it is necessary to consider the portion of the Gibbs free energy that corresponds to the effect of salinity as a function of temperature and pressure, $g(T,p,i)$. This can be expressed as a polynomial in the reduced temperature $\tau$, reduced pressure $\pi$, and reduced salinity $\xi$, which is calculated as follows:

$\displaystyle\frac{ g^i (T,p,i) }{ g_r } = \displaystyle\sum_{j=0}^7 \displaystyle\sum_{k=0}^6 \left( g_{1jk} \xi^2 \ln \xi +\displaystyle\sum_{i=2}^7 g_{ijk} \xi^i \right) \tau ^ j \pi ^ k $

$g_i$
Energía libre de Gibbs molar de la sal
$J/mol$
9342
$g_r$
Energía libre de Gibbs molar de referencia
$J/mol$
9379
$\pi$
Presión reducida
$-$
9348
$\xi$
Salinidad reducida
$-$
9351
$\tau$
Temperatura reducida
$-$
9346

To account for the specific Gibbs free energy of oceanic water, it is necessary to consider the portion of the specific Gibbs free energy that corresponds to the effect of salinity as a function of temperature and pressure, $g(T,p,i)$. This can be expressed as a polynomial in the reduced temperature $\tau$,

$ \tau = \displaystyle\frac{ T - T_0 }{ T_r }$



reduced pressure $\pi$,

$ \pi = \displaystyle\frac{ p - p_0 }{ p_r }$



and reduced salinity $\xi$,

$ \xi = \sqrt{ \displaystyle\frac{ i }{ i_r } }$



which is calculated as follows:

$\displaystyle\frac{ g^i (T,p,i) }{ g_r } = \displaystyle\sum_{j=0}^7 \displaystyle\sum_{k=0}^6 \left( g_{1jk} \xi^2 \ln \xi +\displaystyle\sum_{i=2}^7 g_{ijk} \xi^i \right) \tau ^ j \pi ^ k $

where $g_r$ is the value of the specific Gibbs free energy for the reference temperature and pressure.

ID:(12353, 0)



Gibbs free energy density of ocean water

Equation

>Top, >Model


The specific Gibbs free energy of the ocean can be calculated as the sum of the specific Gibbs free energy of water $g_w(T,p)$ and the specific Gibbs free energy of salt $g_i(T,p,i)$:

$ g(T,p,i) = g^w(T,p) + g^i(T,p,i) $

$g_i$
Energía libre de Gibbs molar de la sal
$J/mol$
9342
$g_w$
Energía libre de Gibbs molar del agua
$J/mol$
9341
$g$
Energía libre de Gibbs molar del océano
$J/mol$
9340

where the latter depends on the salinity $i$.

ID:(12354, 0)



Properties: density

Equation

>Top, >Model


The derivative of Gibbs free energy $G$ with respect to pressure $p$ is equal to volume $V$. Therefore, dividing Gibbs free energy by mass gives us the specific Gibbs free energy $g$. Similarly, when we perform this operation with volume, we obtain the inverse of density $\rho$. Hence, the relationship between the derivative of Gibbs free energy and density is expressed as follows:

$ \rho =\displaystyle\frac{1}{ g_p }$

$\rho$
Densidad
$kg/m^3$
9371
$g_p$
Primera derivada de la energía libre de Gibbs molar en la presión
$m^3/mol$
9364

The derivative of Gibbs free energy $G$ with respect to pressure $p$ is equal to the volume $V$:

$ DG_{p,T} = V $



If we divide the equation by mass, we obtain the same relationship but with the specific Gibbs free energy $g$ and density $\rho$:

$\displaystyle\frac{\partial g}{\partial p}=\displaystyle\frac{1}{M}\displaystyle\frac{\partial G}{\partial p}=\displaystyle\frac{V}{M}=\displaystyle\frac{1}{\rho}$



In other words,

$ \rho =\displaystyle\frac{1}{ g_p }$

ID:(12355, 0)



Properties: specific entropy

Equation

>Top, >Model


The derivative of Gibbs free energy $G$ with respect to temperature $T$ is equal to minus the entropy $S$. Therefore, if we divide Gibbs free energy $G$ by mass $M$, we obtain the specific Gibbs free energy $g$. Similarly, when we perform this operation with entropy, we obtain the specific entropy $s$. Hence, the relationship between the derivative of Gibbs free energy and entropy is expressed as follows:

$ s = - g_T $

$s$
Entropía molar
$J/mol K$
9375
$g_T$
Primera derivada de la energía libre de Gibbs molar en la temperatura
$J/mol K$
9365

The derivative of Gibbs free energy $G$ with respect to temperature $T$ is equal to minus the entropy $S$:

$ DG_{T,p} =- S $



If we divide it by mass, we obtain the same relationship but with the molar Gibbs free energy $g$ and molar entropy $s$:

$\displaystyle\frac{\partial g}{\partial T}=\displaystyle\frac{1}{M}\displaystyle\frac{\partial G}{\partial T}=\displaystyle\frac{S}{M}=s$



In other words,

$ s = - g_T $

ID:(12358, 0)



Properties: specific enthalpy

Equation

>Top, >Model


The specific enthalpy $h$ can be calculated from the specific Gibbs free energy $g$ and its derivative $g_T$ using:

$ h = g - T g_T $

$g$
Energía libre de Gibbs molar
$J/mol$
9356
$h$
Entalpía molar
$J/mol$
9374
$g_T$
Primera derivada de la energía libre de Gibbs molar en la temperatura
$J/mol K$
9365
$T$
Temperatura
$K$
9343

Since the Gibbs free energy is

$ H = U + p V $



we can rearrange it to obtain the enthalpy using

$ G = H - T S $



which gives us

$H = G + TS = G - T\displaystyle\frac{\partial G}{\partial T}$



If we divide the equation by the mass, we obtain the specific version:

$ h = g - T g_T $

where $T$ is the temperature.

ID:(12359, 0)



Properties: specific internal energy

Equation

>Top, >Model


The specific internal energy $u$ can be calculated from the specific Gibbs free energy $g$ and its derivatives with respect to temperature $g_T$ and pressure $g_p$ using:

$ u = g - T g_T - p g_p $

$u$
Energía interna molar
$J/mol$
9373
$g$
Energía libre de Gibbs molar
$J/mol$
9356
$p$
Presión
$Pa$
9347
$g_p$
Primera derivada de la energía libre de Gibbs molar en la presión
$m^3/mol$
9364
$g_T$
Primera derivada de la energía libre de Gibbs molar en la temperatura
$J/mol K$
9365
$T$
Temperatura
$K$
9343

Given that the Gibbs free energy is



we can solve for the internal energy using

$ G = H - T S $



and

$ G =-\displaystyle\frac{1}{ \beta }\ln Z +\displaystyle\frac{ V }{ \beta }\displaystyle\frac{\partial\ln Z }{\partial V }$



which gives us

$U = G + TS + pV = G - T\displaystyle\frac{\partial G}{\partial T} - p\displaystyle\frac{\partial G}{\partial p}$



If we divide the equation by mass, we obtain the specific version:

$ u = g - T g_T - p g_p $

where $T$ is the temperature and $p$ is the pressure.

ID:(12360, 0)



Properties: specific Helmholtz energy

Equation

>Top, >Model


The specific Helmholtz free energy $f$ can be calculated from the specific Gibbs free energy $g$ and its derivative with respect to pressure $g_p$ using:

$ f = g - p g_p $

$g$
Energía libre de Gibbs molar
$J/mol$
9356
$f$
Energía libre de Helmholtz molar
$J/mol$
9372
$p$
Presión
$Pa$
9347
$g_p$
Primera derivada de la energía libre de Gibbs molar en la presión
$m^3/mol$
9364

Since the Gibbs free energy is



we can solve for the Helmholtz free energy using

$ DG_{p,T} = V $



which gives us

$F = G + pV = G - p\displaystyle\frac{\partial G}{\partial p}$



If we divide the equation by the mass, we obtain the specific version:

$ f = g - p g_p $

where $p$ is the pressure.

ID:(12361, 0)



Properties: isobaric specific heat

Equation

>Top, >Model


The second derivative of the molar Gibbs free energy with respect to temperature allows to calculate the specific heat capacity at constant pressure of ocean water using

$ c_p = - T g_{TT} $

$g_{TT}$
Segunda derivada de la energía libre de Gibbs molar respecto de la temperatura
$J/K$
9368
$c_p$
Specific heat at constant pressure
$J/kg K$
9426
$T$
Temperatura
$K$
9343

With the specific Gibbs energy of ocean water given by:

$ g(T,p,i) = g^w(T,p) + g^i(T,p,i) $



and its corresponding second derivative:

$ g_{xy} =\displaystyle\frac{\partial^2 g }{\partial x \partial y }$



it is possible to estimate the specific heat capacity at constant pressure for a given temperature, pressure, and salinity. It is calculated using the following expression:

$ c_p = - T g_{TT} $

ID:(12362, 0)



Properties: coefficient of thermal expansion

Equation

>Top, >Model


The thermal expansion coefficient is calculated as the derivative of volume with respect to pressure, divided by the volume. Since volume is related to the derivative of Gibbs free energy with respect to pressure, we can show that:

$ k_T = \displaystyle\frac{ g_{Tp} }{ g_p }$

$g_p$
Primera derivada de la energía libre de Gibbs molar en la presión
$m^3/mol$
9364
$g_{Tp}$
Segunda derivada de la energía libre de Gibbs molar respecto de la presión y temperatura
$m^3/Pa$
9370
$k_T$
Thermic dilatation coefficient
$1/K$
9361

Since the coefficient of thermal expansion is defined as

$ c ^2=\left(\displaystyle\frac{ \partial p }{ \partial \rho }\right)_ S $



we can calculate it using the relationship

$ DG_{p,T} = V $



The coefficient of thermal expansion can be calculated as

$k_T=\displaystyle\frac{1}{V}D_T V=\displaystyle\frac{D_{pT} G}{D_p G}$



If we multiply and divide the expression by the mass, we can convert the Gibbs free energy into the specific Gibbs free energy, and the relationship becomes

$ k_T = \displaystyle\frac{ g_{Tp} }{ g_p }$

ID:(12363, 0)



Properties: isothermal compressibility

Equation

>Top, >Model


Since the derivative of the molar Gibbs free energy $g$ with respect to pressure $p$ is equal to the volume $V$, we can show that the isothermal compressibility is given by:

$ k_p =- \displaystyle\frac{ g_{pp} }{ g_p }$

$k_p$
Compresividad isotermica
$1/Pa$
9363
$g_p$
Primera derivada de la energía libre de Gibbs molar en la presión
$m^3/mol$
9364
$g_{pp}$
Segunda derivada de la energía libre de Gibbs molar respecto de la presión
$m^3/Pa$
9367

Since the compressibility coefficient is defined as

$ k_p =-\displaystyle\frac{ DV_{p,T} }{ V }$



we can calculate the compressibility coefficient using the relationship

$ DG_{p,T} = V $



The compressibility coefficient can be calculated as

$k_p=\displaystyle\frac{1}{V}D_p V=\displaystyle\frac{D_{pp} G}{D_p G}$



If we multiply and divide the expression by the mass, we can convert the Gibbs free energy into the specific Gibbs free energy, and the relationship becomes

$ k_p =- \displaystyle\frac{ g_{pp} }{ g_p }$

ID:(12364, 0)



Propiedades: compresibilidad isentropica

Equation

>Top, >Model


Con la energía especifica de Gibbs del agua oceánica con energía libre de Gibbs molar de la sal $J/mol$, energía libre de Gibbs molar del agua $J/mol$ and energía libre de Gibbs molar del océano $J/mol$

$ g(T,p,i) = g^w(T,p) + g^i(T,p,i) $



y con la segunda derivada correspondiente con energía libre de Gibbs molar $J/mol$, primera variable termodinámica $-$, segunda derivada de la energía libre de Gibbs molar $J/mol$ and segunda variable termodinámica $-$

$ g_{xy} =\displaystyle\frac{\partial^2 g }{\partial x \partial y }$



con la primera derivada correspondiente con energía libre de Gibbs molar $J/mol$, primera derivada de la energía libre de Gibbs molar $J/mol$ and primera variable termodinámica $-$

$ g_x =\displaystyle\frac{\partial g }{\partial x }$



se puede estimar la compresibilidad isotermal que existe para una temperatura, presión y salinidad dadas. Con energía libre de Gibbs molar $J/mol$, primera derivada de la energía libre de Gibbs molar $J/mol$ and primera variable termodinámica $-$ se calcula mediante

$ k_t = \displaystyle\frac{ g_{Tp} ^2- g_{TT} g_{pp} }{ g_p g_{TT} }$

ID:(12365, 0)



Properties: haline coefficient of contraction

Equation

>Top, >Model


Given that the derivative of the specific Gibbs free energy, denoted as $g$, with respect to pressure $p$ is equal to the inverse of density $\rho$, we can show that the haline contraction coefficient is equal to:

$ k_i =- \displaystyle\frac{ g_{ip} }{ g_p }$

$k_i$
Coeficiente de contracción halina
$-$
9360
$g_p$
Primera derivada de la energía libre de Gibbs molar en la presión
$m^3/mol$
9364
$g_{ip}$
Segunda derivada de la energía libre de Gibbs molar respecto de la presión y salinidad
$m^3/Pa$
9369

Since the haline contraction coefficient is defined by the equation:

$ k_i = -\displaystyle\frac{1}{ \alpha }\left(\displaystyle\frac{ \partial\alpha }{ \partial i }\right)_{ p , T }$



and considering the relationship:

$ \rho =\displaystyle\frac{1}{ g_p }$



we can calculate the compressibility coefficient using:

$k_i=\displaystyle\frac{1}{\alpha}D_p \alpha=\displaystyle\frac{D_{ip} g}{D_p g}$



Therefore, we obtain:

$ k_i =- \displaystyle\frac{ g_{ip} }{ g_p }$

ID:(12368, 0)



Properties: chemical potential

Equation

>Top, >Model


Con la energía especifica de Gibbs del agua oceánica con energía libre de Gibbs molar de la sal $J/mol$, energía libre de Gibbs molar del agua $J/mol$ and energía libre de Gibbs molar del océano $J/mol$

$ g(T,p,i) = g^w(T,p) + g^i(T,p,i) $



y con las derivadas correspondientes con energía libre de Gibbs molar $J/mol$, primera derivada de la energía libre de Gibbs molar $J/mol$ and primera variable termodinámica $-$

$ g_x =\displaystyle\frac{\partial g }{\partial x }$



se puede estimar el potencial químico que existe para una temperatura, presión y salinidad dadas. Con energía libre de Gibbs molar $J/mol$, primera derivada de la energía libre de Gibbs molar $J/mol$ and primera variable termodinámica $-$ se calcula mediante

$ \mu = g + (1- i ) g_i $

ID:(12369, 0)



Propiedades: velocidad del sonido

Equation

>Top, >Model


Con la energía especifica de Gibbs del agua oceánica con energía libre de Gibbs molar de la sal $J/mol$, energía libre de Gibbs molar del agua $J/mol$ and energía libre de Gibbs molar del océano $J/mol$

$ g(T,p,i) = g^w(T,p) + g^i(T,p,i) $



y con la segunda derivada correspondiente con energía libre de Gibbs molar $J/mol$, primera variable termodinámica $-$, segunda derivada de la energía libre de Gibbs molar $J/mol$ and segunda variable termodinámica $-$

$ g_{xy} =\displaystyle\frac{\partial^2 g }{\partial x \partial y }$



con la primera derivada correspondiente con energía libre de Gibbs molar $J/mol$, primera derivada de la energía libre de Gibbs molar $J/mol$ and primera variable termodinámica $-$

$ g_x =\displaystyle\frac{\partial g }{\partial x }$



se puede estimar la compresibilidad isotermal que existe para una temperatura, presión y salinidad dadas. Con energía libre de Gibbs molar $J/mol$, primera derivada de la energía libre de Gibbs molar $J/mol$ and primera variable termodinámica $-$ se calcula mediante

$ c ^2= \displaystyle\frac{ g_p ^2 g_{TT} }{ g_{Tp} ^2- g_{TT} g_{pp} }$

ID:(12366, 0)



Propiedades: tasa de lapso adiabático

Equation

>Top, >Model


Con la energía especifica de Gibbs del agua oceánica con energía libre de Gibbs molar de la sal $J/mol$, energía libre de Gibbs molar del agua $J/mol$ and energía libre de Gibbs molar del océano $J/mol$

$ g(T,p,i) = g^w(T,p) + g^i(T,p,i) $



y con la primera derivada correspondiente con energía libre de Gibbs molar $J/mol$, primera derivada de la energía libre de Gibbs molar $J/mol$ and primera variable termodinámica $-$

$ g_x =\displaystyle\frac{\partial g }{\partial x }$



se puede estimar la compresibilidad isotermal que existe para una temperatura, presión y salinidad dadas. Con energía libre de Gibbs molar $J/mol$, primera derivada de la energía libre de Gibbs molar $J/mol$ and primera variable termodinámica $-$ se calcula mediante

$ \Gamma =- \displaystyle\frac{ g_{Tp} }{ g_{TT} }$

ID:(12367, 0)