Understanding the first law of thermodynamics in a particular situationby fluidistic Tags: situation, thermodynamics 

#1
May2009, 02:58 PM

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Imagine the system Tabledrop 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. 



#2
May2009, 06:15 PM

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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). 



#3
May2009, 06:47 PM

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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. [tex]\Delta U= QW[/tex]. Here Q is worth 0 cal, and I'm almost sure that [tex]\Delta U=0 J[/tex], so [tex]W[/tex] must be [tex]0J[/tex]. 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? 



#4
May2009, 07:21 PM

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Understanding the first law of thermodynamics in a particular situation
You forgot something: there is energy required to change liquid water into vapor. Where does that energy come from/how is it manifest?




#5
May2009, 07:26 PM

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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? 



#6
May2009, 08:19 PM

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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. 



#7
May2009, 08:27 PM

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#8
May2009, 08:51 PM

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Thank you very much to all. 



#9
May2009, 09:01 PM

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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/Interna...nternal_energy you see that we can write E = T (S1 + S2)  P (V1+V2) + mu1(NN2) + 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. 



#10
May2109, 07:43 AM

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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. 



#11
May2109, 08:14 AM

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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. 



#12
May2109, 09:00 AM

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#13
May2109, 11:57 AM

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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. 



#14
May2109, 11:59 AM

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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.




#15
May2109, 05:41 PM

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#16
May2109, 05:51 PM

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#17
May2109, 05:58 PM

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...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. 



#18
May2109, 06:06 PM

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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. 


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