Thermodynamic Functions

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Internal energy and partition function

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The internal energy is allowed to calculate from the partition function as the derivative with respect to \ beta :

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Internal Energy

Equation

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The internal energy ($U$) is with the absolute temperature ($T$), the pressure ($p$), the entropy ($S$) and the volume ($V$) equal to:

$ U = T S - p V $

$T$
Absolute temperature
$K$
5177
$S$
Entropy
$J/K$
5227
$U$
Internal energy
$J$
5228
$p$
Pressure
$Pa$
5224
$V$
Volume
$m^3$
5226

If the absolute temperature ($T$) and the pressure ($p$) are kept constant, the variation of the internal energy ($dU$), which depends on the entropy variation ($dS$) and the volume Variation ($dV$), is expressed as:

$ dU = T dS - p dV $



Integrating this results in the following expression in terms of the internal energy ($U$), the entropy ($S$), and the volume ($V$):

$ U = T S - p V $

ID:(3472, 0)



Internal Energy: differential ratio

Equation

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The dependency of the internal energy differential ($dU$) on the pressure ($p$) and the volume Variation ($dV$), in addition to the absolute temperature ($T$) and the entropy variation ($dS$), is given by:

$ dU = T dS - p dV $

$T$
Absolute temperature
$K$
5177
$dS$
Entropy variation
$J/K$
5225
$p$
Pressure
$Pa$
5224
$dU$
Variation of the internal energy
$J$
5400
$dV$
Volume Variation
$m^3$
5223

As the internal energy differential ($dU$) depends on the differential inexact Heat ($\delta Q$), the pressure ($p$), and the volume Variation ($dV$) according to the equation:

$ dU = \delta Q - p dV $



and the expression for the second law of thermodynamics with the absolute temperature ($T$) and the entropy variation ($dS$) as:

$ \delta Q = T dS $



we can conclude that:

$ dU = T dS - p dV $

.

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Differential of Internal Energy

Equation

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Given that the internal energy ($U$) is a function of the entropy ($S$) and the volume ($V$), the internal energy differential ($dU$) can be expressed as follows:

$ dU = DU_{S,V} dS + DU_{V,S} dV $

$dS$
Entropy variation
$J/K$
5225
$dU$
Internal energy differential
$J$
8736
$DU_{S,V}$
Partial derivative of internal energy with respect to entropy at constant volume
$K$
8735
$DU_{V,s}$
Partial derivative of internal energy with respect to volume at constant entropy
$Pa$
8734
$dV$
Volume Variation
$m^3$
5223

Given that the internal energy ($U$) depends on the entropy ($S$) and the volume ($V$), the internal energy differential ($dU$) can be calculated as follows:

$dU = \left(\displaystyle\frac{\partial U}{\partial S}\right)_V dS + \left(\displaystyle\frac{\partial U}{\partial V}\right)_S dV$



To simplify the notation of this expression, we introduce the derivative of the internal energy ($U$) with respect to the entropy ($S$) while keeping the volume ($V$) constant as:

$DU_{S,V} \equiv \left(\displaystyle\frac{\partial U}{\partial S}\right)_V$



and the derivative of the internal energy ($U$) with respect to the volume ($V$) while keeping the entropy ($S$) constant as:

$DU_{V,S} \equiv \left(\displaystyle\frac{\partial U}{\partial V}\right)_S$



therefore, we can write:

$ dU = DU_{S,V} dS + DU_{V,S} dV $

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Calculo de la derivada parcial de la energía interna en el volumen a entropía constante

Equation

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La derivada de la energía interna en el volumen a entropia constante es

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

$S$
Entropy
$J/K$
5227
$U$
Internal energy
$J$
5228
$DU_{V,s}$
Partial derivative of internal energy with respect to volume at constant entropy
$Pa$
8734
$V$
Volume
$m^3$
5226

ID:(12023, 0)



Calculo de la derivada parcial de la energía interna en la entropia a volumen constante

Equation

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La derivada de la energía interna en la entropia a volumen constante es

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

$S$
Entropy
$J/K$
5227
$U$
Internal energy
$J$
5228
$DU_{S,V}$
Partial derivative of internal energy with respect to entropy at constant volume
$K$
8735
$V$
Volume
$m^3$
5226

ID:(12024, 0)



Internal energy and equation of state at constant entropy

Equation

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

$ DU_{V,S} =- p $

$DU_{V,s}$
Partial derivative of internal energy with respect to volume at constant entropy
$Pa$
8734
$p$
Pressure
$Pa$
5224

The internal energy differential ($dU$) is a function of the variations in the entropy ($S$) and the volume ($V$), as well as the slopes the partial derivative of internal energy with respect to volume at constant entropy ($DU_{V,S}$) and the partial derivative of internal energy with respect to entropy at constant volume ($DU_{S,V}$), which is expressed as:

$ dU = DU_{S,V} dS + DU_{V,S} dV $



When compared with the equation for the internal energy differential ($dU$):

$ dU = T dS - p dV $



it results in the slope of the internal energy ($U$) with respect to the variation in the volume ($V$):

$ DU_{V,S} =- p $

ID:(3535, 0)



Internal energy and equation of state at constant volume

Equation

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Comparing this with the first law of thermodynamics, it turns out that the partial derivative of internal energy with respect to entropy at constant volume ($DU_{S,V}$) is equal to the absolute temperature ($T$):

$ DU_{S,V} = T $

$T$
Absolute temperature
$K$
5177
$DU_{S,V}$
Partial derivative of internal energy with respect to entropy at constant volume
$K$
8735

The internal energy differential ($dU$) is a function of the variations in the entropy ($S$) and the volume ($V$), as well as the slopes the partial derivative of internal energy with respect to volume at constant entropy ($DU_{V,S}$) and the partial derivative of internal energy with respect to entropy at constant volume ($DU_{S,V}$), which is expressed as:

$ dU = DU_{S,V} dS + DU_{V,S} dV $



When compared with the equation for the internal energy differential ($dU$):

$ dU = T dS - p dV $



it results in the slope of the internal energy ($U$) with respect to the variation in the entropy ($S$):

$ DU_{S,V} = T $

ID:(3546, 0)



Internal energy and partition function

Equation

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La energía interna es igual a la energía media calculada con

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



se tiene que con la energía interna es

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

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Internal energy and its relation of Maxwell

Equation

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With the entropy ($S$), the volume ($V$), the absolute temperature ($T$) and the pressure ($p$) we obtain one of the so-called Maxwell relations:

$ DT_{V,S} =- Dp_{S,V} $

$Dp_{S,V}$
Partial derivative of pressure with respect to entropy at constant volume
$K/m^3$
8739
$DT_{V,S}$
Partial derivative of temperature with respect to volume at constant entropy
$K/m^3$
8738

Since the internal energy differential ($dU$) is an exact differential, we should note that the internal energy ($U$) with respect to the entropy ($S$) and the volume ($V$) must be independent of the order in which the function is derived:

$D(DU_{S,V}){V,S}=D(DU{V,S})_{S,V}$



Using the relationship between the slope the partial derivative of internal energy with respect to entropy at constant volume ($DU_{S,V}$) and the absolute temperature ($T$)

$ DU_{S,V} = T $

,

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

$ DU_{V,S} =- p $

,

we can conclude that:

$ DT_{V,S} =- Dp_{S,V} $

ID:(3556, 0)



Calculo de la derivada parcial de la temperatura en el volumen a entropía constante

Equation

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La derivada de la temperatura en el volumen a entropia constante es

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

$T$
Absolute temperature
$K$
5177
$S$
Entropy
$J/K$
5227
$DT_{V,S}$
Partial derivative of temperature with respect to volume at constant entropy
$K/m^3$
8738
$V$
Volume
$m^3$
5226

ID:(12025, 0)



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

Equation

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

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

$S$
Entropy
$J/K$
5227
$Dp_{S,V}$
Partial derivative of pressure with respect to entropy at constant volume
$K/m^3$
8739
$p$
Pressure
$Pa$
5224
$V$
Volume
$m^3$
5226

ID:(12026, 0)



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Video: Internal energy