Cooling Water in a Closed Container

[Electric to be]

I have a bit of confusion about a closed container scenario.

First, I'll start with the open container. Say water is at some temperature, exposed to low humidity atmosphere, and begins to evaporate. The water that evaporates diffuses never to return to the container. To find the final temperature of the water is pretty easy:

Assuming Heat of Vaporization stays somewhat constant ...

(Heat of Vaporization) * (Mass of Water Evaporated) = (Final Temperature * Mass Initial - Initial Temperature * Initial Mass) * (Specific Heat).However, I'm having trouble with the closed container. Assume initially there is no water vapor at all, but still the same atmospheric pressure on the water (just to avoid boiling, though it doesn't really matter). Then assume a certain mass of water quickly evaporated and is now in vapor form so that the vapor is now at saturation.

At this point to find the temperature difference in the liquid water the previous equation should be fine to use, however: won't there now be a small temperature difference between the vapor and the water? The vapor wasn't able to escape and is still in the closed container. So now won't there be some heat exchange in addition to the temperature lost due to evaporation?

How would I quantify this to find the final temperature of the liquid water based on the initial conditions of the water being at a particular temperature and no initial vapor above it?

 

[BvU]

You assume no heat exchange with the container or the outside world, which is fine (but should be mentioned).
At equilibrium you have two conditions: same temperature and same total amount of water. That should be enough...

 

[Electric to be]

You assume no heat exchange with the container or the outside world, which is fine (but should be mentioned).
At equilibrium you have two conditions: same temperature and same total amount of water. That should be enough...

Is the temperature of the liquid water the same though? Shouldn't it be cooler since some of it evaporated?

 

[BvU]

Yes. With 'same temperature' I mean: water and water vapour have the same temperature.

 

[Electric to be]

Yes. With 'same temperature' I mean: water and water vapour have the same temperature.

Okay, well I'm still somewhat confused I guess. I know there will be heat transfer, but I don't know exactly how much because I don't know how much the temperatures will vary prior to reaching equilibrium. Do you think you could help walk me through the energy transfers and math? Starting from the initial point of a certain amount of water at a specific temperature. I know the final amount of vapor will depend on the final temperature as well..

 

[BvU]

First step:
Convert the conditions into equations.

 

[Electric to be]

First step:
Convert the conditions into equations.

Well like I said, my general idea is: (Heat of Vaporization) * (Mass of Water Evaporated) + Heat Transfer between Liquid Water/Gas = (Final Temperature * Mass Initial - Initial Temperature * Initial Mass) * (Specific Heat), but I don't know if this is capturing the full picture..

 

[Chestermiller]

Well like I said, my general idea is: (Heat of Vaporization) * (Mass of Water Evaporated) + Heat Transfer between Liquid Water/Gas = (Final Temperature * Mass Initial - Initial Temperature * Initial Mass) * (Specific Heat), but I don't know if this is capturing the full picture..

Have you studied the first law of thermodynamics? To begin with, you have liquid water and air in the container, with no water vapor in the air. So, this is the same as there being a barrier between the water and the air. Then, you remove the barrier, and let the system re-equilibrate.

So, State 1 is: Liquid water + bone dry air

State 2 is: (less) Liquid water + air saturated with water vapor

If the container is rigid, how much work does the contents of the container do on the rigid container?
If the container is insulated, how much heat is exchanged with the surrounding through the walls of the container?
What is the change in internal energy of the container contents between State 1 and State 2?

 

[Electric to be]

Have you studied the first law of thermodynamics? To begin with, you have liquid water and air in the container, with no water vapor in the air. So, this is the same as there being a barrier between the water and the air. Then, you remove the barrier, and let the system re-equilibrate.

So, State 1 is: Liquid water + bone dry air

State 2 is: (less) Liquid water + air saturated with water vapor

If the container is rigid, how much work does the contents of the container do on the rigid container?
If the container is insulated, how much heat is exchanged with the surrounding through the walls of the container?
What is the change in internal energy of the container contents between State 1 and State 2?

I actually thought of a solution. I'm going from state 1 to state 2 while using heat transfer through a method that isn't exactly what happens in real life, but the initial and final states are the same, and there is also no net heat transfer so that the change in internal energy is the same.

First, I assume a piston is locked over the water leaving basically no volume left for vapor, and so there is negligibly small amount of vapor. The container is completely insulated. The water starts at initial temperature T1. Then I remove heat from the container that brings the water down to temperature T2. This heat exchange is done very quickly so no water has time to evaporate.

So Q-Removed = ( Mass * Specific Heat * T1 - T2)Then, I unlock the piston and place a small weight that provides an external pressure exactly equal to the saturated pressure of the very small amount of vapor. I then, very slowly begin to heat the container. Since the saturated pressure remains constant because of the constant external pressure, the temperature of the heat and water must also be constant P(T2) . The piston however, will be pushed up, and there will also be some energy that goes into Latent heat of evaporation.

So Q-Added = (m * Heat of Vaporization) + P(T2) * V

This solution assumed the volume remains about constant, but it can be easily found as a function of T2 if needed.

Finally, since the original problem description assumed no heat transfer, Q-Added = Q-Removed and T2 can be solved for.

Does this look okay?

 

[Chestermiller]

I actually thought of a solution. I'm going from state 1 to state 2 while using heat transfer through a method that isn't exactly what happens in real life, but the initial and final states are the same, and there is also no net heat transfer so that the change in internal energy is the same.

First, I assume a piston is locked over the water leaving basically no volume left for vapor, and so there is negligibly small amount of vapor. The container is completely insulated. The water starts at initial temperature T1. Then I remove heat from the container that brings the water down to temperature T2. This heat exchange is done very quickly so no water has time to evaporate.

So Q-Removed = ( Mass * Specific Heat * T1 - T2)Then, I unlock the piston and place a small weight that provides an external pressure exactly equal to the saturated pressure of the very small amount of vapor. I then, very slowly begin to heat the container. Since the saturated pressure remains constant because of the constant external pressure, the temperature of the heat and water must also be constant P(T2) . The piston however, will be pushed up, and there will also be some energy that goes into Latent heat of evaporation.

So Q-Added = (m * Heat of Vaporization) + P(T2) * V

This solution assumed the volume remains about constant, but it can be easily found as a function of T2 if needed.

Finally, since the original problem description assumed no heat transfer, Q-Added = Q-Removed and T2 can be solved for.

Does this look okay?

Please precisely define the initial and final states of the system, without any reference whatsoever to how you get from the initial to the final state.

 

[Electric to be]

Please precisely define the initial and final states of the system, without any reference whatsoever to how you get from the initial to the final state.

Well it's exactly what you said:

State 1: Just liquid water in a container at Temperature T1.

State 2: Less Liquid Water, and Vapor in a container at Temperature T2.

 

[Chestermiller]

Well it's exactly what you said:

State 1: Just liquid water in a container at Temperature T1.

State 2: Less Liquid Water, and Vapor in a container at Temperature T2.

So the volume of the container is constant?
So there is space in the container above the liquid water?
So there is air in the space above the liquid water?

 

[Electric to be]

So the volume of the container is constant?
So there is space in the container above the liquid water?
So there is air in the space above the liquid water?

Okay sorry. The initial state has a certain mass of water, with let's say a complete vacuum above it in an isolated container of constant volume at temperature T1.

The final state has slightly less water, with water vapor filling the originally vacuumed volume and both have equilibriated at temperature T2.

I also assume both states have the same internal energy, so no external heat or work.

I also now realize that my original solution is wrong, since there is work being done.

So to begin to solve this I realize I should equate the internal energies of the two states.

Heat Content of Water + Potential Energy of Water = Heat Content of Water + Potential Energy of Water + Heat Content of GasOf these, the heat content of water and gas are straight forward expressions that depend solely on mass and temperature, but I'm not exactly sure how to deal with the potential energy. I would think the latent heat of vaporization has something to do with this potential energy difference, however when water cools due to evaporation it isn't just because of potential energy increases, the gas molecules actually leave with their leftover kinetic energy and take this energy with them.

Basically I'm having trouble with the "energy accounting" of this problem.

 

[Chestermiller]

Okay sorry. The initial state has a certain mass of water, with let's say a complete vacuum above it in an isolated container of constant volume at temperature T1.

The final state has slightly less water, with water vapor filling the originally vacuumed volume and both have equilibriated at temperature T2.

I also assume both states have the same internal energy, so no external heat or work.

I also now realize that my original solution is wrong, since there is work being done.

So to begin to solve this I realize I should equate the internal energies of the two states.

Heat Content of Water + Potential Energy of Water = Heat Content of Water + Potential Energy of Water + Heat Content of GasOf these, the heat content of water and gas are straight forward expressions that depend solely on mass and temperature, but I'm not exactly sure how to deal with the potential energy. I would think the latent heat of vaporization has something to do with this potential energy difference, however when water cools due to evaporation it isn't just because of potential energy increases, the gas molecules actually leave with their leftover kinetic energy and take this energy with them.

Basically I'm having trouble with the "energy accounting" of this problem.

Excellent. This is a major improvement from your original analysis. For the future, it is important to remember to always focus on the initial and final thermodynamic equilibrium states of a system. This technique will serve you well, particularly when you get to the concept of entropy.

When you refer to potential energy, I assume you are not referring to gravitational potential energy, but rather to potential energy of interaction between the molecules. This is included in the internal energy U of the material, so it is not necessary to treat it separately.

So, with this said, you have correctly determined that the internal energy U of the system in State 2 is the same as the internal energy of the system in State 1:$$U_2=U_1$$
Let $T_1$ = temperature of the system in State 1
$T_2$ = temperature of the system in State 2
$V$ = volume of container
$m$ = total mass of water in container
$\Delta m$ mass of liquid that evaporates between States 1 and 2
$v_L$ = Specific volume of liquid water = 0.001 $m^3/kg$

We need to solve for the final temperature $T_2$. To do this we are going to have to determine the final pressure and the amount of water that evaporates. In State 2, the liquid water and water vapor are in equilibrium at the equilibrium vapor pressure:$$P_2=p(T_2)$$where p(T) is the equilibrium vapor pressure at temperature T.
If the total volume of the container is V, and the mass of liquid water remaining in the container in State 2 is $(m-\Delta m)$, what is the volume available for the vapor to fill in State 2?

 

[Electric to be]

Excellent. This is a major improvement from your original analysis. For the future, it is important to remember to always focus on the initial and final thermodynamic equilibrium states of a system. This technique will serve you well, particularly when you get to the concept of entropy.

When you refer to potential energy, I assume you are not referring to gravitational potential energy, but rather to potential energy of interaction between the molecules. This is included in the internal energy U of the material, so it is not necessary to treat it separately.

So, with this said, you have correctly determined that the internal energy U of the system in State 2 is the same as the internal energy of the system in State 1:$$U_2=U_1$$
Let $T_1$ = temperature of the system in State 1
$T_2$ = temperature of the system in State 2
$V$ = volume of container
$m$ = total mass of water in container
$\Delta m$ mass of liquid that evaporates between States 1 and 2
$v_L$ = Specific volume of liquid water = 0.001 $m^3/kg$

We need to solve for the final temperature $T_2$. To do this we are going to have to determine the final pressure and the amount of water that evaporates. In State 2, the liquid water and water vapor are in equilibrium at the equilibrium vapor pressure:$$P_2=p(T_2)$$where p(T) is the equilibrium vapor pressure at temperature T.
If the total volume of the container is V, and the mass of liquid water remaining in the container in State 2 is $(m-\Delta m)$, what is the volume available for the vapor to fill in State 2?

The volume would be $V$ - ($(m -\Delta m)$ *$v_L$). And yes I was referring to potential energy of molecular interactions.

 

[Chestermiller]

The volume would be $V$ - ($(m -\Delta m)$ *$v_L$). And yes I was referring to potential energy of molecular interactions.

Good. So, from the ideal gas law, with this volume, the final temperature $T_2$, and the final pressure $p(T_2)$, what is the final number of moles of water vapor in the head space $n_2$?

 

[Electric to be]

Good. So, from the ideal gas law, with this volume, the final temperature $T_2$, and the final pressure $p(T_2)$, what is the final number of moles of water vapor in the head space $n_2$?

( V - ( (m−Δm)* vL ) ) * p(T2) / (R * T2) = n2

 

[Chestermiller]

( V - ( (m−Δm)* vL ) ) * p(T2) / (R * T2)

Yes. And this is also equal to $\Delta m/M$, where M is the molecular weight of water. So we have:$$\frac{\Delta m}{M}=\frac{( V - (m−Δm)v_L ) p(T_2)} { R T_2}$$ This equation can be used to determine the final mass of vapor $\Delta m$ exclusively in terms of $T_2$. If you solve this for $\Delta m$, what do you get?

 

[Electric to be]

Yes. And this is also equal to $\Delta m/M$, where M is the molecular weight of water. So we have:$$\frac{\Delta m}{M}=\frac{( V - (m−Δm)v_L ) p(T_2)} { R T_2}$$ This equation can be used to determine the final mass of vapor $\Delta m$ exclusively in terms of $T_2$. If you solve this for $\Delta m$, what do you get?

$$\Delta m = \frac {V p(T_2) - m v_L p(T_2)} {R T_2 / M - v_L p(T_2)}$$

I have gotten to this point in previous attempts, and while I know the heat content of both states, how do I deal with a change in potential energy? I know that the two internal energies of the states should be equal, so how do I proceed when I know how much mass has changed?

 

[Chestermiller]

$$\Delta m = \frac {V p(T_2) - m v_L p(T_2)} {R T_2 / M - v_L p(T_2)}$$

I have gotten to this point in previous attempts, and while I know the heat content of both states, how do I deal with a change in potential energy?

Wow. Very impressive. Now, as I said before, the potential energy is embodied in the internal energy, so it doesn't have to be treated separately. I am going to write down the equations for the initial and final internal energies of the system, and let you then ask questions.

$$U_1=mC(T_1-T_R)$$
where $T_R$ is the temperature of the reference (datum) state where U is taken to be zero (usually, zero degrees centigrade) and C is the heat capacity of the liquid water.
$$U_2=mC(T_2-T_R)+\Delta m \Delta u$$
where $\Delta u$ is the internal energy change per unit mass in going from saturated liquid at $T_2$ to saturated vapor at $T_2$. $\Delta u$ is related to the heat of vaporization per unit mass $\Delta h$ at $T_2$ by $$\Delta u=\Delta h-p(T_2)(v_V-v_L)$$, where $v_V$ is the specific volume of the saturated vapor at $T_2$.

 

[Electric to be]

Wow. Very impressive. Now, as I said before, the potential energy is embodied in the internal energy, so it doesn't have to be treated separately. I am going to write down the equations for the initial and final internal energies of the system, and let you then ask questions.

$$U_1=mC(T_1-T_R)$$
where $T_R$ is the temperature of the reference (datum) state where U is taken to be zero (usually, zero degrees centigrade) and C is the heat capacity of the liquid water.
$$U_2=mC(T_2-T_R)+\Delta m \Delta u$$
where $\Delta u$ is the internal energy change per unit mass in going from saturated liquid at $T_2$ to saturated vapor at $T_2$. $\Delta u$ is related to the heat of vaporization per unit mass $\Delta h$ at $T_2$ by $$\Delta u=\Delta h-p(T_2)(v_V-v_L)$$, where $v_V$ is the specific volume of the saturated vapor at $T_2$.

First off, is your expression for U2 supposed to be:

$$U_2= (m- \Delta m) C(T_2-T_R)+\Delta m \Delta u$$

Since the mass of water changed?

EDIT: Also shouldn't there be a component in U2 that accounts for the gas thermal energy of 5/2 nRT?

 

[Chestermiller]

First off, is your expression for U2 supposed to be:

$$U_2= (m- \Delta m) C(T_2-T_R)+\Delta m \Delta u$$

Since the mass of water changed?

No. If you were measuring the change in U for State 2 relative to the reference state, you would first heat all the liquid to T2, and then add additional heat to evaporate the part that becomes vapor.

 

[Electric to be]

No. If you were measuring the change in U for State 2 relative to the reference state, you would first heat all the liquid to T2, and then add additional heat to evaporate the part that becomes vapor.

I'm somewhat confused with Delta U and Delta H. Does the Heat of Vaporization depend on pressure, or is it a constant depending on temperature? Basically here delta H is being treated as a constant, but doesn't delta H depend on pressure and exactly how far a gas has expanded. Are there tables that list values for P,V combinations or something of the sort?

 

[Chestermiller]

I'm somewhat confused with Delta U and Delta H. Does the Heat of Vaporization depend on pressure, or is it a constant depending on temperature? Basically here delta H is being treated as a constant, but doesn't delta H depend on pressure and exactly how far a gas has expanded. Are there tables that list values for P,V combinations or something of the sort?

The definition of the heat of vaporization is the change in enthalpy in going from a specified mass of pure saturated liquid to the same specified mass of pure saturated vapor (either a kg, a gram, or a mole). Since saturated conditions involve a specific temperature and pressure, the change is carried out at the constant temperature and pressure. If the pressure is constant, from the first law, the change in enthalpy is also equal to the heat added at constant pressure to change all the specified mass from liquid to vapor. The change in internal energy between two states and the change in enthalpy between the same two states are related by $\Delta H=\Delta U+\Delta (PV)$. The data on all these state functions for water (saturation temperature and pressure, specific volume, enthalpy, internal energy for both liquid and vapor) is found in the so-called Steam Tables. Just google Steam Tables.

Definition of Heat of Vaporization is another situation where it is helpful to use the State 1 - State 2 formalism:

State 1: 1 kg of liquid water at temperature T and saturation pressure p(T)

State 2: 1 kg of water vapor at temperature T and saturation pressure p(T)

The heat of vaporization (per kg) at T and p(T) is equal to the change in enthalpy of the water between State 1 and State 2.

 

[Electric to be]

The definition of the heat of vaporization is the change in enthalpy in going from a specified mass of pure saturated liquid to the same specified mass of pure saturated vapor (either a kg, a gram, or a mole). Since saturated conditions involve a specific temperature and pressure, the change is carried out at the constant temperature and pressure. If the pressure is constant, from the first law, the change in enthalpy is also equal to the heat added at constant pressure to change all the specified mass from liquid to vapor. The change in internal energy between two states and the change in enthalpy between the same two states are related by $\Delta H=\Delta U+\Delta (PV)$. The data on all these state functions for water (saturation temperature and pressure, specific volume, enthalpy, internal energy for both liquid and vapor) is found in the so-called Steam Tables. Just google Steam Tables.

Definition of Heat of Vaporization is another situation where it is helpful to use the State 1 - State 2 formalism:

State 1: 1 kg of liquid water at temperature T and saturation pressure p(T)

State 2: 1 kg of water vapor at temperature T and saturation pressure p(T)

The heat of vaporization (per kg) at T and p(T) is equal to the change in enthalpy of the water between State 1 and State 2.

This makes a lot of sense thank you. However, as for State 1. In our case, our State 1 is water at temperature T but there's no pressure over it. My thinking is this doesn't matter, since the pressure or lack thereof doesn't affect the internal energy of just water, correct?

In other words, we would use a tabulated vaporization for water in a vacuum, but since this doesn't exist (or is hard to find), we use an existing tabulated enthalpy for the same temperature of water. When you subtract the "work" that was done in this pretend scenario, you get the same change in internal energy as you would had you had the vacuum enthalpy change tabulated. Got it!

 

[Chestermiller]

This makes a lot of sense thank you. However, as for State 1. In our case, our State 1 is water at temperature T but there's no pressure over it. My thinking is this doesn't matter, since the pressure or lack thereof doesn't affect the internal energy of just water, correct?

If the liquid water in State 1 is in contact with the head space, and the head space is a vacuum, then this is not a thermodynamic equilibrium state. In order for State 1 to be a thermodynamic equilibrium state (prior to allowing the system to transition to State 2), we need to temporarily put a barrier between the liquid water and the vacuum, and use the barrier to apply at least the equilibrium vapor pressure to the liquid water at T1. Any pressure higher than the equilibrium vapor pressure will do, since liquid water is essentially incompressible. So, as you said, the internal energy would not be affected. At time zero, we then remove the barrier and allow the system to re-equilibrate.

In other words, we would use a tabulated vaporization for water in a vacuum, but since this doesn't exist (or is hard to find), we use an existing tabulated enthalpy for the same temperature of water. When you subtract the "work" that was done in this pretend scenario, you get the same change in internal energy as you would had you had the vacuum enthalpy change tabulated. Got it!

Not really. It is not necessary to create a scenario for getting from State 1 to State 2. In real life, the change would occur spontaneously. However, I hope you realize that we are doing two separate things in this problem. (1) Using the first law to establish that the change in internal energy of our actual system from State 1 and State 2 is zero, and (2) creating equations for calculating the internal energy for liquid water and water vapor at different temperatures and pressures (and fractions of liquid water), relative to the reference state, irrespective of any process.

To do item 2, we devise a simple and convenient path from the reference state (0 C and 1 atm.) to any other arbitrary state of liquid water or water vapor and use property equations that we know to determine the change in internal energy relative to the reference state. For example, in the case of liquid water, we start with 1 kg of liquid and increase its temperature at 1 atm. from 0 C to temperature T; the internal energy for this change is $C(T-T_R)$. We then drop the pressure on the liquid water from 1 atm. to the equilibrium vapor pressure at temperature T; the internal energy for this change is zero (since no work is done).

In the case of water vapor at temperature T and saturation pressure p(T), we first again start out with liquid water in the reference state, and perform the same two steps again, bringing the liquid water to temperature T and pressure p(T). We then add additional heat at constant temperature T and pressure p(T) to change the liquid water to water vapor; the internal energy change for this step is $\Delta h_{vap}(T)-p(T)\Delta v_{vap}(T)$.

Notice that the changes that we obtain using item 2 do not necessarily bear any resemblance to the changes experienced in our actual process. But, since internal energy is a function only of state (and not path), they will still give the correct values for internal energy U in States 1 and 2.

If we are using Steam Tables to do our calculations, there is no need to implement item 2, since this has already been done for us. Basically, they have derived the same equations that we have derived, and put the results of the item 2 calculations in a convenient table.

 

[Electric to be]

If the liquid water in State 1 is in contact with the head space, and the head space is a vacuum, then this is not a thermodynamic equilibrium state. In order for State 1 to be a thermodynamic equilibrium state (prior to allowing the system to transition to State 2), we need to temporarily put a barrier between the liquid water and the vacuum, and use the barrier to apply at least the equilibrium vapor pressure to the liquid water at T1. Any pressure higher than the equilibrium vapor pressure will do, since liquid water is essentially incompressible. So, as you said, the internal energy would not be affected. At time zero, we then remove the barrier and allow the system to re-equilibrate.

Not really. It is not necessary to create a scenario for getting from State 1 to State 2. In real life, the change would occur spontaneously. However, I hope you realize that we are doing two separate things in this problem. (1) Using the first law to establish that the change in internal energy of our actual system from State 1 and State 2 is zero, and (2) creating equations for calculating the internal energy for liquid water and water vapor at different temperatures and pressures (and fractions of liquid water), relative to the reference state, irrespective of any process.

To do item 2, we devise a simple and convenient path from the reference state (0 C and 1 atm.) to any other arbitrary state of liquid water or water vapor and use property equations that we know to determine the change in internal energy relative to the reference state. For example, in the case of liquid water, we start with 1 kg of liquid and increase its temperature at 1 atm. from 0 C to temperature T; the internal energy for this change is $C(T-T_R)$. We then drop the pressure on the liquid water from 1 atm. to the equilibrium vapor pressure at temperature T; the internal energy for this change is zero (since no work is done).

In the case of water vapor at temperature T and saturation pressure p(T), we first again start out with liquid water in the reference state, and perform the same two steps again, bringing the liquid water to temperature T and pressure p(T). We then add additional heat at constant temperature T and pressure p(T) to change the liquid water to water vapor; the internal energy change for this step is $\Delta h_{vap}(T)-p(T)\Delta v_{vap}(T)$.

Notice that the changes that we obtain using item 2 do not necessarily bear any resemblance to the changes experienced in our actual process. But, since internal energy is a function only of state (and not path), they will still give the correct values for internal energy U in States 1 and 2.

If we are using Steam Tables to do our calculations, there is no need to implement item 2, since this has already been done for us. Basically, they have derived the same equations that we have derived, and put the results of the item 2 calculations in a convenient table.

If you have water at temperature T and a pressure of 1 ATM. How is it possible to lower the pressure to the saturated pressure at T? At pressures above the saturated pressure vapor can't exist, but if the pressure is is lowered to the saturated pressure doesn't that mean that some of the vapor has evaporated, and therefore the temperature of the water dropped below T?

 

[Chestermiller]

If you have water at temperature T and a pressure of 1 ATM. How is it possible to lower the pressure to the saturated pressure at T? At pressures above the saturated pressure vapor can't exist, but if the pressure is is lowered to the saturated pressure doesn't that mean that some of the vapor has evaporated, and therefore the temperature of the water dropped below T?

No. Imagine that you have liquid water contained in a vertical cylinder with a massless, frictionless piston, and you have a set of weights sitting on top of the piston. The cylinder is immersed in a vacuum (so there is no air pressure outside the cylinder acting on the top of the piston), and the sum of the weights divided by the area of the piston is equal to 1 bar. So the pressure on the water is 1 bar, and there is no vapor present in the cylinder above the liquid. Now you begin removing weights from the piston. As you remove weights, the pressure exerted by the piston on the liquid is dropping. Now you remove enough weights that the pressure on the liquid is 0.000001 Pa above the equilibrium vapor pressure of the liquid. So there is still no vapor above the liquid. Even if you removed enough weights to exactly match the vapor pressure, vapor would just be on the verge of forming, but would not form yet.

 

[Electric to be]

No. Imagine that you have liquid water contained in a vertical cylinder with a massless, frictionless piston, and you have a set of weights sitting on top of the piston. The cylinder is immersed in a vacuum (so there is no air pressure outside the cylinder acting on the top of the piston), and the sum of the weights divided by the area of the piston is equal to 1 bar. So the pressure on the water is 1 bar, and there is no vapor present in the cylinder above the liquid. Now you begin removing weights from the piston. As you remove weights, the pressure exerted by the piston on the liquid is dropping. Now you remove enough weights that the pressure on the liquid is 0.000001 Pa above the equilibrium vapor pressure of the liquid. So there is still no vapor above the liquid. Even if you removed enough weights to exactly match the vapor pressure, vapor would just be on the verge of forming, but would not form yet.

Why exactly can't vapor form if there is external pressure being applied of at least the saturated pressure? I thought that the saturated pressure meant that for the specific temperature of vapor, vapor is at a specific pressure and if it exceeds this pressure it will condense into liquid. If the piston is at exactly the saturated vapor pressure won't vapor be perfectly fine being at this pressure?

An example: say you had water and an inert gas in a container. At first there's no water vapor and the pressure of the inert gas happens to be equal to the saturated pressure of water at that temperature. Water will still evaporate in this scenario until the partial pressure of vapor is equal to saturation pressure. Yet the total pressure acting on the liquid water is greater than the saturation pressure when the vapor reached saturation.

What are the differences between these two cases, if there is one?Edit: I figured it out. If water tried to evaporate the temperature of the system would lower, as if we have discovered. However, the vapor would still be constrained to be at the higher pressure set by the piston. Since it can't be at this pressure, the vapor would condens. The result is no vapor at all.

 

[Chestermiller]

Why exactly can't vapor form if there is external pressure being applied of at least the saturated pressure? I thought that the saturated pressure meant that for the specific temperature of vapor, vapor is at a specific pressure and if it exceeds this pressure it will condense into liquid. If the piston is at exactly the saturated vapor pressure won't vapor be perfectly fine being at this pressure?

An example: say you had water and an inert gas in a container. At first there's no water vapor and the pressure of the inert gas happens to be equal to the saturated pressure of water at that temperature. Water will still evaporate in this scenario until the partial pressure of vapor is equal to saturation pressure. Yet the total pressure acting on the liquid water is greater than the saturation pressure when the vapor reached saturation.

What are the differences between these two cases, if there is one?Edit: I figured it out. If water tried to evaporate the temperature of the system would lower, as if we have discovered. However, the vapor would still be constrained to be at the higher pressure set by the piston. Since it can't be at this pressure, the vapor would condens. The result is no vapor at all.

Good. Well done.

Do you want to try some calculations using both the equations we derived and the steam tables to see how the results compare? If so, I will specify a problem that you can solve quantitatively for the temperature decrease.

 

[Chestermiller]

Here's a focus problem for you to work on. You have 1 kg of liquid water, saturated at 50 C, sitting under the barrier in the large, rigid, adiabatic container, with vacuum above the barrier. You remove the barrier and allow the system to re-equilibrate. For the final equilibrium temperature of the resulting liquid water and water vapor mixgture to be 45 C,

1. What is the final mass of water vapor in the head space?
2. What is the total volume of the rigid container?

For using the steam tables to solve this, the applicable equations would be:
$$m_L+m_V=m_0$$
$$m_Lu_L+m_Vu_V=m_0u_{L0}$$
$$m_Lv_L+m_Vv_V=V$$
where $m_0$ is the original mass of liquid water (1 kg), $u_{L0}$ is the original internal energy per kg of the original liquid water at 50 C, V is the total volume of the rigid container, the lower case v's are specific volume (volume per kg) and where lower case symbols without subscript 0's refer to the final state..

Show that these equations can be reduced to:
$$u_L+x(u_V-u_L)=u_{L0}$$
$$v_L+x(v_V-v_L)=\frac{V}{m_0}$$
where x is the equilibrium mass fraction of vapor at the new temperature of 45C.