# Understanding the first law of thermodynamics in a particular situation

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## Main Question or Discussion Point

Imagine the system Table-drop of water on it and air in a closed room, all at the same temperature T. (0°C<T<100°C).
After some time the water drop will evaporate. Shouldn't it make the room pressure increase? If so then I can imagine a system where there's no heat input and the system can do work. (violating the first law of thermodynamics)
So I conclude that as it's impossible, the pressure in the room cannot increase. However I don't understand how/why the pressure of the room does not increase. There are more molecules in the air in an almost same volume. Almost because the water drop is not there anymore. Is it because of the water drop disappearing that the room pressure remains constant all along the process? If so, how can I calculate this? (yeah, it does not convince me at all!)
Thank you very much.

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What is the system that you're imagining?

The evaporation will stop once the partial pressure of water vapour rises high enough. Also, it doesn't violate thermodynamics to perform work without heat input (witness the release of a compressed air tank).

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I'm imagining a perfectly well closed (adiabatic) room (filled with air) with in its center a table and over the table a water drop.
The water will fully evaporates (it won't be able to saturates the air since a single drop cannot do that, considering the big dimensions of the room). All this at a temperature between 0°C and 100°C and initially at a pressure of 1atm.
$$\Delta U= Q-W$$. Here Q is worth 0 cal, and I'm almost sure that $$\Delta U=0 J$$, so $$W$$ must be $$0J$$. Hence the pressure in the room does not increase. How is that possible considering that there are more molecules in the room, temperature doesn't change, and the volume of the room decreases a very bit?

russ_watters
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You forgot something: there is energy required to change liquid water into vapor. Where does that energy come from/how is it manifest?

Mapes
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I'm almost sure that $$\Delta U=0 J$$, so $$W$$ must be $$0J$$.
If you remove energy adiabatically via work ($W>0$), then the energy of the room will decrease ($\Delta U<0$). If you don't, then it won't. What's the problem?

A heat supply isn't required to extract work; you could extract energy in the form of work by letting a weight fall to the floor adiabatically. In that case you extract potential energy. In the case of the evaporating water drop you extract chemical energy: the chemical potential of the water is higher when it's in liquid form. Does this make sense?

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You forgot something: there is energy required to change liquid water into vapor. Where does that energy come from/how is it manifest?
I think it comes from the air's molecules hitting the water drop, exciting water molecules and some of them have enough kinetic energy to leave the drop. But even if the temperature of the drop decreases by a bit and the air temperature increases by a bit, it soon come back to a thermal equilibrium. The process lasts until there's no more liquid water.

A heat supply isn't required to extract work; you could extract energy in the form of work by letting a weight fall to the floor adiabatically. In that case you extract potential energy.
Ah, you're right!
Mapes said:
In the case of the evaporating water drop you extract chemical energy: the chemical potential of the water is higher when it's in liquid form. Does this make sense?
Ah! Yes it makes sense to me.
I should revise my class notes... my physics professor told us that a system cannot produce work if it don't receive heat from its surrounding.

Mapes
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my physics professor told us that a system cannot produce work if it don't receive heat from its surrounding.
Yikes. For some isothermal examples, yes, but not universally.

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Yikes. For some isothermal examples, yes, but not universally.
Ok, he didn't precise so I considered it as universal.
Thank you very much to all.

I think the problem fluidistic is asking about can be formulated better follows: You have a perfectly insulated room kept at constant volume. Heat and work are defined by taking the system boundary to be the boundary of the room, so they are always zero. Then, he concludes that the internal energy must always be constant. This is true, because the total energy inside the room cannot change.

Now, in this case were you have a drop of water that is slowly evaporating, you don't have thermal equilibrium in the room. If you had thermal equilibrium, nothing could change in the room. The state of thermal equilibrium will only be reached if the drop has completely evaporated.

If we want to describe the evaporation process, then merely specifying the macrostate of the room by specifying the total internal energy and volume won't suffice. We want to distinguish between different type of microstates corresponding to how far the drop has evaporated. This can be done by adding extra variables that keeps track of this evaporation process: the number of molecules in the drop of water N1, and the number of water molecules in the vapor phase N2.

Since the evaporation proceeds slowly we can assume that at any given stage of the evaporation process all the microstates for fixed N1 and N2 are equally like, so we have thermal equilibrium in this restricted sense.

The partial derivates of the entropy w.r.t. the number of molecules at fixed energy and volume is defined to be:

dS/dNi = -mu_i/T

The mu_i are called chemical potentials. This then implis that the fundamental thermodynamic relation gets modified to:

dE = T dS - P dV + mu1 dN1 + mu2 dN2

Then, if you read this section from wikipedia:

http://en.wikipedia.org/wiki/Internal_energy#Expressions_for_the_internal_energy

you see that we can write

E = T (S1 + S2) - P (V1+V2) + mu1(N-N2) + mu2 N2

The total volume of V is written as V1 + V2 where V1 is the volume of the drop. N is the total number of water molecules. This equation combined with the ideal gas law for the vapor allows you to compute the pressure and temperature of the room after the drop has evaporated.

Andy Resnick
Imagine the system Table-drop of water on it and air in a closed room, all at the same temperature T. (0°C<T<100°C).
After some time the water drop will evaporate. Shouldn't it make the room pressure increase? If so then I can imagine a system where there's no heat input and the system can do work. (violating the first law of thermodynamics)
So I conclude that as it's impossible, the pressure in the room cannot increase. However I don't understand how/why the pressure of the room does not increase. There are more molecules in the air in an almost same volume. Almost because the water drop is not there anymore. Is it because of the water drop disappearing that the room pressure remains constant all along the process? If so, how can I calculate this? (yeah, it does not convince me at all!)
Thank you very much.
The pressure of the room does not increase. Initially, the system in not in equilibrium- unless there is sufficient water vapor in the room air to set the relative humidity to 100%. So the water drop evaporates to equalize the vapor (partial) pressure. The problem can be simplified if we assume the temperature of the room doesn't change. The change in entropy is then given by both the latent heat of vaporization (the amount of heat required to convert the liquid drop to vapor) and the increased volume the water molecules have to run around. dS = dQ/T + k(V_room/V_drop), where I have been really sloppy- k is some constant that probably depends on T and other stuff, and I think I used the ideal gas law. No coffee = No equations.

Anyhow, the essential point is that your initial configuration is not an equilibrium configuration (again, unless there is 100% relative humidity), so what makes sense is to talk about how the entropy changes and how the free energy changes as equilibrium is approached.

I see now that I made a few mistakes. I forgot to take into account that there is air in the room Also, the last equation I wrote becomes a trivial statement that the total energy is conserved. The whole problem can be simplified as follows. We start with N1 water molecules in the drop, we have N2 water molecules in the vapor and we have N3 air molecules. The total internal energy is always of the form:

E(N1, N2,N3)= N1 e1(T,P) + N2 e2(T,P_vap) + N3 e3(T,P_air)

because the internal energy is an extensive function. P_vap is the vapor pressure, P_air the patial pressure pof the air and P = P_vap + P_air. Conservation of energy and the ideal gas law for the air and the water vapor allows one to solve for the final pressure and temperature after the drop has completely evaporated.

Note that e2 and e3 don't actually depend on their second arguments (the pressure) if we approximate the vapor and the air as ideal gasses.

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Mapes
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The pressure of the room does not increase.
I'm not sure what you're basing this statement on. As water evaporates in a closed room, whether filled with air or not, how can the total pressure not increase?

Andy Resnick
I'm not sure what you're basing this statement on. As water evaporates in a closed room, whether filled with air or not, how can the total pressure not increase?
For the same reason we approximate that the temperature of the room does not change. This is a reasonable simplification for normal-sized rooms and normal-sized drops.

The process can be made more complex (small room, large drop), but the only real change is the number of terms we keep in the free energy change. Again, the essential point is that the starting configuration is not an equilibrium one, but a thermodynamic approach is still valid to model the approach to equilibrium.

Mapes
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OK, I agree that the increase in pressure is small. Since fluidistic originally asked whether it's identically zero, however, I think it's important to emphasize that it's not.

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OK, I agree that the increase in pressure is small. Since fluidistic originally asked whether it's identically zero, however, I think it's important to emphasize that it's not.
It is probably true but I'm not convinced. The volume of the room increases as the water drop evaporates. Using PV=nRT, it's not obvious that if n increases and V increases then P also increases. (It would imply that n increases more than V does. Where n is the number of moles of molecules in the air).

Mapes
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It is probably true but I'm not convinced. The volume of the room increases as the water drop evaporates. Using PV=nRT, it's not obvious that if n increases and V increases then P also increases. (It would imply that n increases more than V does. Where n is the number of moles of molecules in the air).
Why not run the numbers? I'll bet the volume increase in the gas part of the room as the water evaporates is orders of magnitude too little to counteract the pressure increase. (Because the specific volume of water vapor is orders of magnitude more than that of liquid water.)

russ_watters
Mentor
I think it comes from the air's molecules hitting the water drop, exciting water molecules and some of them have enough kinetic energy to leave the drop. But even if the temperature of the drop decreases by a bit and the air temperature increases by a bit...
You have the conclusion in the second sentence backwards. If the air transferrs kinetic energy to the drop (and it does), the temperature of the drop rises while the temperature of the air falls.

...And I hope this part doesn't just add confusion, but since the average temperature of the water is not high enough for boiling, when a water molecule is driven off, the average kinetic energy of the drop decreases, so the temperature of the drop falls again.
...it soon come back to a thermal equilibrium. The process lasts until there's no more liquid water.
Yes.....to a new thermal equilibrium. When water evaporates into a closed room, the resulting temperature of the room is lower than at the beginning.

russ_watters
Mentor
Mapes said:
In the case of the evaporating water drop you extract chemical energy: the chemical potential of the water is higher when it's in liquid form. Does this make sense?
Ah! Yes it makes sense to me.
Makes sense, but it is wrong! The energy of water vapor is higher than that of liquid water. You have to put energy in to water to make it boil/evaporate. You aren't extracting energy here, you are absorbing it.
I should revise my class notes... my physics professor told us that a system cannot produce work if it don't receive heat from its surrounding.
Though you are making some errors here, yes, that is an oversimplification (the professor was probably talking about specific types of thermodynamic systems). But you are also mixing and matching your system boundaries, which is causing confusion.

-In your first post, you defined the system to include the drop of water and the room. The total enthalpy of this system is, in fact, constant.
-Now you are defining the drop of water alone as the system. The enthalpy of the drop of water is not constant. The drop of water extracts thermal energy from the air and converts it to chemical energy as it evaporates.

russ_watters
Mentor
It is probably true but I'm not convinced. The volume of the room increases as the water drop evaporates.
Again, you need to define your system carefully. You talked about a room in your first post. Presumably, the room has walls made of hard materials, not plastic sheeting. Its volume is fixed. That's called a "control volume".
Using PV=nRT, it's not obvious that if n increases and V increases then P also increases. (It would imply that n increases more than V does. Where n is the number of moles of molecules in the air).
With four variables, you can't make statements like that without doing the calculations and I doubt that there is a real-life situation where such a thing as you describe is possible.

Ie, if you have a closed room (as above), the volume is constant but the temperature and pressure both change. Whether the final pressure is higher or lower, I don't know - I suspect it is lower, but I don't know. In any case, solving that requires simultaneous equations - you have the ideal gas law combined with the thermodynamic/chemical process of evaporation.

If you have water an air in a plastic bag, the pressure is constant and the volume is variable, but you again have to do the math to figure out whether the final volume is higher or lower.

That's all described for you in posts #9 & 11.

Mapes
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Makes sense, but it is wrong! The energy of water vapor is higher than that of liquid water.
Russ, I said that the chemical potential of liquid water (i.e., the molar Gibbs free energy G=H-TS=E+PV-TS) is higher than that of (unsaturated) water gas at room temperature. (This can be verified by calculation.) A phase change is spontaneous when a phase with higher chemical potential transforms into a phase with lower chemical potential. Evaporation occurs because it minimizes the total Gibbs free energy.

I just did a fully fledged computation. If you take a room of 4 by 4 by 4 meters and let 1 gram of water evaporate then I find that both the temperature and pressure will drop. I won't post the calculations here as it is quite simple. The important points are:

I assumed ideal gas law for the air and the water vapor. I took the initial temperature to be 20 °C and initial pressure to be 10^5 Pa. To solve the problem we need to use conservation of energy. So, what we need to know is the "absolute" internal energy of water at 20 °C.

The internal energy of the water can be obtained by considering a hypothetical situation were the water would be in equilibrium with the vapor. In that case the specific Gibbs energies of the water and the vapor are the same. This then gives you an equation for the specific internal energy of water in terms of the specific enthalpy of the vapor, the latent heat and the specific volume of the water and the pressure.

The latent heat is roughly 2450 kj/kg at 20 °C . We know that the enthalpy per H20 molecule will be approximately 4 k T, because H2O has 6 degrees of freedom apart from vibrations. So, the internal energy of the water can be easily computed. Internal energy of the air will be 5/2 k T per molecule.

In the final state we only have an ideal gas. Internal energy of the water vapor will be 3 k T per molecule. So, we can easily solve for the final temperature.

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Ok, we cannot simplify the problem as I was doing.
Sorry for all the confusions.
Now I'm aware that the problem is much more complicated than I thought. I'm not qualified enough (yet I hope) to solve it. I didn't even learn about entropy yet but will do soon...
Thank you russ_watters for all the clarifications by the way.

Ok, we cannot simplify the problem as I was doing.
Sorry for all the confusions.
Now I'm aware that the problem is much more complicated than I thought. I'm not qualified enough (yet I hope) to solve it. I didn't even learn about entropy yet but will do soon...
Thank you russ_watters for all the clarifications by the way.

It turns out that the latent heat is the dominant factor in the problem. You can completely ignore the change in volume due to the drop evaporating. And (I have to check this) even the fact that you are adding water molecules to the air.

So, a simplified explanation would be that after the drop evaporates, the internal energy of the gas must become lower because of the huge latent heat of the water. Now internal energy is proportional to N k T. So, N k T will become smaller than what is was. This means that the temperature will become lower. Since N k T = P V, this also means that the pressure will drop.

Mapes
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I just did a fully fledged computation. If you take a room of 4 by 4 by 4 meters and let 1 gram of water evaporate then I find that both the temperature and pressure will drop. I won't post the calculations here as it is quite simple.
I calculate that the gram of water (1/18.02 moles) completely evaporates, contributing an additional partial pressure of 2.11 Pa if the temperature were unchanged. The air in the room (2627 moles) contributes 2.2 kJ of energy towards water vaporization, reducing the temperature by 0.0403°C. The corresponding decrease in pressure is 13.8 Pa, which offsets the increase.

I stand corrected! I figured if evaporating water increased the pressure in a vacuum, it would do the same for an air filled chamber. Of course, it's a little unrealistic to ignore the heat capacity of the walls...

EDIT: Corrected a couple initial errors.

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Mapes
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...In fact, if the container walls are more than a couple millimeters thick, I calculate that the pressure will indeed increase when the drop evaporates. (I assumed plastic walls, specific heat of about 1 kJ/kg-K, density about 1 kg/m3.)