Plot a graph, similar to the one below, with the new variables

(1)/(T)

on the

x

-axis and

LnP

on the

y

-axis. Use the Microsoft Excel program to help you. If the original pressure versus temperature data for the unknown liquid conforms to the behavior predicted by the Claussius-Clapeyron Model, this plot should yield a straight line.\

(4.0*1)/(.9936)(-5472.1)k

\ The slope will yield a numerical value for:\

-(\\\\Delta Hv)/(R)

\ (K)\ The y-intercept will yield a numerical value for:

(\\\\Delta Hv)/(RT_(b))+LnP_(0),

(no units)\ The slope and y-intercept values can be extracted from the graphed data by using the "Trendline" function in the "Microsoft Excel" program. Using the known values for R and

P_(0)

, determine the values for

\\\\Delta H_(V)

and

T_(b)

\ \\\\table[[Parameter,Value],[

\\\\Delta H_(v)

,KiloJoules / mole],[

T_(b)

,],[Correlation Factor,]]\

-5472.1K

\

er=21.899

\

P=33\ y=-5472,1x+21.899\ R^(2)=.9936

\ 33

Plot a graph, similar to the one below, with the new variables 1/T on the x-axis and LnP on the y-axis. Use the Microsoft Excel program to help you. If the original pressure versus temperature data for the unknown liquid conforms to the behavior predicted by the Claussius-Clapeyron Model, this plot should yield a straight line. .993621.894(5472.1)k The slope will yield a numerical value for: RHv (K) The y-intercept will yield a numerical value for: RTbHv+LnP0 (no units) The slope and y-intercept values can be extracted from the graphed data by using the "Trendline" function in the "Microsoft Excel" program. Using the known values for R and P0, determine the values for Hv and Tb 2.1K 1.899 y=5472,1x+21.899R2=.9936