# Chemistry question **URL:** \ **Category:** Factual Questions\ **Created:** [November 3, 2000, 2:21am UTC](https://boards.straightdope.com/t/chemistry-question/39502 "2000-11-03T02:21:20Z")\ **Posts on this page:** 18\ **Page:** 1
**Author:** ![Cardinal](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/cardinal/32/4000_2.png) [@Cardinal](https://boards.straightdope.com/u/Cardinal)\ **Post date:** [November 3, 2000, 2:21am UTC](https://boards.straightdope.com/t/chemistry-question/39502/1 "2000-11-03T02:21:20Z")
I’m tutoring a kid for the SAT II Chemistry, and I don’t get the wording of this question: Which of the previous graphs could be the plot of the average molecular kinetic energy vs the absolute temperature for one mole of an ideal gas? Now, I don’t think 1 mole part has anything to do with it, and I know the av. mol. kin. en. is really temperature, but what is absolute temperature? I don’t recognize this phrase anymore. Any help? I’d appreciate it. He’s coming back tomorrow. Dave ---
**Author:** ![Mr.Child](https://avatars.discourse-cdn.com/v4/letter/m/b19c9b/32.png) [@Mr.Child](https://boards.straightdope.com/u/Mr.Child)\ **Post date:** [November 3, 2000, 2:45am UTC](https://boards.straightdope.com/t/chemistry-question/39502/2 "2000-11-03T02:45:26Z")
I’d say that absoulte temperature is temperature measured in Kelvin (Kelvin scale is defined by absoulute zero and the tripple point of water). Kinetic energy and temperature are the same. Wow, my first post in a really long time, hope I got the answer right! ---
**Author:** ![ellis555](https://avatars.discourse-cdn.com/v4/letter/e/e36b37/32.png) [@ellis555](https://boards.straightdope.com/u/ellis555)\ **Post date:** [November 3, 2000, 2:51am UTC](https://boards.straightdope.com/t/chemistry-question/39502/3 "2000-11-03T02:51:47Z")
As I read the question, it’s asking for which plot has axes KE and T (in K). So, it should have a curve that starts at the origin (KE = 0 at absolute 0), and slopes upward. Unfortunately, you’re not given the P, but it’s safe to assume that it’s constant and probably 1 atm. ``` delta E = (n) * (Cp) * (delta T) ``` n and Cp are constant, so change in KE is directly proportional to change in T. This doesn’t take phase changes into account, but it’ll hopefully help you out. I’ve got to take off now, but I’ll check this thread later tonight if you have any more questions and no one else has come to your aide. -ellis ---
**Author:** ![vandal](https://avatars.discourse-cdn.com/v4/letter/v/858c86/32.png) [@vandal](https://boards.straightdope.com/u/vandal)\ **Post date:** [November 3, 2000, 2:52am UTC](https://boards.straightdope.com/t/chemistry-question/39502/4 "2000-11-03T02:52:03Z")
Is the graph referring to the Law of Charles? Absolute temperature is measured on a scale where absolute zero is always just zero, thus meaning, it’s the absolute temperature. It has something to do with -273C _already_ being the 0 point in the graph, so that’s where the axis’ would cross. Or, something… ---
**Author:** ![bibliophage](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/bibliophage/32/7615_2.png) [@bibliophage](https://boards.straightdope.com/u/bibliophage)\ **Post date:** [November 3, 2000, 2:56am UTC](https://boards.straightdope.com/t/chemistry-question/39502/5 "2000-11-03T02:56:52Z")
It’s been a long time since I studied chemistry, so if I’m wrong, I hope someone will correct me. Energy and Temperature are not _exactly_ the same. There is a specific relation between them. ```auto K[sub] tot[/sub] = (n/2)* RT ``` where R is the universal gas constant (so called), n is the number of moles (one in this example), T is absolute temperature, and n is the number of degrees of freedom of the molecules (3 for monatomic molecules, 5 for diatomic, and 6 for polyatomic). The graph should look like a straight line that passes through the origin. The slope of the line will depend on the values of n and R. ---
**Author:** ![bibliophage](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/bibliophage/32/7615_2.png) [@bibliophage](https://boards.straightdope.com/u/bibliophage)\ **Post date:** [November 3, 2000, 3:00am UTC](https://boards.straightdope.com/t/chemistry-question/39502/6 "2000-11-03T03:00:20Z")
I knew someone would correct me, but I didn’t know it would be me. ```auto K[sub] tot[/sub] = (f/2)*nRT ``` Where K[sub]tot[/sub] is total kinetic energy, f is the number of degrees of freedom, n is the number of moles, R is the universal gas constant, and T is absolute temperature. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [November 3, 2000, 3:06am UTC](https://boards.straightdope.com/t/chemistry-question/39502/7 "2000-11-03T03:06:43Z")
Temperature can, however, be considered as the average kinetic energy per particle. There’s a factor of Boltzman’s constant in there, but most physicists consider k[sub]B[/sub] to be a conversion factor equal to one, like c and hbar. ---
**Author:** ![bibliophage](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/bibliophage/32/7615_2.png) [@bibliophage](https://boards.straightdope.com/u/bibliophage)\ **Post date:** [November 3, 2000, 3:16am UTC](https://boards.straightdope.com/t/chemistry-question/39502/8 "2000-11-03T03:16:16Z")
I’m back to correct myself **again**. Maybe I should stay away from chemistry questions. (f/2)\*nRT is the total internal energy, not all of which is kinetic energy, except for monatomic molecules. There’s some potential energy involved in internal vibrations for diatomic and polyatomic molecules. In any case (3/2)RT is (an approximation of) the total kinetic energy of one mole of monatomic molecules. ---
**Author:** ![Caldazar](https://avatars.discourse-cdn.com/v4/letter/c/8e8cbc/32.png) [@Caldazar](https://boards.straightdope.com/u/Caldazar)\ **Post date:** [November 3, 2000, 3:58am UTC](https://boards.straightdope.com/t/chemistry-question/39502/9 "2000-11-03T03:58:29Z")
Bibliophage has it correct. For an **ideal** gas, the average kinetic energy, KE, is given by: KE = (0.5)\*nRT, per degree of freedom. n=1, R=0.08206, degrees of freedom = 3, so you’d basically this reduced to: KE = (constant)\*T, a linear graph with a positive slope where the Y-intercept=0. ---
**Author:** ![Cardinal](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/cardinal/32/4000_2.png) [@Cardinal](https://boards.straightdope.com/u/Cardinal)\ **Post date:** [November 3, 2000, 10:21pm UTC](https://boards.straightdope.com/t/chemistry-question/39502/10 "2000-11-03T22:21:51Z")
I’m sorry, but I think I’m the least learned in this discussion. I still don’t get how you can have “absolute” temperature. I understand that you can measure temperature in Kelvin, but I don’t understand what T really is in bibliophage’s formula (how you would measure it, or what the units are, and I don’t understand why the f is divided by 2. The answer to the question I mentioned originally is in fact a line starting at the origin, and running at what seems to be a 45 degree angle. This would imply that the two vary directly with each other, where the addition of 1 unit to one adds one unit to the other. But I still don’t get the phrase “absolute temperature.” Is there a more conceptual explanation I can be given? ---
**Author:** ![Dr.Lao](https://avatars.discourse-cdn.com/v4/letter/d/2bfe46/32.png) [@Dr.Lao](https://boards.straightdope.com/u/Dr.Lao)\ **Post date:** [November 3, 2000, 10:38pm UTC](https://boards.straightdope.com/t/chemistry-question/39502/11 "2000-11-03T22:38:03Z")
Absolute temperature just refers to some temperature scale that is always positive and is zero at absolute zero. The absolute temperature scale that everyone uses is the Kelvin scale. The reason an absolute temperature scale is needed is that certain equations using temperature don’t make sense when the temperature is negative. Look at the expression for KE. If Celcius or Farenheit were used then it would be possible, at some temperatures, for the kinetic energy to be negative. Since this is a scalar and not a vector of energy, it must always be positive. Any temperature scale that is not absolute (i.e. has negative values) cannot be used in this expression. ---
**Author:** ![douglips](https://avatars.discourse-cdn.com/v4/letter/d/e68b1a/32.png) [@douglips](https://boards.straightdope.com/u/douglips)\ **Post date:** [November 3, 2000, 10:44pm UTC](https://boards.straightdope.com/t/chemistry-question/39502/12 "2000-11-03T22:44:49Z")
Yes, the definition of absolute temperature is a measure of ‘the average kinetic energy of the molecules of a substance’. So, if you imagine that all the molecules come to rest, their KE is 0 so the absolute temperature is zero. The reason there is such a thing as ‘absolute temperature’ is that people wanting to quantify temperature didn’t really know what it was and used arbitrary origins for the zero point. Thus, any temperature scale which can be measured to be negative is not absolute. (please ignore quantum effects that give less-than-zero absolute temps. Thanks.) So, since essentially absolute temperature is **defined** to be proportional to the average KE of the molecules the graph **must** be a straight line that crosses the origin (assuming reasonable axes.) ---
**Author:** ![Robot\_Arm](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/robot_arm/32/18280_2.png) [@Robot\_Arm](https://boards.straightdope.com/u/Robot_Arm)\ **Post date:** [November 3, 2000, 10:52pm UTC](https://boards.straightdope.com/t/chemistry-question/39502/13 "2000-11-03T22:52:44Z")
There’s just no such thing as -1 degree Kelvin. At 0 Kelvin, all the molecules have completely stopped. Cooling something means to slow the movement of its molecules, and you just can’t get any slower than stopped. If you were to graph molecular kinetic energy vs. temperature in centigrade, the line wouldn’t pass through the origin of the graph. When something is at zero degrees centigrade, the molecules are still moving. The full graph (assuming molecular energy is on the vertical axis) would extend to the right and left. Saying “absolute temperature” implies to me that the zero point on the graph is absolute zero. The horizontal axis doesn’t go any farther to the left because those temperatures have no meaning. Just as the vertical axis doesn’t go below zero because negative energy has no meaning (or is at least beyond the scope of this question). At zero absolute temperature there is zero average molecular kinetic energy. Even without any units on the graph, the line must pass through the origin. ---
**Author:** ![Caldazar](https://avatars.discourse-cdn.com/v4/letter/c/8e8cbc/32.png) [@Caldazar](https://boards.straightdope.com/u/Caldazar)\ **Post date:** [November 3, 2000, 11:32pm UTC](https://boards.straightdope.com/t/chemistry-question/39502/14 "2000-11-03T23:32:29Z")
> [@](#): > > \*Originally posted by Dave Swaney \* > \*\*I’m sorry, but I think I’m the least learned in this discussion. I still don’t get how you can have “absolute” temperature. I understand that you can measure temperature in Kelvin, but I don’t understand what T really is in bibliophage’s formula (how you would measure it, or what the units are, and I don’t understand why the f is divided by 2. > > The answer to the question I mentioned originally is in fact a line starting at the origin, and running at what seems to be a 45 degree angle. This would imply that the two vary directly with each other, where the addition of 1 unit to one adds one unit to the other. But I still don’t get the phrase “absolute temperature.” Is there a more conceptual explanation I can be given? \*\* Absolute temperature simply means that the lowest possible temperature on your scale is zero degrees. The Kelvin scale is the absolute temperature scale typically used. This is the “T” in the equations given above. Regarding the answer that was given, were the units on the T axis and the KE axis explicity numerated? Was one tick mark on either axis stated to be 1 unit (of something)? If not (and usually this is the case), the axis numbers could be anything, and you can’t go by the visual slope of the line. You can’t tell visually that 1 degree K will correspond to a 1 Joule increase in KE from a graph if the graph axis are not labelled with units and numbers. ---
**Author:** ![brad\_d](https://avatars.discourse-cdn.com/v4/letter/b/50afbb/32.png) [@brad\_d](https://boards.straightdope.com/u/brad_d)\ **Post date:** [November 3, 2000, 11:36pm UTC](https://boards.straightdope.com/t/chemistry-question/39502/15 "2000-11-03T23:36:40Z")
This is just a minor nitpick: Kelvins aren’t the only absolute temperature measurement, only the most commonly used. The Rankine temperature scale is the English unit equivalent - 0 Rankine = absolute zero, and a 1-Rankine temperature change is equal to a 1-degree Fahrenheit change. 0 [sup]o[/sup]R = -459.67 [sup]o[/sup]F. In order for any of this to make any sense at all, the concept of _absolute zero_ has to be understood. I realize that those posting replies clearly do, so this is my attempt to answer **Dave Swaney** ’s question. _Absolute zero_ is the temperature at which molecular motion (vibration?) stops completely, like people have said. Temperature, really, is a measure of energy in that motion, so the idea of a temperature below absolute zero simply has no meaning. No more than, say, a stick with negative length. As it turns out, absolute zero is at -459.67 [sup]o[/sup]F, or -273.15 [sup]o[/sup]C. This temperature is defined as 0 K. What good is this? Well, absolute scales make it easier to compare different temperatures. What’s twice as hot (twice the kinetic energy) as, say, the freezing point of water? 2x0 [sup]o[/sup]F=0 [sup]o[/sup]F? Huh? That should sound suspicious, because it’s not right. Similarly, is -2 [sup]o[/sup]F _twice the temperature_ of -1 [sup]o[/sup]F? No, clearly not - it’s colder. When your temperature scale is absolute (zero corresponds to zero), you don’t have these problems. 0 [sup]o[/sup]F = 255.4 K, and since Kelvins are absolute we can simply double that value to get twice the temperature, or twice the kinetic energy: 510.7 K = 459.7 [sup]o[/sup]F. Quite a bit different from 0 [sup]o[/sup]F, and also quite a bit more correct. Hopefully, this is enlightening and not wrong. I’m sure somebody will correct me if I’ve made any errors. ---
**Author:** ![Caldazar](https://avatars.discourse-cdn.com/v4/letter/c/8e8cbc/32.png) [@Caldazar](https://boards.straightdope.com/u/Caldazar)\ **Post date:** [November 4, 2000, 12:02am UTC](https://boards.straightdope.com/t/chemistry-question/39502/16 "2000-11-04T00:02:00Z")
Whoops, didn’t finish up. The conceptual explanations given are correct; temperature is a measure of average kinetic energy of the gas molecules, and hence, is a measure of motion. You can’t have a lower speed than zero. This corresponds to an absolute temperature of zero. As for why in the case of an ideal gas moving in free space (three degrees of freedom) there is a 1/2 term, the 1/2 (3/2’s in this case) is a direct result of the mathematical derrivation of the formula. I won’t post the whole derrivation here, but the short answer is that the 1/2 comes from the formula: KE = (1/2)_m_(v^2). You can usually find the derrivation in any college-level text on the Kinetic Theory Of Gases (look in the sections dealing with Average Kinetic Energy and The Root Mean Squared Speed of gas molecules). If you REALLY want the whole mathematical derrivation, I managed to find this link on the web: [http://www.puc.edu/Faculty/Richard\_Webb/1999/Summer/PHYS112/lectures/lec03.html](http://www.puc.edu/Faculty/Richard_Webb/1999/Summer/PHYS112/lectures/lec03.html) ---
**Author:** ![Cardinal](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/cardinal/32/4000_2.png) [@Cardinal](https://boards.straightdope.com/u/Cardinal)\ **Post date:** [November 5, 2000, 12:34pm UTC](https://boards.straightdope.com/t/chemistry-question/39502/17 "2000-11-05T12:34:35Z")
Thanks, guys. It turns out I did in fact realize what “absolute temperature” was, but it was an example of a question so simple that I couldn’t see what it meant. OF COURSE the Kelvin scale is directly proportional to other scales of measuring temperature (no, there weren’t any units on the axes, but it’s the only straight line with a positive slope to choose from). It didn’t occur to me that they might be asking something so uncomplicated. And all this after I spent almost an entire year two school years helping drill into high schoolers’ heads that using Celcius in many equations can screw up your proportions and give you some _very_ funky numbers when the scale goes negative. ---
**Author:** ![Cardinal](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/cardinal/32/4000_2.png) [@Cardinal](https://boards.straightdope.com/u/Cardinal)\ **Post date:** [November 5, 2000, 12:35pm UTC](https://boards.straightdope.com/t/chemistry-question/39502/18 "2000-11-05T12:35:36Z")
Just in the interest of not looking totally dorky, that’s “two school years _ago_”