Why water evaporates below 100 ºC

[Fernsanz]

Hello everyone.

There is a question I have been thinking about for a time. Let's state it the easy way:

Let's suppose my room is a closed system with constant pressure (1 atm) and temperature (standard 25º C). Under these conditions, if I put a container with water in my room, it will end up empty because water have evaporated. But according to the phase diagram of water, water vapor should not exists as an equilibrium state at those conditions. So, what is wrong?

I understand evaporation (I know the difference with boiling) as a surface phonemenon, but once the system reachs the equilibrium, the water should be liquid as indicated by the phase diagram.

¿Suggestions?

Thanks.

 

[D H]

The phase diagram says water should not exist as a liquid above 100⁰C at 1 atmosphere of pressure. It does not say that water should not exist as a vapor below 100⁰C.

What is happening is that the molecules in the water are "bouncing" around a lot. Some of the molecules have a low velocity while others have a high velocity. The http://en.wikipedia.org/wiki/Maxwell–Boltzmann_distribution" describes the distribution of velocity as a function of temperature. While the average velocity of the molecules in a flask of water at 25⁰C is well below that required to escape the surface of the water, some of the molecules do have sufficient velocity to escape the surface, and that is exactly what they do.

 

[jcsd]

Hello everyone.

There is a question I have been thinking about for a time. Let's state it the easy way:

Let's suppose my room is a closed system with constant pressure (1 atm) and temperature (standard 25º C). Under these conditions, if I put a container with water in my room, it will end up empty because water have evaporated. But according to the phase diagram of water, water vapor should not exists as an equilibrium state at those conditions. So, what is wrong?

I understand evaporation (I know the difference with boiling) as a surface phonemenon, but once the system reachs the equilibrium, the water should be liquid as indicated by the phase diagram.

¿Suggestions?

Thanks.

It's due to the uneven distribution of energy among the water molecules at any given time caused by collisons between water moecules due to their random motion. A water moelcule near the surface can acquire enoguh energy due to these random collisions to escape as water vapour.

Losing the most energetic molecules due to eavporuation will lower the mean energy of the water molecules, meaning temperature of the water will lower which is why evapouration causes the remaining liquid to cool.

 

[Fernsanz]

The phase diagram says water should not exist as a liquid above 100⁰C at 1 atmosphere of pressure. It does not say that water should not exist as a vapor below 100⁰C.

What is happening is that the molecules in the water are "bouncing" around a lot. Some of the molecules have a low velocity while others have a high velocity. The http://en.wikipedia.org/wiki/Maxwell–Boltzmann_distribution" describes the distribution of velocity as a function of temperature. While the average velocity of the molecules in a flask of water at 25⁰C is well below that required to escape the surface of the water, some of the molecules do have sufficient velocity to escape the surface, and that is exactly what they do.

The phase diagram do says that at 1 atm and 25 ºC the equilibrium state of water is in liquid phase.

As for both replys: I know evaporation is a surface phenomena different from boling; however, no matter which process it is, the final state is that we get water vapor at 25 ºC and 1 atm, against predeicted by the phase diagram.

Any more suggestions?

 

[D H]

The phase diagram do says that at 1 atm and 25 ºC the equilibrium state of water is in liquid phase.

You are reading the phase transition diagram a bit too naively. The phase transition diagram says that at 1 atm and 25^oC a vessel that contains H₂O in which the water vapor pressure is at the saturation level will have a liquid phase. It does not say all of the H₂O molecules in the vessel will be in the liquid phase.

 

[Fernsanz]

You are reading the phase transition diagram a bit too naively. The phase transition diagram says that at 1 atm and 25^oC a vessel that contains H₂O in which the water vapor pressure is at the saturation level will have a liquid phase. It does not say all of the H₂O molecules in the vessel will be in the liquid phase.

At 25 ºC and 1 atm in that vessel will exists one and only one state of equilibrium, right? Draw that state of equilibrium for each T and P and you have the phase diagram. Ok, now the phase diagram says that a 25ºC and 1 atm what you have in the vessel is liquid water, not vapor, and, hence it makes no sense to talk about water at the saturation level, simply because it is in liquid zone.

So I insist, how does this reconciliate with the fact that water evaporates at that conditions? Well I think I'm going toward the solution which has to be with the presence of other components in the air. That is, we are dealing with a system with 2 components (H2O and N2) and 2 phases. So chemical potentials will have something to say, but I don't have the correct solution yet.

When people are asked about how can water evaporates at 25 ºC most people tend to give the surface evaporation phenomena explanation. This is, however, not the question. The question has nothing to do with the explanation of mollecular phenomena (Maxwell distribution of velocities of whatever). We are dealing here only with states of equilibrium, no matter how those states are reached. And what should shock a first glance is that the equilibruim state contains water vapor at ambiental conditions, which does not agree with the water state of equilibrium at those conditions. As i said, I suspect that the phase diagram of water as a PURE SUBSTANCE is not directly applicable when more substances are involved.

If any has the closed solution, please let me know.

 

[D H]

At 25 ºC and 1 atm in that vessel will exists one and only one state of equilibrium, right?

By the rest of your post, I assume you mean one and only one phase, and that is simply wrong. Partially fill a vessel with water, vent the air at the top of the vessel to vacuum, and close the vessel. The vacuum portion of the vessel will not remain a vacuum for long. Water will evaporate until the pressure in the chamber reaches the partial pressure for water at 25^oC. The chamber contains water vapor and water liquid. You're persistence in incorrectly interpreting the meaning of the phase diagram is clouding your understanding (pun intended).

As i said, I suspect that the phase diagram of water as a PURE SUBSTANCE is not directly applicable when more substances are involved.

Wrong again. Take my previous example of a vessel comprising water and a vacuum. Now instead of closing the valve immediately, open another valve that let's dessicated air (or helium, or whatever) into the upper part of the vessel. Liquid water will still evaporate until the partial pressure of the H₂O in the gas side of the vessel reaches the temperature-dependent saturation vapor pressure.

 

[Dale]

Let's suppose my room is a closed system with constant pressure (1 atm) and temperature (standard 25º C). Under these conditions, if I put a container with water in my room, it will end up empty because water have evaporated. But according to the phase diagram of water, water vapor should not exists as an equilibrium state at those conditions. So, what is wrong?

I understand evaporation (I know the difference with boiling) as a surface phonemenon, but once the system reachs the equilibrium, the water should be liquid as indicated by the phase diagram.

Rather than looking simply at the phase diagram you need to look at the saturated vapor pressure. The boiling point of water is 100ºC because of the fact that at 100ºC the saturated vapor pressure is 1 atm. However, at lower temperatures there is still a vapor pressure. Assuming your room is a closed system then the container with water will evaporate until the partial pressure of water vapor in the room is equal to the saturated vapor pressure (23.76 mmHg at 25ºC).

http://hyperphysics.phy-astr.gsu.edu/Hbase/kinetic/watvap.html#c1

 

[turbo]

To the OP: Remember that water can evaporate from solid phase directly to vapor without going through the liquid phase, under the right conditions. Look up "sublimation".

 

[pixel01]

I should think the answer is quite simple. Any liquid will have a vapor pressure which depends on only the temperature. Whenever the ambient pressure is equal to the vapor pressure at the given tempeature, then the boiling phenomenon occurs. Water has vapor pressure of 23.7 mmHg at 25oC.
A water tank in a room at 25oC and 1 atm will finally evaporate because the water vapor will condense at any surfaces inside the room. The amount of water vapor in the air is constant (of course if the room is gas tight).
If the water is replaced by motor oil, the same phenomenon happens, but it would take a very long time, which I guess tens of years or even hundreds.

 

[Fernsanz]

You should clarify your understanding of thermodynamic concepts instead of telling my insistence is wrong. It seems to me that you talk more like an engineer than like a physics.

By the rest of your post, I assume you mean one and only one phase, and that is simply wrong. Partially fill a vessel with water, vent the air at the top of the vessel to vacuum, and close the vessel. The vacuum portion of the vessel will not remain a vacuum for long. Water will evaporate until the pressure in the chamber reaches the partial pressure for water at 25^oC. The chamber contains water vapor and water liquid. You're persistence in incorrectly interpreting the meaning of the phase diagram is clouding your understanding (pun intended).

No. What a phase diagram represents is the existing phase(s) at THERMODYNAMICAL EQUILIBRIUM STATES of any substance. It doesn't matter if there exist one, two or n phases, the important is that you take two independent variables (T and P), fix them to a value and the diagram says you that AT THE EQUILIBRIUM STATE corresponding that T and P, the phase(s) corresponding to that substance is this or that (liquid, vapor, liquid-vapor. triple point, etc).

So, take T=300K, P=1 atm and by no means will you get a vessel which contains ONLY water at these conditions to contain vapor. You can get it if you vacuum the vessel, but not at 300K and 1 atm!.

Take a look the images I attach. The first one is the phase diagram for water without concrete values on the axis. The second one tells that at 1 atm and 300K WATER CAN EXIST only as liquid once the equilibrium has been reached.

 

[D H]

So, take T=300K, P=1 atm and by no means will you get a vessel which contains ONLY water at these conditions to contain vapor. You can get it if you vacuum the vessel, but not at 300K and 1 atm!.

Yes, you will. Let's put the empty vessel in a room with completely dessicated air (zero percent humidity). Now partially fill the vessel with liquid water and close the vessel off. The pressure in the vessel will rise to 1.03 atmospheres as some of the water in the vessel evaporates. If you only put a small amount of liquid water in the vessel (i.e., not enough to bring the vapor pressure up to 0.03 atmospheres), all of the water will evaporate, leaving no water in the liquid phase even though the temperature is well below the boiling point.

 

[Fernsanz]

I think the answer goes more or less like this:

For equilibrium to exists, not only must be temperature and pressure definited (uniform trough the system) but also the chemical potential of a component in each phase must be equal (Gibbs rule).

Take a closed room at 300K and 1 atm inside of which there is a open vessel of water. We have two components H2O and N2 (supposing dry air is mostly N2) and two phases: gas and liquid.

Ok, now if there weren't N2 the water would reach the equilibrium state which, accorrding to water phase diagram, would be liquid water. This is what "people" would expect because they don't account for the existence of N2 and, as said in elementary school, water is liquid at 300K.

Now let's account for the existence of N2. As many times in thermodynamics, let's think of a constrained equilibrium state, that is, we imagine vapor water and liquid water separated chemically by mean of a diathermal and mechanical wall which allow mechanical and thermal equilibrium. With this in mind let's answer some questions:

First of all, why can water vapor exist at that pressure and temperature? Well, the key point is that water vapor is surrounded by a pressure of 1 atm, but its parcial pressure is much lower than that; and, according to the phase diagram, water vapor at 300K can exist when the pressure is small enough. So, water vapor exists at that (parcial) pressure.

Now, if we remove the constrain the system will evolve to its global equilibrium state (Zero Law of equilibrium principle of thermodynamics). Note that we come from a constrained equilibrium state, not a global one. That is, the whole system was not at global equilibrium (i.e., not at its maximum possible entropy among constrained equilibrium states) but each part was at equilibrium. Ok, now we know that the chemical potential of vapor water overheated over its saturated level is lower than that of liquid water at 1 atm, so liquid water evaporates in order for chemical equilibrium to exists.

Well, more or less the solution wold have to follow this way, but there is some dark points already, like how to know which chemical potential is actually lower and which higher.

Someone can complete the explanation?

 

[D H]

What do you think the term relative humidity means? Vapor pressure? Partial pressure?

Google these, please. I give up.

 

[Fernsanz]

Yes, you will. Let's put the empty vessel in a room with completely dessicated air (zero percent humidity). Now partially fill the vessel with liquid water and close the vessel off. The pressure in the vessel will rise to 1.03 atmospheres as some of the water in the vessel evaporates. If you only put a small amount of liquid water in the vessel (i.e., not enough to bring the vapor pressure up to 0.03 atmospheres), all of the water will evaporate, leaving no water in the liquid phase even though the temperature is well below the boiling point.

"Dessicated air" -> so you are taking N2 to the experiment also. I said only water, only H2O. Carry out that experiment in a room with contain nothing else than H2O.

As the book of Kestin says, "no matter how small amount of an inert gas you push into a vessel containing vapor-liquid equilibrium water, the water will evaporte." So, what you say is just the experiment of Kelvin-Helmholtz, i.e., perturbation of vapor-liquid equilibrium with a gas. But, if there is no added gas, there will be no vapor water at 300K and 1 atm. These last statement is pretty clear anyway, I don't know if you are trying to find water vapor in any single condition, but at equilibrium state, water alone does not exists as vapor at 300K and 1 atm.

Nevertheless, you are getting closer to the real core of the problem: the existence of another substance (N2) in the room.

 

[D H]

Where in the heck do you think the 1 atmosphere comes from? The ether? Of course water alone does not exist as vapor at 1 atmosphere at 300K. I never said it did. Water does exist as vapor at 0.03 atmospheres at 300K, and liquid water will evaporate to achieve that 0.03 atmospheres if the partial pressure of water is less than 0.03 atmospheres.

Liquid water will continue to evaporate after the partial pressure reaches 0.03 atmospheres, but gaseous water will condense into liquid at the same rate. That is your equilibrium point at 25^oC.

I have given and others have given several examples of and several reasons why your reasoning is wrong.

 

[Fernsanz]

My reasoning is right, I don't have doubts about what I have argued. That is not what worry me.

What I would to like to find is an explanation of WHY the state of equilibrium at 300K and 1 atm is what we all know it is.

You have just given me descriptions of what everyone now that happens, but not the reasons on the ground of rigurous thermodynamics. I want some who say "well, the reason why the system reaches this equilibrium state and water evaporate is because the Gibbs potential of vapor in presence of N2... bla bla bla"

There is a big difference between actual last-causes or theoretical explanations and just enumeration of facts that is what you have provided so far. One have only understood the underlying physics when one is capable to answer "this happens because" and not just "this happens and this other also happen" and so on.

I know what happens when you put water in a vacuum vessel; I know where pressure comes from, I know those thing that anyone not ver intelligent would know. But, up to now, no one has been capable of explain WHY water MUST be in vapor state.

When I read a thermodynamic textbook, a good one such as Kestin, Callen, Atkins or Zemansky one problem could be "Given N2 and H2O in a room, proove that the equilibrium state at 300k will be N2 and H2O vapor" Then, on the ground of the theoretical lessons I would proove the assertion. However, in no place of the book is said "vapor pressure of water in pressence of N2 this or that". You just can deduce it from the very basic principles and mathematics.

Barely could you write a book with the enumeration of facts you have provided. It is nothing against you, I just want a theoretical explanation and not the dumb-engineer recipe. And, I believe strongly that such an explanation is impossibe if you nor even touch the concept of Gibbs chemical potential.

So, I wait for what seem to be a unknown explanation.

By the way, I bet you are an engineer.

 

[D H]

Your first two posts (the OP and #4) illustrated to me and to others an incredible lack of understanding. Freshman level at best, and a poor understanding of freshman-level concepts. We don't know you; we can only infer your level of understanding based on what you post. Given the low caliber of those first few posts, we explained what was going on in simple terms.

If you have a serious question don't come in looking like a fool.

It now appears that you are into psychoceramics and have a hidden agenda. We don't tolerate that here.

 

[Fernsanz]

Ok you are right at to some extent, but not totally.

I admitt that my first posts belied my level of knowledge. I am sorry if that leaded you to think I was at freshman level. Anyway, I didnt mean this to get to you so strongly. In any case I am sorry for that.

However I think it became clear that I am looking for something deeper that just a recipe. So, there is a piece of guilty in you also.

Really, I don't want to bother you and I'm sorry if I have. I hope to keep talking to you sharing our interest in physics.

And, once again, if anyonce could shed light on my question I would be grateful.

 

[Dale]

By the way, I bet you are an engineer.

What is wrong with being an engineer? This is the second time you have said that as though "engineer" were an insult.

You have had essentially the same answer to your question from 4 or 5 different people now, and your responses to DH have become progressively more deliberately ignorant, impolite, and immature.

Do some studying and answer your own questions from now on. It is clear that you need to learn the hard way.

 

[Jax]

What I would to like to find is an explanation of WHY the state of equilibrium at 300K and 1 atm is what we all know it is.

English is probably not your first language; I don't know if I understand exactly what it is you are looking for. I am going to try to provide an explanation as to why there is always some vapor pressure at 25oC. This is going to be an intuitive explanation. If you wanted a mathematical proof then obviously you'll find the explanation not fulfilling.

Let's first consider how our species exist. Your example posits some type of closed system that contains H20 and N2. The molecule of liquid H20 does not stay in constant form. Rather, protons are constantly being swapped between oxygens. This is part of the reason why water, a minimally massed molecule, is even liquid at all at standard temperature. H20 is a polar molecule and does not have a symmetric geometry, with an equilibrium point bond angle of around 104o. This gives water the ability to stick together, and we notice such phenomena as strong surface tension.

N2, on the other hand, IS symmetrical. It contains two nitrogen atoms bonded together with a relatively even charge distribution. This means N2 is terrible at sticking together and the boiling point of liquid nitrogen (which is more massive then water) is very far below 0. This is what makes N2 an inert gas and why it can collide with the surface of water and not get trapped within the liquid.

Now, let's examine your starting conditions: 300K temperature and 1atm pressure. Temperature describes the average kinetic energy of a molecule but not every molecule has exactly 300K. To consider a system at 300K is not enough because it doesn't take into account the variance of kinetic energy. If every molecule of water were exactly 300K then you would not have your dynamic equilibrium situation which includes water vapor. So in reality, we may very well expect differences in molecular velocity to correspond similarly to a Maxwell-Boltzmann distribution.

It is important to consider the differences in velocity in order to answer the question: How does liquid water break from the surface in the first place? Liquid water is held together because of its uneven charge distribution and to break free requires a certain escape speed. Those molecules at the surface with the highest velocities are most likely to break free. It is a cooling effect because once those molecules have left, the remaining average kinetic energy drops, which is what temperature measures.

Now consider the "temperature" of a water vapor molecule just as it escapes the liquid. It is likely higher than the N2 gas around. The N2 gas continues to collide with the surface of the water and if it did not do so there would not be 1 atm pressure. Because it is not polar, N2 remains in the gas phase. However, momenta between N2 and H20 liquid can still be exchanged. Therefore we would expect that the temperature of the N2 gas be LOWER than the newly escaped water vapor.

If hot water vapor continues to rise, the equilibrium of liquid water will never be reached. This explains why unenclosed water volumes continue to evaporate (like a glass of water left outside). In your case, you define the system to be closed. As others have described, dynamic equilibrium is reached when the rate of liquid water evaporating is the same as the rate of water vapor condensing. Pressure is Force per Area. If we consider the system to have a constant volume and a temperature (although varying) that also remains relatively constant, an increase in water vapor pressure would then be indicative of more H20 gas molecules colliding with liquid water surface. Each has a chance to become trapped as part of the liquid again. It is possible for a water vapor molecule to be 300K but can you define one molecule by itself as a liquid?

 

[turbo]

Pathetic.

People are trying to help you. It's not nice to insult them because you can't understand the help. Did you bother Googling on "sublimation"? You seem to have trouble understanding why water can vaporize at temperatures below 100C. How about grasping the fact that water can vaporize at -40C? It happens. Once you accept this, it's easy to grasp how water can vaporize at room temperature.

You have mentioned equilibrium several times. If you put a beaker of water in a sealed room at standard conditions, how will the system (the sealed room) come to equilibrium? Through evaporation and diffusion, the water will eventually be evenly distributed throughout the room. You can do some work to the air and recover the water to once again reduce the entropy of the system and have dry room air and a beaker of water.

 

[Fernsanz]

Hello.

We all know that water at 300K is in vapor form as part of air. Also, we know that at some temperature (dew temperature) liquid water will appear.

Thermodynamically, we are dealing with a system of two components, H2O and N2 -supposing dry air is mostly N2- and two phases.

The precise question is: Could someone explain why water behaves in this way in terms of chemical potentials of water in its different phases?

My reasoning is as follows: in order for a component to be in a single phase, the chemical potential of that component in that phase must be less than in any other fase. Now, in order for a component to be in two phases, the chemical potential of that component in those phases must be equal. This is just the condition of equilibrium of equilaity of chemical potentials. So, if we start at some conditions where only one phase exists and we slowly move to another condition where two phase exist, the point where that second phase appear is just the point where chemical potentials has been done equal. Everything right up to this point.

But now my doubt is: why are chemical potential of water vapor and liquid water equals at dew point? Shouldn't they be different?

Take a closed diathermic vessel which contains water in equilibrium with its vapor (that is, saturated vapor at vapor pressure) at dew point temperature. The vapor pressure will be less than 1 atm. As we have said, the chemical potential of vapor and liquid water are equals in these conditions. Now, to obtain the chemical potencial of liquid water at the same temperature (dew point temperature) and 1 atm, which is what we are interested in, imagine a reversible process to take that water to 1 atm. Well, whichever the process is we have increased its chemical potential. If, in the original conditions inside the vessel both phases had the same chemical potential, now at 1 atm and the same temperature, the liquid water has greater chemical potential. Conclusion: the water should appear only in vapor form.

But the experience is that it condense at that temperature. Where is the flaw of my reasoning?

Thank you very much.

 

[Fernsanz]

Now, to obtain the chemical potencial of liquid water at the same temperature (dew point temperature) and 1 atm, which is what we are interested in, imagine a reversible process to take that water to 1 atm. Well, whichever the process is we have increased its chemical potential. If, in the original conditions inside the vessel both phases had the same chemical potential, now at 1 atm and the same temperature, the liquid water has greater chemical potential. Conclusion: the water should appear only in vapor form.

But the experience is that it condense at that temperature. Where is the flaw of my reasoning?

Ok ok, I have found what the flaw is. For liquids and solids the chemical potential is practically constant with pressure. So, for air surrounded water at the dew point and 1 atm we have the same chemical potential that for water at its vapor pressure and the same temperature. As for the vapor, since the chemical potential of a gas depends only of his partial pressure (or, more generally, membrane pressure) and, in both circunstances we have the vapor at the same partial pressure, it also has the same chemical potential in both circunstances.

This is why we can use the pure-water phase-diagram to make predictions about phase changes of water surrounded by air, even though we can't use it for general predictions, as evidenced by the fact that pure-water phase-diagram predicts liquid water at 300K. This is the rigurous thermodynamic explanation which no one seemed to notice. We wouldn't have been able to explain this if it were not by mean of the chemical potential concept. I can't even imagine how can some discuss about phase changes without a mention to chemical potential.

So, perhaps someone has realize that the question was deeper than they were able to see; and so is the explanation. And hopes anyone has gained insight in what they thought they fully understood.

 

[monish]

I defy any of the know-it-alls who make fun of Fernsanz to find a point on the phase diagram of water where the liquid is at 1 atm 300K and the vapor is at .03 atm 300K.
Yes, the phase diagram shows liquid at .03 atm being in equilibrium with vapor at .03 atm when the temperature is 300K, but that's not the situation here.

His question was about how to read the phase diagram and nobody answered it. And the hypocrisy of those who berated Fernsanz for being insulting to others is unbelievable.

Marty

 

[Fernsanz]

Thank you very very much for your support Marty.

You have summed up the problem perfectly well and I am glad that you have fully understood my question.

Once again, thank you very much.

 

[Fernsanz]

The post from Marty has encouraged me to better restate my question which I have been chewing over.

Lets suppose the partial pressure of water in the air is 0.03 atm and we are at 30ºC. If I asked you how much have I to cool the air to get liquid water, most of you would say 25 ºC. If I asked you why, your answer would be "because if look at the phase diagram of water at 25ºC and a vapor partial pressure of 0.03 atm the water just start to being in equilibrium with liquid water".

Well, but there is a little point here you have passed over: if look at the phase diagram of water at 25ºC and a vapor partial pressure of 0.03 atm the water just start to being in equilibrium with liquid water at those same 0.03 atm, not at 1 atm! So, how could be your reasoning right? In fact it is not generally right.

That was what didn't fit in my mind; and I recognize that perhpas I didn't know to state it clearly, but the post from Marty has made clear to me that I could have been understood.

Thanks and thanks Marty.

 

[Fernsanz]

So, the dew point is, strictly talking, less than 25ºC.

But how much less? Well, we have to make liquid and vapor chemical potentials equal to find the dew point:

$$\mu_v(T_{dew}, 0.03 \mbox{ atm})=\mu_l(T_{dew}, 1 \mbox{ atm})$$​

where l stands for liquid and v for vapor. As I have pointed out in a previous reply, at 1 atm and 25ºC the liquid water chemical potential is greater than that of the vapor at 0.03 atm and 25ºC by an amount - molar volume times pressure difference in right units- $$v \cdot (\Delta P) = 18 \cdot 10^{-6} \cdot (101125-101125 \cdot 0.03)=1.766 J/mol$$, where $$18 \cdot 10^{-6} m^3/mol$$ is the molar volume of liquid water; i.e.

$$\mu_l(25 \mbox{C}, 1 \mbox{ atm}) = \mu_v(25 \mbox{C}, 0.03 \mbox{ atm}) + 1.766$$​

If we want to compensate for this difference we can lower the temperature. The contribution of a difference in temperature to the chemical potential is, if the temperature difference is not to big, $$\mu(T_2, P) = \mu(T_1, P) -s \cdot (\Delta T)$$, where s is the molar entropy and $$\Delta T=T_2-T_1$$ is the temperature difference. The molar entropy for water vapor is 188.83 and for liquid water is 69.91. So, the balance in order for liquid water at 1 atm and water vapor at 0.03 atm to have equals chemical potential starting from the temperature $$T_1$$=25ºC is

$$\mu_v(T_{dew}, 0.03 \mbox{ atm})=\mu_l(T_{dew}, 1 \mbox{ atm}) \hspace{20} \Rightarrow \hspace{20} 1.766 - 69.91·(T_{dew}-T_1) +188.83 \cdot (T_{dew}-T_1) = 0$$​

Solving for $$T_{dew}-T_1$$ we have $$T_{dew}-T_1=-0.015$$ ºC. That is, the dew point is practically 25ºC.

So, numerical calculations show us that the true dew point is practically the same that the dew point you would arrive by the wrong way. However this is just a consequence of the numbers. If you apply the wrong reasoning you will obtain wrong results in general. And, what is more important by far, you will show a lack of understanding of the physics which is going on and will transmit misconceptions to those who receive ur answer.

______________________
"Thermodynamics is a funny subject. The first time you go through it, you don't understand it at all. The second time you go through it, you think you understand it, except for one or two small points. The third time you go through it, you know you don't understand it, but by that time you are so used to it, so it doesn't bother you any more". Arnold Sommerfeld

Anyone is in that second stage?

Regards.