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Jump to The vapor pressure, P, of a certain liguid was measured at two temperatu

ID: 544640 • Letter: J

Question

Jump to The vapor pressure, P, of a certain liguid was measured at two temperatures, If you were going to graphk ally determine the enthalpy vaporizaton·dHmp, for this liquid, what points would you plot? The data is shown in this table. Number Numbar T (K) (kPa) 225 3.01 675 6.42 To avoid rounding errors, use three significant figures in the x vaiues and four significant figures in the y values point 1:11 Number Number NOTE: Keep the pressure units in kPa. point [D Determine the rise, run, and slope of the line formed by these points. rise run slope Number Number Number What is the enthalpy of vaporization of this liquid? Number J/mol Exit Check Answer O Next O Previous Give Up & View Solution Hint

Explanation / Answer

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)

Where

P2,P1 = vapor pressure at point 1 and 2

dH = Enthalpy of vaporization, typically reported in kJ/mol, but we need to use J/mol

R = 8.314 J/mol K

T1,T2 = Saturation temperature at point 1 and 2

Therefore, we need at least 4 variables in order to solve this.

Substitute all known data:

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

now...

get

x/y point 1

ln(3.01) = 1.1019

1/T = 1/225 = 0.00444

x/y point 2

ln(6.42) = 1.859

1/T = 1/675= 0.00148

now..

rise = y2-y1 = 1.859 - 1.1019 = 0.7571

run = x2-x1 = 0.00148-0.00444 = -0.00296

slope = dy/dx = 0.7571/-0.00296

slope = -255.777

HVap/R = -slope

HVap = 255.777*8.314 = 2126.5 J/mol