Question

A certain liquid has a vapor pressure of 92.0 Torr at 23.0 °C and 347.0 Torr at 45.0 °C. Calculate the value of ΔH°vap for this liquid. calculate the normal boiling point of this liquid.

Answer 1

using the clasius -claperyan equation , we have

ln(p2/p1) = Hvap / R ( 1/ T1 - 1/ T2 )

putting all the values , we get

ln(347.0 / 92.0 ) =  Hvap / 8.314 J/Kmol ( 1/ 296 K - 1 / 318 )

ln 3.77 =  Hvap / 8.314 * ( 1/ 296 K - 1 / 318 )

  Hvap = 35571.82 J/mol * ln 3.77

= 47206.47 J/mol

= 47.2 kJ/mol

and at normal boiling point P = 760 torr , therefore

ln(p2/p1) = Hvap / R ( 1/ T1 - 1/ T2 )

ln( 760 / 347.0 ) = 47206.47 J/mol/ 8.314 J/Kmol ( 1/ 318 K - 1 / T2 )

ln( 760 / 347.0 ) = 47206.47 J/mol / 8.314 J/Kmol ( 1/ 318 K - 1 / T2 )

( 1/ 318 K - 1 / T2 ) = 0.000138

1/T2 = 0.00314 - 0.000138

T2 = 1 / 0.0030067

T2 = 332.59 K

T2 = 59.59 degree celcius