Proportionality and Absolute Temperature
Date: 06/06/99 at 18:38:32 From: Sean Subject: Variation and polynomial equation word problems I don't understand how to do the word problem below: Absolute temperature is measured in degrees Kelvin (K). A degree on the Kelvin scale is the same size as the Celsius degree, but 0 K is -273 C (Celsius). [This is a part that puzzles me; please explain]. The volume of a fixed amount of gas kept at constant pressure varies directly as its absolute temperature. If the gas occupies 100 liters at -13 C what is its volume at 26 C? I have tried to do -13/100 is equal to 26C/x as a proportion but I have failed to reach the answer: 115 liters. I was given a clue to convert the Kelvin to Celsius, but have failed to see how. (That is why I wrote - I don't understand the puzzling part above.) Thank you for taking the time to solve this. My thanks to you are square root pi x infinity.
Date: 06/07/99 at 09:22:59 From: Doctor Rick Subject: Re: Variation and polynomial equations word problems Hi, Sean. I'm glad you are persistent - that's the way to learn. Here is what is going on with Celsius and Kelvin. I'll draw a Celsius thermometer and a Kelvin thermometer: A B C +-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+----- -273 -223 -173 -123 -73 -23 27 77 127 177 227 A B K +-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+----- 0 50 100 150 200 250 300 350 400 450 500 You see that 0 K = -273 C (actually, I think it's -273.15 C or something like that, but we're rounding). The difference between A and B is 27 - (-23) = 50 C. On the Kelvin thermometer the difference between A and B is also 50 K (300-250). That's why the problem can state that a degree on the Kelvin scale is the same size as the Celsius degree. All you do to convert Celsius to Kelvin is to add 273 degrees; for instance: A = -23 C = -23 + 273 K = 250 K The problem stated that the volume of a fixed amount of gas kept at constant pressure varies directly as its absolute temperature. Absolute temperature is the temperature on the Kelvin scale, or another scale that has the same zero point. (The Rankine scale, for example, is to the Fahrenheit scale what Kelvin is to Celsius: its degrees are the same size as Fahrenheit degrees, but 0 R = 0 K.) So, convert the two Celsius temperatures to Kelvin before you work out the proportionality. -13 C = -13 + 273 = 260 K 26 C = 26 + 273 = 299 K x L 299 K ----- = ----- = 1.15 100 L 260 K x = 1.15 * 100 L = 115 L Think about WHY we must express the temperatures in Kelvin (or Rankine) for this proportionality to work. Let's invent a new scale for measuring people's height, the Sean scale. This scale uses feet and inches, but we call a person who is 5 feet tall "0 feet on the Sean scale." I am 6 feet tall on the "absolute" height scale, but I would be called 1 foot tall on the Sean scale (because I'm 1 foot more than 5 feet tall). A child 3 feet tall would be called -2 feet S. Sounds bizarre, doesn't it? But that's what Celsius temperature is like. Could you tell that I (1 foot S) am twice as tall as that -2 foot S child? You would have to convert both of our heights to "absolute height" by adding 5 feet: I am 1 + 5 = 6 feet (absolute) and the child is -2 + 5 = 3 feet (absolute). Then you can divide and find that the ratio of our heights is 6/3 = 2. When the common scales (Fahrenheit and Celsius) were developed, no one knew yet what the lowest possible temperature was, or even that there was a lowest temperature. The proportionality rule stated in the problem wasn't known yet. It seemed that one could choose a zero point arbitrarily, the way a zero point of longitude was chosen arbitrarily at the meridian of the Royal Observatory in Greenwich, England. Indeed, Fahrenheit and Celsius chose different zero points. But when experiments with gases produced the law that is used in this problem, the zero point was no longer arbitrary. That's why the Kelvin scale was developed. I hope this helps you understand the why and the how of absolute temperatures and proportions. If not, keep at it! - Doctor Rick, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2015 The Math Forum