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Thermodynamic Functions

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ID:(172, 0)



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 :

ID:(11723, 0)



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
dU = T * dS - p * dV U = T * S - p * V U =-d ln Z /d beta DU_VS =- p DU_SV = T DT_VS=- Dp_SV dU = DU_SV * dS + DU_VS * dV DU_VS = dU / dV DU_SV = dU / dS DT_VS = dT / dV Dp_SV = dp / dS TbetaSdSZUUdUDU_SVDU_VSDp_SVDT_VSpdUVdV

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
dU = T * dS - p * dV U = T * S - p * V U =-d ln Z /d beta DU_VS =- p DU_SV = T DT_VS=- Dp_SV dU = DU_SV * dS + DU_VS * dV DU_VS = dU / dV DU_SV = dU / dS DT_VS = dT / dV Dp_SV = dp / dS TbetaSdSZUUdUDU_SVDU_VSDp_SVDT_VSpdUVdV

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

.

ID:(3471, 0)



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
dU = T * dS - p * dV U = T * S - p * V U =-d ln Z /d beta DU_VS =- p DU_SV = T DT_VS=- Dp_SV dU = DU_SV * dS + DU_VS * dV DU_VS = dU / dV DU_SV = dU / dS DT_VS = dT / dV Dp_SV = dp / dS TbetaSdSZUUdUDU_SVDU_VSDp_SVDT_VSpdUVdV

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

ID:(8185, 0)



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
dU = T * dS - p * dV U = T * S - p * V U =-d ln Z /d beta DU_VS =- p DU_SV = T DT_VS=- Dp_SV dU = DU_SV * dS + DU_VS * dV DU_VS = dU / dV DU_SV = dU / dS DT_VS = dT / dV Dp_SV = dp / dS TbetaSdSZUUdUDU_SVDU_VSDp_SVDT_VSpdUVdV

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
dU = T * dS - p * dV U = T * S - p * V U =-d ln Z /d beta DU_VS =- p DU_SV = T DT_VS=- Dp_SV dU = DU_SV * dS + DU_VS * dV DU_VS = dU / dV DU_SV = dU / dS DT_VS = dT / dV Dp_SV = dp / dS TbetaSdSZUUdUDU_SVDU_VSDp_SVDT_VSpdUVdV

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
dU = T * dS - p * dV U = T * S - p * V U =-d ln Z /d beta DU_VS =- p DU_SV = T DT_VS=- Dp_SV dU = DU_SV * dS + DU_VS * dV DU_VS = dU / dV DU_SV = dU / dS DT_VS = dT / dV Dp_SV = dp / dS TbetaSdSZUUdUDU_SVDU_VSDp_SVDT_VSpdUVdV

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
dU = T * dS - p * dV U = T * S - p * V U =-d ln Z /d beta DU_VS =- p DU_SV = T DT_VS=- Dp_SV dU = DU_SV * dS + DU_VS * dV DU_VS = dU / dV DU_SV = dU / dS DT_VS = dT / dV Dp_SV = dp / dS TbetaSdSZUUdUDU_SVDU_VSDp_SVDT_VSpdUVdV

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}

ID:(3534, 0)



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
dU = T * dS - p * dV U = T * S - p * V U =-d ln Z /d beta DU_VS =- p DU_SV = T DT_VS=- Dp_SV dU = DU_SV * dS + DU_VS * dV DU_VS = dU / dV DU_SV = dU / dS DT_VS = dT / dV Dp_SV = dp / dS TbetaSdSZUUdUDU_SVDU_VSDp_SVDT_VSpdUVdV

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
dU = T * dS - p * dV U = T * S - p * V U =-d ln Z /d beta DU_VS =- p DU_SV = T DT_VS=- Dp_SV dU = DU_SV * dS + DU_VS * dV DU_VS = dU / dV DU_SV = dU / dS DT_VS = dT / dV Dp_SV = dp / dS TbetaSdSZUUdUDU_SVDU_VSDp_SVDT_VSpdUVdV

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
dU = T * dS - p * dV U = T * S - p * V U =-d ln Z /d beta DU_VS =- p DU_SV = T DT_VS=- Dp_SV dU = DU_SV * dS + DU_VS * dV DU_VS = dU / dV DU_SV = dU / dS DT_VS = dT / dV Dp_SV = dp / dS TbetaSdSZUUdUDU_SVDU_VSDp_SVDT_VSpdUVdV

ID:(12026, 0)



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