Question

c. Fill in the table below by calculating the corresponding value of 1/T for each temperature and the log(P) for each vapor pressure Temperature (K) 1/T

263 ______________ 273 ______________ 283 ______________ 293 ______________ 303 ______________ 313 ______________

Vapor pressure (P, in torr)

80.1 133.6 213.3 329.3 495.4 724.4

log (P)

_____________ _____________ _____________ _____________ _____________ _____________

d. Using the calculated data from the table, graph log(P) vs 1/T. e. What is the difference between the two graphs?

f. If you wanted to estimate the vapor pressure of the liquid at a different temperature (e.g., 253 K or 298 K) whichgraphwouldprovideamoreaccurateresult? Why?

Answer #1

C)

Graphing plot:

Recall that in equilibrium; especially in vapor-liquid equilibriums, we can use Clasius Clapyeron combination equation in order to relate two points in the same equilibrium line.

The equation is given as:

ln(P2/P1) = -dHvap/R*(1/T2-1/T1)

then, plot ln(P) vs. T(K)

f)

The best graph is the one which include both points

that is

T = 253 K; 298K

ln(P) = -dHvap/R*(1/T) + dS/R

y = m*x + b

slope = -dHvap/R

-3622.1 = -dHVap/8.314 J/molK

dHvap = 3622.1 *8.314 = 30114.1394 J/mol = 30.11 kJ/mol

once you get all values, simply plug T = and get P

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