Boiling point elevation and phase diagram

[Tokspor]

I can understand how lowering the vapor pressure leads to an elevation in the boiling point. Adding a nonvolatile solute makes it more difficult for the pure solvent's vapor pressure to reach 1 atm so a greater increase in temperature is needed.

But I'm having a hard time reconciling this with what the phase diagram seems to indicate. I attached a quick illustration of what I'm trying to explain.

If you start at any point within the liquid region of the diagram, suppose you decrease P at constant T but remain within the liquid region of the graph. In any case, you are still brought to a point within the liquid phase where you are much closer to that liquid/gas equilibrium line and thus closer to becoming a gas. Doesn't this indicate that lowering the pressure lowers the temperature at which a substance boils?

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[G1SG2]

I can understand how lowering the vapor pressure leads to an elevation in the boiling point. Adding a nonvolatile solute makes it more difficult for the pure solvent's vapor pressure to reach 1 atm so a greater increase in temperature is needed.

But I'm having a hard time reconciling this with what the phase diagram seems to indicate. I attached a quick illustration of what I'm trying to explain.

If you start at any point within the liquid region of the diagram, suppose you decrease P at constant T but remain within the liquid region of the graph. In any case, you are still brought to a point within the liquid phase where you are much closer to that liquid/gas equilibrium line and thus closer to becoming a gas. Doesn't this indicate that lowering the pressure lowers the temperature at which a substance boils?

I think that is the phase diagram for CO2, yes? When you add a solute to a pure liquid, you would shift the liquid/gas boundary line to the right for a higher temperature required to reach the gas phase. You lower the vapor pressure by adding the solute but I think that inevitably requires an increase in temperature as you know need more energy to get the molecules to escape into the gas phase (hence the shifting of the liquid/gas boundary line to the right).

Does this answer your question?

 

[Tokspor]

I think that is the phase diagram for CO2, yes? When you add a solute to a pure liquid, you would shift the liquid/gas boundary line to the right for a higher temperature required to reach the gas phase. You lower the vapor pressure by adding the solute but I think that inevitably requires an increase in temperature as you know need more energy to get the molecules to escape into the gas phase (hence the shifting of the liquid/gas boundary line to the right).

Does this answer your question?

Yes, that's for CO2. Thanks for your response. It didn't occur to me that the curve actually shifts due to, I'm supposing, a change in the conditions of the system (i.e. adding a solute). I'm a bit confused though about the two examples you gave. The first, by adding a solute, the curve shifts to the left. The second, by lowering the vapor pressure, the curve shifts to the right. Isn't adding a solute (if it's nonvolatile) and lowering the vapor pressure the same thing? Would both lead to a shift to the right since that way, it takes a higher temperature to reach the liquid/gas boundary?

 

[G1SG2]

Yes, that's for CO2. Thanks for your response. It didn't occur to me that the curve actually shifts due to, I'm supposing, a change in the conditions of the system (i.e. adding a solute). I'm a bit confused though about the two examples you gave. The first, by adding a solute, the curve shifts to the left. The second, by lowering the vapor pressure, the curve shifts to the right. Isn't adding a solute (if it's nonvolatile) and lowering the vapor pressure the same thing? Would both lead to a shift to the right since that way, it takes a higher temperature to reach the liquid/gas boundary?

Np, but I don't think I said anything about the curve shifting to the left? It shifts to the right upon addition of the nonvolatile solute...

 

[Tokspor]

Np, but I don't think I said anything about the curve shifting to the left? It shifts to the right upon addition of the nonvolatile solute...

Oh, I misread that then. Oops.

What led me to ask this was a practice question I had to answer about a substance that was a gas under standard conditions, thereby making its boiling point hard to determine. It asked what could be done to determine the boiling point.

I chose an answer that basically said to lower the pressure, determine the boiling point there, and then extrapolate the curve to a plot with P at 1 atm. My reasoning was that lowering the vapor pressure increases the boiling point.

But the correct answer was to increase the pressure so that the substance boils at higher temperatures. I still can't make sense of this, and having gone over this now makes it less convincing of an answer. Is there something I'm missing?

 

[G1SG2]

Oh, I misread that then. Oops.

What led me to ask this was a practice question I had to answer about a substance that was a gas under standard conditions, thereby making its boiling point hard to determine. It asked what could be done to determine the boiling point.

I chose an answer that basically said to lower the pressure, determine the boiling point there, and then extrapolate the curve to a plot with P at 1 atm. My reasoning was that lowering the vapor pressure increases the boiling point.

But the correct answer was to increase the pressure so that the substance boils at higher temperatures. I still can't make sense of this, and having gone over this now makes it less convincing of an answer. Is there something I'm missing?

Was it asking you in relative terms? Maybe the answer means that if you increased the pressure, the substance will boil at THE higher temperature? It may be a weird wording type of problem, do you have the exact wording? If you increased the pressure, it would be relatively easier to boil. Like if you had two solutions and one had a higher vapor pressure than the other, then it would boil faster.

 

[Tokspor]

Was it asking you in relative terms? Maybe the answer means that if you increased the pressure, the substance will boil at THE higher temperature? It may be a weird wording type of problem, do you have the exact wording? If you increased the pressure, it would be relatively easier to boil. Like if you had two solutions and one had a higher vapor pressure than the other, then it would boil faster.

Ah, thank you for clarifying this. I realized the question did not mention anything about adding solutes to decrease the vapor pressure, so the phase diagram doesn't shift. Going by the phase diagram (which should look like CO2, right? This is a Kaplan pseudo-discrete question which just refers to the compound as a low-molecular weight hydrocarbon), increasing the pressure in that case would likely return the compound to the liquid phase. It could then be boiled at a higher temperature. Hope that does make sense.

 

[housemd?]

I can understand how lowering the vapor pressure leads to an elevation in the boiling point. Adding a nonvolatile solute makes it more difficult for the pure solvent's vapor pressure to reach 1 atm so a greater increase in temperature is needed.

But I'm having a hard time reconciling this with what the phase diagram seems to indicate. I attached a quick illustration of what I'm trying to explain.

If you start at any point within the liquid region of the diagram, suppose you decrease P at constant T but remain within the liquid region of the graph. In any case, you are still brought to a point within the liquid phase where you are much closer to that liquid/gas equilibrium line and thus closer to becoming a gas. Doesn't this indicate that lowering the pressure lowers the temperature at which a substance boils?

from your original post -- are you mixing up vapor pressure with pressure? lowering pressure is not same as lowering vapor pressure.

 

[Tokspor]

from your original post -- are you mixing up vapor pressure with pressure? lowering pressure is not same as lowering vapor pressure.

Well, I know the vapor pressure would be the pressure at which the rate of evaporation is equal to the rate of condensation, correct? And so would pressure refer to the force exerted by the molecules over the walls of the container? I'm a bit confused now--how would a change in vapor pressure then create a shift in the phase diagram?

 

[G1SG2]

Well, I know the vapor pressure would be the pressure at which the rate of evaporation is equal to the rate of condensation, correct? And so would pressure refer to the force exerted by the molecules over the walls of the container? I'm a bit confused now--how would a change in vapor pressure then create a shift in the phase diagram?

Think about adding salt to a pot of hot water and pasta. You want to raise the boiling point so your pasta can cook longer/cook well.

What happens is that the molecules exert a pressure above the liquid as they try to escape into the gas phase. This is the vapor pressure. Now, if you add NaCl, it will block some of the molecules from escaping at the surface since NaCl will also occupy some room. This lowers the vapor pressure, since you don't have that many molecules crashing up at the surface trying to escape. By LOWERING the vapor pressure, you elevated the boiling point because remember, boiling occurs when the vapor pressure=atmospheric pressure. Thus, by doing so, you will have shifted the liquid/gas boundary (evaporation point) to the RIGHT, meaning at an increased temperature (since T is on the x axis). Silly way to make you understand:

liquid molecules attempting to escape into the gas phase: muahaha, let's become gas molecules! We're creating more vapor pressure by hanging out at the surface and trying to escape into the gas phase. That pasta won't be in there for long!

nonvolatile solute: not so fast! Let me block your way for a while.

liquid molecules attempting to escape into the gas phase: dammit! You just lowered the vapor pressure. We worked so hard! Now this pasta will have to stay in here longer because we have to create more pressure to equal to the atmospheric pressure. We almost had it! Dang.

pasta: shut up, I'm cooking here.

water: you're all lame.

The end.

Sorry if I'm corny, but does this help? lol.

 

[loveoforganic]

For a more accurate less analagous description 🙂P I'm not knockin it, if it works it works) ions form stronger intermolecular forces with water than water forms with itself. So the water is bound more tightly to its container and is less likely to boil off because instead of having water-water interaction it has stronger water-ion interaction.

 

[Tokspor]

Now this pasta will have to stay in here longer because we have to create more pressure to equal to the atmospheric pressure.

Thank you for that analogy. I was wondering about the part I quoted above--do you mean we have to create more vapor pressure, instead of just pressure, to equal the atmospheric pressure? The phase diagram seems to indicate that even if there is a shift to the right, we can remain at a constant P and just increase the temperature to get to the liquid/gas boundary.

 

[G1SG2]

Thank you for that analogy. I was wondering about the part I quoted above--do you mean we have to create more vapor pressure, instead of just pressure, to equal the atmospheric pressure? The phase diagram seems to indicate that even if there is a shift to the right, we can remain at a constant P and just increase the temperature to get to the liquid/gas boundary.

I'm a bit confused by what you mean by "pressure"-do you mean the atmospheric pressure? Vapor pressure is independent of atmospheric pressure. I don't see anywhere in the diagram where you would have to remain at constant pressure to get to where you want to be in terms of temperature (for the liquid/gas boundary, that is). If you lowered the vapor pressure by adding a nonvolatile solute, now you have to start building up that pressure up again. Think about it this way: suppose you have five runners that have to hit the 20 meter mark before they can stop running. Now, suppose that the runners just passed the 14 meter mark (they have to run 6 more meters until they can stop) when you stopped them, put them in your car, drove back to the 8 meter mark and let them out. Now they have to run 12 more meters to get to the 20 meter mark since you drove them back. They have to work longer. It doesn't change their 20 meter cut off-that's still the deal. They just have to run longer because you removed them from their 14 meter mark to the 8 meter mark.

Same goes for vapor pressure-you lowered the vapor pressure by adding the nonvalatile solute. Now the liquid molecules have to build that vapor pressure again, but they still have that cutoff of equaling the local atmospheric pressure for boiling to occur.