Relationship between Gas T and Pvap.

[Rhino1000]

Edit: My goodness. I meant liquid T.

Q1: Pvap increases exponentially with T. Can anyone provide some intuition for this phenomenon? Specifically for the fact that it is an exponential relationship.

Q2: Why does changing atmospheric pressure change the boiling point of a liquid, but not the vapor pressure? (TBR says, according to my notes) that substance x will have a higher Pvap if it has a lower (normal) boiling point than y. Vapor pressure only depends on the solution temperature and the enthalpy of vaporization.

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[gettheleadout]

Q1: I believe if you plot P_vap as a function of T you should see logarithmic increase in P_vap. This should be the result of the Clausius-Clapeyron relationship such that (∂P / ∂T) = (1 / T). Mathematically this is a little outside my realm of expertise but I believe I'm interpreting it correctly. In any case the point is that you should expect P_vap to increase and approach P_atm smoothly; that is, P_vap should relatively asymptotically approach P_atm, since it can't go any higher.

Q2: By definition b.p. is the value of T where f(T) = P_vap = P_atm. Thus, when you lower P_atm you change the value of T necessary for the P_vap function f to yield a value equal to P_atm. It is true that for two substances at the same temperature, the one with the lower normal b.p. will have the higher P_vap at the given temperature (we specify normal b.p. so that our reference frame is the same for both b.p.'s). This is because while the temperatures are the same, the differences in substance characteristics (e.g. mass, intermolecular interactions) are captured by their enthalpies of vaporization.

 

[Rhino1000]

Q1: I believe if you plot P_vap as a function of T you should see logarithmic increase in P_vap. This should be the result of the Clausius-Clapeyron relationship such that (∂P / ∂T) = (1 / T). Mathematically this is a little outside my realm of expertise but I believe I'm interpreting it correctly. In any case the point is that you should expect P_vap to increase and approach P_atm smoothly; that is, P_atm should relatively asymptotically approach P_atm, since it can't go any higher.

Q2: By definition b.p. is the value of T where f(T) = P_vap = P_atm. Thus, when you lower P_atm you change the value of T necessary for the P_vap function f to yield a value equal to P_atm. It is true that for two substances at the same temperature, the one with the lower normal b.p. will have the higher P_vap at the given temperature (we specify normal b.p. so that our reference frame is the same for both b.p.'s). This is because while the temperatures are the same, the differences in substance characteristics (e.g. mass, intermolecular interactions) are captured by their enthalpies of vaporization.

I knew you'd reply 😀. I was even thinking about @gettheleadout ing you. But I think in the second to last clause of your first paragraph, you mean that P_vap should relatively asymptotically approach P_atm, not that P_atm approaches itself.

I am not sure about the Clausius-Clapeyron relationship. I'm just saying I never heard of it. That equation doesn't seem to be consistent with the given situation though, because if you increase temperature, it would seem that you should get an increase in the dP/dT, rather than a decrease (as predicted by the equation, if done correctly), because vapor Pressure increases exponentially with Temperature (at least according to TBR)... I think it would make sense for it to increase logarithmically. Perhaps I misinterpreted the equation. Plus, the Pvap doesn't seem to be reaching any sort of maximum pressure in any case (it increases exponentially as per TBR; I) as far as its relationship with a liquids temperature. Of course, the liquid can only reach a certain temperature before it no longer is a liquid.

I wonder if TBR itself misinterpreted the equation, because your logarithmic interpretation makes sense. They definitely do graph and describe Pressure as increasing exponentially with Temperature, though.

I'm going to try to contemplate your reply to the second question soon here.

Besides, what IS your area of expertise? What's your major (if you care to share)?

 

[gettheleadout]

I knew you'd reply 😀. I was even thinking about @gettheleadout ing you. But I think in the second to last independent clause of your first paragraph, you mean that P_vap should relatively asymptotically approach P_atm, not that P_atm approaches itself.

Ah yes, I accidentally said P_atm twice there, I'll fix that.

I am not sure about the Clausius-Clapeyron relationship. I'm just saying I never heard of it.

My first thought was the Clausius-Clapeyron equation, and rather than derive the P vs. T relationship myself I found it here.

That equation doesn't seem to be consistent with the given situation though, because if you increase temperature, it would seem that you should get an increase in the dP/dT, rather than a decrease (as predicted by the equation, if done correctly), because vapor Pressure increases exponentially with Temperature. Perhaps I misinterpreted the equation. Plus, the Pvap doesn't seem to be reaching any sort of maximum pressure in any case (it increases exponentially) as far as its relationship with a liquids temperature. Of course, the liquid can only reach a certain temperature before it no longer is a liquid.

No, dP/dT should definitely decrease, plot an asymptotic approach of P to P_atm on a plot where P is on the y-axis and T is on the x-axis. The limit is:

lim(T -> inf.) P(T) = P_atm

Which means that since P_atm is constant the slope of P(T) at T = infinity is zero, whereas the slope is a positive number while P is increasing initially. The derivative dP/dT must decrease, since the change in P per change in T gets smaller and smaller until it is zero.

I'm going to try to contemplate your reply to the second question soon here.

Besides, what IS your area of expertise? What's your major (if you care to share)?

Haha I'll PM you.

 

[Rhino1000]

Ah yes, I accidentally said P_atm twice there, I'll fix that.


My first thought was the Clausius-Clapeyron equation, and rather than derive the P vs. T relationship myself I found it here.


No, dP/dT should definitely decrease, plot an asymptotic approach of P to P_atm on a plot where P is on the y-axis and T is on the x-axis. The limit is:

Which means that since P_atm is constant the slope of P(T) at T = infinity is zero, whereas the slope is a positive number while P is increasing initially. The derivative dP/dT must decrease, since the change in P per change in T gets smaller and smaller until it is zero.


Haha I'll PM you.

Therefore I say that you and TBR are in direct disagreement, and they have the incorrect interpretation. I edited my first post to see if you could confirm this. TBR does describe Pvap as having an exponential rather than logarithmic relationship with Temperature. The logarithmic relationship is sounding much more logical.

By the way, I wasn't saying that the Clausius-Clapeyron equation would show an exponential relationship, I was merely suggesting that the Clausius-Clapeyron equation predicts a relationship inconsistent with TBR (I was accepting their interpretation as truth, rather than the equation, which is why I said that the equation must be inconsistent with the given situation); so I was actually stating that the two are inconsistent with each other.

 

[gettheleadout]

Therefore I say that you and TBR are in direct disagreement, and they have the incorrect interpretation. I edited my first post to see if you could confirm this. TBR does describe Pvap as having an exponential rather than logarithmic relationship with Temperature. The logarithmic relationship is sounding much more logical.

By the way, I wasn't saying that the Clausius-Clapeyron equation would show an exponential relationship, I was merely suggesting that the Clausius-Clapeyron equation predicts a relationship inconsistent with TBR (I was accepting their interpretation as truth, rather than the equation, which is why I said that the equation must be inconsistent with the given situation); so I was actually stating that the two are inconsistent with each other.

Ah I see. Yes, I'm fairly confident this is the case, since if you solve the ODE from the Clausius-Clapeyron relation you find that P(T) = ln(T) + C.