Question

IPUIN. 6.61 The vapor pressure of toluene is 8.00 kPa at 40.3 °C and 2.67 kPa at 18.4°C. Calculate (a) the heat of vaporization and (6) the vapor pressure at 25°C. 66 AOC i hoorho 222.5 1.-1in molti

Answer 1

(a) Heat of vaporization is = 38.032374 kilo joules per moles

1Q-6.61 vapor pressure of any liquid substance is defined as the preuve et which vaporizaron and condensation rates are same. vapor pressure has direct relation with temperature, As temperature Vores) increases, its vapor pressure increases, vapor pressure 14 inversely proportional to Heat of vaporization, The Heat of vaporization is a measure o required energy for breaking the intermolecular forces between liquid molecules before they vaporize. To find vapor pressure at given Temperature, A Relation given by clausius- clapeyson as: vapor pressure 2+ given temperature 25°C = 25+273 = 298 K can be delermind by using equation o consider p is vapor Presslehe at T = 298K Since Heat of vaporization Romain same, og combining One combination PSP, T = 273K H = 38032-34 TM P2 = 2.67% 103 Pa Tz = 29114k, en como ] = 38032.374 Tato 3148] en p - ln (2670) = 0.3477 o emp = 7.8898 + 0.3477 en p = 8.2375 en - Hvap [ -] -0 R e (B-2375) Þ p = 3780.078 Pascal or vapor pressure at 25°C - 3780 kPa Here I and Tz are two differente Temperatures in Kelvin, P, P2 are vapor pressure in Pas Cel, BHvap is reat of vaporization, R is gas constant (a) To determine heat of vaporization we use equation ④ P = BkPa = BX10 P. P = 2.67 kPa = 2.67813 Pa R = 8.3144 J K mot!, T = 40.3°C = 31313 Kelvin T2 = 18.4°C = 184+273 = 291.4 Kelving en is natural logarithm. en en 12.614103 = Strap rint ex103 7 813144 291,4 3/2,77 or or or Hvap = 8.314 X 2914 x 313,3 x ln (2.996) (31313-291.4) shvap = 38032.374 I MOR! Hear of vaporizatim 4H : 38.032374 K Tmel?

(b) vapor pressure is = 3.780m kilo pascal