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Clausius-Clapeyron equation to estimate sublimation pressure of water

  • #1

Homework Statement:

Using the Clausius-Clapeyron equation and the triple-point data of water, estimate the sublimation pressure of water at ##-30°C## and compare to the value in your Thermodynamic Tables (Cengel and Boles).

Homework Equations:

##\int_1^2 (\frac {dP} P)_{sat}=\frac {\Delta h} {R} \int_1^2(\frac {dT} {T^2})_{sat}##
Before this question, the questions were about the Clapeyron equation, and how to estimate ##\Delta s##. I'm completely put off by this question however, and don't know where to start.

I've found that the triple point of water is at ##0.01°C##, and there is indeed data in the table for saturated water, for that temperature. So I know I've got this data available to me, but what next? I know also that sublimation is a state transition directly from a solid to a gas. How is sublimation linked to the triple point, and how do I manipulate this equation to find a single pressure?

After integration, the equation involves two pressures and two temperatures: what are states ##1## and ##2##, and in what state are we taking ##\Delta h## at?
 

Answers and Replies

  • #2
19,679
3,989
You are trying to determine the equilibrium vapor pressure of ice at -30 C, knowing the equilibrium vapor pressure and the molar heat of sublimation at 0.01 C (assuming that the heat of sublimation does not change significantly over the intervening temperature range).
 
  • #3
After reading your response, I've had an attempt at the solution, and can't seem to reach the right answer.

I took phase ##1## to be at ##-30°C##, and phase ##2## at ##0.01°C##. Manipulating the equation to find the pressure at phase ##1##:

##ln(P_1)=ln(P_2)-\frac {\Delta h} R (\frac 1 {T_1} - \frac 1 {T_2})_{sat}##

##R=8.31~JK^{-1}mol^{-1}## >> (assuming this is the same as ##8.31~JK^{-1}kg^{-1}##

##T_1=243.15~K##

##T_2=273.16~K##

##P_2## I took to be ##saturation~pressure## at the triple point. From the tables, this pressure is ##0.6117~kPa##.

##\Delta h##, initially, I took from the table. The table value for ##\Delta h## at ##0.01°C## is ##2500.9~kJ/kg##. However, this value of ##\Delta h## I believe is the enthalpy of vaporisation, ##h_{fg}##.

Using this first value of ##\Delta h## provides an answer that couldn't possibly ever be correct. That's ##2500(1000)~J##.

So, I went online, and from Wikipedia I got a value for the ##enthalpy~of~sublimation## at ##273.15~K## (close enough). This value is ##51.1~kJ/mol## which I assume again, is the same as ##51.1~kJ/kg##.

Using these numbers in the equation, and multiplying the enthalpy by a thousand, yields the result:

##P_2=0.0380~kPa##.

The answer given is: ##P_2=0.03824~kPa##.

So there is a very small difference; insignificant, I would think. I'm just a little worried that I took a value off of the internet, rather than using the thermodynamic tables provided. I don't see however, where else I would get that value from. There is no such value for enthalpy of sublimation in the tables.
 
  • #4
19,679
3,989
The heat of vaporization of water at 273 is 2501 kJ/kg and the heat of melting of water at 273 is 335 kJ/kg. So the heat of sublimation is 2501 + 335 = 2836 kJ/kg = 51048 J/mole. So your answer is correct: 38.1 Pa
 
Last edited:
  • #5
After a little thinking, I believe I've made a mistake using the universal gas constant, ##8.31~J K^{-1}mol^{-1}##. I honestly don't know how I got the correct answer doing the calculation how I did. From the thermodynamic tables, the gas constant, ##R##, for steam, is ##0.4615~kJ kg^{-1} K^{-1}##. This means back in the equation, I no longer have to multiply by a thousand.

That's the gas constant sorted. Next, the enthalpy of sublimation:

##\Delta h_{sub}=51.1~\frac {kJ} {mol}(\frac {1~mol} {18.02~g})(\frac {1000~g} {1~kg})=2836~kJ kg^{-1}~(to~4~s.f.)##

Well, NOW after a little ##more## experimenting, it appears the fraction ##\frac {\Delta h} R## doesn't really change much between using these pairs of values. I don't know why. Using these new values however yields the answer: ##P_2=0.03809~kPa##, which I guess is a little closer. Is it wrong the way I did it before?

And I've just noticed my value for the enthalpy of sublimination is exactly the same as the one you've just stated.

Your latest comment I believe answers all my questions. In the tables, I'm given a value for the latent heat of fusion and vaporisation; summing these yields the latent heat of sublimination. I still think I'm missing something as the answers aren't ##exactly## the same, but I can refer to my lecturer for that. For the sake of completeness:

Enthalpy of vaporisation at ##0.01°C## is ##2500.9~kJ/kg##

Enthalpy of fusion for water is ##333.7~kJ/kg##

Enthalpy of sublimation: ##2500.9+333.7=2834.6~kJ/kg##

This value has been obtained using only what's given in the thermodynamic tables. Using this value for the enthalpy in the main equation gives a final answer of:

##P_2=0.03813~kPa##

Closest value yet, and I also believe this to be the most 'correct' answer as it's using only what's given.

Thank you for your help Chestermiller! Is there any way to mark questions as, 'solved'. I know there was before, though I can't seem to find how to do it anymore (if you still can).
 
  • #6
19,679
3,989
After a little thinking, I believe I've made a mistake using the universal gas constant, ##8.31~J K^{-1}mol^{-1}##. I honestly don't know how I got the correct answer doing the calculation how I did. From the thermodynamic tables, the gas constant, ##R##, for steam, is ##0.4615~kJ kg^{-1} K^{-1}##. This means back in the equation, I no longer have to multiply by a thousand.

That's the gas constant sorted. Next, the enthalpy of sublimation:

##\Delta h_{sub}=51.1~\frac {kJ} {mol}(\frac {1~mol} {18.02~g})(\frac {1000~g} {1~kg})=2836~kJ kg^{-1}~(to~4~s.f.)##

Well, NOW after a little ##more## experimenting, it appears the fraction ##\frac {\Delta h} R## doesn't really change much between using these pairs of values. I don't know why. Using these new values however yields the answer: ##P_2=0.03809~kPa##, which I guess is a little closer. Is it wrong the way I did it before?

And I've just noticed my value for the enthalpy of sublimination is exactly the same as the one you've just stated.

Your latest comment I believe answers all my questions. In the tables, I'm given a value for the latent heat of fusion and vaporisation; summing these yields the latent heat of sublimination. I still think I'm missing something as the answers aren't ##exactly## the same, but I can refer to my lecturer for that. For the sake of completeness:

Enthalpy of vaporisation at ##0.01°C## is ##2500.9~kJ/kg##

Enthalpy of fusion for water is ##333.7~kJ/kg##

Enthalpy of sublimation: ##2500.9+333.7=2834.6~kJ/kg##

This value has been obtained using only what's given in the thermodynamic tables. Using this value for the enthalpy in the main equation gives a final answer of:

##P_2=0.03813~kPa##

Closest value yet, and I also believe this to be the most 'correct' answer as it's using only what's given.

Thank you for your help Chestermiller! Is there any way to mark questions as, 'solved'. I know there was before, though I can't seem to find how to do it anymore (if you still can).
You did it correctly. You just gotta make sure the units match and cancel properly.
 

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