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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

so my problem is:

1.

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

3.

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

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.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

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