Question

This is a question for homework in an undergraduate Physical Chemistry class:

- Please use 273.15 when converting between Kelvin and Celsius temperatures.

QUESTION 5 Between 298.2 K to its normal boiling point 353.24 K the vapor pressure of a certain organic compound can be fitted to the following expression: 3229.86 In P = 16.725 -- - 118345 T2 Where P is measure in torr and T in K. For this substance, the molar enthalpy of vaporization at 353.24 K is 30.8 kJ mol-1 and the molar volume of its liquid phase at the same temperature is 96.0 cm3 mol-1. Calculate the molar volume (L mol-1) of the vapor at its equilibrium pressure at 353.24 K.

- Use 96485 for Faraday Constant

- In problems that require fitting of a curve or straight line use linear regression. Do not use trend line. Sometimes the results can be different.

Answer 1

At the bolling point of this organic compound the vapour pressure will be egnal to the atmospheric pressun a Lalom. T = 353.24K lir - vap According to claussion clapeyson egnition 3- ver: 35. Tauran . Jurga Juage-rein "" di dupa 15.725 - 3229.26 - 19345 TL Inp = 0 + 3229.96 2x118345 T2 211934 $ (one) to 3229.PE p ( 389.03 - 2x ) T A Hund 1 AVV - p/ 3219.8% 42 dT 77

& D Avay my latom. / 3229-82 tot die lalo 2x1183457 T2352, T? tabr. 19.14 +1.89 - 11.037 abr. It way 4 Juap? 0.101325kg ž 16it-abon 30.8kg , * 30.8 11.03zatori 30.say7.107325 = 303.9 Elites 303.97 lit-ala mo.. 11.037 alm 1. = 96 cm3 mln done a1.c.c. 2 96 mb molt! :D - 96x10-3 lit mit = 0.097 litma l Jump - Veiq = 27.46 lit mot = 27.464 0.096 bit mn! 7 Jump = 27.55 hat minn