Natural Log of Vapor Pressure?
Given just mmHg values, how would I find ln P? I have the following information:
Temperature in Celsius:
90,91,92,93,94,95,96,97,98,99,100 (I will convert these to Kelvin)
mmHg:
525, 546, 567, 588, 611, 634, 658, 682, 706, 734, 760
I am supposed to do a linear regression on my calculator (which I know how to do, I just need help with the numbers) and graph the equation.
Update:Using the Clausius-Claeyron Equation, it asks for P1 and P2. Which numbers would I plug into which?
Comments
CC equation...
ln(P1 / P2) = -ΔHvap / R) x (1/T1 - 1/T2)
rearranging..
ln(P1) - ln(P2) = (-ΔHvap / R) x (1/T1) + (ΔHvap / R) x (1/T2)
rearranging...
ln(P1) = (-ΔHvap / R) x (1/T1) + (ΔHvap / R) x (1/T2) + ln(P2)
and if we assume (T2,P2) = normal bp of the liquid..
i.e.. it's a FIXED VALUE
i.e.. it's a constant for the particular liquid
then.. if we call the normal bp To, Po... and let P1,T1 vary.. and call them "P" and "T"... then we can write
ln(P) = (-ΔHvap / R) x (1/T) + [ (ΔHvap / R) x (1/To) + ln(Po) ]
and that is of the form y = mx + b
IF.. y = ln(P)
IF.. m = (-ΔHvap / R)
IF.. x = 1 / T
and
IF.. b = [ (ΔHvap / R) x (1/To) + ln(Po) ]
(notice... "m" is a constant.. dHvap/R is a constant right?.. also notice "b" is a constant.. dHvap/R is a constant.. To and Po are constants so "b" is a constant)
Therefore...
a plot of ln(P) on the y-axis and 1/T (in K) on the x-axis will have...
slope = (-ΔHvap / R)
intercept = [ (ΔHvap / R) x (1/To) + ln(Po) ]
follow all that?
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I plugged the data into excel (note the T in K !)
made the 1/T vs ln(P) plot and got this equation
y = -5003.9x + 20.044.... r^2 = 1
the r^2 is an indication of how linear the data is.. i.e.. how well the equation fit the actual data points. r^2 = 1 is a perfect fit. This shows the clausius clapyron equation fit the data perfectly! great job measuring!
so.. this means
-ΔHvap / R = -5003.9 K... . (the units of the slope are K)
and
(ΔHvap / R) x (1/To) + ln(Po) = +20.044... (dimensionless)
we already know R = 8.314 J/molK
and Po = 760mmHg
right?
(To is the normal bp of the substance.. ie.. the bp for P = 760mmHg)
so.. let's find -ΔHvap and To.
from
-ΔHvap / R = -5003.9 K
ΔHvap = +5003.9 K x 8.314 J/molK x (1kJ/1000J) = 41.6 kJ / mol
from
(ΔHvap / R) x (1/To) + ln(Po) = +20.044
(ΔHvap / R) x (1/To) = +20.044 - ln(Po)
To = (ΔHvap / R) / [ +20.044 - ln(Po) ]
solving..
To = (5003.9 K) / (20.044 - ln(760)) = 5003.9K / 13.411 = 373 K
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recap..
ΔHvap = 41.6 kJ/mol
normal bp = 373 K
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were you working with water?
and btw.. when I converted to K, I added 273.15.. not just 273.
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how much of this did you follow?
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