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Reply
| | From:  Abalo3 (Original Message) | Sent: 1/9/2008 7:12 PM |
1. Record your data table of temperature vs. volume here.
Temperature(Deg C) Volume(ml)
100 126.87
95.86 125.16
90.88 123.77
85.21 121.70
80.91 120.25
75.31 118.35
70.06 116.58
65.18 114.92
60.67 113.40
55.78 111.83
50.98 110.20
45.87 108.40
40.97 106.74
2. Determine the slope of the line and record its value here. Be sure to include the 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 temperature?
Answer for verification purposes
After I sketched the slope I found:
slope m= y2-y1/x2-x1
=-126.87-(-106.74)/9100-40.97)
= 0.341
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| | | From: ·Steve· | Sent: 1/9/2008 8:19 PM |
2. Determine the slope of the line and record its value here. Be sure to include the units for this slope value. The slope looks correct but what are its units? 6. Is the volume of air proportional to its temperature?
You can see the general relationship that as T decreases, the volume decreases also. They are using the word "proportional" rather loosely here, but clearly there is a relationship between the temperature of the gas and its volume. If the volume is directly proportional to the temperature, then you would have a relationship V = CT, where C is a constant. Then if you double the temperature, the volume will double also. If you triple the temperature, the volume will triple. If you halve the temperature, the volume will be 1/2 its previous value as well. This is normally what we mean by "proportional." A graph of V vs. T would give a straight line with a y-intercept of zero. So, check out your graph. If you double the temperature, does the volume likewise double? If so, then the volume is directly proportional to the temperature. Or, does your data fit a slightly different relationship; does your graph fit the y = mx + b formula of a straight line? Using V for y and T for x, the relationship in this case is V = mT + b, where m is the slope and b is the y-intercept. In this case, V is not directly proportional to T. If have this kind of relationship, pick a V and T value and calculate b. Then, just for the fun of it, use the V = mT + b formula with your values of m and b to calculate T when V = 0. You will get a pleasant surprise.  Steve |
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| | | From: ·Steve· | Sent: 1/9/2008 8:27 PM |
FYI �?Here is a common definition of "proportional" in the mathematical sense: b. (of a first quantity with respect to a second quantity) a constant multiple of: The quantity y is proportional to x if y = kx, where k is the constant of proportionality. |
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Reply
| | | From: Abalo3 | Sent: 1/10/2008 2:51 AM |
slope m= y2-y1/x2-x1
=-126.87ml - (-106.74ml) / 9100 - 40.97)
= 0.341ml/Vol
A graph of V vs. T would give a straight line with a y-intercept of zero. From the graph we can see that the volume is directly proportional to the temperature. As the volume increases so does the temperature.
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| | | From: ·Steve· | Sent: 1/10/2008 4:21 AM |
slope m= y2-y1/x2-x1 Yep.  =-126.87ml - (-106.74ml) / 9100 - 40.97)
Yep. = 0.341ml/Vol <–�?/STRONG> Nope!  A graph of V vs. T would give a straight line with a y-intercept of zero. From the graph we can see that the volume is directly proportional to the temperature. Just to confirm, is the unit of temperature for this graph °C or K? Depending on which, your answer is or ! |
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| | | From: Abalo3 | Sent: 1/10/2008 4:40 AM |
I meant to say:
slope m= y2-y1/x2-x1
=-126.87ml - (-106.74ml) / (100 - 40.97)
= 0.341ml/ Deg C (Sorry about that)
The unit of temperature for this graph °C. |
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| | | From: ·Steve· | Sent: 1/10/2008 5:02 AM |
= 0.341ml/ Deg C Yes! The unit of temperature for this graph °C. Then, V is not proportional to T, because the data do not fit the formula V = mT, required for a proportional relationship. "m" is the slope of the line (I called it "C" earlier) and the y-intercept b has to be zero. From the equation of a straight line, y = mx + b, which is V = mT + b in this experiment if your volumes are on the y-axis and your temperatures are on the x-axis. If b = 0, V is proportional to T. If b �?nbsp;0, V is not proportional to T. This is according to the strict definition of "proportional." Guess what the result is when you use Kelvin units for the temperature? |
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| | | From: Abalo3 | Sent: 1/10/2008 2:51 PM |
| I guess the result is thermodynamic temperature. |
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| | | From: Abalo3 | Sent: 1/11/2008 2:39 PM |
| If the Celsius scale is used, as in my graph, this temperature is Deg C. |
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