Find the Relationship: An Eercise in Graphing Analsis Computer 5 In several laborator investigations ou do this ear, a primar purpose will be to find the mathematical relationship between two variables. For eample, ou might want to know the relationship between the pressure eerted b a gas and its temperature. In one eperiment ou do, ou will be asked to determine the relationship between the volume of a confined gas and the pressure it eerts. A ver important method for determining mathematical relationships in laborator science makes use of graphical methods. OBJECTIVE In this eperiment, ou will determine several mathematical relationships using graphical methods. EXAMPLE 1 Suppose ou have these four ordered pairs, and ou want to determine the relationship between and : 2 6 3 9 5 15 9 27 The first logical step is to make a graph of versus. Evaluation cop Chemistr with Vernier 5-1
Computer 5 Since the shape of the plot is a straight line that passes through the origin (0,0), it is a simple direct relationship. An equation is written showing this relationship: = k. This is done b writing the variable from the vertical ais (dependent variable) on the left side of the equation, and then equating it to a proportionalit constant, k, multiplied b, the independent variable. The constant, k, can be determined either b finding the slope of the graph or b solving our equation for k (k = /), and finding k for one of our ordered pairs. In this simple eample, k = 6/2 = 3. If it is the correct proportionalit constant, then ou should get the same k value b dividing an of the values b the corresponding value. The equation can now be written: = 3 ( varies directl with ) EXAMPLE 2 Consider these ordered pairs: 1 2 2 8 3 18 4 32 First plot versus. The graph looks like this: Since this graph is not a straight line passing through the origin, ou must make another graph. It appears that increases as increases. However, the increase is not proportional (direct). Rather, varies eponentiall with. Thus might var with the square of or the cube of. The net logical plot would be versus 2. The graph looks like this: Since this plot is a straight line passing through the origin, varies with the square of, and the equation is: = k 2 2 5-2 Chemistr with Vernier
Find the Relationship: An Eercise in Graphing Analsis Again, place on one side of the equation and 2 on the other, multipling 2 b the proportionalit constant, k. Determine k b dividing b 2 : k = / 2 = 8/(2) 2 = 8/4 = 2 This value will be the same for an of the four ordered pairs, and ields the equation: = 2 2 ( varies directl with the square of ) EXAMPLE 3 2 24 3 16 4 12 8 6 12 4 A plot of versus gives a graph that looks like this: A graph with this curve alwas suggests an inverse relationship. To confirm an inverse relationship, plot the reciprocal of one variable versus the other variable. In this case, is plotted versus the reciprocal of, or 1/. The graph looks like this: Since this graph ields a straight line that passes through the origin (0,0), the relationship between and is inverse. Using the same method we used in eamples 1 and 2, the equation would be: = k(1/) or = k/ To find the constant, solve for k (k = ). Using an of the ordered pairs, determine k: k = 2 X 24 = 48 1/ Chemistr with Vernier 5-3
Computer 5 Thus the equation would be: = 48/ ( varies inversel with ) EXAMPLE 4 The fourth and final eample has the following ordered pairs: A plot of versus looks like this: 1.0 48.00 1.5 14.20 2.0 6.00 3.0 1.78 4.0 0.75 Thus the relationship must be inverse. Now plot versus the reciprocal of. The plot of versus 1/ looks like this: 1/ Since this graph is not a straight line, the relationship is not just inverse, but rather inverse square or inverse cube. The net logical step is to plot versus 1/ 2 (inverse square). The plot of this graph is shown below. The line still is not straight, so the relationship is not inverse square. 5-4 Chemistr with Vernier
Find the Relationship: An Eercise in Graphing Analsis 1/ 2 Finall, tr a plot of versus 1/ 3. Aha! This plot comes out to be a straight line passing through the origin. 1/3 This must be the correct relationship. The equation for the relationship is: = k(1/ 3 ) or = k/ 3 Now, determine a value for the constant, k. For eample, k = 3 = (6)(2) 3 = 48. Check to see if it is constant for other ordered pairs. The equation for this relationship is: = 48/ 3 ( varies inversel with the cube of ) MATERIALS computer Logger Pro PROCEDURE 1. Obtain three problems from our instructor to solve using the graphical method described in the introduction. Follow the procedure in Steps 2-9 to find the mathematical relationship for the data pairs in each problem. 2. Begin b opening the file 05 Find the Relationship from the Chemistr with Vernier folder of Logger Pro. 3. Enter the data pairs in the table. a. Click on the first cell in the column in the table. Tpe in the value for the first data pair. b. Click on the first cell in the column. Tpe in the value for the first data pair. c. Continue in this manner to enter the remaining data pairs. Chemistr with Vernier 5-5
Computer 5 4. Eamine the shape of the curve in the graph. If the graph is curved (varies inversel or eponentiall), proceed as described in the introduction of this eperiment. To do this using Logger Pro, it is necessar to create a new column of data, ^n, where represents the original column in the Table window, and n is the value of the eponent: a. Choose New Calculated Column from the Data menu. b. Enter a Name that corresponds to the formula ou will enter (e.g., ^2, ^ 1 ). Use an eponent of 2 or 3 for a power that increases eponentiall, 1 for the reciprocal of n, 2 for inverse square, or 3 for inverse cube. Leave the Short Name and Unit boes empt. c. Enter the correct formula for the column, (^n) in the Equation edit bo. To do this, select from the Variables list. Following in the Equation edit bo, tpe in ^, then tpe in the value for the eponent, n, that ou used in the previous step. Click. According to the eponent of n ou entered, a corresponding set of calculated values will appear in a modified column in the table. d. Click on the horizontal-ais label, select ^n. You should now see a graph of vs. ^n. To autoscale both aes starting with zero, double-click in the center of the graph to view Graph Options, click the Ais Options tab, and select Autoscale from 0 from the scaling menu for both aes. Click. 5. To see if ou made the correct choice of eponents: a. If a straight line results, ou have made the correct choice proceed to Step 6. If it is still curved, double-click on the calculated column, ^n, heading. Decide on a new value for n, then edit the value of n that ou originall entered (in the ^n formula in the Equation edit bo). Change the eponent in the Name. Click. b. A new set of values for this power of will appear in the modified column; these values will automaticall be plotted on the graph. c. You should now see a graph of vs. ^n. If necessar, autoscale both aes starting from zero. d. If the points are in a straight line, proceed to Step 6. If not, repeat the Step-5 procedure until a straight line is obtained. 6. After ou have obtained a straight line, click the Linear Fit button,. The regression line is calculated b the computer as a best-fit straight line passing through or near the data points, and will be shown on the graph. 7. Since ou will need to use the original data pairs in Processing the Data, record the and values, the value of n used in our final graph, and the problem number in the Data and Calculations table (or, if directed b our instructor, print a cop of the table). 8. To print our linear graph of vs. ^n: a. Label the curves b choosing Tet Annotation from the Insert menu, and tping the number of the problem ou just solved in the edit bo (e.g., Problem 23). Drag the bo to a position near the curve. Adjust the position of the arrowhead b clicking and dragging it. b. Print a cop of the graph. Enter our name(s) and the number of copies of the graph. 9. To confirm that ou made the right choice for the eponent, n, ou can use a second method. Instead of a linear regression plot of vs. n, ou can create a power regression curve on the original plot of vs.. Using the method described below, ou can also calculate a value for a and n in the equation, = A ^n. a. To return to the original plot of vs., click on the horizontal-ais label, and select. Remove the linear regression and annotation floating boes. 5-6 Chemistr with Vernier
Find the Relationship: An Eercise in Graphing Analsis b. To return to the original plot of vs., click on the horizontal-ais label, and select. Remove the linear regression and annotation floating boes. c. Click the Curve Fit button,. d. Choose our mathematical relationship from the list at the lower left: Use Variable Power ( = A^n). To confirm that the eponent, n, is the same as the value ou recorded earlier, enter the value of n in the Power edit bo at the bottom. Click. e. A best-fit curve will be displaed on the graph. The curve should match up well with the points, if ou made the correct choice. If the curve does not match up well, tr a different power and click again. When the curve has a good fit with the data points, then click. f. (Optional) Print a cop of the graph, with the curve fit still displaed. Enter our name(s) and the number of copies of the graph ou want, then click. 10. To do another problem, reopen 05 Find the Relationship. Important: Click on the No button when asked if ou want to save the changes to the previous problem. Repeat Steps 3-9 for the new problem. PROCESSING THE DATA 1. Using,, and k, write an equation that represents the relationship between and for each problem. Use the value for n that ou determined from the graphing eercise. Write our final answer using onl positive eponents. For eample, if = k -2, then rewrite the answer as: = k / 2. See the Data and Calculations table for eamples. 2. Solve each equation for k. Then calculate the numerical value of k. Do this for at least two ordered pairs, as shown in the eample, to confirm that k is reall constant. See the Data and Calculations table for eamples. 3. Rewrite the equation, using,, and the numerical value of k. OPTIONAL: TWO-POINT FORMULA A two-point formula is one that has variables for two ordered pairs, 1, 1, and 2, 2. To derive a two-point formula, a constant, k, is first obtained for a direct or inverse formula. All ordered pairs for a particular relationship should have the same k value. For eample, in a direct relationship ( = k ), first solve for k (k = /). Thus 1 / 1 = k, and 2 / 2 = k. Since both k values are the same: 1 = 2 In another eample of an inverse square relationship, = k/ 2 or k = 2. If k = 1 1 2 and k = 2 2 2, then: 1 1 2 = 2 2 2 Derive a two-point formula for each of the three problems ou have been assigned. Show the final answer in the space provided in the Calculation Table. 1 2 Chemistr with Vernier 5-7
Computer 5 DATA AND CALCULATIONS TABLE Problem Number Problem Number Problem Number X Y X Y X Y n = n = n = Problem Number Equation (using,, & k) Solve for k (find the value of k for two data pairs) Final Equation (,, and value of k) k = 2 eample = k/ 2 k = (4)(2) 2 = 16 k = (1)(4) 2 = 16 = 16/ 2 5-8 Chemistr with Vernier
Vernier Lab Safet Instructions Disclaimer THIS IS AN EVALUATION COPY OF THE VERNIER STUDENT LAB. This cop does not include: Safet information Essential instructor background information Directions for preparing solutions Important tips for successfull doing these labs The complete Chemistr with Vernier lab manual includes 36 labs and essential teacher information. The full lab book is available for purchase at: http://www.vernier.com/cmat/cwv.html Vernier Software & Technolog 13979 S.W. Millikan Wa Beaverton, OR 97005-2886 Toll Free (888) 837-6437 (503) 277-2299 FAX (503) 277-2440 info@vernier.com www.vernier.com