Introduction.

Objective:  Understanding the equilibrium between liquid and vapor phases is very important for such practical applications as distillation and vapor pressure calculation. The purpose of this lab is:

• to learn the relation between enthalpy of vaporization, DHvap,  and temperature variation in vapor pressure for different pure liquids and its relation to liquids' structure.
• to learn about relation between vapor pressure above mixture of two liquids

The heat of vaporization of a liquid is a useful thermodynamic quantity because it allows calculation of the vapor pressure of a liquid at any temperature.

In this experiment the vapor pressure of four pure liquids (water, methanol (or ethanol), benzene and CCl4) will be measured as a function of temperature and their heats of vaporization, DHvap, will be calculated.

Vapor Pressure of Pure Liquids.

Two phases (a and b) in equilibrium at constant pressure and temperature have the same Gibbs free energy:

Ga = Gb

Recalling that dG = VdP - SdT yields:

Va dP - Sa dT = Vb dP - Sb dT

or   dP/dT = (Vb - Va)/(Sb - Sa) = DS/DV

For a phase transition occurring at constant temperature and pressure, the definition of entropy dS = dq/T implies that DStransition = DHtransition/T.  Thus,  the variations of pressure and temperature along the phase coexistence line are linked via the Clapeyron equation:
(1)

Again,  P and T here are the pressure and the temperature, respectively; Sx and Vx are the molar entropies and molar volumes for corresponding phases and DS and DV are their changes; DH is the change in molar enthalpy, i.e. it is the molar enthalpy of vaporization, DHvap, for a liquid-vapor transition.

Assuming that the molar volume of liquid can be neglected (explain why) and vapor behaves as an ideal gas, the general Clapeyron equation can be rewritten as the Clasius- Clapeyron equation:

d (ln P)/d(1/T) = - DHvap/RZ        (2)

Here the compressibility factor:

Z = PVgas/RT,                          (3)

was introduced, which for the vapor is very close to 1. Then after integrating the Clasius- Clapeyron equation (2), for temperature independent DHvap and Z = 1, one gets:

ln (P/Po) = - DHvap/R *(1/T - 1/To)     (4)

This dependence predicts linear slope for ln P when plotted as a function of inverse absolute temperature, 1/T. Some assumptions we've made are not strictly correct:

a) Nonideality of the vapor makes Z to differ from unity and become temperature dependent (explain how), which should introduce some curvature to the dependence (4).

b) Since enthalpy of both liquid and vapor changes with temperature, the difference between them, i.e. the enthalpy of vaporization, DHvap, is also temperature dependent. After all, above critical temperature it should zero. Temperature dependence of enthalpy can be estimated based on heat capacity,  Cp, which, for simplicity can be taken as a constant. Then one can write:

DHvap (T2) =   DHvap (T1) +  DCp (T2 - T1)           (5)

where   DCp  =  Cg - Cl                                   (6)

c) Neglecting molar volume of liquid causes less than 1% error. I suggest you to evaluate it yourself.

It turns out that the effects a) and b) are each quite significant; however, they very nearly cancel each other out1 and Clausius-Clapeyron plots tend to be quite linear.

Vapor Pressure of Solutions.

The vapor pressure of an ideal solution of a nonvolatile solute in a volatile liquid should be given by Raoult's law:

Pl = xl Pl*                           (7)

where xl is the mole fraction of the volatile liquid and Pl* is the vapor pressure of the pure volatile liquid. Thus, for an ideal solution:
a) vapor pressure should follow Raoult'slaw, Eq(7),
b)  DHvap for the solution should be the same as DHvap for the volatile solvent.

You will study mixtures between ethanol (or methanol) and glycerol (in proportions 1:1, 2:1 and 3:1) on the same setup and see if the Raoult's law is valid for these solutions.