The vapor pressure, P, of a certain liquid was measured at two temperatures, T.
ID: 1057760 • Letter: T
Question
The vapor pressure, P, of a certain liquid was measured at two temperatures, T. The data is shown in this table.
T(K) P(kPa)
275 3.64
575 5.28
If you were going to graphically determine the enthalpy of vaporizaton, ?Hvap, for this liquid, what points would you plot?
x y
point 1 x axis in 3 significant figures and y axis in 4 significant figures _____ _______
pont 2 x axis in 3 significant figures and y axis in 4 significant figures _____ _______
Determine the rise, run, and slope of the line formed by these points.
rise _______________ run ________________ slope ______________
What is the enthalpy of vaporization of this liquid?
_________________ J/mol
Explanation / Answer
It is just a form of Clausius-clapyeron equation
Let temperature be x and pressure be y
at point 1 x1 = 1 / 275 = 3.636*10-3
y1 = ln(3.64) = 1.2919
at point 2 x2 = 1 /575 = 1.739*10-3
y2 = ln(5.28) = 1.6639
Run = x2 - x1 = (1.739*10-3 - 3.636*10-3) = -1.897 *10-3
Rise = y2 - y1 = (1.6639 - 1.2919) = 0.372
slope = m = y2 - y1 / x2 - x1 = 0.372 / (-1.897 *10-3) = -196.099
Enthalpy of vaporization of liquid = Hvap = m * R = -196.099 * 8.314 = 1630.3679 J/mol
Note: Do not consider the minus sign as the enthalpy changes of vaporization are always positive
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