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Question of enthalpy of vaporization the diference from ln to log

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hi, i'm new around here but i really need your help this question is similar to one already solved here some years ago
Vapour pressure of a liquid in the temperature range of 200Kto 260K is given by this expression

ln (p/Torr) = 16.255 - 2501.8(T/K)

Calculate the enthaly of vaporization of hte liquid

since this is the liquid vapour boundary



and

so then the ratio of Ln p to ln p* would yield the expression for chi which i cna then solve for delta H but it doesnt yield that same answer

what am i doing wrog here can you push (or shove) me iin the right direction

i have an exam today thus i need to answer this once and for all

thank you for help!
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:
Originally Posted by stunner5000pt
Vapour pressure of a liquid in the temperature range of 200Kto 260K is given by this expression

ln (p/Torr) = 16.255 - 2501.8(T/K)

Calculate the enthalpy of vaporization of the liquid

First of all it is important to get the question right if you want people to help you. Your expression is unintelligible as it is. The expression must be:

where P is the ambient pressure in Torr (mm/hg) and T is in Kelvins. Now it makes sense.

The expression for vapour pressure is given by:

where is the Heat or Enthalpy of vaporization in J/mol.

From the expression for this gas, it is apparent that:




AM
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:

Answers and Replies

GCT
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State all of the relevant equations not just the equations that you were given.
 
26
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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:
GCT
Science Advisor
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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.
 
GCT
Science Advisor
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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)
 
26
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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... :)
 
GCT
Science Advisor
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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?
 
26
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yeah the result on solutions is that 34,075
 
GCT
Science Advisor
Homework Helper
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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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