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Reply
| | From:  Abalo3 (Original Message) | Sent: 1/10/2008 3:29 PM |
1.
Air Volume(ml) Pressure(atm) 1/Pressure
0 1.000 1
20.00 1.154 0.866
40.00 1.364 0.733
60.00 1.667 0.600
80.00 2.143 0.467
100.00 3.000 0.333
2. Construct a graph by hand or using a spreadsheet program, plotting the values for (1 / pressure) on
the x-axis and air volume on the y-axis.
3. Draw a straight line through the points.
5. Determine the slope of this line and record its value here. Be sure to include units for this slope value. If you are using a spreadsheet program, use its built-in function to find the slope of the straight line. In Excel, use the function SLOPE (y-values, x-values).
6. Is the volume of air proportional to its "reciprocal pressure", i.e., inversely proportional to its pressure?
Answers to the questions
After constructing the graph, the slope will be:
m= y2 - y1/x2-x1
= (100-0) / (0-1)
= 100 ml/-1atm
= -100 ml/atm
6. We see that as the pressure decreases the volume increases. From the graph, we can tell that the pressure is not proportional to the volume.
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| | | From: ·Steve· | Sent: 1/10/2008 8:23 PM |
This time, you are looking at Boyle's law. You are right, V is not proportional to P (this graph is not even a straight line), but V is proportional to 1/P. A plot of V (y-axis) vs. 1/P (x-axis) will be a straight line with a positive slope. Did you use the reciprocals of the pressure values, 1/P? You do not want to use P = 0 because 1/P would be undefined. Since the instructions want you to give the units of the slope, what are the units of 1/P? Then, what are the units of the slope? It is also best to get the slope from the complete graph, not just from two points. But if those two points lie exactly on the best fit straight line through all the points, that would be OK. |
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Reply
| | | From: Abalo3 | Sent: 1/11/2008 2:45 PM |
| Yes, I use the reciprocal of the pressure which is 1/P. The unit of 1/P is atm. The graph is a straight line with middle above and below the line which is fine. V is proportional to 1/P as you mentioned it in your message. |
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Reply
| | | From: ·Steve· | Sent: 1/11/2008 7:38 PM |
OK, but: m= y2 - y1/x2-x1
= (100-0) / (0-1)
You have a zero value for x2. But none of the pressures give 1/P = 0. The unit of 1/P is atm. Nope! The unit of P is atm, but this is "one over P," 1/P.  |
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| | | From: ·Steve· | Sent: 1/12/2008 1:33 AM |
Sorry, "nope" again. Just stick with atmosphere units. If the unit of P is atm, the unit 1/P is _____. This all leads up to answering the question, "What are the units of the slope of the line that results from the graph of V vs. 1/P?" Seems it is a good question! |
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| | | From: ·Steve· | Sent: 1/12/2008 4:05 AM |
YES!! You can call it a "one over atmosphere unit" or "reciprocal atmosphere unit," which can also be written atm�?. Now, what is the unit of the slope of the line (V vs. 1/P graph)? |
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| | | From: ·Steve· | Sent: 1/12/2008 4:12 AM |
>> The unit will be V.atm << Oops, you beat me to the draw. That is right, the units of the slope of the line are going to be volume x pressure, but you need to give the volume unit, such as mL. Slope units = mL = mL·atm 1/atm |
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