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Vapor Pressure and Enthalpy of Vaporization of Water
Kyle Miller
November 19, 2006
1 Purpose
The purpose of this experiment is to calculate the enthalpy of vaporization of water by finding the vapor pressure of water over a range of temperatures.
2 Procedure
In an inverted 10mL graduated cylinder, a sample of air is trapped. The cylinder is submerged in a 1L beaker of water. The distance between the surface of the water in the beaker to the top of the water in the cylinder is then measured and shall be denoted as h. The beaker is heated to 80◦C and the volume of the air is measured. While stirring the water in the beaker to be sure of even heat distribution, let the water cool and record the volume of the gas every 5◦C. After the temperature of the beaker reaches 50◦C, the beaker is cooled to near 0◦C by adding ice. The volume of air and the near-zero temperature is measured.
3 Data
The following data were collected:
Patmosphere = 747.81 mmHg
h = 74.5 mm
Table 1: Measured values T (◦C) V (mL) 17 5. 73.1 10. 70 9. 65 8. 60 8. 55 7. 50 7. 1.0 5.
4 Calculations
- Tk is in K so it is T + 273. For example, 17◦C + 273 = 290K.
- To correct for inverted menisci, 0.20mL is subtracted from each volume reading. So, Cor- rected V = V − 0 .20mL. For example, corrected V = 10.00mL − 0 .20mL = 9.80mL
- The pressure on the cylinder is Pcylinder = Patmosphere + h · 113.^00. 6 mmHg. Substituting in the actual values, Pcylinder = (747.81 + 7413 ..^56 )mmHg = 753mmHg
- The number of moles of trapped gas is nair = PR^ ··VT where R = 62. 4 mmHgmol·K·L. Substituting
values where T is near zero, nair = 753 · 10005.^20
4 · 274 mol = 0.000229mol
The partial pressure of air in the cylinder for each temperature is Pair = nairV^ · R·T. For example, when T = 346, Pair = 0.^0002299 .· 8062.^4 ·^346 1000
= 505mmHg
- The vapor pressure is Pwater = Pcylinder − Pair. For example, when T = 1, Pwater = (753 − 753)mmHg = 0mmHg
- The natural logarithm of P column is simply ln Pwater. For example, when T = 73.1, ln P = ln 248 = 5. 51
6.1. The (^) T^1 k is, for example, 3461 = 0. 00289
Using this data, and a linear regression algorithm, we can find the relation
ln Pwater = − 5190 ·
T
The slope, −5190, is − ∆H Rvap , so ∆Hvap = 5190 · R = 5190 · 8. (^31) molJ = 43100 (^) molJ =
- (^1) molkJ
Compared to the accepted value of 40. (^656) molkJ , the error is 43.^140 −.^407.^7 = 5.90%
- For fitting to two points, the Clausius-Clapeyron equation is
ln
Pvap,T 1 Pvap,T 2
∆Hvap R
T 2
T 1
The “graphical” method (though it should rather be called the “linear regression” method as a graph is not needed for the calculations) is more accurate than the “two point” method since it takes into account more than two points. The more points there are, a truer correlation can be made as there can be measurement error that goes by unnoticed with only two points. Statistically, the quality of a linear approximation is proportional to the percentage of accurate measurements, and when there are more points, the linear approximation is better.
6 Sources of error
As was mentioned above, some error was due to the assumptions made, which amounted to less than a percent for 0mmHg as the partial pressure at 0◦C (which could compound and be more or less of an error later on, though) and an unknown amount for letting h be constant. Another source of error was that we did not use distilled water for the experiment. The dissolved ions would have lowered the measured partial pressure of the water. For the calculations, a (hopefully) small error could have arisen due to rounding after every step to 3 or 4 significant figures (depending on the numbers in the equations). Finally, Steven could have been a source of error because he kept taking a little water out of the beaker to boil on the hot plate. This would change the pressure the water exerts on the air and could skew the volume measurements.
