Question

A. Vapor Pressure Measurements Vapor Pressure (mm Hg) *53.5 3.51 623 4.41 Temp.(°C) Temp.(K) Gol 219.3 11.0 292.2 23.2 296.4 26.0 30112 _ 1/T 00368 .00342 .60331 60332 .00323 - 4.61 lib. 214.2 4.75 5.367 36.0 309.2 Sample Calculations: Oct 213,15=k 6.1 +218,15 = 214,3 279300368 k' Use your data above to prepare an Excel or LoggerPro graph of In P vs. 1/T. Have Excel or LoggerPro draw the best fit straight line through the data points and determine the regression equation and R2 value for the line. Determine the slope of this line and calculate the molar heat of vaporization from the slope. Slope = molar heat of vaporization = kJ/mol

Answer 1

ANSWER

According to Claousius-Claproyron Equation:- The relationship between vapour pressure of the liquid and temperature.

Mathematically, ln P = -∆Hvap(1/T) /R + C ............... (1)

A plot of ln P vs. 1/T has a form of the straight line.

y = mx + C(General Regression eqn)........... (2)

Let's calculate slope of the eqn(1),

Slope = -∆Hvap/R Here, R= 8.31 J/K.mol

-∆Hvap/R = -9/0.0026 = 3461.53

5.34 4.75 4.61 + 4.61 3.51 ya 0.0099 7 000332 0.00337 0.00342 0.00368 (ky Y₂ y (K) e 3.51, y = 4.41 M2=0.00368 -M, 50.00342 Calculate the slope - Youy, M-u, - 3.51-4.41 0100368 -0.00342 -org 0, 00026 0.0026

$Slope =$$Slope = 3461.53$

∆Hvap = 3461.53 × R = 28765.384J = 28765384KJ/mol

  $Molar Heat of Vaporisation = 28765384KJ/mol$