Chem20Graphing - Module 20 — Graphing Calculations in...

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Unformatted text preview: Module 20 — Graphing Calculations in Chemistry Modules 19 and above have been re-numbered. Module 19 – Graphing is now Module 20. Module 20 – Spectra is now Module 21 If you are looking for Spectra topics, check Module 21 ***** Module 20 — Graphing Module 20 – Graphing .................................................................................................. 495 Lesson 20A: Lesson 20B: Lesson 20C: Lesson 20D: Lesson 20E: Lesson 20F: Graphing Fundamentals................................................................................... 495 The Specific Equation for a Line...................................................................... 504 Graphing Experimental Data........................................................................... 514 Deriving Equations From Linear Data ........................................................... 521 Linear Equations Not Directly Proportional ................................................. 532 Graphing Inverse Proportions......................................................................... 539 ©2010 ChemReview.net v. 1m Page 417 Module 20 — Graphing Module 20 — Graphing Timing: Begin this module when you are asked to make a line graph of data as part of a lab report or other assignment. Prerequisites: It will be helpful to do Lessons 17A, 18A and 18B prior to this module. Provisions: You will need a pencil, eraser, and about 5-10 sheets of graph paper (quarterinch grid preferred), but you can begin the lessons without graph paper. Pretests: If you feel confident about your graphing ability, try the last graph in the problem set at the end of each lesson. If you can do the last graph easily, you should not need to do the others. If the last graph is difficult, complete the lesson. ***** Lesson 20A: Graphing Fundamentals Graphs P vs. V for Trapped Air A graph is a way to display numbers visually. 12 Pressure (atm) There are many types of graphs, including bar graphs, pie charts, and histograms. In these lessons, our interest will be limited to line graphs: a type of graph often used to display experimental results. An example of a line graph is at the right. 14 10 8 6 4 2 0 0 20 40 60 80 100 120 Volume (L) Computer software can also create graphs. However, in order to create a proper software graph, it is important to be able to do the basic graphing operations without a computer. Graphing Exercises In the following lessons you will “learn by doing” several types of line graphs. The exercises will proceed from relatively easy to difficult. The early examples can be solved in easier steps, but the steps we practice on simple cases will make more complex and computer graphs easier. Please try the rules suggested here. Tip #1: Graph In Pencil When making a graph by hand, use a pencil and eraser. Simple or rough graphs may be sketched in ink, but complex graphs may require draft numbers that are later erased. ©2010 ChemReview.net v. 1m Page 495 Module 20 — Graphing Graphing Numbers Near 0,0 In science experiments that study two variables at a time, the data are generally a series of pairs of numbers and their units. A relationship between two variables can often be determined from a graph of the points in two dimensions. In such experiments, two measurements are recorded after each change in conditions. To simplify our initial study, we will begin with “math graphs” of numbers without units. Table 1 at the right lists an example of these data points. Our focus will be two-dimensional graphs in Cartesian coordinates with a horizontal x-scale and a vertical y-scale. The x-axis is the heavy line perpendicular to the y-scale at y = 0. The y-axis is drawn perpendicular to the x-scale at x = 0. Graphs with x and y-scales may or may not include the xand/or y-axis. Table 1 Data Points x y 20 297 10 177 4 105 ─3.5 15 8 153 In science, data for related variables usually plot as points that fall close to smooth curves or straight lines. Based on the shape and characteristics of the graph, our goal is to develop an equation that describes the relationship between the two variables. To graph data, use the following steps. Complete these steps using the data in Table 1. 1. Decide which column will be plotted on the x-scale. For this data, that’s been done. 2. Fill-in a range chart for each scale. In Cartesian coordinates, each scale is • numbered to start lower than the lowest data value on that scale and end higher than the highest value, and • evenly numbered so that the lines along each scale increase by the same value. To number the scales, begin by filling in the range chart below. Use the data in Table 1. The lowest value in the x column goes in the first blank. Range Chart: x-scale: Low #: __________ High #: __________ Minor unit: ___ Major: ____ y-scale: Low #:___________ High #: __________ Minor unit: ___ Major: ____ 3. Consider adding zero to each range if the range does not include 0. Some graphing exercises will specify a range to use on the scales. If not, you will need to decide if the origin (0,0) should be shown on the graph. On most graphs, the answer will be yes. Including (0,0) will help you to recognize the type of equation that the data represents. For a range does not include zero, to include zero in the range, change one of the numbers in the range to zero. Change the number that increases the range. The number changed to zero will usually be the lower number (but if both of the range numbers are negative, the higher value is changed to zero). ©2010 ChemReview.net v. 1m Page 496 Module 20 — Graphing For this current graph, include zero on both scales. Apply the rule above to the numbers in the range chart, then check your answer below. ***** ( means cover below the * * * * * , do the work above, then check below.) x-scale: Low #: ─3.5 y-scale: Low #: 15 High #: 20 0 Minor Unit: ____ Major: ____ High #: 297 Minor Unit: ____ Major: ____ The x-scale range already includes 0, so no change is needed. For the y-scale, changing the 15 to 0 increases the range of the numbers on the scale. 4. Mark the boundaries of the plot area on your graph paper. Decide how much space will be used to plot your points. In science, graphs must include a title, the numbers on each scale, and a label for each axis. We will include room for those science labels on this math graph even though the labels of the two current columns are simply x and y. Graphs may be done in the portrait or landscape mode. Using the steps for graphing suggested here, data can be plotted to use all of a sheet of graph paper, or a half or quarter sheet. How large to make a graph depends on how it will be used. A small graph is useful for a quick check of relationships. Using an entire sheet of paper is better for calculating an accurate slope of a line, or if you want to include zero on a scale but the data is clustered away from zero. For this problem, either graph on the sample layout provided at the right, or on a copy of this page, or use a half-sheet of your own graph paper. If using your own paper, try to use the same number of squares (12 up and 16 across) as at the right. ©2010 ChemReview.net v. 1m Page 497 Module 20 — Graphing Lightly sketch the borders of the box of gridlines which you will plot the points inside. Leave room at the top, bottom, and left for the title, numbers and scale labels. 5. Calculate the scale minor unit. Round UP. The minor unit is the spacing number for the scale. When numbering the scale, we increase the value of each line by this number. To number each scale so that all of the data will fit, use this equation. Scale minor unit = (High # on scale) minus (Low # on scale) The count of the grid lines on the scale ─2 Then round UP to the next easy number to count by and divide into pieces. The ─2 is a factor that helps to assure that the data will fit if we write the numbers on every other line along the scale. To write numbers less often, this factor would need to have a larger number after the negative sign. What does the round up to the next easy number rule mean? Counting by 2’s means 2, 4, 6, 8, 10, …. 1’s, 2’s, 4’s, 5’s, 10’s, 20’s, 50’s, and 100’s are easy to count by and divide into parts. You could count by 3’s, but in graphing, we need to be able to plot values between numbered lines. Estimating where 4.8 falls on a line between 4 and 6 is probably easier than between 3 and 6. On occasion, the division in the equation above results exactly in a round number that is easy to count by and count between. If that is the case, use that round number as the minor unit. If it does not, round up. For the current graph, on your paper, calculate the x-scale minor unit. Use the equation above. • In the numerator, enter the numbers from the range chart for the x-scale. • In the denominator, count the number of lines on the scale in the box you sketched to plot in. Start with 0 on the bottom left, and count the lines on the graph paper that cross that scale, including the line at the far border. For example, in the sample layout above, count of grid lines along the x-scale. (The xscale is the horizontal scale (“the horizon is horizontal” = -------- .) ***** The count of the x-scale grid lines is 16. Finish the math of the x-scale minor unit equation, then check your answer below. ***** x-scale minor unit = = 23.5 = 1.68 20 ─ (─3.5) 16 grid lines ─ 2 14 Round up to 2. Counting by 2’s is easy, and plotting points between 0 and 2 is easy. ©2010 ChemReview.net v. 1m Page 498 Module 20 — Graphing 6a. Decide the major unit. The major unit is the value by which you increase the numbers that show on the scale. In these lessons, we will write numbers on the scale at every other line. This means the major unit will be double the minor unit. In the x-scale range chart above, write the minor and major unit for the scale. ***** x-scale: Low #: ─3.5 High #: 20 Minor: 2 Major: 4 Now calculate the y-scale minor unit, then check your answer below. ***** = 29.7 Round up to 40. y-scale minor unit = 297 ─ 0 12 lines ─ 2 40 is familiar to count by. (You could round to 50, but for this example choose 40.) In the range chart above, for the y-scale, fill in the minor and the major unit. 6b. Make both ranges slightly wider to be evenly divisible by the major unit. Change the numbers in the range chart using these rules. a. Do not change a 0. Treat zero as evenly divisible by all numbers. b. Do not change numbers that are already evenly divisible by the major unit for that scale. c. Cross out numbers that are not zero and not evenly divisible by the major unit. Increase the range by substituting the next number that is evenly divisible by the major unit for that scale. Make the high number higher and the low number lower. For this current graph, try that step for the x-scale, then check below. ***** Range Chart: x-scale: Low #: ─3.5 ─4 High #: 20 Minor: 2 Major: 4 The major unit for the x-scale is 4. ─3.5 is not evenly divisible by 4, so change ─3.5 to the next number lower that is evenly divisible by 4. Since the high value of 20 is already evenly divisible by 4, no change is needed. Try those steps above for the y-scale, then check below. ***** y-scale: Low #: 15 0 High #: 297 320 Minor: 40 Major: 80 The minor unit on y is 40; make the range wider and divisible by 80. Do not change 0. Change 297 to the next higher number that is evenly divisible by 80, which is 320. (Making the higher number divisible by the major unit may not change how scales are numbered in graphs done by hand, but it’s good preparation for software graphing.) ©2010 ChemReview.net v. 1m Page 499 Module 20 — Graphing 7. Number and label each scale. When graphing by hand, numbering the scales may involve some trial and error, but the following rules will minimize erasing. • The y-axis is a line perpendicular to x = 0. • The x-axis is a line perpendicular to y = 0. If the x- or y-scale includes an axis, we number next to the axis. If not, we number on the left or bottom edge of the grid. Some graphs will have one or both axes in the middle of the plot. Some graphs will omit x = 0, y = 0, or both. A relatively easy way to number the scales is a. At the bottom left corner of the grid, start with the lowest number in the range for each scale. Lightly number the x-scale on the bottom edge, and the y-scale on the left edge. b. If a scale includes zero, draw the axis line all the way across the graph perpendicular to that zero. Put the tick marks and numbers below an x-axis or scale, and to the left of a y-axis or scale. Then move the numbers to the axis, writing firm, final numbers with the same spacing as the light numbers. Erase the light numbers. c. If a scale does not include an axis, simply replace the light numbers on the bottom or left edge for that scale with heavier final numbers. d. For each line on the scale, increase the count by the scale minor unit, but write the count on every other line. The difference in value between the lines is the minor unit, and the difference in value between the numbered lines is the major unit. These steps will be more clear with an example. Using the range chart for Table 1 and rules above, start with the x-scale. Number either the sample grid provided on a previous page or the graphing box on your own paper. Then check your answer below. ***** Since the x-scale includes zero, draw the heavy line perpendicular to zero. Now number the y-scale. Since you already have the y-axis marked, write the numbers to the left of the axis, instead of on the left edge of the grid. If the y-scale includes y = 0, draw the perpendicular heavy line for the x-axis. Replace the light x-scale numbers with final numbers below the axis. Then, label each scale in words. Use the labels for the column plotted on the scale. Add a title to the graph. Base the graph title on the title of the data table for the graph. 240 200 160 120 80 40 0 -4 0 4 8 12 16 20 24 28 ***** ©2010 ChemReview.net v. 1m Page 500 Module 20 — Graphing Your result should look like the graph at the right. ***** 8. Plot the points. Use your own graph paper or the grid at the right. 300 250 200 150 y Using the numbers in Table 1, go out the xscale by the number in the first column, then up by the number in the 2nd column, and make a dot. For visibility, either make the dot somewhat thick or circle it. Table 1 -- Data Points 100 50 0 -4 0 4 8 12 16 20 24 28 x Add a point for each row in the data table. Do that step, and then check your answer below. ***** 9. Draw the function. For natural phenomena, two related quantities will often graph to produce a smooth curve or a straight line. Add a line or curve to your graph that represents where you think all measurements for the two variables would fall if the data were without experimental error. In drawing the line or curve, do not change the data based on where the points plot. You may, however, want to check the plotting of points that are far off the line or curve. 400 320 240 y Do not “connect the dots.” The function drawn should be a smooth curve or line that passes close to most of the points, through or between them, representing the best fit between the data and a smooth function. Table 1 -- Data Points 480 160 80 0 -4 0 4 8 12 16 20 24 28 x If the points plot close to a line, use a straight edge to add the best-fit line to the graph. Complete step 9, then check your answer in the next lesson. ***** ©2010 ChemReview.net v. 1m Page 501 Module 20 — Graphing Summary of Initial Rules: Graphing In Two Dimensions 1. Decide which variable to plot on x. 2. Write the range chart for each scale. x-scale: Low #: _________ High #: __________ Minor Unit: ___ Major: ___ y-scale: Low #: _________ High #: __________ Minor Unit: ___ Major: ___ 3. Consider adding 0 to each range, increasing the range. In most cases, if a range does not include zero, change one number in the range to zero. Change the number that increases the range. 4. Mark the boundaries of the grid plot on the graph paper. 5. Calculate spacing number for each scale. Round UP. Use this equation: Scale minor unit = (High # on scale) minus (Low # on scale) (The count of the grid lines on the scale) ─ 2 and then round UP to the next easy number to count by and count between. 6. Decide the major unit for each scale, then make both ranges slightly wider and evenly divisible by the major unit for that scale. To number every second line on a scale, make the major unit double the minor unit. Add the minor and major units to the range chart. 7. Number the scales or axes. Label the scales. Title the graph. 8. Plot the points. 9. Draw the function: either a smooth curve or straight line that “best fits” the points. Practice: For more practice in graphing, do these. 1. Graph the data in Table 2. Include zero on both scales. Use your own graph paper. Plot on a grid that includes 15 lines across the page and 8 lines up, counting from zero and including the lines on the far borders of the grid. 2. Graph the data in Table 3. Include zero on both scales. Table 3 Table 2 Data Points Data Points x y x y 1.5 4.1 100 0.030 4.2 5.2 225 0.0050 6.3 6.0 325 ─0.015 2.7 4.6 ─0.045 475 Use your own graph paper. Plot on a grid that includes 16 lines across the page and 10 lines up, counting from zero and including the lines on the far borders of the grid. ©2010 ChemReview.net v. 1m Page 502 Module 20 — Graphing ANSWERS 1. At Step 3, the range chart is x-scale: Low #: 1.5 0 y-scale: Low #: 4.1 High #: 6.3 Major: ____ High #: 6.0 0 Minor unit: ____ Minor unit: ____ Major: ____ At Step 5: x-scale minor unit = 6.3 ─ 0 15 lines on x-scale ─ 2 y-scale minor unit = 6.0 ─ 0 =1 8 grids on y-scale ─ 2 = 6.3 = 0.48 13 Round up to 0.5 1 is round. Use it. At Step 6, the range chart is x-scale: Low #: 1.5 0 y-scale: Low #: 4.1 0 7 Minor unit: 0.5 Minor unit: 1 High #: 6.0 Major: 1 Major: 2 Table 2 -- Data Points 8 6 y If the major unit is double the minor unit, number every other line. Change each range number to increase the range, if needed, to be evenly divisible by the major unit. Zero is always evenly divisible. High #: 6.3 4 2 For this graph, at Step 9, all points should fall exactly on a straight line. 0 0 1 2 3 x 4 5 6 7 2. At Step 3, the range chart is x-scale: Low #: 100 0 y-scale: Low #: ─0.045 High #: 475 Minor unit: ____ Major: ____ High #: 0.030 Minor unit: ____ Major: ____ The y-scale, in going from negative to positive, already includes zero. ***** Count the grid lines starting from 0. At Step 5, the spacing numbers are: x-scale minor unit = 475 ─ 0 16 lines ─ 2 = 475 = 33.9 14 y-scale minor unit = 0.030 ─ (─ 0.045) 10 lines ─ 2 ©2010 ChemReview.net v. 1m Round up to 40 = 0.075 = 0.0094 8 Round up to 0.01 Page 503 Module 20 — Graphing At Step 6, the range chart is x-scale: Low #: 100 0 High #: 475 y-scale: Low #: ─0.045 ─0.060 At Step 9 for this graph, all points should fall very close to a straight line. Minor unit: 40 Minor unit: 0.01 High #: 0.030 0.040 Major: 80 Major: 0.02 Table 3 -- Data Points 0.04 0.02 0 0 y To write numbers on every other line, make the major unit is double the minor unit. Each number in a range is moved away from the other, if needed to be evenly divisible by the major unit. Zero is always evenly divisible. 480 80 160 240 320 400 480 560 640 -0.02 -0.04 -0.06 x Lesson 20B: The Specific Equation For a Line Timing: Do this lesson if you are asked to derive an equation from graphed data in which the data points fall close to a straight line. Pretest: If you can do the last problem in this lesson, you may skip the lesson. ***** Graphs of Straight Lines If two variables graph as a straight line, a specific equation relating the variables can be written using the general equation for a line (also called the slope-intercept formula): y = m x +b The equation y = mx + b has two variables ( y and x ) and two constants ( m and b ): y = the value of the data point on the y-scale x = the value of the data point on the x-scale m = the slope of the line, and b = the y-intercept = the value of y at x = 0 (where the line crosses the y-axis). Calculating the Slope On a straight line, the slope is the same between any two points. The slope can therefore be calculated between any two points. Each point is identified by its coordinates (x, y). The point with the lower x value is designated (x1, y1), and the other point (x2, y2). ©2010 ChemReview.net v. 1m Page 504 Module 20 — Graphing The equation that defines slope must be memorized. m = slope = rise = change in y = Δ y = y2 ─ y1 run change in x Δx x2 ─ x1 The symbol Δ means change in. When an equation is needed to solve a problem, the steps are: Write the equation, make a data table using the symbols in the equation, then solve for the WANTED symbol. Using those rules, solve the following problem and then check your answer below. Assume “math” numbers without units are exact; round if needed to 2 or 3 places. Q. Calculate the slope between the points (2, 3) and (5, 18). ***** 1. WANTED: m When a slope is needed, first write the equation for slope. Writing an equation each time you need it helps to set up your data table and helps to store the equation in your long-term memory. m = slope = rise = change in y = Δ y = y2 ─ y1 run change in x Δx x2 ─ x1 2. DATA: m= ? x1 = ________ y1 = ________ x2 = ________ y2 = _________ To solve for the slope, the point with the lower x value is designated (x1, y1), and the other point as (x2, y2). Identify x1, then complete the data table above. 3. SOLVE: Substitute into the equation and solve for the WANTED unit. If needed, finish those steps and then check your answer below. ***** Answer WANTED: m DATA: m = slope = rise = change in y = Δ y = y2 ─ y1 run change in x Δx x2 ─ x1 m= ? SOLVE: x1 = 2 y1 = 3 m = y2 ─ y1 = x2 ─ x1 ©2010 ChemReview.net v. 1m x2 = 5 18 ─ 3 = 1 5 5 ─2 3 y2 = 18 = (the lower x is x1 = 2) 5 =m Page 505 Module 20 — Graphing The Slope of a Line on a Graph A slope can be a positive number, negative number, zero, or infinity. When a straight line on a graph is • shaped / , the slope is a positive number; • shaped \ , the slope is a negative number; • shaped ── , the slope is zero, and • shaped │ , the value of the slope is infinity (∞). Work this example, then check your answer below. Table 1 Q. In the previous lesson, for the data in Table 1 at the right, all points plotted were on a straight line. Data Points a. Calculate the slope between the points in Table 1 that have the highest and lowest x values. x y 20 297 b. Calculate the slope between any other two points in Table 1. 10 177 4 105 ─3.5 15 8 153 ***** WANTED: m DATA: m = slope = rise = change in y = Δ y = y2 ─ y1 run change in x Δx x2 ─ x 1 a. m = ? lower x = x1 = ─3.5 y 1 = 15 x 2 = 20 297 ─ 15 = 282 = SOLVE: m = y2 ─ y1 = x2 ─ x1 20 ─ (─3.5) 23.5 y2 = 297 12 = m for the Table 1 data b. Between any other two points, the slope must be the same, allowing for uncertainty. ***** Calculating the y-Intercept In the slope-intercept formula y = mx + b , the value of the y-intercept (b), a constant, can be found in two ways: • By reading the graph and estimating the value of y at the point where the line crosses the y-axis, or • By calculating the y-intercept using the slope value and the coordinates of any one point on the line. ©2010 ChemReview.net v. 1m Page 506 Module 20 — Graphing At the right is a graph for the Table 1 data. Looking at the graph, estimate a value for the y-intercept. Table 1 -- Data Points 480 400 b ≈ _______ 320 y The value of the y-intercept can also be calculated mathematically. The general equation for a line is 240 160 y = mx + b 80 0 The values for y and x will vary -4 0 4 8 12 16 20 24 28 for each point on the line, but m x and b are constants: they stay the same for all points on the line. If you know the values for y and x for any two points on the line, you can find the slope (m) of the line, as was done in the calculations above. Once the value for the slope (m) of the line and the coordinates (x,y) for any one point on the line are known, you know values for three of the four symbols in y = mx + b. When values are known for all of the symbols in an equation except one, the missing value, in this case the y-intercept (b), can be found using algebra. For the graph of the Table 1 data, all points fell exactly on the line. In that case, any points in the data can be used to calculate numeric values for m and b. Use the slope for the Table 1 data that was calculated above, plus the data for the highest x value in the data, calculate a value for b, then check your answer below. ***** WANT: DATA: b Writing the WANTED symbol helps to identify the possible equations that may solve the problem. At this point, we only know one equation that uses b. Write that equation, then make a data table using the symbols in the equation. ***** y = mx + b m = 12 (solved above) For the point with the highest x value in the Table 1 data, x = 20 and y = 297 Solve y = mx + b b= ? for the WANTED symbol. ***** b = ? = y ─ mx = 297 ─ (12)(20) = 297 ─ 240 = 57 = b Compare the calculated b value to your b value estimated from the graph above. Are they close? Which is likely to be the more accurate value? ©2010 ChemReview.net v. 1m Page 507 Module 20 — Graphing The Specific Equation for the Line The general equation for a line is y = mx + b . The specific equation for a line is obtained by substituting into y = mx + b the numeric values for the two constants. The specific equation for the line on the graph above is y = 12x + 57 The specific equation can be used to calculate what the values would be at other points on the line. For any y, using the specific equation, you can calculate what x must be. For any x, you can calculate y. In science, this is a key benefit of graphing data. Based on a data sample that you find by experiment, if you can find an equation that fits the data, you can predict results for other cases without having to test every case. Testing the Equation To be certain that you have solved for the specific equation correctly, test the equation: substitute into the specific equation for the line either the x or y value for a point on the line where both x and y are known. Then use the specific equation to calculate a value for the other variable, and compare this equation prediction to the known data. Try it. Substitute x = 4 into the specific equation for the graphed line above. Calculate a value for y. Compare that answer to the y value for x = 4 in Table 1. ***** y = 12x + 57 = 12(4) + 57 = 48 + 57 = 105 = y That value matches the value for y in the table at x = 4. Try another. Predict the value for x when y = 15. ***** y = 12x + 57 ; 15 = 12x + 57 ; ─ 42 = 12x ; x = ─ 3.5 This matches the x value at y = 15 in the data table. Summary 1. The general equation for a line is y = mx + b , where m and b are constants. The slope of the line is m, b is the y-intercept, and x and y are variables that are the coordinates of the points on the line. 2. m = slope = rise = change in y = Δ y = y2 ─ y1 run change in x Δx x2 ─ x1 3. Substitute the values of the constants m and b into the general equation This gives the specific equation for the line. If any one of the x or y coordinates for a point on the line is substituted into the specific equation, the equation will predict the value for the other coordinate. ©2010 ChemReview.net v. 1m Page 508 Module 20 — Graphing 4. To find the specific equation for a line, a. Find the value of the constant slope (m) between any two points on the line. b. Find the x and y coordinates of any point on the line, substitute those values plus the value for m into y = mx + b , and solve for the constant value of b. c. Substitute the numeric values of m and b into the general equation y = mx + b . Practice: On Problems 1 and 2, do every other part. On the rest, do the odd problems. Do the evens if you need more practice. Assume these “math” numbers are exact; round answers to 2 or 3 places. 1. Calculate the slope between these points. Try these without a calculator. a. (40., 20.) and (10., 30.) b. (20., 50.) and the origin. 2. Two points determine a line. Calculate the specific equation for the line that passes through the two points a. in Problem 1a. b. In Problem 1b. 3. If the slope of a line is ─2 and one point on the line is (1, 7), a. calculate the y-intercept. b. Write the specific equation for the line. c. What is the value for y on the line at x = 12 ? 4. Look at the two graphs below labeled Problem 5 and Problem 6. In which case will the calculated slope be a negative number? Problem 5 5. Try this problem without a calculator. 6 4 For the graph at the right, 2 a. Fill in this chart. y1 = _____ x2 = 4 y2 = _____ 0 y x1 = 0 -2 0 1 2 3 4 -4 b. Calculate the slope. -6 c. Read the y-intercept. -8 x d. Write the specific equation for the line. e. On the line, at x = 25, y = ? ©2010 ChemReview.net v. 1m Page 509 Module 20 — Graphing 6. For the graph at the right, Problem 6 a. Calculate the slope. 60 b. Calculate the y-intercept. 56 d. On this line, at y = 78, x = ? 52 y c. Write the specific equation for the line. 48 44 40 36 32 28 1 2 7. The data in Table 2 at the right was graphed in the problems at the end of the previous lesson. The data graphed as a straight line, with each data point on the line. a. Calculate the slope for the line. b. Calculate the value of the y-intercept. c. Write the specific equation for the line. d. Test the equation by calculating a predicted value of x when y = 5.2 . Compare that calculated value to the value for x at y = 5.2 in the data table. 8. The data in Table 3 at the right was graphed in the problems at the end of the previous lesson. The data graphed as a straight line, with each data point very close to the line. 3 4 x 5 Table 2 Data Points x y 1.5 4.1 4.2 5.2 6.3 6.0 2.7 4.6 Table 3 Data Points b. Test the equation by calculating a predicted value of y when x = 225 . ©2010 ChemReview.net v. 1m x y 100. 0.030 225 0.0050 325 ─0.015 475 a. Use the data in the table or the graph to calculate the specific equation for the line. ─0.045 Page 510 Module 20 — Graphing ANSWERS 1. WANT m. The equation that uses m and data for two points is m = slope = rise = change in y = Δ y = y2 ─ y1 run change in x Δx x2 ─ x1 1a. Define the lower x value as x1, so x1 = 10. , y1 = 30., x2 = 40., y2 = 20. Plug those numbers into the equation for m and solve. ***** m = slope = rise = change in y = Δ y = y2 ─ y1 = 20. ─ 30. run change in x Δ x x2 ─ x1 40. ─ 10. = ─ 10 = ─ 0.33 30 1b. Define the lower x as x1, so x1 = 0 , y1 = 0 , x2 = 20. , y2 = 50. m = slope = rise = change in y = Δ y = y2 ─ y1 = run change in x Δx x2 ─ x1 50. ─ 0 20. ─ 0 = 2.5 2. To find the equation for a line, first calculate the slope (m), then use the coordinates of any point on the line plus the slope to find the y-intercept (b), using y = mx + b . To write the specific equation, substitute the values for m and b into y = mx + b . a. For problem 1a, the slope is ─ 0.33 (see above). Two points were used to determine the line, so by definition both points are exactly on the line. y = mx + b is the only equation we know that uses b. Any point (x,y) exactly on the line can be used to calculate b. If we use the first point, y = 30. m = ─ 0.33 x = 10. Solving in symbols first b = y ─ m x = 30 ─ ( ─ 0.33(10) ) = 30 + 3.3 b = 33.3 The specific equation for the line is y = ─ 0.33(x) + 33.3 b. Only two points were used to determine the line, so by definition both points are exactly on the line. Any point exactly on the line can be used to calculate b. If we use the second point, x = 20. , y = 50. m = 2.5 (see 1b). y = mx + b 50 = 2.5(20) + b 50 = 50 + b b = 0 The specific equation for the line is y = 2.5(x) The value for b can also be solved by inspection. The y-intercept is the value of y when x = 0 . For any line that passes through the origin, b = 0. ©2010 ChemReview.net v. 1m Page 511 Module 20 — Graphing 3a. WANT: b The only equation we know that uses b is DATA: y = mx + b m = ─2 , x = 1 , y = 7 . SOLVE: y = mx + b ; 7 = ─2(1) + b ; 7 = ─2 + b ; b = 9 3b. WANT: specific equation for line. Use y = mx + b with the values substituted for the constants m and b. From 3a above, b = 9 , m = ─2 , y = ─2(x) + 9 3c. WANT: y DATA: x = 12 SOLVE: from part b, y = ─2(x) + 9 = ─2(12) + 9 = ─24 + 9 = ─15 = y 4. Problem 4. If the data plots to give a \ slope, the slope (m) must have a negative value. In the graph for problem 5, the / slope means a positive m value. 5a. x1 = 0 y1 = 5 x2 = 4 y2 = ─7 5b. m = slope = rise = Δ y = y2 ─ y1 = ─ 7 ─ 5 = ─ 12 = run Δx x 2 ─ x1 4 ─0 4 ─3 = m The slope is negative, consistent with the slant of the graphed line. 5c. The line crosses the y-axis at +5 . b = 5 5d. To get the specific equation, substitute the specific values for the constants m and b into y = mx + b From above, m = ─3 , b = + 5 , y = ─3(x) + 5 5e. The specific equation calculates values for points on the line. On this line, at x = 25, ? = y = ─3(x) + 5 6a. WANT: m = ─3(25) + 5 = ─70 = y Write the equation for m, and a data table with the symbols needed to solve m. m = slope = rise = change in y = Δ y = y2 ─ y1 run change in x Δx x2 ─ x1 DATA: If the lowest and highest x values on the graph are used, m=? x1 = 1 y1 = 34 x2 = 5 m = slope = rise = Δ y = y2 ─ y1 = run Δx x2 ─ x1 y2 = 50 50 ─ 34 = 16 5─1 4 = 4.0 The slope is positive, consistent with the / slant of the graphed line. 6b. The graph does not show x = 0, so the y-intercept does not show on the graph and must be calculated. WANTED: b Write the one equation that you know that uses b: y = mx +b . ©2010 ChemReview.net v. 1m Page 512 Module 20 — Graphing DATA: b = ? At x = 1, y = 34 ; y = mx +b ; 34 = 4.0(1) + b ; b = 30 6c. WANT: specific equation for line. Use y = mx + b with the values substituted for the constants m and b. From above, m = 4 , b = 30 , y = 4x + 30 6d. Substitute y = 78 into the specific equation: y = 4x + 30 ; 78 = 4x + 30 ; 4x = 48 ; x = 12 7a. WANT: m Write the equation for m, and a data table with the symbols needed to solve m. m = slope = rise = change in y = Δ y = y2 ─ y1 run change in x Δx x2 ─ x1 DATA: If we use the lowest and highest x values in the data table, m=? x1 = 1.5 y1 = 4.1 x2 = 6.3 y2 = 6.0 m = slope = rise = Δ y = y2 ─ y1 = 6.0 ─ 4.1 = 1.9 run Δx x2 ─ x1 6.3 ─ 1.5 4.8 The slope is positive, consistent with the slant of the graphed line. 7b. WANTED: b = 0.40 Write the one equation that you know that uses b: y = mx +b DATA: b = ? , m = 0.40, If we use point x = 2.7, y = 4.6, y = mx + b 4.6 = 0.40(2.7) + b 4.6 = 1.08 + b b = 3.52 = 3.5 7c. To write the specific equation, substitute the calculated values for m and b into y = mx + b . The specific equation for the line is y = 0.40(x) + 3.5 7d. Substitute y = 5.2 into the specific equation y = 0.40(x) + 3.5 to find x . 5.2 = 0.40(x) + 3.5 1.7 = 0.40(x) ; x = 1.7/0.40 = 0.425 = 4.2 = x This answer matches Table 2 for y = 5.2 8a. WANT: the specific equation for the line. Start from the general equation for a line: y = mx + b . To find the specific equation, • calculate the slope (m) between any two points on the line, then • use the coordinates of any point on the line plus m to find the y-intercept (b) using y = mx + b ; • substitute the values for m and b into y = mx + b . All of the points plotted very close to the line, so any two points may be used. If we use the lowest and highest x values in the table, x1 = 100. , y1 = 0.030 , x2 = 475 , y2 = ─ 0.045 ©2010 ChemReview.net v. 1m Page 513 Module 20 — Graphing m = slope = rise = y2 ─ y1 = ─ 0.045 ─ 0.030 = ─ 0.075 = ─ 0.00020 = ─ 2.0 x 10─4 = m run x2 ─ x1 475 ─ 100. 375 Note that the slope is negative, consistent with the \ slant of the graphed line. Any point on the line can be used to calculate b. If we use the data point x = 325, y = ─ 0.015, y = mx + b ─ 0.015 = (─2.0 x 10─ 4)(325) + b ─ 0.015 = ─ 0.065 + b b = + 0.050 The specific equation for the line is y = ─ 2.0 x 10─ 4(x) + 0.050 8b. WANT: y Substitute x = 225 into y = (─2.0 x 10─ 4) (x) + 0.050 y = ( ─ 2.0 x 10─ 4) (225) + 0.050 = ─ 0.045 + 0.050 = + 0.005 = y This answer matches Table 3 for x = 225 . ***** Lesson 20C: Graphing Experimental Data Pretest: If you can do the last problem at the end of this lesson, you may skip the lesson. ***** Graphs in Science Graphs of experimental data collected in laboratory experiments are similar to the “math” graphs studied in the previous two lessons, but graphs in science differ in some respects. • Most math graphs are assumed to have exact numbers. In science, graphs of experimental data must allow for uncertainty in the data. • Math graphs are usually done with numbers. Graphs in science are usually done using measurements: numbers with units and labels. Graphs in science explore how the physical world works. When the rules for our environment can be expressed as equations, predictions about the behavior of natural phenomena become easier, and the results can be applied to improve the quality of our lives. Studies of Two Variables In science, to investigate the relationships among a number of variables, one strategy is to design an experiment that holds all but two variables constant. For the two remaining variables, we measurably change one variable and measure what happens to the other. ©2010 ChemReview.net v. 1m Page 514 Module 20 — Graphing The two variables studied in experiments can often be classified as dependent and independent. • An independent variable is one that is changed in a controlled way or is measured at a regular interval. • A dependent variable responds to the change in the independent variable. For example, in experiments we often measure a variable at specific points of time. Time is then the independent variable: we control the interval at which we record measurements. However, for an experiment in which we change a solution concentration by regular amounts, then measure the time a reaction takes, time is the dependent variable. Which quantities are dependent and independent depend on the experimental design. In many experiments, there will be no controlled or regular interval at which the value of a variable is measured, and neither variable will be considered to be dependent or independent. Data Tables A graph must be accompanied by the data table upon which it is based. In a data table, it is easier to see the numeric measurements recorded in an experiment. Graphs assist in finding the relationships among the variables. A data table must have the following elements. • A title that identifies what quantities are being studied; • columns of measurements; and • labels for each column showing the quantity being measured and its consistent units. In some cases, the quantity is clear from its unit and is omitted. If one of the variables in an experiment is independent, it is generally listed in the first column and graphed on the x-scale. Graphs of Data Near 0,0 In an experiment, for a collection of solid (not hollowed) cylinders that are all made of the same metal, mass and volume is recorded for each cylinder. The data is at the right. We would like to know: are the numbers mathematically related? Can we answer questions such as: if a similar cylinder had a mass of 8,532 grams, what would be its volume? To answer these questions, begin by graphing the data. The steps listed below are similar to the steps used in math graphs As we solve, we will address how science graphs may differ. 1. Decide which variable to plot on x. In Table 4, no variable is more controlled than the other or recorded at a regular interval. Either unit can therefore be plotted on the x-scale. For this example, plot mL on x. ©2010 ChemReview.net v. 1m Table 4 Mass and Volume for Metal Cylinders Volume in mL Mass in grams 2.7 21.2 4.0 29.5 9.2 72.4 12.6 99.2 7.6 61.9 Page 515 Module 20 — Graphing 2. Fill in the range chart for each scale. The data in science graphs includes units. All of the data on the same scale must have the same units. In the range chart, including the units is optional, but doing so will serve as a check that the numbers are placed correctly. Fill in the range chart using the data in Table 4. x-scale: Low #: __________ High #:__________ Minor unit: ___ Major: ___ y-scale: Low #:___________ High #:__________ Minor unit: ___ Major: ___ 3. Consider adding 0 to each range if the range is not specified in the problem. For this graph, add the origin to the range, then check your answer below. ***** x-scale: Low #: 2.7 mL 0 mL y-scale: Low #: 21.2 g 0 g High #: 12.6 mL Minor Unit: _____ Major: _____ High #: 99.2 g Minor Unit: _____ Major: _____ For the origin (0,0) to show on the graph, zero must be a part of both ranges. For Table 4, to include the origin in the graph, zero will need to be added to both ranges. 4. Mark the boundaries of the plot area on your graph paper. For this problem, either graph on the sample grid provided below or on a printed copy of this page, or use a half-sheet of your own graph paper. Leave room at the top, bottom, and left for the title and scales. Use the same number of boxes up and across that are shown in the sample grid below. 5. Calculate the scale minor unit. Round UP. Scale minor unit = (High # on scale) minus (Low # on scale) (The count of the grid lines on the scale) ─ 2 and then round UP to the next easy number to count by and count between. Calculate the scale minor unit for the x-scale, then check your answer below. ***** x-scale minor unit = = 12.6 = 1.05 Round up to 2. 12.6 ─ 0 14 grid lines ─ 2 12 Calculate the minor unit for the y-scale, then check your answer below. * * * ** y-scale spacing = 99.2 ─ 0 12 grid lines ─ 2 = 99.2 = 9.92 10 Round up to 10. Enter both minor units to the range chart. Enter the major units: double the minor. 6. Make both ranges slightly wider and evenly divisible by the scale major unit. For this example, first adjust the range for the x-scale, then check below. ***** ©2010 ChemReview.net v. 1m Page 516 Module 20 — Graphing x-scale: Low #: 2.7 mL 0 mL High #: 12.6 mL 16 Minor Unit: 2 Major: 4 Zero is always evenly divisible. Make the high number higher. Move 12.6 to the next higher number divisible by the scale major unit (4), which is 16. Try that step for the y-scale, then check below. ***** y-scale: Low #: 21.2 g 0 g High #: 99.2 g 100 Minor Unit: 10 Major: 20 Adjust the y-scale high number higher, from 99.2 to 100, since 100 is divisible by 20. 7. Number and label each scale. The low number on each scale is 0. That results in the most familiar type of scale, where the bottom left corner is the origin (0,0). For this example, number each scale, then add a label to the scale. For the label, use the heading of the data-table column that you are plotting on that scale. Add a title at the top of the graph, based on the title of the data table. 8. Plot the points. As one indication that the data has uncertainty, either make each point somewhat thick, or circle it. 120 100 80 60 40 20 0 0 4 ©2010 ChemReview.net v. 1m 8 12 16 20 24 28 Page 517 Module 20 — Graphing 9. Draw the function. Add a line or curve to the graph. In natural phenomena, two related quantities will usually graph to produce a smooth curve or a straight line. The function should be a smooth curve or line that passes close to most of the points, representing your best estimate of where the points would be if there were no experimental error (do not “connect the dots”). Do those steps and then check your answer on the second page of the next lesson. ***** The position of the points on your graph, relative to the line, should match the graph below. Note that the data does not fit exactly on a straight line. Should the line for this graph go through the origin? Let’s work through the logic of the experiment. • If the volume of the metal cylinders approaches zero, what should their mass approach? If the mass of the cylinders approaches zero, what should their volume approach? ***** If the mass of a solid metal cylinder approaches zero, its volume must approach zero as well. The line showing the relationship of mass to volume for the cylinders should approach the origin (0,0). However, including (0,0) as a point on the line would be questionable science. We did not record data at (0,0), and if the instruments used in measurements had a systematic error, or the scale on the instrument was read with a consistent error, the line might not pass though (0,0). Graphing can help us to find such systematic errors. That said, if the line representing the smooth function passes through (0,0), it would fit the theoretical results for this experiment. ***** ©2010 ChemReview.net v. 1m Page 518 Module 20 — Graphing Practice If unsure about an answer, check it after each part. Answers are at the end of the lesson. 1. In an experiment, a hollow glass container is filled with dry air and attached to a pressure gauge. The system is then sealed so that no air can enter or escape. The glass container is placed into beakers of ice water, boiling water, and liquid water at various temperatures. The temperature of the water baths and the pressure exerted by the air in the container is reported in Table 5. a. Based on PV=nRT, what are the symbols for the variables held constant in this experiment? b. What are the symbols for the two variables that can change in this experiment? c. In this experiment, which is the independent variable? d. Assuming we will plot the independent variable on x, write a range chart for the data. For this graph, do not modify the ranges to add zero. Table 5 Sealed Fixed Volume of air: Temp. vs. Pressure ºC kPa 0.0 202 100. 276 58 248 82 262 20 250 12 210 44 230 e. Based on the grid below, calculate the scale minor unit for each scale. f. Adjust the ranges in the range chart based on the scale major unit that is double the minor unit. g. Plot the graph for Table 5. If you use your own graph paper, use the same number of grid lines as the grid below. Add the function to the graph. h. Are there any outlier points in which errors in measurement may have been made? tem eratu in d p re egrees F Kelvin vs. Fahrenheit temperature temperature in K ©2010 ChemReview.net v. 1m Page 519 Module 20 — Graphing ANSWERS 1a. The variables V and n are held constant. (R is a constant, not a variable, in PV=nRT.) 1b. P and T vary in this experiment (though temperature is recorded in degrees Celsius instead of kelvins.) 1c. Temperature is the more controlled variable, and is therefore the more independent, because if we place the cylinder into ice water, or boiling water at standard pressure, we know and control what the temperature will be. Once the system is sealed, we do not control what the pressure readings will be: the pressure is dependent on the controlled temperature. 1d. The problem says to plot the independent variable on x. x-scale: Low #: 0.0 ºC Minor unit: ______ Major: ______ y-scale: Low #: 202 kPa 1e. High #: 100. ºC High #: 276 kPa Minor unit: ______ Major: ______ x-scale minor unit = 100 ─ 0 = 100 = 5 22 grid lines ─ 2 20 Don’t round. If the division results in a round number that is easy to count by, use it. y-scale spacing = 276 ─ 202 = 10 grid lines ─ 2 74 8 = 9.25 Round up to 10. x-scale: Low #: 0.0 ºC High #: 100. ºC Minor unit: 5 y-scale: Low #: 202 kPa 200 1f. High #: 276 kPa 280 Minor unit: 10 Major: 10 Major: 20 The x-scale numbers are divisible by 10. Do not change them. For the y-scale, expand the range at both ends. Make both ends divisible by 20, making the low number lower and high number higher. 1g. The graph should result in a smooth curve or a straight line. The data does not fit exactly on a line, but most of the points fit close to a straight line. The function should be a best estimate of where the data would be without experimental Sealed, Fixed Volume: P vs. temp. error. 1h. The point at 20 ºC would seem to be an outlier (a point not consistent with the other data). This often represents an error in measurement, but on occasion may be accurate -and interesting. 300 Pressure in kPa 280 ***** ©2010 ChemReview.net v. 1m 260 240 220 200 0 10 20 30 40 50 60 70 80 90 100 110 temperature in degrees C Page 520 Module 20 — Graphing Lesson 20D: Deriving Equations From Linear Data Timing: Do this lesson if you are asked to derive an equation from a graph in which experimental data points fall close to a straight line. Pretest: If you can do the last problem in this lesson, you may skip the lesson. ***** Developing Equations From Graphs of Straight Lines In investigating two related quantities in science, a key goal is to develop a mathematical equation that accurately predicts how one quantity will change when the other is changed. With such an equation, we can answer “what if” questions by solving an equation, without having to experiment to determine every result. If two variables graph as a straight line, the equation relating the variables can be determined using y = m x +b Reading the Coordinates of Points on a Line In determining the equation for a line, a key step is determining the slope. When graphing experimental data, often the points will be “roughly linear,” falling close to, but not exactly on, a straight line. For roughly linear data, we try to draw the line “through the middle” of the points: where we think the data points would be if there were no error. To describe and predict the relationship between two variables, the slope of a line “through the middle” of the points should result in a more accurate equation than a slope calculated between two points that are not exactly on the line. For this reason, calculating the slope from experimental data requires reading the coordinates of two points on the line, rather than the values of data points that are not exactly on the line. Determining the coordinates of points on a line requires estimation. The graph below represents the Table 4 relationship found in the previous lesson. For each of the cases below, knowing the value for one variable, estimate the value for the other variable on the graphed line. a. At mass = 20.0 grams, volume ≈ _______________. ***** Your answer should be close to 2.5 mL. You may want to use an index card or sticky note as a “T-square” to line up with the line and the two scales. Putting a “tick mark” on scale with the WANTED unit may help in estimating the answer. b. Try this one. At volume = 8.0 mL, mass ≈ __________________ ***** ©2010 ChemReview.net v. 1m Page 521 Module 20 — Graphing Mass and Volume for Metal Cylinders 120 Mass in grams 100 80 60 40 20 0 0 4 8 12 16 20 24 28 Volume in mL Your answer should be close to 63 grams. c. At volume = 4.0 mL, estimate the mass based on the graph = __________________ In Table 4 at the beginning of the prior lesson, the mass at 4.0 mL = ________ , but the line is slightly above the point. Do the following by estimating, based on the graphed line above. d. At volume = 12.0 mL, mass ≈ __________________ e. If mass = 40.0 grams, volume ≈ _______________. f. At mass = 90.0 grams, volume ≈ _______________. ***** Your answers should be close to these. d. At volume = 12.0 mL, mass ≈ 94 g e. If mass = 40.0 grams, volume ≈ 5.1 mL f. At mass = 90.0 grams, volume ≈ 11.4 mL How can you improve your estimates? Making a larger graph, using graph paper with finer lines, using a numbering scale that allows for easy estimation, and using widely separated points all help. However, in reading the coordinates of a point on a line, your answers will have uncertainty. ©2010 ChemReview.net v. 1m Page 522 Module 20 — Graphing Calculating the Slope of a Line When calculating the slope of a line graphed from experimental data, the two points chosen on the line should be widely separated to reduce the impact of uncertainty in reading the point coordinates. Try the following example, then check your answer below. Q. For the line in the graph of the metal cylinders data above, calculate the slope of the line between the two points where mass = 80.0 grams and mass = 0 grams. ***** WANTED: m DATA: m = slope = rise = change in y = Δ y = y2 ─ y1 run change in x Δx x2 ─ x1 ***** At y = 80.0 g , reading the graph, the x value is about 10.2 mL. At y = 0 g , since this line passes through the origin, assume x = 0 mL. Solve for slope. Include units. ***** y1 = 0 g (assumption) m = ? x 1 = 0 mL x2 = 10.2 mL (estimate) y2 = 80.0 g ***** 80.0 g ─ 0 g = 7.85 g m = y2 ─ y1 = x2 ─ x1 10.2 mL ─ 0 mL mL In calculations based on a line through the origin, (0,0) is considered to be exact. The 0 values for the origin therefore have infinite sf and do not limit the sf in an answer. Because points on the line that are not the origin can be difficult to read accurately, the slope should be calculated a second time using two different widely separated points. On the graph above, estimate the values for grams at mL = 2.0 and mL = 12.0 . Then calculate the slope between those two points, and check your answer below. ***** Your estimates for y values should be close to these. x1 = 2.0 mL y1 = 15.6 g (estimate) x2 = 12.0 mL y2 = 94.3 g (estimate) ***** m = slope = rise = change in y = Δ y = y2 ─ y1 = 94.3 g ─ 15.6 g = 7.87 g run change in x Δx x2 ─ x1 12.0 mL ─ 2.0 mL mL The slope between any two points on a line should be the same. In practice, two slope calculations based on coordinate estimates must be close, but due to uncertainty often will not be exactly the same. What value should we use for the slope of the line? If the two slopes are close, average the two answers. ©2010 ChemReview.net v. 1m Page 523 Module 20 — Graphing If your two slope calculations are not close, check your reading of the point coordinates and/or your slope calculation. Based on the calculations above, write the value for the slope. Include units. ***** The calculated slope is the average of the two calculations: m = 7.86 g/mL. Practice A 1. Calculate the slope between these points. Include units. a. Points (10. cm, ─40. ºC) and (─20. cm, 50 ºC) b. (─20 ºF, ─50 cm) and the origin. 2. Using the graph of P vs. ºC at the end of the answers to the previous lesson, a. Estimate ºC at 270 pKa b. Estimate P at 5 ºC. c. Using those two points, find m. Writing the Specific Equation For A Line Through the Origin The equation y = mx + b has two variables ( y and x ) and two constants ( m and b ). For the math graphs in Lesson 20B, to write the specific equation for a line on a graph, we left the variables as y and x, but substituted values for m and b. For graphs in science, when writing the specific equation for a straight line on a graph, • In place of y and x, write the quantities and units plotted on the scale (such as mass in grams or volume in mL) and • In place of m and b, write the numeric value of each constant with its units. Based the calculations above for the metal-cylinders graph above, do those substitutions into y = mx + b and write the specific equation for the line. ***** For this graph, y = mass in grams (the quantity plotted on the y-axis) m = 7.86 g/mL calculated above x = volume in mL (the quantity plotted on the x-axis) b = 0 , since the line goes through the origin. Substitute those terms in y = mx +b. ***** The specific equation for the graphed line and the data is (mass in grams) = 7.86 g • (volume in mL) + 0 mL ©2010 ChemReview.net v. 1m Page 524 Module 20 — Graphing Testing the Equation Once a specific equation is written, it should be tested. • Pick a point in the data table that plotted close to the line. • Table 4 Substitute one of the two coordinates for that point into the specific equation and calculate a value for the other coordinate. • Compare the calculated to the table value. If the two values are close, the equation works. Try it. For this example, use V = 9.2 mL, plug it into the specific equation, and calculate a predicted mass. Compare the prediction to Table 4 at that volume. Mass and Volume for Metal Cylinders Volume in mL Mass in grams 2.7 21.2 4.0 29.5 9.2 72.4 12.6 99.2 7.6 61.9 ***** Specific equation: (mass in grams) = 7.86 g • (volume in mL) + 0 mL Substituting V = 9.2 mL: (mass in grams) = 7.86 g • (9.2 mL) = 72 grams mL The table value at 9.2 mL is 72.4 g. The equation predicts, allowing for uncertainty, the measured value. Allowing for uncertainty, the equation should predict the volume given any mass in the table, and predict the mass given any volume in the table. The specific equation describes the line on the graph, and The specific equation predicts the relationship between the two variables plotted on the graph and written in the equation. Extrapolating From the Data If an equation correctly predicts the experimental results for several small samples, it be able to be used for predictions on larger samples. Use the specific equation above to answer this question. Q. If a cylinder of the same material had a volume of 1,000. mL, what would be its mass? ***** (mass in grams) = 7.86 g/mL • (1,000 mL) = 7,860 g However, though that is the predicted value, the prediction may not be accurate. For experimental data, predicting values on the line between data points (interpolating) is usually safe, but predicting results beyond the range of the measured data (extrapolating) is risky. For example, predictions for real gases based on ideal gas law break down as gases approach conditions at which they condense into liquids or solids. When evaluating experimental results, conclusions should be drawn cautiously and limited to the conditions under which the experiment was run. ©2010 ChemReview.net v. 1m Page 525 Module 20 — Graphing Explaining the Specific Equation In Words The specific equation for the metal cylinders data above can be written as (mass in grams) = 7.86 g/mL • (volume in mL) This equation matches the form y = (constant) • x which is the general equation for a direct proportion. In Lesson 18A, we discussed statements and equations that can be written when a graph results in a straight line through the origin. Review those rules if needed, then express the meaning of the data in Table 4, and the graph of that data, in words and in equations. ***** If a graph results in a straight line through the origin, the variables plotted on y and x are directly proportional. In general, for a graph of a line through the origin, y = (slope) • x which means y = (constant) • x which means y/x is constant. Based on the specific equation that fits the data for the cylinders, • Mass in grams for these cylinders (y) is directly proportional to their volume (x): Mass of each cylinder = (constant) (Volume of each cylinder) • The ratio of mass to volume for the cylinders is constant, and is equal to the slope of the line on the graph. Mass of cylinders = a constant value = the slope of the graphed line Volume of cylinders This last statement can be tested by dividing the second column by the first for any point in the data. Try that calculation for any two points in Table 4. Fill in the table with each result. * * * ** Does the mass/volume ratio equal the slope that you found above? At the end of a laboratory experiment, you are often asked to speculate on the possible meaning of the results. Take a moment to ponder: in this experiment, why would the g/mL ratio be constant? Table 4 Mass and Volume for Metal Cylinders Volume in mL Mass in grams 2.7 21.2 4.0 29.5 9.2 72.4 12.6 99.2 g/mL 7.6 61.9 * * * ** The problem stated that the cylinders were solid (not hollow) and that all were made of the same metal. The data show that these cylinders all had the same mass to volume ratio. The ratio of mass to volume for an object is its density, so all of these cylinders have the same density. The data suggests that solid samples of this metal have a constant density. Will the density identify the metal? If you are near a computer, try an online search on densities of metals. See if the results allow you to tentatively identify the metal used in the cylinders for this experiment. ©2010 ChemReview.net v. 1m Page 526 Module 20 — Graphing ***** To our prior list of steps to use when graphing data, let us add step 10. If the graph is a straight line, a. write the specific equation for the line. Calculate the slope using two points on the line. Find b using x and y values for any one point on the line and y = mx+b. Substitute values for m and b, and quantities and units for y and x, into y = mx+b. b. Test your specific equation. For a point in the graph close to the line, pick one coordinate from the data table. Use the specific equation to predict the value for the other coordinate. The calculated and table value should be close. c. Explain the equation in words. If the graphed line goes through the origin, write the statements and equations that can be applied to direct proportions. Practice B Volume vs. Absolute Temperature 1. For the graph at the right, 30 a. If x1 = 0 K , y1 = _____ Liters If x2 = 300. K , y2 ≈ _____ b. Calculate the slope. 20 10 c. Read the y-intercept. d. Write the specific equation for the line. 0 0 e. Translate the equation into words. f. 100 200 kelvins 300 At x = 400. K, y = ? 2. On the grid at the right, complete the graph of distance versus time represented by the following equation. Distance in miles = (25 miles/hr) (time in hours) Draw the function. Add labels for the title and the axes. ©2010 ChemReview.net v. 1m 100 75 50 25 0 0 1 2 3 Page 527 Module 20 — Graphing 3. For a collection of solid metal cylinders, mass and volume measurements are recorded for each cylinder. The data is at the right. a. Graph this data using the steps in the summary above. Use your own graph paper or the sample below. b. What might explain the difference between the data in Table 6 at the right and the data for the similar experiment in Table 4? Table 6 Mass and Volume for Metal Cylinders – Set 2 Volume in mL Mass in grams 3.5 9.5 7.1 55.9 4.2 33.1 c. Label the third column g/mL. Calculate a value for that ratio for each data point, including units. 5.6 15.1 13.1 35.4 d. Explain the results in the third column. 2.6 20.5 tem perature in degrees F Kelvin vs. Fahrenheit temperature temperature in K ©2010 ChemReview.net v. 1m Page 528 Module 20 — Graphing ANSWERS Practice A 1a. Define the lower x as x1: x1 = ─20. cm y1 = 50. ºC x2 = 10. cm y2 = ─40. ºC m = slope = rise = change in y = Δ y = y2 ─ y1 = ─40. ºC ─ (50. ºC) = ─90 ºC = ─ 3.0 ºC run change in x Δ x x2 ─ x1 10. cm ─ (─ 20. cm) 30 cm cm 1b. If numbers are on the same scale, they must have the same units, so the units of the 0 at the origin match the units of the other numbers on that scale. m=? x1 = ─20 ºF y1 = ─50 cm x 2 = 0 ºF m = slope = rise = change in y = Δ y = y2 ─ y1 = run change in x Δ x x2 ─ x1 2a. At 270 pKa, ºC ≈ 92 ºC y2 = 0 cm 0. cm ─ (─50. cm) 0 ºF ─ (─ 20. ºF) = 50 = 2.5 cm 20 ºF b. At 5 ºC, kPa ≈ 206 kPa 2c. WANTED: m = slope = rise = Δ y = y2 ─ y1 run Δx x2 ─ x1 DATA: Make the data table using the symbols needed to solve the equation. Set x1 as the lower x value in the two points. ***** Since ºC is plotted on x, set x1 = 5 ºC m = ? x1 = 5 ºC , y1 = 206 kPa , x2 = 92 ºC , y2 = 270 kPa SOLVE: m = y2 ─ y1 = x2 ─ x1 270 kPa ─ 206 kPa 92 ºC ─ 5 ºC = 64 kPa = 0.74 kPa 87 ºC ºC Practice B 1a. x1 = 0 K y1 = 0 L x2 = 300. K y2 = 24.6 L (your answer should be close) 1b. WANT = m Using the points above, m = y2 ─ y1 = 24.6 L ─ 0 L = 0.0820 L x2 ─ x1 300. K ─ 0 K K In practice, it may be difficult to calculate a slope on a graph to more than two sf. 1c. For a line through the origin, b = 0 1d. Into y = mx+ b , substitute the above values for m and b and the quantities and units on y and x . Specific equation: (Volume in L) = 0.0820 L ( temperature in kelvins) K 1e. Possible statements include: For this data, • volume in liters and temperature in kelvins are directly proportional. • The ratio of volume in liters to temperature in kelvins is 0.0820 L/K . 1f. Substitute the value into the specific equation. Specific equation: (Volume in L) = 0.0820 L ( 400. K) = 32.8 L K ©2010 ChemReview.net v. 1m Page 529 Module 20 — Graphing 2. The equation is in the form y = (constant)(x) , which is the form of a direct proportion. Data that is directly proportional plots as a straight line through the origin. 100 75 Miles Based in the equation, the slope must be 25 mi/hr. You can use that slope to draw the line. The rise is 25 miles when the run is one hour, and the line has a constant slope. Distance vs. Time 50 25 0 3a. Remember that in science, data 0 often plots as straight lines or smooth curves. Be on the lookout especially for straight lines that extend through the origin. Do not “connect the dots.” ***** The graph is shown after the summary below. 1 Hours 2 3 3b The data in Table 4 was found to fit one metal with constant density. This table may represent data for two different metals, each with constant density. 3c. All of the values should be about 7.9 g/mL or 2.7 g/mL. 3d. The results are consistent with data for two different metals, one with a density of about 7.9 g/mL, the other about 2.7 g/mL. An online search for metal densities may identify familiar metals fitting these results. Summary of Initial Rules: Graphing In Two Dimensions 1. Decide which variable to plot on x. Often, this is the independent variable. 2. Write the range chart for each scale. x-scale: Low #: _________ High #: __________ Minor Unit: ___ Major: ___ y-scale: Low #: _________ High #: __________ Minor Unit: ___ Major: ___ 3. Consider adding zero to each range, increasing the range. In most cases, if a range does not include zero, change the number to zero that increases the range. 4. Mark the boundaries of the plot on the graph paper. 5. Calculate the minor unit for each scale. Round UP. Use this equation: Scale minor unit = (High # on scale) minus (Low # on scale) (The count of the grid lines on the scale) ─ 2 and then round UP to the next easy number to count by and count between. 6. Decide the major unit for each scale, then make both ranges slightly wider and evenly divisible by the major unit for that scale. To number every second line on a scale, make the major unit double the minor unit. Add the minor and major units to the range chart. ©2010 ChemReview.net v. 1m Page 530 Module 20 — Graphing 7. Number the scales based on the minor unit. Label the scales. Title the graph. 8. Plot the points. 9. Draw the function: either a smooth curve or straight line near most points. 10. If the graph is a straight line, a. write the specific equation for the line. • Calculate m using two points on the line. • Find b using y = mx+b , x and y for any one point on the line, and m. • Substitute values for m and b, and quantities and units for y and x, into y = mx+b. b. Test the specific equation. For a point on the graph close to the line, pick one coordinate from the data table. Use the specific equation to predict the value for the other coordinate. Compare the predicted and the data table values. c. Explain the equation in words. If the graphed line goes through the origin, explain the relationship using the statements and equations that apply to direct proportions. ***** Mass vs. Volume - Metal Cylinder Set 2 80 70 Mass (g) 60 50 40 30 20 10 0 0 2 4 6 8 10 12 14 16 18 20 Volume (mL) ©2010 ChemReview.net v. 1m Page 531 Module 20 — Graphing Lesson 20E: Linear Data Not Directly Proportional Pretest: If you can do the last problem at the end of this lesson, you may skip the lesson. ***** Data Clustered Away from Zero The following table shows the correlation between temperatures in the Fahrenheit and Kelvin scales. In this data, the Kelvin temperatures are clustered far from zero. When the numbers on a scale are clustered far from zero, you have a choice. You can number the scale to include zero, or not. To recognize the type of equation that the data represents, including (0,0) will help. However, if the data on a scale is clustered away from zero, choosing not to include zero will help in calculating an accurate slope for the line. For this graph, do not change the range to include zero. Table 7 Fahrenheit versus Kelvin temperatures K ºF 373 212 293 68 273 32 233 ─ 40. For this data, no variable is more controlled than the other. We will arbitrarily choose to graph the Kelvin-scale temperatures on the x-scale. Graph this data using your own half-sheet of graph paper or the sample grid below. temperature in degrees F Kelvin vs. Fahrenheit temperature temperature in K ©2010 ChemReview.net v. 1m Page 532 Module 20 — Graphing Use the ten graphing steps. If you get stuck, check the answers below. ***** Step 2. x-scale: Low #: 233 High #: 373 High #: 212 y-scale: Low #: ─ 40 Minor Unit: ___ Major: _____ Minor Unit: ___ Major: _____ ***** x-scale minor unit = = 140 373─ 233 18 x grid lines ─ 2 16 y-scale minor unit = Step 5. 212 ─ (─40) 16 y grid lines ─ 2 = 8.75 = 252 14 Round up to 10. = 18 Round up to 20. ***** Step 6. x-scale: Low #: 233 220 y-scale: Low #: ─ 40 High #: 373 380 Minor Unit: 10 Major: 20 High #: 212 240 Minor Unit: 20 Major: 40 In adjusting the ends of the range to be divisible by the major unit, make the lower numbers lower and the higher higher. If a number is already divisible, such as ─ 40, no change is needed. ***** 9. See the graph on the following page. ***** 10. If you get a straight line, a) write the specific equation for the line. Substitute into y = mx + b the quantities and units for the two variables and the values for the two constants. For the variables in Table 7, y = temperature in ºF x = temperature in K To calculate the constant slope, we must read coordinates of two points on the line, not in the data, since the line may not pass through the points. But for this data, which is defined rather than experimental, all of the points fall very close to the line. Calculate the slope. For this graph, use the highest and lowest x scale values in the data, and then check your answer below. ***** ©2010 ChemReview.net v. 1m Page 533 Module 20 — Graphing Kelvin vs. Fahrenheit temperature temperature in degrees F 280 240 200 160 120 80 40 0 220 -40 240 260 280 300 320 340 360 380 400 temperature in K Using the lowest x (233, ─40) and highest x (373, 212), x1 = 233 K , y1 = ─40 ºF , x2 = 373 K , y2 = 212 ºF m = slope = rise = Δ y = y2 ─ y1 = 212 F ─( ─40 F) = 252 F = x2 ─ x1 run Δx 373K ─ 233K 140. K 1.80 ºF = m K Using m, calculate b. * * * ** b = the y intercept = the value of y where the line crosses the y-axis = the value of y when x = 0 . You cannot read the y intercept if a graph does not show x = 0, and this graph does not. To find b, we can either redo the graph, this time including 0 on both scales, or we can calculate b mathematically. Let’s try the latter method. If you have not already solved for b, use the data table point at x = 293 K, plus the value for m above, and solve for y = mx +b for b. ***** ©2010 ChemReview.net v. 1m Page 534 Module 20 — Graphing WANT: DATA: b The only equation we know that uses b is y = mx + b m = (1.80 ºF/K) At x = 293 K, y = 68 ºF ? = b = y ─ mx = ( 68 ºF ) ─ [(1.80 ºF/K)( 293 K)] = ( 68 ºF ) ─ (527 ºF ) = ─ 459 ºF = b Now write the specific equation for the line. ***** Substitute the variable quantities and their units in place of y and x, and substitute the numerical constants calculated above for m and b. The specific equation for the line is ( temp. in ºF ) = ( 1.80 ºF/K )( temp in K ) ─ 459 ºF b. Test your equation. For a point in the data table that is on or very near the line, plug one coordinate (x or y) into the specific equation, and see if the equation correctly predicts the other. Using the specific equation, solve for the predicted ºF at 273 K. ***** (temp. in ºF = (1.80 ºF/K)(temp in K) ─ 459 ºF (temp. in ºF = (1.80 ºF/K)(273 K) ─ 459 ºF = 491 ºF ─ 459 ºF = 32 ºF Compare the calculated to the table value. ***** The equation predicts the data. Does this answer make sense? Fill in these blanks based on our temperature scale definitions: 273 K = standard temperature for gas measurements = ________ ºC = ________ ºF Do Step 10c. Explain the equation in words. ***** The specific equation for the line supplies us with an equation that converts between temperature measurements in kelvins and in degrees Fahrenheit. The data in this problem was based on calculations rather than measurements, and the results are more certain than would be the case for data obtained by experiments. Calculations based on graphs of experimental data must allow for a higher level of uncertainty. ***** ©2010 ChemReview.net v. 1m Page 535 Module 20 — Graphing Practice 1. For the graph at the right, a. at 0.20 mol/L, s ≈ ______ Time vs. Concentration 60 At 55 s , mol/L ≈ _______ Seconds b. Using those two points, calculate the slope. c. Read the y-intercept. d. Calculate the y-intercept based on the data at 0.50 mol/L. 40 20 0 0 e. Write the specific equation for the line. f. 0.2 0.4 0.6 Mol/L At 80. seconds, the predicted mol/L = ? 2. Using the graph of the Table 5 data at the right, Sealed, Fixed Volume: P vs. temp. 300 a. estimate the y-intercept. c. Write the specific equation for the line. d. On this line, calculate the ºC at 0 kPa. 280 Pressure in kPa b. If the slope is 0.74 kPa/ºC, calculate the y-intercept. 260 240 220 200 0 10 20 30 40 50 60 70 80 90 100 110 temperature in degrees C e. For an ideal gas, what should be the answer to part d? f. Is the graphed relationship a direct proportion? Why or why not? g. What change could be made that would result in the graph plotting as a direct proportion? ©2010 ChemReview.net v. 1m Page 536 0.8 Module 20 — Graphing ANSWERS 1a. At 0.20 mol/L , seconds = 30. s b. At 55 s , mol/L = close to 0.58 M 1b. WANT = m = slope = rise = Δy = y2 ─ y1 run Δx x2 ─ x1 DATA: Set x1 as the lower x value in the two points. m=? SOLVE: x1 = 0.20 M, y1 = 30. s , x2 = 0.58 M , y2 = 55 s m = y2 ─ y1 = 55 s ─ 30 s = x2 ─ x1 0.58 M ─ 0.20 M 25 s = 66 s 0.38 M M Your answer should be close: 63-70 s/M 1c. Close to 16.5 s. 1d. WANT: DATA: b The only equation we know that uses b is y = mx + b m = 66 s/M , x = 0.50 mol/L , y = ~ 50. sec (reading graph) SOLVE: y = mx + b ; 50. s = (66 s/M) (0.50 M) + b ; 50. s = 33 s + b ; b = 17 s (your answer should be close: 15-19 seconds) 1e. Specific equation: (Time in seconds) = (66 s/M)(concentration in mol/L) + 17 seconds 1f. WANT: mol/L DATA: 80. s Equation: (Time in seconds) = (66 s/M)(concentration in mol/L) + 17 seconds or (s) = (66 s/M)(? mol/L) + 17 s (80 s) = (66 s/M)(? mol/L) + 17 seconds 63 sec = (66 s/M)(? mol/L) (63 sec)/(66 s/M) = (? mol/L) ? Mol/L = 0.95 M Your answer should be close. However, this answer is based on extrapolation beyond the data, and extrapolated predictions may not be reliable. 2a. Should be close to the part b answer below. 2b. WANT: DATA: b The only equation we know that uses b is y = mx + b Make a data table using those symbols. m = 0.74 kPa/ºC Knowing m, b can be calculated from the coordinates of any point on the line. ©2010 ChemReview.net v. 1m Page 537 Module 20 — Graphing An easy point to use is the highest point, since it is on the line. Using the data in Table 5, x = 100. ºC , y = 276 kPa SOLVE: y = mx + b ; 276 kPa = (0.74 kPa/ºC) (100. ºC) + b ; 276 kPa = 74 kPa + b ; 2c. b = 202 kPa (your answer should be close) Pressure in kPa = (0.74 kPa/ºC)(temperature in ºC) + 202 kPa 2d. Substitute 0 kPa into the specific equation: 0 kPa = (0.74 kPa/ºC) (? ºC) + 202 kPa ; ─202 kPa/0.74 kPa/ºC = ? ºC = ─270 ºC A closer answer would be absolute zero (─273 K) , but graphed experimental data has uncertainty. 2e. For an ideal gas, the pressure, measured in any units, should be zero at absolute zero. 2f. The graphed relationship is not a direct proportion. A direct proportion must graph through the origin: b must equal zero. 2g. If all of the temperatures were converted to an absolute temperature scale, such as the Kelvin scale, a graph of pressure versus temperature values should go through the origin. For ideal gases, P is directly proportional to T (absolute temperature). ***** ©2010 ChemReview.net v. 1m Page 538 Module 20 — Graphing Lesson 20F: Graphs of Inverse Proportions Pretest: If you can do the last problem at the end of this lesson, you may skip the lesson. ***** Graph of Gas Pressure Vs. Volume In an experiment, a large syringe with a scale marked in mL is filled with air. The tip is sealed, and the syringe is then clamped so that it is vertical, with its moveable plunger pointed upward. A series of identical chemistry textbooks is placed on top of the plunger one at a time. As the air in the syringe is compressed, the volume of air inside the syringe decreases. The books are then removed one at a time. In Table 8, the data for books and the average volume of air is recorded. Graph the data in Table 8 on the grid below or on your own paper. Though the more controlled variable is the number of books exerting the pressure, for this graph plot books on the y-scale. Include zero on both scales. Table 8 Books on Syringe With Trapped Air Pressure In Books Volume In mL 1 80.0 2 62.5 3 50.6 4 43.0 5 37.3 6 32.9 If you get stuck, take a look at the answers below. ***** ©2010 ChemReview.net v. 1m Page 539 Module 20 — Graphing Step 5. x-scale minor unit = 80 ─ 0 18 lines ─ 2 y-scale minor unit = 6─0 14 lines ─ 2 = 80 = 5 16 = 6 12 Leave at 5. = 0.50 Leave at 0.5. ***** Step 6. x-scale: Low #: 32.9 0 y-scale: Low #: 1 High #: 80.0 High #: 6 0 Minor Unit: 5 Major: 10 Minor Unit: 0.5 Major: 1 ***** ***** P and V are varied; n and T are held constant. The relationship between these two variables should be Boyle’s law: Books on Syringe with Trapped Air 7 Pressure (books) In this experiment, which ideal gas variables are varied? Which are held constant? What should the relationship be? Which historic gas law is this? PV = c at constant n and T 6 5 4 3 2 1 0 0 10 What kind of proportion exists between these two variables, and how should it plot? 20 30 40 50 60 70 80 90 Volume (mL) ***** P and V should be inversely proportional, and inverse proportions plot as part of a hyperbola. The shape of the graph appears to be a part of a hyperbola. When y and x are inversely proportional, y and 1/x are directly proportional, and a plot of y and 1/x should result in a straight line through the origin. The equation for the line is y = (constant slope)(1/x) Let’s test this hypothesis. In the graph above, mL was plotted on the x-scale. For the data in Table 8B, calculate 1/x, in this case 1/mL, and enter the result in the last column. Check your answers on the next page. Table 8B Books on Syringe With Trapped Air Pressure In Books Volume In mL 1 80.0 2 62.5 3 50.6 4 43.0 5 37.3 6 32.9 1/mL ***** ©2010 ChemReview.net v. 1m Page 540 Module 20 — Graphing When 1/mL is calculated, the unit of the answer is written as mL─1. Now graph pressure on the y-scale versus 1/volume on the x-scale. Include zero on both scales. Table 8B Books on Syringe With Trapped Air ***** ©2010 ChemReview.net v. 1m 1/Volume In mL─1 80.0 1 0.0125 62.5 2 0.0158 50.6 3 0.0200 4 0.0233 5 0.0268 32.9 Check your answers below. Pressure In Books 37.3 Use the grid below or your own graph paper. Volume In mL 43.0 Set the initial range on the y-scale from ─6 to +6 books. 6 0.0304 Page 541 Module 20 — Graphing Step 5. x-scale minor unit = 0.0304 ─ 0 18 lines ─ 2 y-scale minor unit = 6 ─ (─6) 14 lines ─ 2 = 0.0304 16 = 12 12 = 0.0019 =1 Round up to 0.002 Leave at 1 ***** Step 6. x-scale: Low #: 0.0089 0 High #: 0.0304 0.032 Minor Unit: 0.002 Major: 0.004 y-scale: Low #: ─6 High #: 6 Minor Unit: 1 Major: 2 ***** The graph is shown on the next page. The graph is not as predicted. By Boyle’s Law, we know that P and V are inversely proportional. The plot of P versus 1/V should therefore plot as a straight line through the origin. In this case, the graph is a straight line, but not through the origin. Another concern: because by Boyle’s law (PV = c) , P times V should equal a constant value. Multiply P times V in the chart at the right for several cases. Is the result close to constant? Table 8C Books on Syringe With Trapped Air Volume In mL Pressure In Books 80.0 1 62.5 2 50.6 3 43.0 4 37.3 5 In science, relationships tend to be simple, such as a direct proportion that graphs as a straight line through the origin. When a straight line does not go through the origin, it often means that we have defined zero poorly. P times V 32.9 6 Consider this data and graph. When the syringe had one book piled on top, was there only one book of pressure on the gas in the syringe? What are we forgetting? Write an answer to that question, then check below. ***** For the gas in the syringe, the pressure exerted by the gas inside equals the pressure exerted on the plunger outside. What we tend to take for granted is atmospheric pressure. We live on the bottom of an ocean of air. That air exerts a considerable pressure on the surfaces around us that are in contact with the atmosphere. How much pressure is atmospheric pressure? Let’s assume that the line on the graph passes through true zero pressure. That way, pressure and 1/volume graph as a direct proportion, as they should. Let’s keep the same size unit of pressure on the y-axis: books. Estimate the y-intercept on the graph: ____________ books. Now take that y-intercept number on the axis and make it true zero. Keeping the same size pressure unit, what number for P does the old zero become? ______ ***** ©2010 ChemReview.net v. 1m Page 542 Module 20 — Graphing Books on Syringe, P vs. 1/V 8 Pressure (books) 6 4 2 0 0.000 0.004 0.008 0.012 0.016 0.020 0.024 0.028 0.032 0.036 -2 -4 -6 1/Volume (1/mL) Answers The y-intercept should be close to ─ 2.5 books. Numbering up from the new zero, using the same size P unit, the old zero becomes +2.5 books. The difference between the old zero books pressure and the true zero pressure is the atmospheric pressure: 2.5 books. Table 8D At one book on the original scale, what is the value for the true pressure? _______ books ***** The original 1 book on the yaxis becomes 1 + 2.5: a true pressure of 3.5 books. Books on Syringe With Trapped Air Volume In mL Pressure In Books 80.0 1 62.5 2 50.6 3 43.0 4 37.3 5 32.9 True Pressure True P times V 6 To find the true pressure for each number of books piled on the syringe, 2.5 books (atmospheric pressure) is added to each original reading for the pressure. Do that addition in the chart on the above, then complete the additional columns. Is true pressure times volume constant? ***** ©2010 ChemReview.net v. 1m Page 543 Module 20 — Graphing SF: Assuming that the y-intercept is only accurate to tenths, the true pressure is accurate to tenths, and the PV products will have two sf. Table 8D Conclusion: The experimental data indicate that true pressure times volume is constant for gases, if moles and temperature are held constant. Volume In mL Pressure In Books True Pressure True P times V 80.0 1 3.5 280 62.5 2 4.5 280 50.6 3 5.5 280 43.0 4 6.5 280 37.3 5 7.5 280 32.9 6 8.5 280 Summary When searching for relationships among two variables, one strategy is to graph the data, and from the shape of the graph to develop a hypothesis about the type of type of equation that the graph represents. Books on Syringe With Trapped Air In the example above, the initial graph of the data seemed to be in the shape of a portion of a hyperbola, indicating an inverse proportion between the two variables. The data is then treated by functions so that, if the hypothesis is correct, the data is in the form y = mx + b and graphs as a straight line. If the data is linear, an equation for a line that predicts the data and relates the variables can be calculated. If the straight line does not pass through the origin, it is often an indication that a better definition for zero exists for one or both of the variables than the ones chosen in the experiment. In measuring quantities, the most useful scales are those that define zero in a way that results in direct proportions between one quantity and other quantities taken to a power, such as P = cV─1 , which is one way to write Boyle’s law. Let’s add to our list of science graphing steps. 10d. If a straight line does not go through the origin, try to explain why it does not, and/or consider a different definition for zero. 11. If the graph is a smooth curve but not a straight line, write the equation that you think fits the general equation for the curve. Then, adjust the equation to fit the format y = (constant) x , and plot the adjusted variables. If the graph is linear, do step 10. If not, try a different equation to explain the curve. Practice 1. For the linear graph in this lesson , using the original y-scale pressure values, a. estimate these values by reading the graph. At x = 0.0320 mL─1 , y = ____________ At y = ─ 1 book, x = ____________ b. Using those two points, calculate the slope of the line. c. Write the specific equation for the line. ©2010 ChemReview.net v. 1m Page 544 Module 20 — Graphing d. Change the format of the specific equation so that it matches the format for a direct proportion. e. Compare the part b answer to the last column in Table 8D. Why are they related? 2. In an experiment, liquids A and B are mixed together and stirred. The reaction mixture changes color after time. To study the reaction rate, the time required after mixing for different mixtures to change color is recorded. [A] is held constant, and [B] is varied. Based on the resulting data in Table 9, a. Calculate 1/seconds and fill in the chart. b. Graph [B] on the y-axis versus 1/seconds. Include the origin on the graph. c. Calculate the slope of the line, with units. d. Estimate the y-intercept by reading the graph. e. Write the specific equation for the line. f. Explain the results in words. g. What is the inherent relationship between the rate of a process and time it takes? Table 9 [B] versus Time of Reaction [B] in M seconds 0.100 25 0.080 30. 0.050 48 0.040 57 0.020 1/seconds 120 h. Write a relationship between [B] and rate. 0.14 0.12 0.1 0.08 0.06 0.04 0.02 0 0.0000 ©2010 ChemReview.net v. 1m 0.0200 0.0400 0.0600 Page 545 Module 20 — Graphing ANSWERS Your answers are correct if they are close to these. When interpreting a graph, allowances must be made for estimation and uncertainty. 1a. At x = 0.0320 mL─1 , y should be close to 6.5 books. At y = ─ 1 book , x should be close to 0.0054 mL─1. 1b. WANT = m = slope = rise = Δy = y2 ─ y1 run Δx x2 ─ x1 DATA: m = ? x1 = 0.0054 mL─1, y1 = ─ 1 book , x2 = 0.032 mL─1 , y2 = 6.5 books SOLVE: 1c. 7.5 books = 280 book • mL = m = y2 ─ y1 = 6.5 books ─ (─ 1 book) ─1 ─ 0.0054 mL─1 x2 ─ x1 0.0266 mL─1 0.0320 mL (Original P reading in books = (280 book • mL) (1/Volume in mL─1) ─ 2.5 books 1d. (Original P reading in books + 2.5 books) = (280 book • mL) (1/Volume in mL─1) or (True pressure) = (280 book • mL) (1/Volume in mL─1) This equation matches the format y = mx , a formula for a direct proportion between y and x . This equation will therefore graph as a straight line through the origin. By adding the value + 2.5 books to the original pressure, this new form of the equation reflects the adjustment to the original pressure needed to calculate true pressure, so that P = m(1/V) . 1e. Since the line equation is True P = m(1/V) , then (True P)(V) = m . 2a. See table at right. 2b. See next page. Table 9 2c. See below, under answer 2h. 2d. b = 0 [B] versus Time of Reaction 2e. [B] = (2.4 M • s) (1/s) [B] in M seconds 1/seconds 0.100 25 0.040 Based on the graph, [B] = (constant slope)(1/s) 0.080 30. 0.033 This means that [B] and (1/time) are directly proportional. The above equation can be rewritten as 0.050 48 0.021 0.040 57 0.018 0.020 120 0.0083 2f. As the [B] goes down, the time required for the reaction goes up proportionally. [B] • (s) = constant = graph slope = 2.4 M • s This format matches the xy = constant form of an inverse proportion. Therefore, [B] and the time of the reaction are inversely proportional. 2g. As the rate of a process increases, the time the process takes decreases. You can write: Rate = c (1/time) ©2010 ChemReview.net v. 1m and Rate in seconds = c (1/time in seconds) Page 546 Module 20 — Graphing 2h. Since [B] = (constant slope)(1/s) above, [B] = (a different constant) (Rate) , or in words: the reaction rate is proportional to [B]. 2c. WANT = m = slope = rise = Δy = y2 ─ y1 run Δx x2 ─ x1 Read any two widely separated points on the line. If we estimate DATA: m = ? x1 = 0.0 s─1, y1 = 0 M , x2 = 0.050 s─1 , y2 = 0.120 M SOLVE: m = y2 ─ y 1 = x2 ─ x1 0.120 M ─ 0 M 0.050 s─1 ─ 0 s─1 = 0.120 M = 2.4 M • s = m 0.050 s─1 Timed Reaction: [B] vs. 1/s 0.14 0.12 [B] 0.1 0.08 0.06 0.04 0.02 0 0.000 0.020 0.040 0.060 1/seconds ©2010 ChemReview.net v. 1m Page 547 Module 20 — Graphing SUMMARY: Graphing In Two Dimensions 1. Decide which variable to plot on x. Often, this is the independent variable. 2. Write the range chart for each scale. x-scale: Low #: _________ High #: __________ Minor Unit: ___ Major: ___ y-scale: Low #: _________ High #: __________ Minor Unit: ___ Major: ___ 3. Consider adding zero to each range, increasing the range. In most cases, if a range does not include zero, change the number to zero that increases the range. 4. Mark the boundaries of the plot on the graph paper. 5. Calculate the minor unit for each scale. Round UP. Use this equation: Scale minor unit = (High # on scale) minus (Low # on scale) (The count of the grid lines on the scale) ─ 2 and then round UP to the next easy number to count by and count between. 6. Make both ranges slightly wider and evenly divisible by the major unit for that scale. To write a number at every second line on a scale, set the major unit as double the minor unit. 7. Number the scales based on the minor unit. Label the scales. Title the graph. 8. Plot the points. 9. Draw the function: either a smooth curve or straight line near most points. 10. If the graph is a straight line, a. write the specific equation for the line. • Calculate m using two widely spaced points that are on the line. • Find b using y = mx+b , x and y for any one point on the line, and m. • Substitute values for m and b, and quantities and units for y and x, into y = mx+b. b. Test the specific equation. For a point on the graph close to the line, pick one coordinate from the data table. Use the specific equation to predict the value for the other coordinate. Compare the predicted and the data table values. c. Explain the equation in words. If the graphed line goes through the origin, write the statements and equations for direct proportions. d. If a straight line does not go through the origin, try to explain why it does not do so, and/or consider a different definition for zero. 11. If the graph is a smooth curve but not a straight line, write the equation that you think fits the general equation for the curve. Then, adjust the equation to fit the format y = (constant) x , and plot the adjusted variables. If the graph is linear, do step 10. If not, try a different equation to explain the curve. ©2010 ChemReview.net v. 1m Page 548 ...
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This note was uploaded on 11/16/2011 for the course CHM 1025 taught by Professor J during the Summer '09 term at Santa Fe College.

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