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?

    ********

    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

    *******

    recap..

    ΔHvap = 41.6 kJ/mol

    normal bp = 373 K

    *******

    were you working with water?

    and btw.. when I converted to K, I added 273.15.. not just 273.

    *******

    *******

    how much of this did you follow?

  • RE: chemistry graph ln(ok) vs a million/T? In graph of ln(ok) vs. a million/T, the slope is a unfavorable promptly line. yet what does it signify? I'm attempting to locate the modifications in entropy and enthalpy.

Sign In or Register to comment.