Answer to Question #156615 in Physical Chemistry for ffffw

Question

The pressure was exerted with a piston on molted naphthalene (C₁₀H₁₈) at 95 ^oC. The vapour pressure of naphthalene under 1.0 bar is 2.0 kPa and its density is 0.962 g cm^-3, calculate the normal boiling point.

Answer 1

Standard vaporization enthalpy of naphthalene is 53.4 kJ/mol (table data). Let's assume enthalpy does not change with temperature.

Normal boiling point means vapour pressure is equal to the standard atmospheric pressure (1 atm or 1.01325 bar or 101325 Pa)

ln(p1/p2) = $\Delta$_vapH/R*(1/T2-1/T1)

p2 is 2kPa and T2 is 95⁰C or 368.15K

ln(101325/2000) = 53400/8.314*(1/368.15-1/T_n.b.p.)

T_n.b.p. = 475.02K or 201.87⁰C