Entalpía

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Se obtienen mediante la función partición las distintas funciones y relaciones termodinámicas.

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Enthalpy and partition function

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The enthalpy can be calculated from the partition function if it is remembered that this is equal to the internal energy and the pressure times the volume:

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Enthalpy $H(S,p)$

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If we need to take into account the energy required to form the system in addition to the internal energy, we must consider the enthalpy ($H$).

the enthalpy ($H$) [1] is defined as the sum of the internal energy ($U$) and the formation energy. The latter corresponds to the work done in the formation, which is equal to $pV$ with the pressure ($p$) and the volume ($V$).

Therefore, we have:

$ H = U + p V $

$H$
Enthalpy
$J$
$U$
Internal energy
$J$
$p$
Pressure
$Pa$
$V$
Volume
$m^3$



the enthalpy ($H$) is a function of the entropy ($S$) and the pressure ($p$).

An article that can be considered as the origin of the concept, although it does not include the definition of the name, is:

[1] "Memoir on the Motive Power of Heat, Especially as Regards Steam, and on the Mechanical Equivalent of Heat," written by Benoît Paul Émile Clapeyron (1834).

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Differential Enthalpy Relationship

Equation

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Since the enthalpy ($H$) is a function of the internal energy ($U$), the pressure ($p$), and the volume ($V$) according to the equation:

$ H = U + p V $



and this equation depends solely on the entropy ($S$) and the pressure ($p$), we can show that its partial derivative with respect to the differential enthalpy ($dH$) is equal to:

$ dH = T dS + V dp $

$T$
Absolute temperature
$K$
$dH$
Differential enthalpy
$J$
$dS$
Entropy variation
$J/K$
$dp$
Pressure Variation
$Pa$
$V$
Volume
$m^3$

If we differentiate the enthalpy function:

$ H = U + p V $



we obtain:

$dH = dU + Vdp + pdV$



With the differential of the internal energy:

$ dU = T dS - p dV $



we can conclude:

$ dH = T dS + V dp $



where the entropy variation ($dS$), the pressure Variation ($dp$), and the absolute temperature ($T$) are also considered.

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Calculo de la derivada parcial de la entalpia en la entropia a presión constante

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La derivada de la entalpia en la entropia a presión constante es

$ DH_{S,p} =\left(\displaystyle\frac{\partial H }{\partial S }\right)_ p $

$H$
Enthalpy
$J$
$S$
Entropy
$J/K$
$DH_{S,p}$
Partial derivative of enthalpy with respect to entropy at constant pressure
$K$
$p$
Pressure
$Pa$

ID:(12028, 0)



Calculo de la derivada parcial de la entalpia en la presión a entropía constante

Equation

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La derivada de la entalpia en la presión a entropia constante es

$ DH_{p,S} =\left(\displaystyle\frac{\partial H }{\partial p }\right)_ S $

$H$
Enthalpy
$J$
$S$
Entropy
$J/K$
$DH_{p,S}$
Partial derivative of enthalpy with respect to pressure at constant entropy
$m^3$
$p$
Pressure
$Pa$

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Enthalpy differential

Equation

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Given that the enthalpy ($H$) is a function of the entropy ($S$) and the pressure ($p$), we can express the differential enthalpy ($dH$) as follows:

$dH=\left(\displaystyle\frac{\partial H}{\partial S}\right)_pdS+\left(\displaystyle\frac{\partial H}{\partial p}\right)_Sdp$



This allows us to define the differential enthalpy ($dH$) in terms of the slopes the partial derivative of enthalpy with respect to entropy at constant pressure ($DH_{S,p}$) and the partial derivative of enthalpy with respect to pressure at constant entropy ($DH_{p,S}$):

$ dH = DH_{S,p} dS + DH_{p,S} dp $

$dH$
Differential enthalpy
$J$
$dS$
Entropy variation
$J/K$
$DH_{S,p}$
Partial derivative of enthalpy with respect to entropy at constant pressure
$K$
$DH_{p,S}$
Partial derivative of enthalpy with respect to pressure at constant entropy
$m^3$
$dp$
Pressure Variation
$Pa$

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Enthalpy and equation of state at constant pressure

Equation

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The differential enthalpy ($dH$) is a function of the variations in the entropy ($S$) and the pressure ($p$), as well as the slopes the partial derivative of enthalpy with respect to entropy at constant pressure ($DH_{S,p}$) and the partial derivative of enthalpy with respect to pressure at constant entropy ($DH_{p,S}$), which is expressed as:

$ dH = DH_{S,p} dS + DH_{p,S} dp $



Comparing this with the first law of thermodynamics, it turns out that the partial derivative of enthalpy with respect to entropy at constant pressure ($DH_{S,p}$) is equal to minus the volume ($V$):

$ DH_{S,p} = T $

$T$
Absolute temperature
$K$
$DH_{S,p}$
Partial derivative of enthalpy with respect to entropy at constant pressure
$K$

ID:(3548, 0)



Enthalpy and equation of state at constant entropy

Equation

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The differential enthalpy ($dH$) is a function of the variations in the entropy ($S$) and the pressure ($p$), as well as the slopes the partial derivative of enthalpy with respect to entropy at constant pressure ($DH_{S,p}$) and the partial derivative of enthalpy with respect to pressure at constant entropy ($DH_{p,S}$), which is expressed as:

$ dH = DH_{S,p} dS + DH_{p,S} dp $



Comparing this with the first law of thermodynamics, it turns out that the partial derivative of enthalpy with respect to pressure at constant entropy ($DH_{p,S}$) is equal to minus the absolute temperature ($T$):

$ DH_{p,S} = V $

$DH_{p,S}$
Partial derivative of enthalpy with respect to pressure at constant entropy
$m^3$
$V$
Volume
$m^3$

ID:(3538, 0)



Enthalpy and partition function

Equation

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La entalpía se logra calcular de la función partición si se recuerda que esta es igual a la energía interna y a la presión por el volumen que con enthalpy $J$, internal energy $J$, pressure $Pa$ and volume $m^3$ es:

$ H = U + p V $



Como la energía interna es con igual a

$U=-\displaystyle\frac{\partial\ln Z}{\partial\beta}$



y con la presión es

$\bar{p}=\displaystyle\frac{1}{\beta}\displaystyle\frac{\partial\ln Z}{\partial V}$



se tiene que con es

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

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Enthalpy and its relation of Maxwell

Equation

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Since the enthalpy ($H$) is an exact differential, it means that you can first vary the entropy ($S$) and then the pressure ($p$), or in the reverse order, and the result will be the same. This can be expressed by taking derivatives of slopes in different orders, and there will be no difference:

$D(DH_{S,p})_{p,S}=D(DH_{p,S})_{S,p}$



If you replace the differential with the corresponding variable, you obtain the relationship involving the absolute temperature ($T$) and the volume ($V$):

$ DT_{p,S} = DV_{S,p} $

$DT_{p,S}$
Partial derivative of temperature with respect to pressure at constant entropy
$K/Pa$
$DV_{S,p}$
Partial derivative of volume with respect to entropy at constant pressure
$K/Pa$

Since the differential enthalpy ($dH$) is an exact differential, we should note that the enthalpy ($H$) with respect to the entropy ($S$) and the pressure ($p$) must be independent of the order in which the function is derived:

$D(DH_{S,p}){p,S}=D(DH{p,S})_{S,p}$



Using the relationship between the slope the partial derivative of enthalpy with respect to entropy at constant pressure ($DH_{S,p}$) and the absolute temperature ($T$)

$ DH_{S,p} = T $

,

and the relationship between the slope the partial derivative of enthalpy with respect to pressure at constant entropy ($DH_{p,S}$) and the volume ($V$)

$ DH_{p,S} = V $

,

we can conclude that:

$ DT_{p,S} = DV_{S,p} $

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Calculo de la derivada parcial de la temperatura en la presión a entropía constante

Equation

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La derivada de la temperatura en la presión a entropia constante es

$ DT_{p,S} =\left(\displaystyle\frac{\partial T }{\partial p }\right)_ S $

$T$
Absolute temperature
$K$
$S$
Entropy
$J/K$
$DT_{p,S}$
Partial derivative of temperature with respect to pressure at constant entropy
$K/Pa$
$p$
Pressure
$Pa$

ID:(12030, 0)



Calculo de la derivada parcial del volumen en la entropia a presión constante

Equation

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La derivada el volumen en la entropia a presión constante es

$ DV_{S,p} =\left(\displaystyle\frac{\partial V }{\partial S }\right)_ p $

$S$
Entropy
$J/K$
$DV_{S,p}$
Partial derivative of volume with respect to entropy at constant pressure
$K/Pa$
$p$
Pressure
$Pa$
$V$
Volume
$m^3$

ID:(12029, 0)



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Video: Enthalpy