PHYSICAL CHEMISTRY IN BRIEF

CHAP. 7: PHASE EQUILIBRIA [CONTENTS] 192

Solution

We use equation (7.16) into which we substitute the following values:T 1 = 272. 7 K, ps(T 1 ) =

101325 Pa, ps(T 2 ) = 32000Pa.

ln

32000

101325

=

22400

8. 314

( 1

272. 7

−

1

T 2

)

→ T 2 = 244. 2 K → t=− 28. 9 ◦C.

Butane would not boil at Mount Everest at− 30 ◦C, it would be in the liquid state of matter.

For the dependence of the saturated vapour pressure on the temperature, a number of
empirical relations have been suggested, e.g. theAntoine equation

lnps=A−

B

T+C

, (7.17)

whereA, B, Crepresent adjustable constants.

Example

The Antoine equation constants for toluene areA= 14. 01415 , B= 3106. 4 ,C=− 53. 15 (for

pressure given in kPa). Calculate the normal boiling temperature and the enthalpy of vaporization

at this temperature.

Solution

According to7.1.5, the normal boiling temperature is the temperature at a saturated vapour

pressure of 101 325 Pa. By substituting in (7.17) we get

T=

B

A−lnps

−C=

3106. 46

14. 01415 −ln 101. 325

+ 53.15 = 384. 77 K

We then calculate the enthalpy of vaporization from equation (7.14)

∆vapH = RT^2

d lnps

dT

=RT^2

B

(T+C)^2

= 8. 314 × 384. 772

3106. 46

(384. 77 − 53 .15)^2

= 34770J mol−^1.