Alcohol Evaporation

by
Ted Cook
Susan English
Katie Lanier

Abstract:

The purpose of our experiment was to measure the rate of evaporation of anhydrous isopropyl alcohol. The experiment involved measuring the amount of alcohol that evaporated from a glass, funnel-shaped container every twelve hours. These values were then plotted and the resulting equation was found to be y=101.38e-.0026x.

Introduction:

When alcohol is exposed to air, it evaporates. The more concentrated the alcohol, the more rapidly evaporation should occur. When the surface area of the alcohol exposed to the air is reduced, the evaporation occurs slower as a function of time. The rate of

evaporation should obey the differential equation:

dy/dx = -ky (1)

If one solves the differential equation above for Y, one finds the solution to be:

Y = Ae-kt (2)

In attempt to prove this theory, we exposed one hundred milliliters of anhydrous isopropyl alcohol to the atmosphere in a container in which the surface area varies as a function of time.





Description of Experiment:

One hundred milliliters of anhydrous isopropyl alcohol (at least 99% alcohol) were poured into a clean, dry, funnel-shaped glass container. The room temperature was kept relatively constant at 23 degrees Celsius, so that the temperature would not be a variable of the solution. The container had clear milliliter markings on it to make taking measurements easier. Measurements of the amount of alcohol evaporated were taken at 7:00 a.m. and 7:00 p.m. each day for 12 and a half days.

Results and Discussion:

After obtaining all of our data, we fed the data into a computer using the spreadsheet program Microsoft Excel. To find our constant k for our solution:

Y = Ae-kt (2)

for our differential equation:

dy/dx =-kx , (1)

we first took the natural log of our data. Knowing from previous experience in our differential equations class and the required book, Exploring Differential Equations via Graphics and Data, we prepared a graph of the ln(amount) VS. time. This line that is shown on the graph has a slope m that is equal to our constant k. We then plotted the amount of alcohol VS. time to achieve the second graph below. Using a feature allowed by the program Excel, we obtained the equation:

Y(t) = -.0026t + 4.6189 (3)

for the plotted data. The data and graphs follow on the next page.

Data
TIME(HRS) AMT (in mL) LN(AMT)
0 100 4.60517
12 97 4.574711
24 94 4.543295
36 92 4.521789
48 90 4.49981
60 87 4.465908
72 85 4.442651
84 82 4.406719
96 79 4.369448
108 77 4.343805
120 74 4.304065
132 71 4.26268
144 70 4.248495
156 69 4.234107
168 66 4.189655
180 63 4.143135
192 61 4.110874
204 59 4.077537
216 57 4.043051
228 56 4.025352
240 54 3.988984
252 52 3.951244
264 51 3.931826
276 49 3.89182
288 47 3.850148



Conclusion:

The differential equation that was expected for the evaporation of alcohol was Y= -kt (equation 1). A solution to this equation was found for the graph we derived from our data points is y=101.38e-0.0026x (equation 3). Our experimental data revealed a graph that was more linear than expected. This could be attributed to the experimental error in measuring the alcohol that evaporated. Another factor could be the amount of time elapsed between measurements. Future experimentation could possibly correct this error by obtaining data every six hours instead of twelve.