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The graphs in this problem should be made with a graphing program like Excel. Us

ID: 857075 • Letter: T

Question

The graphs in this problem should be made with a graphing program like Excel. Use
linear regression to obtain the appropriate slopes and intercepts. Given the following
data for ethanol, and assuming Clausius-Clapeyron behavior where gaseous ethanol behaves
perfectly, plot ln p vs. 1/T (use Pa for p) and fit the data with a straight line to obtain the
slope and intercept and estimate the molar ?Hvap [kJ/mol]. To your plot, add theoretical
curves for ethanol assuming a molar ?Hvap = 38.56 kJ/mol (the value at 352 K) and a molar
?Hvap = 42.08 kJ/mol (the value at 298 K) for comparison:

Vapor Pressure (mm Hg) 1 10 40 100 400 760 T/degrees C -31.3 -2.3 19 34.9 63.5 78.4

Explanation / Answer

Vapor pressure (mm Hg), P P (in Pa) T/degree C T/ in Kelvin,K lnP 1/T Bitmap 1 1.33E-08 -31.3 241.85 -18.13 0.0041 10 1.33E-07 -2.3 270.85 -15.83 0.0037 40 5.33E-07 19 292.15 -14.44 0.0034 100 1.33E-06 34.9 308.05 -13.53 0.0032 400 5.33E-06 63.5 336.65 -12.14 0.0030 760 1.01E-05 78.4 351.55 -11.50 0.0028 760 mm Hg = 1.013E-5 Pa 760 mm Hg = 1 atm 1 atm = 1.013E-5 Pa lnP = -delH/ RT +constant Slope 0.0002 Intercept 0.0006 Extrapolate this graph for 352 K to add theoritical curve for ethanol