Question

2. The vapor pressure of liquid 1-butene in a temperature range of -198 and 398 K is expressed by the following equation: 475 - - 1343.516 -- 167.515 x 10-5T Calculate the normal boiling point and the vaporization enthalpy of the liquid 1-butane fluid.

Answer 1

Claurius – Clapeyron equation can be expnessed as — hP = -afvap te ---- Page 1 whene, P-pressure Strap - enttialpy of vaporization T - absolute temperature R- universal gas constant C = countant At normal boiling point, pat atm. Given, in ) - 5.475 – 1343.516 – 167.515X10 T ----CÜ > linke - 5.475 - 1343,516 - 167.515*10* 0 - 5.475 - 1343.516 -167.515 x10 T 1343.516 + 167,515x105 = 5.475 T T → 1343.516 + 167.515 x1072 = 5.475 T ~ 167.515 x10° T2-5.475 T + 1343.516 = 0 solving we get, T -65.475) £165,475)=-4x 167.515x16 5x1343.516 2 x 16 7.515 x10-5 5.475 4.579 2x 167.515 x10-5 = 267.44 K or 3000,93K = -5,56 °C M 2727.93 °C

Now, as the temperature 3000.93 K is out of the given mange of -198 K to 398 K , it can't be the amwer, So the normal boiling point is 267.44 Kor - 5.56 °C do Comporing eq. cuj aw eq; cij we get- - 1343.516 - t Strap RX → AH rap = ( 1343,516) (8.314) J moll - 11169.99 J moni | 11.169 kJ moi (page-2