In dynamic-R/RTM: Reaction Transport Modelling: Course Tutorials and Exercises
Introduction
Consider a closed container filled with water and air (Figure 1). The volume (in $m^3$) and amount (in $mol$) of water in the liquid phase are $V_{liq}$ and $N_{liq}$, respectively. The gas phase has a volume $V_{gas}$ and contains a mixture of air and water vapor with the corresponding amounts $N_{air}$ and $N_{vap}$. The liquid water can evaporate to form water vapor, and vice versa, water vapor can precipitate to form liquid water. These processes occur across the liquid-gas boundary whose area is $A_{liq}$, and they can be denoted by the following reversible reaction:
$$
H_2O\,(liq) \mathrel{\mathop{\rightleftarrows}^{\mathrm{k_f}}_{\mathrm{k_b}}} H_2O\,(gas)
$$
{width=30%}
In the first approximation, and as long as the liquid water is present, evaporation is described by zero-order kinetics, i.e., the rate is independent of how much vapor is in the gas phase. In contrast, precipitation is described by first-order kinetics, i.e., the rate is linearly proportional to the concentration of water vapor molecules in the gas phase, $N_{vap}/V_{gas}$. Because the exchange of water molecules occurs across the interface between the liquid and gas phase, the overall rate is proportional to the area of this interface, $A_{liq}$. Taken together, the numbers of water molecules in the liquid and vapor form are changing over time according to the following differential equations:
\begin{equation}
\label{eq1}
\begin{array}{rcr}
\displaystyle\frac{dN_{liq}}{dt} & = &
\displaystyle
- k_f\cdot A_{liq}
+ k_b\cdot A_{liq} \cdot \frac{N_{vap}}{V_{gas}} \[5mm]
\displaystyle
\frac{dN_{vap}}{dt} & = &
\displaystyle
k_f\cdot A_{liq}
- k_b\cdot A_{liq} \cdot \frac{N_{vap}}{V_{gas}}
\end{array}
\end{equation}
In these equations, the rate constants $k_f$ (in $mol\,m^{-2}\,s^{-1}$) and $k_b$ (in $m\,s^{-1}$) correspond to the forward (evaporation) and backward (precipitation) process, respectively. Note that because the system is closed, the total amount of water molecules in the system is conserved, as follows from the fact that the two rates have the same magnitude but an opposite sign ($dN_{liq}/dt = -dN_{vap}/dt$).
The system reaches an equilibrium when the time derivatives are equal to zero. Equation 1 implies that this situation occurs when the concentration of vapor molecules is equal to the ratio between the rate constants for the forward and backward reaction. Using an asterisk ($$) to denote equilibrium values, we can therefore write
\begin{equation}
\label{eq3}
\mathrm{equilibrium}!!:\qquad \frac{dN_{liq}}{dt} = \frac{dN_{vap}}{dt} = 0 \quad\implies\quad
\left( \frac{N_{vap}}{V_{gas}}\right)^ = \frac{k_f}{k_b}.
\end{equation}
This notation allows us to rewrite Eq. 1 according to:
\begin{equation}
\label{eq4}
\frac{dN_{liq}}{dt} = - \frac{dN_{vap}}{dt} = k_b\cdot A_{liq} \cdot
\left[ \left(\frac{N_{vap}}{V_{gas}}\right) -
\left(\frac{N_{vap}}{V_{gas}}\right)^{!!}\, \right].
\end{equation}
The net process of evaporation and precipitation can therefore be thought of as being driven by a physical-chemical disequilibrium: the net rate of exchange of water molecules between the liquid and gas phase is proportional to the distance* from the equilibrium, which in this particular case is equal to the difference between the instantaneous and equilibrium concentrations of vapor molecules above the liquid.
Reformulation of the mass balance equations
In the literature, the driving force in the mass balance of water molecules (Eq. \ref{eq4}) is often formulated in terms of differences in vapor pressure or relative humidity instead of differences in the concentration of water vapor molecules. As can be expected, these formulations are valid under certain assumptions, and they are equivalent except for differences in the values (and units) of the parameters used. Here we summarize basic concepts that allow us to switch between the different formulations.
Vapor pressure
If water vapor behaves as an ideal gas, the concentration of water vapor molecules is related to the vapor pressure, $P_{vap}$ in Pascal ($Pa$), through the ideal gas law:
\begin{equation}
\label{eq5}
\frac{N_{vap}}{V_{gas}} = \frac{P_{vap}}{R\cdot T},
\end{equation}
where $R=8.314\,J\,K^{-1}\,mol^{-1}$ is the gas constant and $T$ is the vapor temperature (in $K$). Using this relationship, Eq. \ref{eq4} can be rewritten in the form
\begin{equation}
\label{eq6}
\frac{dN_{liq}}{dt} = -\frac{dN_{vap}}{dt} = \frac{k_b}{R\cdot T} \cdot A_{liq}\cdot \left( P_{vap} - P_{vap}^ \right),
\end{equation}
where $P_{vap}^$ denotes the equilibrium vapor pressure calculated from the equilibrium concentration $(N_{vap}/V_{gas})^$ using the ideal gas law (Eq. \ref{eq5}). The formulation in Eq. \ref{eq6} is equivalent to the empirical relationship developed by Carrier (1918):
\begin{equation}
\label{eq7}
\frac{dN_{vap}}{dt} = \frac{c_1 + c_2\cdot v}{\Delta\bar{H}{vap}} \cdot A{liq}\cdot \left( P_{vap}^ - P_{vap} \right),
\end{equation}
where $\Delta \bar{H}_{vap}$ is the molar enthalpy of vaporization of water ($\approx 40.66~kJ~mol^{-1}$), $v$ is the wind velocity (in $m~s^{-1}$), and $c_1 = 0.0888~m~s^{-1}$ and $c_2=0.0783$ are empirical constants.
By combining Eqs. \ref{eq3}, \ref{eq6} and \ref{eq7}, we see that the evaporation ($k_f$) and precipitation ($k_b$) rate constants are related to the empirical constants $c_1$ and $c_2$ according to
\begin{equation}
\label{eq8}
\begin{array}{l}
\displaystyle
k_f = \frac{P_{vap}^*}{\Delta \bar{H}{vap}}\cdot (c_1+c_2\cdot v),\[5mm]
\displaystyle
k_b = \frac{R\cdot T}{\Delta \bar{H}{vap}}\cdot (c_1+c_2\cdot v).
\end{array}
\end{equation}
Equilibrium vapor pressure
Equation \ref{eq8} shows that the value of the equilibrium vapor pressure, $P_{vap}^$, is required to calculate $k_f$. This quantity depends on temperature and can be estimated based on the Clausius-Clapeyron equation, which relates equilibrium vapor pressures at different temperatures:
\begin{equation}
\label{eq9}
\ln\left[\frac{P_{vap}^(T_1)}{P_{vap}^(T_2)}\right] =
-\frac{\Delta \bar{H}{vap}}{R}\cdot \left(\frac{1}{T_1} - \frac{1}{T_2}\right).
\end{equation}
This relationship follows from equilibrium thermodynamics and assumes that the molar enthalpy of vaporisation of water, $\Delta \bar{H}{vap}$, is temperature-independent and the molar volume of liquid water is negligible in comparison with the molar volume of water vapor. Because we know that water boils at a temperature of $100\,^\circ C$ when the external pressure is $1~atm$, the equilibrium vapor pressure at a temperature of $T_{boil} = 373.15\,K$ is $P_{vap}^(T_{boil}) = 1\,atm = 101325\,Pa$. The Clausius-Clapeyron equation allows us to calculate the equilibrium vapor pressure at any other temperature, $T$, according to:
\begin{equation}
\label{eq10}
P_{vap}^(T) = P_{vap}^(T_{boil}) \cdot\exp\left[
-\frac{\Delta \bar{H}{vap}}{R}\cdot \left(\frac{1}{T} - \frac{1}{T{boil}}\right)\right].
\end{equation}
Molar fraction and partial pressure
Because we are interested in modeling evaporation and precipitation of water surrounded by atmospheric air, we must consider that air is a mixture of multiple gaseous components, including major components such as $N_2$ and $O_2$ and minor components such as $Ar$, $CO_2$, $CH_4$, and water vapor.
When dealing with mixtures, properties of individual components in the mixture are described by partial molar quantities. Molar fraction, typically denoted by $x$, is one of the most basic partial molar quantities. For a component $A$, this quantity is defined as the ratio between the amount of component $A$ and the total amount of the mixture (both amounts are in moles):
\begin{equation}
\label{eq11}
x_A = \frac{N_A}{N_{tot}}\quad (mol_A~mol^{-1}_{tot}).
\end{equation}
When dealing with gas mixtures such as atmospheric air, individual components are typically characterized by another quantity: partial pressure. If a mixture behaves as an ideal gas, partial pressures are additive. Thus, partial pressure of component $A$ is equal to the product of the total pressure of the gas, $P_{tot}$, and the molar fraction of the component: $P_A=x_A\cdot P_{tot}$. Because moist air at typical conditions behaves as an ideal gas, partial pressure of water vapor is related to the total air pressure, $P_{air}$, according to
\begin{equation}
\label{eq12}
P_{vap} = x_{vap}\cdot P_{air},
\end{equation}
where $x_{vap}$ is the molar fraction of water vapor in air. For example, if the total air pressure is $1~Atm$ and 1 in every 1000 molecules of air is water, the molar fraction of water vapor is $x_{vap}=0.001$ and the corresponding partial pressure of water vapor is $P_{vap}= 1~mAtm$.
Relative air humidity
Rather than using the molar fraction, water content of a moist air is very often expressed by a quantity called relative humidity, $\phi$. By definition, this quantity is equal to the ratio between the vapor pressure and the equilibrium vapor pressure $\phi = P_{vap}/P_{vap}^$. Thus, if the relative humidity of air is known, the vapor pressure is calculated according to
\begin{equation}
\label{eq13}
P_{vap} = \phi \cdot P_{vap}^.
\end{equation}
Because the molar fraction of water vapor in air and the relative humidity of air are relative quantities, they are often expressed in %. This can cause confusion. We emphasize here that these two quantities are fundamentally different: while the molar fraction of water vapor in air is the ratio between the (partial) vapor pressure and total air pressure (Eq. \ref{eq12}), relative air humidity is the ratio between the (partial) vapor pressure and the equilibrium vapor pressure (Eq. \ref{eq13}).
Tasks
Solve the following tasks using the theoretical background explained in the previous section.
Task 1
Calculate the equilibrium vapor pressure, $P_{vap}^*$, for temperatures of $10\,^\circ C$, $20\,^\circ C$ and $50\,^\circ C$. What are the corresponding vapor pressures, $P_{vap}$, and molar fractions, $x_{vap}$, if the total air pressure is $0.95\,Atm$ and the relative air humidity is 80%?
Task 2
Consider a closed container filled with liquid water and moist air (Figure 1). Initially, the volume of the liquid water is $V_{liq} = 1\,L$, the total volume of the container ($V_{liq}+V_{gas}$) is $1\,m^3$, and the relative air humidity is 0% (dry air). The system is kept at a constant temperature of $20\,^\circ C$ for $0.5\,hr$. Then, the system is cooled and kept at a constant temperature of $10\,^\circ C$ for another $0.5\,hr$. At all times, the total pressure in the gas phase is kept constant at $1\,Atm$.
-
Write a model that calculates the amount of water vapor above the liquid water as a function of time. Express this amount in moles, as a molar fraction, and as a relative air humidity.
-
Evaluate water budgets, i.e., the gross and net amounts of water transferred between the liquid and gas phase during the first and second half-hour intervals.
Assumptions
-
Area of the liquid water surface is constant ($A_{liq} = 1~m^2$).
-
Air is stagnant. Thus, you can assume the wind velocity of $v=0$ (see Eq. \ref{eq8}).
-
Dissolution of air components such as $N_2$ or $O_2$ in water can be ignored.
-
The air and water vapor behave as an ideal gas. Since the total pressure of the gas phase and the temperature of the entire system are kept constant, the evaporation and precipitation processes as well as the isobaric cooling phase will lead to a volume change for the liquid water, the gas phase (air + water vapor), as well as the total volume (liquid + gas phase). Except for the change in volume of the liquid water during cooling, these changes cannot be neglected. You can assume that the walls of the container are flexible to allow for such changes to occur without the need to perform an extra work.
-
To net evaporate $n$ moles of water at a constant pressure and temperature, a certain amount of heat is required ($q=n\cdot \Delta\bar{H}{vap}$). You can assume that this heat is supplied externally such that the pressure and temperature of the system can remain constant during this process. Similarly, you can assume that the heat produced during net precipitation of $n$ moles of water ($q=-n\cdot\Delta\bar{H}{vap}$) is transferred to the surrounding. During the phase where the system is cooled from $20\,^\circ C$ to $10\,^\circ C$, the required amount of heat is also transferred to the surrounding.
-
The liquid and gas phases are well mixed (i.e., homogeneous). Thus, you can ignore transport processes and base your model entirely on mass balances (Eqs. \ref{eq1}) and the ideal gas law (Eq. \ref{eq5}).
dynamic-R/RTM documentation built on Feb. 28, 2025, 1:23 p.m.
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Introduction
Consider a closed container filled with water and air (Figure 1). The volume (in $m^3$) and amount (in $mol$) of water in the liquid phase are $V_{liq}$ and $N_{liq}$, respectively. The gas phase has a volume $V_{gas}$ and contains a mixture of air and water vapor with the corresponding amounts $N_{air}$ and $N_{vap}$. The liquid water can evaporate to form water vapor, and vice versa, water vapor can precipitate to form liquid water. These processes occur across the liquid-gas boundary whose area is $A_{liq}$, and they can be denoted by the following reversible reaction: $$ H_2O\,(liq) \mathrel{\mathop{\rightleftarrows}^{\mathrm{k_f}}_{\mathrm{k_b}}} H_2O\,(gas) $$
{width=30%}
In the first approximation, and as long as the liquid water is present, evaporation is described by zero-order kinetics, i.e., the rate is independent of how much vapor is in the gas phase. In contrast, precipitation is described by first-order kinetics, i.e., the rate is linearly proportional to the concentration of water vapor molecules in the gas phase, $N_{vap}/V_{gas}$. Because the exchange of water molecules occurs across the interface between the liquid and gas phase, the overall rate is proportional to the area of this interface, $A_{liq}$. Taken together, the numbers of water molecules in the liquid and vapor form are changing over time according to the following differential equations: \begin{equation} \label{eq1} \begin{array}{rcr} \displaystyle\frac{dN_{liq}}{dt} & = & \displaystyle - k_f\cdot A_{liq} + k_b\cdot A_{liq} \cdot \frac{N_{vap}}{V_{gas}} \[5mm] \displaystyle \frac{dN_{vap}}{dt} & = & \displaystyle k_f\cdot A_{liq} - k_b\cdot A_{liq} \cdot \frac{N_{vap}}{V_{gas}} \end{array} \end{equation} In these equations, the rate constants $k_f$ (in $mol\,m^{-2}\,s^{-1}$) and $k_b$ (in $m\,s^{-1}$) correspond to the forward (evaporation) and backward (precipitation) process, respectively. Note that because the system is closed, the total amount of water molecules in the system is conserved, as follows from the fact that the two rates have the same magnitude but an opposite sign ($dN_{liq}/dt = -dN_{vap}/dt$).
The system reaches an equilibrium when the time derivatives are equal to zero. Equation 1 implies that this situation occurs when the concentration of vapor molecules is equal to the ratio between the rate constants for the forward and backward reaction. Using an asterisk ($$) to denote equilibrium values, we can therefore write \begin{equation} \label{eq3} \mathrm{equilibrium}!!:\qquad \frac{dN_{liq}}{dt} = \frac{dN_{vap}}{dt} = 0 \quad\implies\quad \left( \frac{N_{vap}}{V_{gas}}\right)^ = \frac{k_f}{k_b}. \end{equation} This notation allows us to rewrite Eq. 1 according to: \begin{equation} \label{eq4} \frac{dN_{liq}}{dt} = - \frac{dN_{vap}}{dt} = k_b\cdot A_{liq} \cdot \left[ \left(\frac{N_{vap}}{V_{gas}}\right) - \left(\frac{N_{vap}}{V_{gas}}\right)^{!!}\, \right]. \end{equation} The net process of evaporation and precipitation can therefore be thought of as being driven by a physical-chemical disequilibrium: the net rate of exchange of water molecules between the liquid and gas phase is proportional to the distance* from the equilibrium, which in this particular case is equal to the difference between the instantaneous and equilibrium concentrations of vapor molecules above the liquid.
Reformulation of the mass balance equations
In the literature, the driving force in the mass balance of water molecules (Eq. \ref{eq4}) is often formulated in terms of differences in vapor pressure or relative humidity instead of differences in the concentration of water vapor molecules. As can be expected, these formulations are valid under certain assumptions, and they are equivalent except for differences in the values (and units) of the parameters used. Here we summarize basic concepts that allow us to switch between the different formulations.
Vapor pressure
If water vapor behaves as an ideal gas, the concentration of water vapor molecules is related to the vapor pressure, $P_{vap}$ in Pascal ($Pa$), through the ideal gas law: \begin{equation} \label{eq5} \frac{N_{vap}}{V_{gas}} = \frac{P_{vap}}{R\cdot T}, \end{equation} where $R=8.314\,J\,K^{-1}\,mol^{-1}$ is the gas constant and $T$ is the vapor temperature (in $K$). Using this relationship, Eq. \ref{eq4} can be rewritten in the form \begin{equation} \label{eq6} \frac{dN_{liq}}{dt} = -\frac{dN_{vap}}{dt} = \frac{k_b}{R\cdot T} \cdot A_{liq}\cdot \left( P_{vap} - P_{vap}^ \right), \end{equation} where $P_{vap}^$ denotes the equilibrium vapor pressure calculated from the equilibrium concentration $(N_{vap}/V_{gas})^$ using the ideal gas law (Eq. \ref{eq5}). The formulation in Eq. \ref{eq6} is equivalent to the empirical relationship developed by Carrier (1918): \begin{equation} \label{eq7} \frac{dN_{vap}}{dt} = \frac{c_1 + c_2\cdot v}{\Delta\bar{H}{vap}} \cdot A{liq}\cdot \left( P_{vap}^ - P_{vap} \right), \end{equation} where $\Delta \bar{H}_{vap}$ is the molar enthalpy of vaporization of water ($\approx 40.66~kJ~mol^{-1}$), $v$ is the wind velocity (in $m~s^{-1}$), and $c_1 = 0.0888~m~s^{-1}$ and $c_2=0.0783$ are empirical constants.
By combining Eqs. \ref{eq3}, \ref{eq6} and \ref{eq7}, we see that the evaporation ($k_f$) and precipitation ($k_b$) rate constants are related to the empirical constants $c_1$ and $c_2$ according to \begin{equation} \label{eq8} \begin{array}{l} \displaystyle k_f = \frac{P_{vap}^*}{\Delta \bar{H}{vap}}\cdot (c_1+c_2\cdot v),\[5mm] \displaystyle k_b = \frac{R\cdot T}{\Delta \bar{H}{vap}}\cdot (c_1+c_2\cdot v). \end{array} \end{equation}
Equilibrium vapor pressure
Equation \ref{eq8} shows that the value of the equilibrium vapor pressure, $P_{vap}^$, is required to calculate $k_f$. This quantity depends on temperature and can be estimated based on the Clausius-Clapeyron equation, which relates equilibrium vapor pressures at different temperatures: \begin{equation} \label{eq9} \ln\left[\frac{P_{vap}^(T_1)}{P_{vap}^(T_2)}\right] = -\frac{\Delta \bar{H}{vap}}{R}\cdot \left(\frac{1}{T_1} - \frac{1}{T_2}\right). \end{equation} This relationship follows from equilibrium thermodynamics and assumes that the molar enthalpy of vaporisation of water, $\Delta \bar{H}{vap}$, is temperature-independent and the molar volume of liquid water is negligible in comparison with the molar volume of water vapor. Because we know that water boils at a temperature of $100\,^\circ C$ when the external pressure is $1~atm$, the equilibrium vapor pressure at a temperature of $T_{boil} = 373.15\,K$ is $P_{vap}^(T_{boil}) = 1\,atm = 101325\,Pa$. The Clausius-Clapeyron equation allows us to calculate the equilibrium vapor pressure at any other temperature, $T$, according to: \begin{equation} \label{eq10} P_{vap}^(T) = P_{vap}^(T_{boil}) \cdot\exp\left[ -\frac{\Delta \bar{H}{vap}}{R}\cdot \left(\frac{1}{T} - \frac{1}{T{boil}}\right)\right]. \end{equation}
Molar fraction and partial pressure
Because we are interested in modeling evaporation and precipitation of water surrounded by atmospheric air, we must consider that air is a mixture of multiple gaseous components, including major components such as $N_2$ and $O_2$ and minor components such as $Ar$, $CO_2$, $CH_4$, and water vapor.
When dealing with mixtures, properties of individual components in the mixture are described by partial molar quantities. Molar fraction, typically denoted by $x$, is one of the most basic partial molar quantities. For a component $A$, this quantity is defined as the ratio between the amount of component $A$ and the total amount of the mixture (both amounts are in moles): \begin{equation} \label{eq11} x_A = \frac{N_A}{N_{tot}}\quad (mol_A~mol^{-1}_{tot}). \end{equation}
When dealing with gas mixtures such as atmospheric air, individual components are typically characterized by another quantity: partial pressure. If a mixture behaves as an ideal gas, partial pressures are additive. Thus, partial pressure of component $A$ is equal to the product of the total pressure of the gas, $P_{tot}$, and the molar fraction of the component: $P_A=x_A\cdot P_{tot}$. Because moist air at typical conditions behaves as an ideal gas, partial pressure of water vapor is related to the total air pressure, $P_{air}$, according to \begin{equation} \label{eq12} P_{vap} = x_{vap}\cdot P_{air}, \end{equation} where $x_{vap}$ is the molar fraction of water vapor in air. For example, if the total air pressure is $1~Atm$ and 1 in every 1000 molecules of air is water, the molar fraction of water vapor is $x_{vap}=0.001$ and the corresponding partial pressure of water vapor is $P_{vap}= 1~mAtm$.
Relative air humidity
Rather than using the molar fraction, water content of a moist air is very often expressed by a quantity called relative humidity, $\phi$. By definition, this quantity is equal to the ratio between the vapor pressure and the equilibrium vapor pressure $\phi = P_{vap}/P_{vap}^$. Thus, if the relative humidity of air is known, the vapor pressure is calculated according to \begin{equation} \label{eq13} P_{vap} = \phi \cdot P_{vap}^. \end{equation}
Because the molar fraction of water vapor in air and the relative humidity of air are relative quantities, they are often expressed in %. This can cause confusion. We emphasize here that these two quantities are fundamentally different: while the molar fraction of water vapor in air is the ratio between the (partial) vapor pressure and total air pressure (Eq. \ref{eq12}), relative air humidity is the ratio between the (partial) vapor pressure and the equilibrium vapor pressure (Eq. \ref{eq13}).
Tasks
Solve the following tasks using the theoretical background explained in the previous section.
Task 1
Calculate the equilibrium vapor pressure, $P_{vap}^*$, for temperatures of $10\,^\circ C$, $20\,^\circ C$ and $50\,^\circ C$. What are the corresponding vapor pressures, $P_{vap}$, and molar fractions, $x_{vap}$, if the total air pressure is $0.95\,Atm$ and the relative air humidity is 80%?
Task 2
Consider a closed container filled with liquid water and moist air (Figure 1). Initially, the volume of the liquid water is $V_{liq} = 1\,L$, the total volume of the container ($V_{liq}+V_{gas}$) is $1\,m^3$, and the relative air humidity is 0% (dry air). The system is kept at a constant temperature of $20\,^\circ C$ for $0.5\,hr$. Then, the system is cooled and kept at a constant temperature of $10\,^\circ C$ for another $0.5\,hr$. At all times, the total pressure in the gas phase is kept constant at $1\,Atm$.
-
Write a model that calculates the amount of water vapor above the liquid water as a function of time. Express this amount in moles, as a molar fraction, and as a relative air humidity.
-
Evaluate water budgets, i.e., the gross and net amounts of water transferred between the liquid and gas phase during the first and second half-hour intervals.
Assumptions
-
Area of the liquid water surface is constant ($A_{liq} = 1~m^2$).
-
Air is stagnant. Thus, you can assume the wind velocity of $v=0$ (see Eq. \ref{eq8}).
-
Dissolution of air components such as $N_2$ or $O_2$ in water can be ignored.
-
The air and water vapor behave as an ideal gas. Since the total pressure of the gas phase and the temperature of the entire system are kept constant, the evaporation and precipitation processes as well as the isobaric cooling phase will lead to a volume change for the liquid water, the gas phase (air + water vapor), as well as the total volume (liquid + gas phase). Except for the change in volume of the liquid water during cooling, these changes cannot be neglected. You can assume that the walls of the container are flexible to allow for such changes to occur without the need to perform an extra work.
-
To net evaporate $n$ moles of water at a constant pressure and temperature, a certain amount of heat is required ($q=n\cdot \Delta\bar{H}{vap}$). You can assume that this heat is supplied externally such that the pressure and temperature of the system can remain constant during this process. Similarly, you can assume that the heat produced during net precipitation of $n$ moles of water ($q=-n\cdot\Delta\bar{H}{vap}$) is transferred to the surrounding. During the phase where the system is cooled from $20\,^\circ C$ to $10\,^\circ C$, the required amount of heat is also transferred to the surrounding.
-
The liquid and gas phases are well mixed (i.e., homogeneous). Thus, you can ignore transport processes and base your model entirely on mass balances (Eqs. \ref{eq1}) and the ideal gas law (Eq. \ref{eq5}).
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