Question

ethanol has a heat vaporization of 38.56

8.314 J/mol K 782.4+ 27. All work must be show to receive credit. Remember, significant figures are important! P-AH/ TT, "P, R T ,T, - 25.0+ 2 1. (4 points) Ethanol has a heat or vaporization of 38 56 kJ/mol and a normal boiling point of 78.4°C. a. What is the vapor pressure of ethanol at 25 0 C? MP2 = - À HVED (T.. T2 ) - 38.56 kJlnit 78.4°C -25.06 0.0821 atmo b. What is the boiling temperature of ethanol on Pikes Peak, where the atmo- spheric pressure iş 540 tort? 540 torr tum = 0.7102 760 torr

Answer 1

Answer: 6 Given that en (972), P = 0·0821 4 28:314 g/molok @ Ethanol has a heat of rapourizations, AH = 8 38.56 kJ/mol = 38.56X103 g/mol Boiling point, T, = 78.4°C = 273 + 78.4) K = 351.4K. We have to find vapour pressure (P2) at tempt Te = 2500 = (273 +25)* = 298K At boiling point vapour pressure is equal to the atmospheric pressure. 1 {f, = latm = 760 torr. / Then Then /351-4k-298k P2 =- 38:56 8.314 Zimola (351.4k X 298k 760 torr => ln 2 - 38560 K x 53.4K -38,314 x 3h14X298 K =) in P2o toro = -2.365 P = (769 X 2-2 345 tr >P2 = 71.4 torn / . The vapour pressure at 25°C tempo is 71.4 torr

6 Given that Also, P = 760 fort, Ti = 351.4K We have to find Tz at P = 540 torr. Then the above egn beume - In 540 tort - - 38.56X10 /mol (351.4K-T21 " 760 torr - 8.314 I.mot. (351.44 TL => + 01 342 = + 13.129 ( 35746-73) → (5.6k 1) =%a172 - 42 = 0.026 = 351.43 = 1+0,026 = 1.026 >> Ty 3 357.4K 1026 => 152 = 342.5W => T2 = (3425–273) => 12 = (69.5°C) Hence the boiling temps of 540 torr is 69.5°C ethanol at