OBJECTIVES/OVERVIEW

This experiment is intended to help the student become familiar with the computers that will be used in several lab exercises during the semester. It is also intended to demonstrate that graphing is an important and powerful means of determining the relationship between variables. We will use two variables in this lab, initially identified as "X" and "Y". Later you will analyze a set of experimental data from an actual chemical reaction. A primary purpose in several laboratory investigations this semester will be to find the mathematical relationship between two variables. An example is kinetics where you will be asked to determine the relationship between the concentration of a reactant and the time elapsed during the reaction. In this exercise, you will use a computer and the Graphical Analysis software to help determine several of these relationships.

EXAMPLE 1
Suppose you have these four ordered pairs, and you want to determine the relationship between x and y:

x	y
2	6
3	9
5	15
9	27

The first step is to make a graph of y versus x. The y-axis is used for values of the dependent variable while the x-axis is used for the independent variable.

Since the shape of the plot is a straight line with x increasing as y increases, it is a simple direct relationship. An equation can be written showing this relationship: y = k•x. This is done by writing the variable from the vertical axis (dependent variable) on the left side of the equation, and then equating it to a proportionality constant, k, multiplied by x, the independent variable. The constant, k, can be determined either by finding the slope of the graph or by solving the equation for k (k = y/x), and substituting data for one of the ordered pairs. In this simple example, k = 6/2 = 3. If it is the correct proportionality constant, then you should get the same k value by dividing any of the y values by the corresponding x value. The equation can now be written: y = 3•x. That is y varies directly with x with a constant (slope) of 3.

EXAMPLE 2
Consider these ordered pairs:

x	y
1	2
2	8
3	18
4	32

First plot y versus x. The graph looks like this

Since this graph is not a straight line, you must make another graph. It appears that y increases as x increases. However, the increase is not directly proportional. Rather, y might vary exponentially with x. Thus y might vary with the square of x or the cube of x, so the next logical plot would be y versus x2. The graph looks like this:

Since this plot is a straight line, y varies with the square of x, and the equation is:

y = k•x²

Again, place y on one side of the equation and x² on the other, multiplying x² by the proportionality constant, k. Determine k by dividing y by x²:

k = y/x² = 8/(2)² = 8/4 = 2

This value will be the same for any of the four ordered pairs, and yields the equation:

y = 2•x² (y varies directly with the square of x)

EXAMPLE 3:

x	y
2	24
3	16
4	12
8	6
12	4

A plot of y versus x gives a graph that looks like this:

A graph with this curve shape always suggests an inverse relationship, because Y decreases as X increases. To confirm an inverse relationship, plot the reciprocal of one variable versus the other variable. In this case, y is plotted versus the reciprocal of x, or 1/x. (In the Graphical Analysis software, this is indicated as a power or exponent of X. This would be X^-1. The software would use the symbols X^-1 to indicate this. The "^" symbol means exponent. The graph looks like this:

Since this graph yields a straight line , the relationship between x and y is inverse. Using the same method we used in examples 1 and 2, the equation would be:

y = k(1/x) or y = k/x

To find the constant, solve for k (k = y•x). Using any of the ordered pairs, determine k:

k = 2 X 24 = 48

Thus the equation would be:

y = 48/x (y varies inversely with x)

EXAMPLE 4:

The fourth and final example has the following ordered pairs

x	y
1.0	48.00
1.5	14.20
2.0	6.00
3.0	1.78
4.0	0.75

A plot of y versus x looks like this:

Thus the relationship appears to be inverse. Now plot y versus the reciprocal of x. The plot of y versus 1/x looks like this:

Since this graph is not a straight line, the relationship is not just inverse, but rather inverse with a higher power, such as inverse square (X^-2) or inverse cube (X^-3).

The next logical step is to plot y versus 1/x² (inverse square). The plot of this graph is shown below. The line still is not straight, so the relationship is not inverse square.

Finally, try a plot of y versus 1/x³. Yes! This plot comes out to be a straight line and thus

must be the correct relationship. The equation for the relationship is:

y = k(1/x³) or y = k/x³

Now, determine a value for the constant, k. For example, k = y•x³ = (6)(2)³ = 48. Check to see if it is constant for other ordered pairs. The equation for this relationship is:

y = 48/x³ (y varies inversely with the cube of x)