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Question of enthalpy of vaporization the diference from ln to log

  1. Apr 24, 2008 #1
    hi, i'm new around here but i really need your help this question is similar to one already solved here some years ago
    so my problem is:

    1.The vapour pressure of benzene between 10°C and 30°C fits the
    expression log(p/Torr) = 7.960 - 1780/(T/K). Calculate (a) the enthalpy of
    vaporization and (b) the normal boiling point of benzene.


    this exercise is in atkins 8th edition physical chemistry page 133 for someone who could have it


    3.i try to solve the problem using the same idea as the you said for the exercise i quoted that was:
    seem that something changes from ln to log if it helps the solution is supposed to be around +34,08Kj mol^-1
    the b) is not important if you could help me in a) i would thank a lot

    that question of enthalpy of
    vaporization will appear in a next test so i will await you're help thanks

    thin i have said all necessary any necessary chnage say it please
    but it's making me crazy how a simple change from a ln to log can make so much difference :S

    PS: i tried to put the link to the post instead of the quote but i was not allowed
     
    Last edited: Apr 24, 2008
  2. jcsd
  3. Apr 24, 2008 #2

    GCT

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    State all of the relevant equations not just the equations that you were given.
     
  4. Apr 24, 2008 #3
    complementation to the first post

    State all of the relevant equations

    well usually when we learn about clausius clapeyron equation we learn with ln
    [tex]ln(P) = constant - \frac{\Delta H_{vap}}{RT}[/tex]

    however the problem i shown has log instead of ln
    instead of this shot thing i think the resolution must be something like using this [tex]\frac{\Delta H_{vap}}{R} = 1780[/tex] i'm not sure
    1780 is the number i give in the question log(p/Torr) = 7.960 - 1780/(T/K), but maybe my error starts just here :S

    but now i don't know how the change from ln to log can influence the method cause the original equation is in ln not in log

    i think is this what was missing,
    can a admin add this to original post please?
     
    Last edited: Apr 24, 2008
  5. Apr 24, 2008 #4

    GCT

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    Solve the Clasius Clapeyron equation for P and also the equation that you were given for P - set these two equations equal to each other then utilize some math to relate the two sides in a linear fashion - someone else here may be able to give you a simpler solution. The equations should be the same except for the constants however you are simply concerned with the term that has the delta Enthalpy value so it should not be a problem.
     
  6. Apr 24, 2008 #5

    GCT

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    That is set the two equations for P and then set the independent portions equal to each other. Next take the Ln of both sides - use a special rule for the Ln to simplify one of the sides into linear form - what happens is that you get your equation for Ln (P) and it turns out that it is nearly the same as your Log (P) except for an additional constant which is - Ln (10) .


    in value terms

    Ln (P) = -1780/T - -7.960 -Ln (10)
     
  7. Apr 25, 2008 #6
    that ln10 seem to be a big key in this situation
    cause with this i was cheking and found the following propertie that i really didn't know that
    substituting
    [tex]\frac{\ln n}{ln 10}= constant - \frac{\Delta H_{vap}}{RT}[/tex]

    so passing the ln10 to the other side became:
    [tex]ln n= constant*ln 10 - \frac{\Delta H_{vap}}{RT}ln 10[/tex]

    so using the expression and not forgeting the multiplication with ln10
    [tex]\frac{\Delta H_{vap}}{R}ln10 = 1780[/tex]
    and puting in order to [tex]\Delta H_{vap}[/tex]
    [tex]\Delta H_{vap} = 1780*R*ln10[/tex]

    the last pass is making the product of the ln10 with the result of the last expression

    i've tried and as R=8,314
    [tex]\Delta H_{vap} = 1780*8,314*ln 10 [/tex]

    the result is 34075J =34,075KJ the solution

    is this reasoning right? sometimes the result is right but the method can be wrong

    i would never remember to work the log to obtain the ln, if i didn't put this question here... :)
     
  8. Apr 25, 2008 #7

    GCT

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    You would actually have the ln(10) on the denominator when solving for the Enthalpy - so 34.075kJ is the right answer?
     
  9. Apr 25, 2008 #8
    yeah the result on solutions is that 34,075
     
  10. Apr 25, 2008 #9

    GCT

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    Yeah I just found that log to ln relation myself - your work seems fine forget what I mentioned about multiplying and dividing - it seems that the only value with units is R ; were you given the units for the value of 1780? I would be just concerned with how you show the steps in the cancellation of the units - that's all.
     
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