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GilSE
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- I've made some calculations and found that equilibrium vapor pressure depends on air pressure above the liquid. Is it true?
It’s usually being assumed that points of equilibrium liquid – its vapor is given by a curve in the (P,T) coordinates, and this curve doesn’t change no matter is there another gas in the system or not. For example: if water is put in the empty volume, it will obviously vaporize, filling the emptiness, until vapor pressure reach equilibrium value (at this temperature). If the volume hadn’t been empty but filled with air, the water still would vaporize until the equilibrium and the equilibrium vapor pressure would be the same as in the “vacuum case”. But this assumption looks not obvious for me, I had tried to prove it using thermodynamics, but got an unexpected result.
My reasoning:
The system contains of two phases: the gas one and the liquid one. The liquid one is represented by water with the molar volume v_{w} and the molar Gibbs energy \varphi_{w} . The gas one is represented by water vapor and by air. Pressure which is produced by molecules of air is p_{a} and doesn’t change no matter what happens, partial pressure of water vapor is p_{v} . The molar Gibbs energy of the gas phase is \varphi_{g} , the molar volume of the gas phase is v_{g}. No water molecules can leave the system.
If the pressure and the temperature is the same everywhere in the system, an expression for the total Gibbs energy of the system would be ( \nu in this expression stands for corresponding amount of substance):
\Phi(p, T) = \varphi_{w}(p, T)\nu_{w} + \varphi_{g}(p, T)(\nu_{v} + \nu_{a})
When vaporization or condensation occur, the total Gibbs energy will change because \nu_{w} and \nu_{v} are changing. Amount of air in this processes obviously doesn't change.
So, change of the total Gibbs energy when small change of liquid/vapor amount has occurred:
d\Phi = \varphi_{w}d\nu_{w} + \varphi_{g}d\nu_{v}
Also, if amount of vapor has increased by d\nu, amount of liquid should decrease by d\nu and vice versa:
d\nu_{v} = -d\nu_{w}
Taking this in account:
d\Phi = (\varphi_{w} - \varphi_{g})d\nu_{w}
With such conditions in the system, processes must decrease Gibbs energy. Thus, if \varphi_{w} is not equal to \varphi_{g} phase transitions will occur. Equilibrium is possible only if \varphi_{w} = \varphi_{g} .
Everywhere at the equilibrium curve \varphi_{w} must be equal to \varphi_{g}, hence, when moving along the curve, Gibbs energy of gas phase and Gibbs energy of liquid phase must change identically: d\varphi_{w} = d\varphi_{g} , what could be rewritten as:
-s_{w}dT+v_{w}d(p_{v}+p_{w}) = -s_{g}dT+v_{g}d(p_{v}+p_{a})
Air pressure doesn’t change:
-s_{w}dT+v_{w}dp_{v} = -s_{g}dT+v_{g}dp_{v}
Let’s transform the expression above with the notice that change of an entropy during an isothermal process is equal to the received heat divided by the temperature, \Delta s = s_{g}-s_{w} = \frac q T , where q is the heat consumed to transform 1 mol of liquid water to vapor.
(s_{g}-s_{w})dT = (v_{g}-v_{w})dp_{v}
\frac {dp_{v}} {dT} = \frac q {T(v_{g}-v_{w})}
Now, let’s write down the ideal gas law for 1 mol of the gas phase and express v_{g} from it.
p_{v}+p_{a} = \frac {RT} {v_{g}}
v_{g} = \frac {RT} {p_{v}+p_{a}}
Then, substitute it into the previous equation, also assuming that v_{w} \ll v_{g} .
\frac {dp_{v}} {dT} = \frac q {Tv_{g}}
\frac {dp_{v}} {dT} = \frac {q(p_{v}+p_{a})} {RT^2}
Solve this differential equation:
\frac {dp_{v}} {p_{v}+p_{a}} = \frac {qdT} {T^2}
\ln(p_{v}+p_{a}) + \ln\left(\frac 1 C\right) = -\frac q T
p_{v}+p_{a} = Ce^{-q/T}
Let’s it’s known that when T = T_0 and p_{v} = p_{v_0} the system is in an equilibrium state.
p_{v_0}+p_{a} = Ce^{-q/T_0}
C = (p_{v_0}+p_{a})e^{q/T_0}
p_{v}+p_{a} = (p_{v_0}+p_{a})e^{q(1/T_0-1/T)}
p_{v} = (p_{v_0}+p_{a})e^{q(1/T_0-1/T)}-p_{a}
p_{v} in this expression – vapor pressure in an equilibrium state. We can see that this pressure depends on air pressure, so at the same temperature but at the different air pressure equilibrium curves will differ.
P.S. I hope TeX language in this post will render correctly. I did my best to format it like in other posts on the forum but for some reason it still doesn't render in preview.
My reasoning:
The system contains of two phases: the gas one and the liquid one. The liquid one is represented by water with the molar volume v_{w} and the molar Gibbs energy \varphi_{w} . The gas one is represented by water vapor and by air. Pressure which is produced by molecules of air is p_{a} and doesn’t change no matter what happens, partial pressure of water vapor is p_{v} . The molar Gibbs energy of the gas phase is \varphi_{g} , the molar volume of the gas phase is v_{g}. No water molecules can leave the system.
If the pressure and the temperature is the same everywhere in the system, an expression for the total Gibbs energy of the system would be ( \nu in this expression stands for corresponding amount of substance):
\Phi(p, T) = \varphi_{w}(p, T)\nu_{w} + \varphi_{g}(p, T)(\nu_{v} + \nu_{a})
When vaporization or condensation occur, the total Gibbs energy will change because \nu_{w} and \nu_{v} are changing. Amount of air in this processes obviously doesn't change.
So, change of the total Gibbs energy when small change of liquid/vapor amount has occurred:
d\Phi = \varphi_{w}d\nu_{w} + \varphi_{g}d\nu_{v}
Also, if amount of vapor has increased by d\nu, amount of liquid should decrease by d\nu and vice versa:
d\nu_{v} = -d\nu_{w}
Taking this in account:
d\Phi = (\varphi_{w} - \varphi_{g})d\nu_{w}
With such conditions in the system, processes must decrease Gibbs energy. Thus, if \varphi_{w} is not equal to \varphi_{g} phase transitions will occur. Equilibrium is possible only if \varphi_{w} = \varphi_{g} .
Everywhere at the equilibrium curve \varphi_{w} must be equal to \varphi_{g}, hence, when moving along the curve, Gibbs energy of gas phase and Gibbs energy of liquid phase must change identically: d\varphi_{w} = d\varphi_{g} , what could be rewritten as:
-s_{w}dT+v_{w}d(p_{v}+p_{w}) = -s_{g}dT+v_{g}d(p_{v}+p_{a})
Air pressure doesn’t change:
-s_{w}dT+v_{w}dp_{v} = -s_{g}dT+v_{g}dp_{v}
Let’s transform the expression above with the notice that change of an entropy during an isothermal process is equal to the received heat divided by the temperature, \Delta s = s_{g}-s_{w} = \frac q T , where q is the heat consumed to transform 1 mol of liquid water to vapor.
(s_{g}-s_{w})dT = (v_{g}-v_{w})dp_{v}
\frac {dp_{v}} {dT} = \frac q {T(v_{g}-v_{w})}
Now, let’s write down the ideal gas law for 1 mol of the gas phase and express v_{g} from it.
p_{v}+p_{a} = \frac {RT} {v_{g}}
v_{g} = \frac {RT} {p_{v}+p_{a}}
Then, substitute it into the previous equation, also assuming that v_{w} \ll v_{g} .
\frac {dp_{v}} {dT} = \frac q {Tv_{g}}
\frac {dp_{v}} {dT} = \frac {q(p_{v}+p_{a})} {RT^2}
Solve this differential equation:
\frac {dp_{v}} {p_{v}+p_{a}} = \frac {qdT} {T^2}
\ln(p_{v}+p_{a}) + \ln\left(\frac 1 C\right) = -\frac q T
p_{v}+p_{a} = Ce^{-q/T}
Let’s it’s known that when T = T_0 and p_{v} = p_{v_0} the system is in an equilibrium state.
p_{v_0}+p_{a} = Ce^{-q/T_0}
C = (p_{v_0}+p_{a})e^{q/T_0}
p_{v}+p_{a} = (p_{v_0}+p_{a})e^{q(1/T_0-1/T)}
p_{v} = (p_{v_0}+p_{a})e^{q(1/T_0-1/T)}-p_{a}
p_{v} in this expression – vapor pressure in an equilibrium state. We can see that this pressure depends on air pressure, so at the same temperature but at the different air pressure equilibrium curves will differ.
P.S. I hope TeX language in this post will render correctly. I did my best to format it like in other posts on the forum but for some reason it still doesn't render in preview.
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