Storyboard

Normally the pressure of the real water vapor is lower than the maximum it can withstand before starting to condense it. Therefore, the concept of relative humidity is introduced.

>Model

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Concept

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Concept

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The gaseous phase, which in our case corresponds to water vapor, is the phase in which atoms can move relatively freely.

In this phase, there is only minimal interaction that can affect the behavior of atoms without significantly confining them.

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Concept

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The liquid phase, which in our case corresponds to water, is the phase in which atoms can move relatively freely while maintaining their unity and adapting to the shape that contains them.

In this phase, no specific structure is observed.

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Concept

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The solid phase, which in our case corresponds to ice, is the phase in which atoms cannot move relatively and can only oscillate around their equilibrium point.

In this phase, one can observe a structure that is often crystalline and, therefore, regular.

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Concept

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One of the most relevant phase diagrams for our planet is that of water. This diagram exhibits the three classical phases: solid, liquid, and gas, along with several phases featuring different crystalline structures of ice.

The significant distinction compared to other materials is that within a pressure range spanning from 611 Pa to 209.9 MPa, the solid phase occupies a greater volume than the liquid phase. This characteristic is reflected in the phase diagram as a negative slope along the boundary line separating the solid phase (hexagonal ice) and the liquid phase (water).

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Concept

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The measurement of the heat of vaporization is carried out by heating a sample, causing it to evaporate, while simultaneously measuring the heat delivered to the sample. Then, the vapor is cooled and condensed back, and the mass that originally evaporated is measured.

In this way, we can estimate the energy required to evaporate a given mass, which corresponds to latent Heat ($L$) measured in joules per kilogram (J/kg) or joules per mole (J/mol).

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Concept

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The gaseous phase of water corresponds to what is known as water vapor. It is created as water molecules acquire enough kinetic energy to escape from the liquid phase and begin to move through the space above the liquid. Periodically, the molecules in the gaseous state collide with the liquid surface again and are captured, returning to the liquid state.

As the number of molecules in the gaseous state increases, so does the number that returns to the liquid. This process continues until an equilibrium is reached between the molecules leaving the liquid and those being reabsorbed. In this situation, it is said that the space above the liquid is saturated.

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Concept

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Selected parameter

Calculations

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Equation

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The first law of thermodynamics states that energy is conserved, meaning that the internal energy differential ($dU$) is always equal to the differential inexact Heat ($\delta Q$) supplied to the system (positive) minus the differential inexact labour ($\delta W$) done by the system (negative).

Therefore, we have:

$ dU = \delta Q - \delta W $

Conditions

Variables

$\delta W$

Differential inexact labour

$J$

$\delta Q$

Variation of heat

$J$

$dU$

Variation of the Internal Energy

$J$

Relation

While the exact differential does not depend on how the variation is executed, the inexact differential does. When referring to a differential without specifying that it is inexact, it is assumed to be exact.

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Equation

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In analogy to the definition of work $dW$ in mechanics:

which is defined in terms of force $F$ and displacement $dx$, in thermodynamics, we work with the expression of work in terms of pressure $p$ and volume change $dV$:

$ \delta W = p dV $

Conditions

Variables

Relation

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Equation

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The second law of thermodynamics states that it is not possible to fully convert the energy the internal energy differential ($dU$) into useful work the differential inexact labour ($\delta W$). The difference between these quantities is associated with the unutilized energy the differential inexact Heat ($\delta Q$), which corresponds to the heat generated or absorbed in the process the absolute temperature ($T$).

In the case of the differential inexact labour ($\delta W$), there exists a relationship between the intensive variable the pressure ($p$) and the extensive variable the volume ($V$), expressed as:

An intensive variable is characterized by defining the state of the system and not depending on its size. In this sense, the pressure ($p$) is an intensive variable, as it describes the state of a system independently of its size. On the other hand, an extensive variable, such as the volume ($V$), increases with the size of the system.

In the case of the differential inexact Heat ($\delta Q$), an additional extensive variable, denoted as the entropy ($S$), is needed to complement the intensive variable the absolute temperature ($T$) in order to define the relationship as follows:

$ \delta Q = T dS $

Conditions

Variables

$T$

Absolute temperature

$K$

$dS$

Entropy variation

$J/K$

$\delta Q$

Variation of heat

$J$

Relation

This new variable, which we shall call the entropy ($S$), is presented here in its differential form (the entropy variation ($dS$)) and models the effect that not all of the energy the internal energy differential ($dU$) can be entirely converted into useful work the differential inexact labour ($\delta W$).

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Equation

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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:

we can replace the differential inexact Heat ($\delta Q$) with the expression from the second law of thermodynamics in terms of the absolute temperature ($T$) and the entropy variation ($dS$), resulting in the expression for the internal energy differential ($dU$):

$ dU = T dS - p dV $

Conditions

Variables

$T$

Absolute temperature

$K$

$dS$

Entropy variation

$J/K$

$p$

Pressure

$Pa$

$dU$

Variation of the Internal Energy

$J$

$dV$

Volume Variation

$m^3$

Relation

Development

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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Equation

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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 $

Conditions

Variables

$H$

Enthalpy

$J$

$U$

Internal energy

$J$

$p$

Pressure

$Pa$

$V$

Volume

$m^3$

Relation

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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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:

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 $

Conditions

Variables

$T$

Absolute temperature

$K$

$dH$

Differential enthalpy

$J$

$dS$

Entropy variation

$J/K$

$dp$

Pressure Variation

$Pa$

$V$

Volume

$m^3$

Relation

Development

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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Equation

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As the gibbs free energy ($G$) [1,2] depends on the enthalpy ($H$), the entropy ($S$), and the absolute temperature ($T$):

The dependence on the gibbs free energy ($G$) with respect to the pressure ($p$) is obtained, and from the absolute temperature ($T$), we obtain the differential:

$ dG =- S dT + V dp $

Conditions

Variables

$S$

Entropy

$J/K$

$dp$

Pressure Variation

$Pa$

$dT$

Temperature variation

$K$

$dG$

Variation of Gibbs Free Energy

$J$

$V$

Volume

$m^3$

Relation

Development

The gibbs free energy ($G$) as a function of the enthalpy ($H$), the entropy ($S$), and the absolute temperature ($T$) is expressed as:

$ G = H - T S $

The value of the differential of the Gibbs free energy ($dG$) is determined using the differential enthalpy ($dH$), the temperature variation ($dT$), and the entropy variation ($dS$) through the equation:

$dG=dH-SdT-TdS$

Since the differential enthalpy ($dH$) is related to the volume ($V$) and the pressure Variation ($dp$) as follows:

$ dH = T dS + V dp $

It follows that the differential enthalpy ($dH$), the entropy variation ($dS$), and the pressure Variation ($dp$) are interconnected in the following manner:

$ dG =- S dT + V dp $

[1] "On the Equilibrium of Heterogeneous Substances," J. Willard Gibbs, Transactions of the Connecticut Academy of Arts and Sciences, 3: 108-248 (October 1875 May 1876)

[2] "On the Equilibrium of Heterogeneous Substances," J. Willard Gibbs, Transactions of the Connecticut Academy of Arts and Sciences, 3: 343-524 (May 1877 July 1878)

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Equation

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The differential of the Gibbs free energy ($dG$) can be calculated using the pressure Variation ($dp$) and the temperature variation ($dT$) in conjunction with the entropy ($S$) and the volume ($V$):

This equation remains constant along the lines separating phases during a phase change. This allows us to demonstrate the Clausius-Clapeyron Law with the assistance of the latent Heat ($L$), the absolute temperature ($T$), and the volume variation in phase change ($\Delta V$) [1,2,3]:

$\displaystyle\frac{ dp }{ dT }=\displaystyle\frac{ L }{ \Delta V T }$

Conditions

Variables

$T$

Absolute temperature

$K$

$L$

Latent Heat

$J/kg$

$dp$

Pressure Variation

$Pa$

$dT$

Temperature variation

$K$

$\Delta V$

Volume variation in phase change

$m^3$

Relation

Development

If the differential of the Gibbs free energy ($dG$) is constant, it means that for the pressure Variation ($dp$) and the temperature variation ($dT$), the values of the entropy ($S$) and the volume ($V$) in phase 1

$dG = -S_1dT+V_1dp$

and the entropy ($S$) and the volume ($V$) in phase 2

$dG = -S_2dT+V_2dp$

yield

$\displaystyle\frac{dp}{dT}=\displaystyle\frac{S_2-S_1}{V_2-V_1}$

The change in the entropy ($S$) between both phases corresponds to the latent Heat ($L$) divided by the absolute temperature ($T$):

$S_2 - S_1 =\displaystyle\frac{ L }{ T }$

So, with the definition of the volume variation in phase change ($\Delta V$)

$\Delta V \equiv V_2 - V_1$

we obtain

$\displaystyle\frac{ dp }{ dT }=\displaystyle\frac{ L }{ \Delta V T }$

[1] "Über die Art der Bewegung, welche wir Wärme nennen." (On the Nature of Motion We Call Heat), Rudolf Clausius, Annalen der Physik und Chemie, 155(6), 368-397. (1850)

[2] "Ueber eine veränderte Form des zweiten Hauptsatzes der mechanischen Wärmetheorie" (On a Modified Form of the Second Law of the Mechanical Theory of Heat), Rudolf Clausius, Annalen der Physik, 176(3), 353-400. (1857)

[3] "Mémoire sur la puissance motrice de la chaleur." (Memoir on the Motive Power of Heat), Benoît Paul Émile Clapeyron, Journal de l'École Royale Polytechnique, 14, 153-190. (1834)

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Equation

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In many cases the latent molar heat is not available but the latent heat that is expressed, for example, in Joules per kilogram (J/Kg). Since the vapor pressure equation works with the latent molar heat we must convert the latent heat into latent molar heat. Since the latter is per mole, it is enough to divide the latent heat L by the molar mass M_m to obtain the latent molar heat

$ l_m \equiv\displaystyle\frac{ L }{ M_m }$

Conditions

Variables

$L$

Latent Heat

$J/kg$

$l_m$

Molar Latent Heat

$J/kg$

$M_m$

Molar Mass

$kg/mol$

Relation

In the case of water, the latent heat of evaporation is of the order of L\ sim 2,256\ times 10^6J/kg while the molar mass is M_m\sim 18,g/cm^3 = 0.018,kg/m^3, resulting in a latent molar heat of L_m\sim 40600,J/mol = 40.6,kJ/mol.

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Equation

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The Clausius-Clapeyron equation [1,2,3], which depends on the pressure Variation ($dp$), the temperature variation ($dT$), the latent Heat ($L$), the volume variation in phase change ($\Delta V$), and the absolute temperature ($T$), is expressed as:

This equation can be rewritten in terms of the molar Latent Heat ($l_m$) and the variation of molar volume during phase change ($\Delta v_m$) as:

$\displaystyle\frac{ dp }{ dT }=\displaystyle\frac{ l_m }{ \Delta v_m T }$

Conditions

Variables

$T$

Absolute temperature

$K$

$l_m$

Molar Latent Heat

$J/kg$

$p$

Pressure

$Pa$

$\Delta v_m$

Variation of molar volume during phase change

$m^3/mol$

Relation

Development

With the Clausius-Clapeyron law, which depends on the pressure Variation ($dp$), the temperature variation ($dT$), the latent Heat ($L$), the volume variation in phase change ($\Delta V$), and the absolute temperature ($T$), expressed as:

$\displaystyle\frac{ dp }{ dT }=\displaystyle\frac{ L }{ \Delta V T }$

and the definition of the molar Latent Heat ($l_m$), where the latent Heat ($L$) is related to the molar Mass ($M_m$) as follows:

$ l_m \equiv\displaystyle\frac{ L }{ M_m }$

and the variation of molar volume during phase change ($\Delta v_m$), where the volume variation in phase change ($\Delta V$) is related to the molar Mass ($M_m$) as follows:

$\Delta v_m =\displaystyle\frac{ \Delta V }{ M_m }$

we obtain:

$\displaystyle\frac{ dp }{ dT }=\displaystyle\frac{ l_m }{ \Delta v_m T }$

[1] "Über die Art der Bewegung, welche wir Wärme nennen" (On the Nature of Motion Which We Call Heat), Rudolf Clausius, Annalen der Physik und Chemie, 155(6), 368-397. (1850)

[2] "Über eine veränderte Form des zweiten Hauptsatzes der mechanischen Wärmetheorie" (On a Modified Form of the Second Fundamental Theorem in the Mechanical Theory of Heat), Rudolf Clausius, Annalen der Physik, 176(3), 353-400. (1857)

[3] "Mémoire sur la puissance motrice de la chaleur" (Memoir on the Motive Power of Heat), Benoît Paul Émile Clapeyron, Journal de l'École Royale Polytechnique, 14, 153-190. (1834)

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Equation

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When undergoing a phase change from a liquid to a gas, the change in volume can be expressed as:

$\Delta V = V_{\text{gas}} - V_{\text{liquid}}$

Since the volume of the gas is much greater than that of the liquid ( ($$)), we can approximate it as:

$\Delta V \approx V_{\text{gas}}$

As water vapor exhibits behavior similar to that of an ideal gas, we can state that with values the universal gas constant ($R$), the number of moles ($n$), the absolute temperature ($T$), and the water vapor pressure unsaturated ($p_v$):

$ \Delta V =\displaystyle\frac{ n R T }{ p_v }$

Conditions

Variables

$T$

Absolute temperature

$K$

$\Delta V$

Amount of Water Vapor

$m^3$

$n$

Número de Moles

$-$

$R$

Universal gas constant

8.4135

$J/mol K$

$p_v$

Water vapor pressure unsaturated

$Pa$

Relation

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Equation

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Using the Clausius-Clapeyron equation for the gradient of the pressure ($p$) with respect to the absolute temperature ($T$), which depends on the molar Latent Heat ($l_m$) and the variation of molar volume during phase change ($\Delta v_m$):

and the ideal gas law, we can calculate the the pressure saturated water vapor ($p_s$) with respect to the reference pressure ($p_{ref}$) and the universal gas constant ($R$):

$ p_s = p_{ref} e^{- l_m / R T }$

Conditions

Variables

$T$

Absolute temperature

$K$

$l_m$

Molar Latent Heat

$J/kg$

$p_s$

Pressure saturated water vapor

$Pa$

$p_{ref}$

Reference pressure

3.65e+10

$Pa$

$R$

Universal gas constant

8.4135

$J/mol K$

Relation

Development

Using the Clausius-Clapeyron equation for the gradient of the pressure ($p$) with respect to the absolute temperature ($T$), which depends on the latent Heat ($L$) and the volume variation in phase change ($\Delta V$):

$\displaystyle\frac{ dp }{ dT }=\displaystyle\frac{ L }{ \Delta V T }$

In the case of the phase change from liquid to gas, we can assume that the change in volume is approximately equal to the volume of the vapor. Therefore, we can employ the ideal gas equation with the number of moles ($n$), the volume ($V$), the universal gas constant ($R$), and the water vapor pressure unsaturated ($p_v$):

$ \Delta V =\displaystyle\frac{ n R T }{ p_v }$

Since the Clausius-Clapeyron equation can be written as:

$\displaystyle\frac{dp}{dT}=\displaystyle\frac{L}{n}\displaystyle\frac{p}{R T^2}$

Where the molar Latent Heat ($l_m$) ($l_m = L/n$) corresponds to the change in enthalpy during the phase change h (the energy required to form water), we finally have:

$\displaystyle\frac{dp}{dT}=l_m\displaystyle\frac{p}{RT^2}$

If we integrate this equation between the pressure saturated water vapor ($p_s$) and the pressure at point p_0 at temperature T_0, we obtain:

$p_s=p_0e^{l_m/RT_0}e^{-l_m/RT}$

If we evaluate this expression with the data at the critical point:

$p_{ref}=p_0e^{l_m/RT_0}$

We finally have:

$ p_s = p_{ref} e^{- l_m / R T }$

the reference pressure ($p_{ref}$) can be calculated using data from both the triple point and the critical point. In the first case, the temperature is 0.01°C = 273.16K, and the pressure is 611.73 Pa. If the molar latent heat is 40.65 kJ/mol (for water), then the reference pressure ($p_{ref}$) has a value of 3.63E+10 Pa.

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Equation

>Top, >Model

The relationship between the concentration of water vapor molecules ($c_v$) and saturated water vapor concentration ($c_s$) is referred to as the relative humidity ($RH$). In other words, when a relative humidity of 100% is reached, the existing concentration will be equal to the saturated concentration.

$ RH =\displaystyle\frac{ c_v }{ c_s }$

Conditions

Variables

$c_v$

Concentration of water vapor molecules

$mol/m^3$

$RH$

Relative humidity

$-$

$c_s$

Saturated water vapor concentration

$mol/m^3$

Relation

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Equation

>Top, >Model

When the pressure ($p$) behaves as an ideal gas, satisfying the volume ($V$), the number of moles ($n$), the absolute temperature ($T$), and the universal gas constant ($R$), the ideal gas equation:

can be rewritten in terms of the molar concentration ($c_m$) as:

$ p = c_m R T $

Conditions

Variables

$T$

Absolute temperature

$K$

$c_m$

Molar concentration

$mol/m^3$

$p$

Pressure

$Pa$

$R$

Universal gas constant

8.4135

$J/mol K$

Relation

Development

When the pressure ($p$) behaves as an ideal gas, satisfying the volume ($V$), the number of moles ($n$), the absolute temperature ($T$), and the universal gas constant ($R$), the ideal gas equation:

$ p V = n R T $

and the definition of the molar concentration ($c_m$):

$ c_m \equiv\displaystyle\frac{ n }{ V }$

lead to the following relationship:

$ p = c_m R T $

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Equation

>Top, >Model

The relationship between the concentration of water vapor molecules ($c_v$) and saturated water vapor concentration ($c_s$) is called the relative humidity ($RH$):

and it can be expressed in terms of the water vapor pressure unsaturated ($p_v$) and the pressure saturated water vapor ($p_s$) as follows:

$ RH =\displaystyle\frac{ p_v }{ p_s }$

Conditions

Variables

$p_s$

Pressure saturated water vapor

$Pa$

$RH$

Relative humidity

$-$

$p_v$

Water vapor pressure unsaturated

$Pa$

Relation

Development

The relationship between the relative humidity ($RH$) with the concentration of water vapor molecules ($c_v$) and saturated water vapor concentration ($c_s$) is expressed as:

$ RH =\displaystyle\frac{ c_v }{ c_s }$

and by relating the pressure ($p$) with the molar concentration ($c_m$), the absolute temperature ($T$), and the universal gas constant ($R$), we obtain:

$ p = c_m R T $

This applies to the vapor pressure of water, where:

$p_v = c_v R T$

and the saturated vapor pressure of water:

$p_s = c_s R T$

resulting in the following equation:

$ RH =\displaystyle\frac{ p_v }{ p_s }$

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