Question

The normal boiling point of liquid bromoethane is 311 K. Assuming that its molar heat of vaporization is constant at 27.1 kJ/mol, the boiling point of CH3Br when the external pressure is 1.35 atm is

Answer 1

() Other Ct =) Pi is the vapour pressure at e Vapour pressure at a temperature T is the vapour pressure at a temperature Tz. Ti,To are in Kelvin A Huap enthalpy of vapourization 'R universal gas constant = 8.314 J. mot kl

P = latm Tr= 3118 P₂ = 1.35 atm Ta=? pH = 27.18 (03/mal et 311k In ( 135 ) : 31-4X to smert 8-3145-mol!x- 8:3143 vie in (1.35) x 8.314 J. mod 27-1X 103 7. met 31 - 2 / 2 -2 9 -20689883x669 30 T to = - 9-20684883 x 10-5 I To = 0.0031233651 To = 320.1675013913