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Thermodynamics, Clausius-Clapeyrons equation

  1. Mar 6, 2013 #1


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    1. The problem statement, all variables and given/known data
    Find the standard enthalpy of vaporization, standard entropy of vaporization and standard free energy of vaporization.

    Given are two pairs of temperature and vapor pressure for Arsine, AsH3.
    Vapor pressure is 35 Torr at -111.95C and 253 Torr at -83.6C.
    T [K] = T [C] + 273.15

    2. Relevant equations
    ln(P2/P1) = H*((1/T1) - (1/T2))/R
    H is the standard entropy of vaporization and R the gas constant.

    3. The attempt at a solution
    To find the standard enthalpy I simply solved for H and substituted all the values. The pressures as they are, since the units cancel, and the temperature in kelvin. I got H to be about 17.7 kJ/mol but is supposed to be 28.3 kJ/mol according to solutions.
  2. jcsd
  3. Mar 7, 2013 #2
    I think, there in the red is the problem. That should be "L", not H. The difference is that L gives latent phase change in total mass. You will have to divide it by moles. How will you find moles ? I think you should use gas laws.

    And yet again, H is enthalpy, not entropy.
  4. Mar 7, 2013 #3


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    In my book, Chemical Principles by Atkins, H is in units joules per moles and has a small circle at the top right to indicate that it's for standard states of the substance. H, in the book, is called the standard enthalpy of vaporization and it is this same thing that is asked for in the problem.

    You might be right though, the quotient between the enthalpy given in the solution and the one I calculated is about 1.6. Which could possibly be the number of moles.

    What's confusing is that, to my knowledge, the standard enthalpy is always in joules per moles.
  5. Mar 7, 2013 #4
    Is the circle drawn on it ? Then that means ΔH is measured at 298 K temperature and 1 atm pressure and is known as standard enthalpy change.

    Lets check dimensionally. You know the dimensions of R and T. ln(P2/P1) is dimensionless. So right hand of the side should also be dimensionless. That will let you know, whether in formula H is J/mole or simply J.
  6. Mar 7, 2013 #5


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    Yes, it has the circle. It has subscript "vap" for vaporization as well. I checked the units for the gas constant, it's joules per kelvin per mole. The temperature units cancel, therefore the enthalpy change has to have units joules per moles for the units to cancel.

    Then we know that standard enthalpy change of vaporization is in units joules per moles. The book defines enthalpy of vaporization as the difference in molar enthalpy between the vapor and liquid states. This is the enthalpy that's being used in the equation above (in standard conditions), and is what's being asked for in the problem (in standard conditions as well).

    I'm guessing the solution is wrong then?
  7. Mar 8, 2013 #6
    Yes, wrong.

    The question is asking for standard enthalpy of vaporization, and that will be at 298 K temperature and 1 atm pressure. So this problem is not, I guess, a simple formula applying one. Also note that, standard enthalpy of vaporization is written as ΔvapHo and not just Ho. If you write just H, then it will be absolute enthalpy and question is not asking for it anywhere.

    So I suggest you to do the following,

    Take one pair at a time. 35 Torr at -111.95C. These will be P1 and T1. Temperature should be in kelvin. Now, P2 and T2 will be 1 atm (that will be how many torrs ?) and 298 K. Then solve for H. Do that for other pair also.
  8. Mar 8, 2013 #7
    You have received some faulty information in some of the responses.

    In the form of the clausius clapyron equation that you have written, H* is not supposed to be the heat of vaporization under standard conditions. It is supposed to be the average heat of vaporization over the temperature interval T1 to T2. In addition, this relationship inherently assumes ideal gas behavior, and neglects the volume of the liquid (compared to the vapor) in determining the volume change during vaporization. Look up the derivation of the clausius clapyron equation to refresh your memory. The value you have calculated for the average heat of vaporization over the temperature interval is correct: 17.7 kJ/mol. Therefore, the answer, "according to the solutions", must be incorrect (unless there is a mistake in the input data).

  9. Mar 8, 2013 #8


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    I could have called the standard enthalpy of vaporization Q if I wanted, what's important is that I defined it.

    Since it's the standard enthalpy of vaporization used in the equation above, how is it not a simple application of the formula?

    1 atm is 760 Torr. Doing what you're saying will get me two standard vaporization enthalpies, in the answer there should only be one. Combining the two equations will get me the same enthalpy I would've gotten by using the values given from the beginning since enthalpy is a function of state.

    Plus, it's not certain that the substance in this problem has a vapor pressure of 1 atm at 298 K temperature.

    What the book says: In the equation above, it's the standard enthalpy of vaporization that is used, I called it H for simplification (but that doesn't make it the absolute enthalpy).
  10. Mar 8, 2013 #9


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    Thanks for the clarification. I didn't like the way they derived the equation in the book. Do you know of a good derivation online somewhere?
  11. Mar 8, 2013 #10

    I myself was confused, but Chestermiller cleared it up slightly. Here are other sites where Clausius–Clapeyron relation is derived.


    Does that help ?
  12. Mar 8, 2013 #11
    Even if you don't understand every last detail of the derivation of the equation, your fallback position should be to at least know what each of the parameters in the equation stands for. I explained the latter in may previous response. A fairly decent derivation of the Clausius Clapeyron equation is presented on pages 180 - 182 of Introduction to Chemical Engineering Thermodynamics by Smith and van Ness.

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