**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.