Question

"Vapor Pressure of Water" assignment:

VAPOR PRESSURE OF WATER INTRODUCTION All liquids exist in equilibrium with the vapor state. As temperature increases, this equilibrium shifts more and more in favor of the vapor until one reaches the boiling point, at which all liquid is converted to gas. The pressure exerted by the vapor over the liquid is known as its vapor pressure and is a function of temperature. Boiling occurs at the temperature at which the vapor pressure is equal to the external pressure. The normal boiling point of a liquid is the temperature at which the liquid boils when the external pressure is one atmosphere or 760 mm Hg. For instance, the normal boiling point of water is 100°C. However, it is well known that water will boil at a lower temperature in the mountains. This is because the atmospheric pressure is lower at higher elevations. The relationship between vapor pressure and temperature is shown in equation 1: (Eq. 1) In P=AHRT) +A where P is the vapor pressure, H is the heat of vaporization of the liquid, R is the gas constant expressed in units of energy, 8.314 J/mol K, and A is a constant unique to each gas. This type of In Y vs. 1/X behavior is common to many types of physical processes and is known as Arhennius type behavior. You will notice that equation I can be written as the equation of a line, where Y is equal to In P, X is equal to 1/T, A is the intercept, and AHR is the slope. In this experiment, you will determine the vapor pressure of water at several temperatures. You will then determine the heat of vaporization of water by plotting the natural log of the vapor pressure, measured in atmospheres, versus 1/T in Kelvin. The slope of this line will be AHR EXPERIMENTAL The Heat of Vaporization of Water: Obtain a special one hole stopper, as shown in figure 1, and a test tube. Fill the test tube to just below the rim with deionized water and firmly insert the special one-hole rubber stopper into the neck of the test tube. NOTE: IT IS IMPORTANT THAT THERE ARE NO AIR BUBBLES IN THE TEST TUBE Figure 1: Special One-Hole Rubber Stopper with Stopcock 79

Answer 1

1. we make ln P dimensionless by divi

v.p. (mmHg)	ln (P /1mm Hg)	(T/K)	1/(T/K)
40	3.689	305	0.003279
60	4.094	313	0.003195
80	4.382	320	0.003125
100	4.605	325	0.003077

(a) Slope of line is = -4504 K

(b) slope = -$\Delta$vap H /R = -4504 K

$\Delta$vap H = 4504 K * 8.314 J/K-mol = 37446 J/mol = 37.45 kJ/mol

(c) Normal boiling point : P = 1 atm or 760 mmHg :

we have : ln (P /1mm Hg) =  -4504 / (T/K)+A (intercept) = -4504 / (T/K)+18.5

ln 760 = 6.63 =  -4504 / (T/K)+18.5

T = Tb = 379.5 K = 106.5 C

(d) " Substance melts at T when vapor pressure of the substance in its solid phase is equal to its vapor pressure in the liquid phase" P = ?

In P against 17T A In (P/1mm Hg) = -4504*x + 18.5 R2 = 0.999 5.000 4.000 3.000 In( P/1mm Hg) 2.000 1.000 0.000 0.003100 0.003150 0.003200 0.003250 1/(TIK)