Question

At 213.1 K a substance has a vapor pressure of 45.77 mmHg. At 243.7 K it has a vapor pressure of 193.1 mm Hg. Calculate its heat of vaporization in kJ/mol.

Answer 1

`20.3125` kJ/mol

Step-by-step explanation:

`P_(i)` = initial vapor pressure = 45.77 mm Hg

`P_(f)` = final vapor pressure = 193.1 mm Hg

`T_(i)` = initial temperature = 213.1 K

`T_(f)` = final temperature = 243.7 K

`H` = Heat of vaporization

Using the equation

`ln\left ( (P_(f))/(P_(i)) \right ) = \left ( (-H)/(R) \right )\left ( (1)/(T_(f)) - (1)/(T_(i))\right)`

`ln\left ( (193.1)/(45.77) \right ) = \left ( (-H)/(8.314) \right )\left ( (1)/(243.7) - (1)/(213.1)\right)`

`H = 20312.5` J/mol

`H = 20.3125` kJ/mol