Does Humidity Affect Entropy in Climate-Controlled Environments?

[Dreebs]

Homework Statement

During the fall, the outside air's temperature is comfortable but its humidity is too high for direct use inside the cabaret. The air feels clammy and damp. So your climate control system chills the outdoor air to extract some of its moisture and then reheats that air back up to its original temperature for use inside the cabaret. When all this is done, you have drier air in the cabaret and water running down the drain. Which arrangement has more entropy (more disorder): a cabaret full of outside air (moist but at the right temperature) or a cabaret full of dried air (drier and at the right temperature) plus the extracted water in the drain?
Select one:
a. There is more entropy in the cabaret full of outside air because if you allow the dried air and extracted water to recombine, they will do so spontaneously.
b. There is more entropy in the cabaret full of outside air because it contains more total energy than does the cabaret full of dried air plus the extracted water.
c. There is more entropy in the cabaret full of dried air plus the extracted water because electrical energy was converted to thermal energy while producing those two separated materials.
d. There is more entropy in the cabaret full of dried air plus the extracted water because electrical energy was transformed into entropy while separating the moisture from the air.

Homework Equations

N/A

The Attempt at a Solution

Either A or B, but I am leaning towards B because moist air full of both water and air particles will be higher in entropy than the two parts separated from each other.

 

[haruspex]

Either A or B, but I am leaning towards B because moist air full of both water and air particles will be higher in entropy than the two parts separated from each other.

Seems to me that is the same as the reason given in A. B mentions energy, but your justification of B does not.

 

[Dreebs]

Seems to me that is the same as the reason given in A. B mentions energy, but your justification of B does not.

But will the water and air spontaneously recombine if allowed to? That didn't make sense to me, or maybe I just don't quite understand.

 

[haruspex]

But will the water and air spontaneously recombine if allowed to? That didn't make sense to me, or maybe I just don't quite understand.

Will they be at the same temperature initially?
What will happen if they are at the same temperature and the air is less than 100%RH?

 

[Dreebs]

Will they be at the same temperature initially?
What will happen if they are at the same temperature and the air is less than 100%RH?

They will be different temperatures because the water is extracted and the air reheated.

 

[haruspex]

They will be different temperatures because the water is extracted and the air reheated.

Right, so what will happen over time with the two in contact?

 

[Chestermiller]

Entropy is a function of state, and has nothing to do with any process for achieving the final state, starting from the initial state.

The initial state is a mixture of air and water vapor at temperature T, both with partial pressures less than 1 atm.

The final state is (presumably) dry air at 1 atm and temperature T, and liquid water at a temperature equal to or lower than T.

From Gibbs' Theorem, the entropy of a component in an ideal gas mixture is the same as the entropy of that same pure component at the temperature of the mixture and at the partial partial pressure in the mixture. So the entropy of the air in the final state is less than that in the initial state, because its partial pressure has increased from whatever it was in the mixture to 1 atm (and the entropy of a pure ideal gas decreases with increasing pressure).

The liquid water also has a lower entropy than the original mixture because the heat of vaporization has been removed. So both the air and the water have lower entropies in the final state than in the initial state. And their sum, of course, is also lower.

 

[Dreebs]

Entropy is a function of state, and has nothing to do with any process for achieving the final state, starting from the initial state.

The initial state is a mixture of air and water vapor at temperature T, both with partial pressures less than 1 atm.

The final state is (presumably) dry air at 1 atm and temperature T, and liquid water at a temperature equal to or lower than T.

From Gibbs' Theorem, the entropy of a component in an ideal gas mixture is the same as the entropy of that same pure component at the temperature of the mixture and at the partial partial pressure in the mixture. So the entropy of the air in the final state is less than that in the initial state, because its partial pressure has increased from whatever it was in the mixture to 1 atm (and the entropy of a pure ideal gas decreases with increasing pressure).

The liquid water also has a lower entropy than the original mixture because the heat of vaporization has been removed. So both the air and the water have lower entropies in the final state than in the initial state. And their sum, of course, is also lower.

So the entropy would be greater in the room with dried air due to the sum of the water and air's entropy being lower than at the start?

 

[Chestermiller]

So the entropy would be greater in the room with dried air due to the sum of the water and air's entropy being lower than at the start?

Sure. This change in entropy can be precisely calculated knowing the partial pressure of the air and the water vapor in the initial state.

 

[Dreebs]

Sure. This change in entropy can be precisely calculated knowing the partial pressure of the air and the water vapor in the initial state.

So this would be due to D, where electrical energy is converted to entropy during the separation?

 

[Chestermiller]

So this would be due to D, where electrical energy is converted to entropy during the separation?

What did I say about the entropy change being independent of any specific process?

In my judgment, all the choices offered are inappropriate (i.e., suck). The choice that comes closest to being correct (again, in my judgment) is option (a).