Vapor Pressure of a Pure Liquid
Objective
Measure the vapor pressure of a liquid as a function of temperature and to u this information to determine the heat of vaporization and the compressibility factor.
Theory
Consider a system compod of a pure liquid that is in coexistence with its vapor at a fixed temperature and (total) pressure. Assume that the substance itlf caus the pressure of the system and, thus, that there are no other gas prent in the vapor pha. The pressure in this ca is referred to as the orthobaric pressure . Becau liquid and vapor coexist, the rate of vaporization of the liquid is equal to the rate of condensation of the vapor, and this dynamic
equivalence corresponds to liquid-vapor equilibrium for the substance. Although this system can be treated using kinetic molecular theory, we will consider the process from a macroscopic
perspective bad on the equivalence of the chemical potentials of the two phas in coexistence.
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Suppo pure substance A is prent in both the liquid and gaous phas, A(l) and A(g), respectively, at a temperature T and pressure p . We have already assumed that the only vapor prent in the system is that produced by the substance A (i.e., p = p A ). We call p A the saturation vapor pressure of A at the temperature T . The constraint of equilibrium between the two phas of A can be stated as: µA liquid (T ,p )=µA vapor (T ,p ) [1]
where µA liquid =µA vapor are the chemical potentials of liquid and vapor at T and p . If the temperature is changed infinitesimally to T + dT , and the pressure produced by A changes
accordingly to p + dp in order to maintain equilibrium between the two phas, then at the new values for T and p , we must have
southamptonµA liquid (T +dT ,p +dp )=µA vapor (T +dT ,p +dp ) [2]
This can be rewritten in terms of the differentials of the chemical potentials:在线英语发音器
µA liquid (T ,p )+d µA liquid =µA vapor (T ,p )+d µA vapor [3]
Inspecting equations [1] and [2], we e the following relation must hold true:
d µA liquid =d µA vapor
[4]
The differential of µ for any pure material is d µ=V m dp "S m dT , where V m and S m are the molar volume and molar entropy of the substance. Substituting this relation into equation [4], we obtain
V m liquid dp !S m liquid dT =V m vapor dp !S m vapor dT
[5]
Equation [5] is a fundamental equation for any one-component pha equilibrium and can be rewritten as dp dT =!S m !V m大幅度
[6]
造价员考试用书Since we are at a pha equilibrium, ΔH m = T ΔS m . Substituting this into equation [6] we obtain dp dT =!H m T !V m
[7]
This result is the Clapeyron equation and is valid for all first-order transitions between two
phas. For equilibria involving a gas pha and a condend pha (liquid or solid), the change in molar volume will be dominated 1 by the vapor:
!V m =V m vapor "V m condend #V m
gravvapor [8]
And thus equation [7] becomes: dP dT =!H vap TV m vapor [9]
Using the mathematical relationships dlnp = dp/p and d(1/T) = -dT/T 2 we can express Equation
[9] as: dln p d(1/T)=!"H vap R RT pV m vapor [10]
The cond term on the right hand side of the equation, RT/pV m vapor , is equal to the reciprocal of the compressibility factor Z for the vapor (Z =pV m vapor /RT ) and thus Equation [10] can be rewritten as: dln p d(1/T)=!"H vap RZ
[11]
If we assume that the vapor pha is reasonably ideal at the conditions, and thus Z = 1, and parate the differentials we obtain
d ln p ()=!"H vap R
d 1T # $ % & ' ( [10]
This equation implies that a plot of ln p (the equilibrium vapor pressure) versus 1/T will have a slope equal to the negative of the enthalpy of vaporization of the liquid divided by R . We will u this equation to determine the enthalpy of vaporization. Note that, in general, the plot of ln p versus 1/T will not necessarily be linear but may be reasonably so over a small temperature range.
chine food1 This approximation breaks down if the pressure is too high or if the critical point is approached.
You will be assigned one of the following liquids: n-hexane, n-heptane, methanol, acetone, 2-butanone, or cyclohexane. You must obtain vapor pressure as a function of temperature data for this liquid as described below. From your data you will calculate the enthalpy of vaporization for your liquid and compare your experimental value to the literature value.
We will u a device called an isoteniscope to determine the equilibrium vapor pressure as a function of temperature. The isoteniscope containing the liquid is placed in a beaker of water on a stirring hotplate. The isoteniscope is connected to a ballast bottle and an vacuum gauge that is ud to measure the vapor pressure.
First, the dissolved air is removed from the liquid by gentle pumping using an aspirator pump. Then, an external pressure is applied to the system so that the liquid levels of the isoteniscope (acting esntially as a null manometer) are made equal. Under this condition the external pressure is equal to the pressure in the sample bulb (i.e., the vapor pressure of the substance). This method is particularly nsitive becau the density of the liquid is quite low.
1.First t up a water bath by filling a large beaker with water and placing it on a stirring hot
plate. Add a stir bar and allow it to rotate slowly to ensure thermal equilibrium in the bath.
Suspend a thermometer in the bath so that it is immerd deeply enough into the water to provide accurate readings but is kept away from the side of the bath. Add ice to lower the bath temperature to approximately 10˚C.
2.Place the liquid to be studied in the isoteniscope, so that the bulb is about two-thirds full.
Then place the isoteniscope in the water bath and connect it to the ballast bulb and vacuum gauge (See Figure 1). Clamp the isoteniscope in place using a ring stand.
3.Remove the air from the bulb by carefully and slowly reducing the pressure in the ballast
bulb until the liquid boils very gently. (For some liquids, this may not be possible. In such cas, pumping on the system for about 10 minutes with a good vacuum al should be
sufficient). Continue pumping for about 4 minutes, but be careful to avoid evaporating too much of the liquid.
4.Tilt the isoteniscope so that some liquid from the bulb is transferred into the U-tube.
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Carefully admit air to the ballast bulb through stopcock until the levels of the liquid in both arms of the liquid in both arms of the U-tube are the same. Record the temperature of the bath and the pressure on the vacuum gauge. To establish that all the air has been removed from the isoteniscope, carefully reduce the pressure in the ballast bulb until a few bubbles are obrved passing through the U-tube liquid. Then redetermine the equilibrium pressure reading. Repeat, if necessary, until successive readings are in good agreement.triste
5.Once all the air is removed and a good reading is obtained at ~10˚C, heat the thermostat一起英文
bath about 5°C (Caution, this must be done slowly!). Keep the liquid levels in the U-tube equal at all times. When the temperature has stabilized, adjust the pressure in the ballast bulb until the levels in the U-tube are equal and record both temperature and pressure.
6.Take readings every 5 degrees until you reach the normal boiling point.
1.Plot ln p versus 1/T. If the data appear linear, obtain the slope and determine ΔH vap. If the
plot is nonlinear, fit with a polynomial and u the slope at 25°C to determine ΔH vap.
2.Compare your enthalpy of vaporization to the literature value. Make a table of the enthalpy
of vaporization for each of the liquids listed in the procedure and comment on the factors that influence the vaporization of liquids.
3.Determine the compressibility factor for the vapor pha of your liquid
Things to include on your ONE-PAGE report.
•The graph you ud to determine the enthalpy of vaporization.
• A table of data that contains the temperature, pressure, Ln(p), 1/T, your experimental value for the enthalpy and entropy of vaporization, the literature values for the enthalpy, entropy of vaporization, and the compressibility factor.
•All appropriate errors.
