VAPOR PRESSURE AND HEAT OF VAPORIZATION

A) Theoretical
	We study the liquid-vapor phase equilibrium of a one-component system in this measurement. Such a system is of one degree of freedom as the phase rule of Gibbs points out:

			F = C - P + 2  =  1 - 2 + 2 = 1					1)

One of the intensities (potentials) is independent and the equilibrium values of the others are determined by it. Setting the temperature determines the equilibrium vapor pressure and setting the pressure determines the equilibrium temperature. The relative quantities of the two equilibratted phases can not be determined by any way of setting intensities, it depends on extensities as total number of  moles and volume.
	The two phases, liquid and vapor, are in equilibrium if the rate of vaporization and that of condensation are compensated by each other:

								2)

and consequently, the quantity of both phases remains unchanged. Since both vaporization and condensation are driven by some potential, condition 2) induces an other condition: these potentials must have the same value. The equilibratted potentials must be chemical potentials in this case as we are dealing with molar fluxes and the amount of material posesses chemical potential as conjugated intensity:

										3)

Chemical potentials must be unchanged in equilibrium, so Eq. 3) holds true of differential form, too:

									4)

	Since we are given one-component phases, chemical potentials in Eq's 3) and 4) can be replaced by molar Gibbs free energies (isotherm-isobar ones, of course):

									5)
and
									6)
,respectively.

Substitution of the expression of the differential of Gibbs free enegy in Eq. 6):

					7)

yields the Clapeyron equation:

									8)

Here we have taken into account that dH=TdS under isobar condition and, as our condition is isotherm as well, its intgral is H=TS.
			Of course, the Clapeyron equation holds true of  the  phase equilibrium 			of any other one-component two-phase system, not just liquid-				vapor.
	Vaporization is considered to be positive in the differences of Eq. 8):

									9)
									10)
								11)

Heat of vaporization is the enthalpy difference in Eq. 11).
	 Integration of the Clapeyron equation may become a difficult task as, in the most general case, both heat of vaporization and change of molar volumes depend on temperature and pressure. The problem may be simplified by assuming that molar volume of the liquid can be neglected and the vapor obeys the perfect gas law:

									12)

Substituting this expression to Eq. 8) and assuming the heat of vaporization to be constant one gets a readily integralable differential equation:

									13)

whose integral yields the original form of the Clausius-Clapeyron equation:

									14)
(C is the integration constant.)

	A better agreement with experimental results can be achieved while mantaining the simple form if one modifies the perfect gas law by use compressibility factors to calculate the molar volumes of vapor and liquid:

						15)

Replacing change of molar volumes from Eq. 15) to Eq. 8) and assuming that  is independent of temperature one gets the ordinarily used form of the Clausius-Clapeyron equation:

									16)
						,where				17)

 defined in Eq. 17) is the apparent heat of vaporization.
			The heat of vaporization may not be independent of  temperature since it 			must become zero at the critical temperature, consequently, it must be its 			decreasing function. Sm, and Vm and, by virtue of the latter,  Z have the 			same tendency. It is quite reasonable to expect a considerably milder 				temperature dependence for virtual heat of vaporization than for heat of 			vaporization. Experience supports this expectation.
	According to Eq. 16) ln p is a linear function of 1/T.  The apparent heat of vaporization can be given by multiplying the slope of this plot with -R.

B) Equipment
	We use a U-shaped manometer filled with mercury to measure the vapor pressure. (See the attached figure.)
	It consists of a closed vessel 1 and an open one 3 connected to each other by a flexible tube 2. Column of mercury keeps liquid and vapor closed in vessel 1 and it meets atmospheric pressure in the open vessel 3. A larger vessel 4 may be filled with water of different temperature to thermostate the content of vessel 1. The open vessel is attached to the end of a tape measure 5 with whose aid one can change its position. The position of the open vessel can be fixed with a locking 6.
	Since the level of mercury can be read only in the right side of the manometer the level in the left side must be fixed. The O-ring 8 on the closed vessel serves this goal. Before reading the manometer one has to lift or sink the open vessel until mercury-level in the closed vessel meets the O-ring. As the bridle 7 of the locking should have identical position with the O-ring one has to read the levels at the bridle (h1) and at the mercury surface in the open vessel (h2) on the tape measure. In order to know the atmospheric pressure (b) there is a mercury barometer in the laboratory. In an ideal case, the vapor pressure, in mm's of mercury (Hgmm) is

									18)

	I n reality some corrections are to be made on Eq. 18).
	A) Bridle correction: With the aid of a builder's level (or at least a ruler) we must check if the levels of O-ring and bridle are the same. If not, the difference must be added to (or subtracted from) h1 .
	B) Hydrostatic pressure correction: The vapor pressure in Eq. 18) contains the hydrostatic pressure of the liquid, too, which should be subtracted from it. The height of the liquid-column (usually 2-3cm) can be measured with a ruler, the density of the liquid can be gained from tables of handbooks. Since density of  mercury exceeds that of an organic liquid at least by an order of magnitude, this correction is 1-2 Hgmm. Its negligence causes an error less than 1% in most cases (however this error is systematic).	C) Temperature correction: Density of mercury decreases as temperature increases, consequently, a given h1-h2 means less pressure-difference at higher temperature than at lower one. There is a table at the barometer containing additive correction factors a as function of temperature and pressure. This correction can be practically connected to conversion of unit of pressure to Pa.:

			19)

	D) Air correction:  Despite of all care, some amount of air may be present in the closed vessel. It causes the measured vapor pressure to be dependent on the vapor/liquid-volume ratio: the larger the latter is the smaller it is. In order to determine the volume of vapor phase some assumptions are needed:
		a) air does not dissolve in the liquid;
		b) the volume of the liquid is constant;
		c) both air and vapor obey the perfect gas law;
		d) by lifting of the open vessel one can compress air in the closed vessel 			to a bubble whose volume is negligible comparing with volume of 			the vapor phase during measurements.
	As the volume of the flexible tube is constant, changes in the mercury-volumes in the closed and that in the open vessel, apart from opposite signs, must be equal if the conditions above are fulfilled. The pressure measurable under Eq. 18) is the sum of the vapor pressure and the partial pressure of air. It writes for measurements 1 and 2 carried out on the same temperature:

						20a)
						20b)

(Subscript "corr" designates bridle correction at h1 and conversion/correction according to Eq. 19) at the right bracket.)
	Subtracting Eq. 20a) from Eq. 20b):

				21)

and by virtue of condition c):

							22)
		where  is the mercury-level in the open vessel when air is compressed 		to its minimum volume.
	After substitution of the partial pressure of air in measurement  2 () to Eq. 21) from Eq. 22) one can express its value in measurement 1:

								23)

	This is the value of air correction we have to subtract from the pressure calculated with Eq. 19), assuming that the level of mercury was at the O-ring during measurement 1. It is to be converted for other temperatures with the perfect gas law.

	C) Experimental
	1. Measure bridle correction
	2. Read the barometer and its thermometer. Decide the correction factor from the table.
	3. Determine the air correction by measuring the levels h1 and h2 when the mercury-level in the closed vessel is 
		at the O-ring (mesurement 1);
		between the O-ring and the maximum (measurement 2);
		at its maximum (air compressed to its minimum, measurement 3).
Do not forget to measure the level-difference of the O-ring and the mercury-surface when measuring . One can use a ruler to do that. Measuring  is not needed.
	4. Lift up the open vessel to its maximum position and fill up vessel 4 with water whose temperature almost reaches the atmospheric boiling point of the measured liquid. After 1-2 minutes and some mixing the equilibrium has been settled. Read h1, h2 and the thermometer of vessel 4. Set a new temperature by making use of natural cooling or adding cooler (or warmer) water to vessel 4.
	5. Calculate the vapor pressure for each temperature using Eq.18) and 19).
	6. Calculate the air correction for room temperature by using Eq.21) and 23). Convert its value for other temperatures with the perfect gas law. Subtract the air corrections from the vapor pressures calculated under 5. 
	7. Draw two plots:
			C  
		and	  
Read the slope of the lattest and calculate the apparent heat of vaporization. 

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