Wikipedia:Reference desk/Archives/Science/2011 October 22

= October 22 =

a fixed volume of gas is heated from 273K to 546K ... does free energy and entropy increase or decrease?
Pressure and temperature both increase in this case (they are doubled). I don't get the last two equations in Entropy ... how would you use the heat capacity at constant volume if volume is changing? Or are those supposed to be the result of two reversible changes ? elle vécut heureuse à jamais  (be free) 01:31, 22 October 2011 (UTC)
 * In this case, the pressure is changing due to the change in temperature, so the volume remains constant. 67.169.177.176 (talk) 02:53, 22 October 2011 (UTC)
 * But that doesn't make sense. Nature can't distinguish between causes in change in pressure....and furthermore, entropy is a state function, not a path function. elle vécut heureuse  à jamais  (be free) 02:59, 22 October 2011 (UTC)
 * I don't understand what's your difficulty here. You can always use either formula for any path because as you said, entropy is a state function. So what's the problem again? Dauto (talk) 03:02, 22 October 2011 (UTC)
 * As far as the volume is concerned, remember the gas equation, PV = nRT? In this case, P(final) = 2P(initial), and T(final) = 2T(initial); this means that the 2's cancel, and you're left with the same volume as before. 67.169.177.176 (talk) 03:07, 22 October 2011 (UTC)


 * I remember that the Gibbs free energy increases with pressure, but decreases with temeperature, all other things being constant, -- but then again, it was not a constant volume relation. elle vécut heureuse  à jamais  (be free) 03:41, 22 October 2011 (UTC)
 * Increases. Dualus (talk) 08:07, 22 October 2011 (UTC)



\begin{align} dE &= T dS - P dV\longrightarrow\\ dG &= - S dT + V dP = - S dT + V\left(\frac{\partial P}{\partial T}\right)_{V}dT + V\left(\frac{\partial P}{\partial V}\right)_{T}dV \end{align} $$

So, the partial derivative of G w.r.t. T at constant V is:
 * $$\left(\frac{\partial G}{\partial T}\right)_{V} = V\left(\frac{\partial P}{\partial T}\right)_{V} - S$$

For an ideal gas this is N k - S which is smaller than zero. You can claculate G exactly using statistical mechanics, this yields:


 * $$G = N k T\left[1 -\log\left(\frac{V}{\Lambda^{3}}\right)\right]$$

where $$ \Lambda = \frac{h}{\sqrt{2\pi m k T}} $$ is the so-called thermal de-Broglie wavelength Count Iblis (talk) 23:20, 22 October 2011 (UTC)

Where can I get some brown or grey
lightbulbs for Halloween? μηδείς (talk) 02:53, 22 October 2011 (UTC)


 * I don't think you can. 'Brown' and 'grey' aren't so much colours, as less-intense shades of other colours - basically 'red'(mostly) and 'white'. If you want 'grey' light, you just need less light. AndyTheGrump (talk) 02:58, 22 October 2011 (UTC)


 * I don't think Andy is right about the lights. However, I am no physicist. You can get purple and amber from here and make your own brown by overlapping. Bielle (talk) 03:01, 22 October 2011 (UTC)


 * I think you're both right. In this screenshot of a color picker, brown is a range of colors located somewhere between bright orange (at the top of the rainbow panel) and grey (at the bottom), with the lightness slider moved towards the middle. That is to say, it has low saturation and low lightness. Low saturation means that the color tends towards white, grey or black, which is your mixing (Color mixing, Colorfulness). Low lightness is what Andy's saying. You can't have a light which shines on a surface and forces it to be darker, so you can't have a brown light, but you can have a pale reddish-orange light (a peachy pink), which will look brown when it isn't very bright. Card Zero  (talk) 03:34, 22 October 2011 (UTC)


 * Overlapping of red and green lights could sometimes give a dark brownish-yellow amber color. (This is a common problem with three-color VASIs as used in the former USSR -- when the green light ("on slope") and the red light ("too low") overlap, the resulting brownish-amber color can be misinterpreted as a yellow light ("too high"), causing the pilot to correct the wrong way.) 67.169.177.176 (talk) 03:52, 22 October 2011 (UTC)
 * Sure, green and red make yellow. I question the part about "dark", though, unless the green and red lights were dark to start with. Card Zero  (talk) 04:06, 22 October 2011 (UTC)
 * Quote from NOAA's publication "Instrument Approach Procedures (Charts)", legend L2 (Approach Lighting System -- United States) (given in Van Sickle's Modern Airmanship) regarding the "Tri-color Visual Approach Slope Indicator" (chart symbol VT): "CAUTION: When the aircraft descends from green to red, the pilot may see a dark amber color during the transition". I can't speak from personal experience about it -- so far I've never seen any tri-color lightboxes, only the standard red-and-white variety.  AFAIK tri-color VASIs are not used in the US to any significant extent, but I've talked to a couple of Russian pilots and they say that the tri-color system is the most common there.  FWIW 67.169.177.176 (talk) 04:28, 22 October 2011 (UTC)
 * Oh, of course -- in the three-color VASI, the red and green lights overlap with the very edges of their beams, so they would be much fainter than in the middle of the beam. That explains why the amber light is dark and said to be brownish-looking. 67.169.177.176 (talk) 04:32, 22 October 2011 (UTC)
 * You can, but for red brown it's one spectrum and for green brown it's another. Dualus (talk) 08:23, 22 October 2011 (UTC)


 * The 'brownest' brown light I've ever seen was in old dark rooms with the safelight on. It is to the the human eye, a monochromatic brown light.    Whites show up brown but other colours  are  completely muted - for obvious reasons. Other editor who  did B&W printing should be able to put in their $0.02 on this.--Aspro (talk) 21:21, 23 October 2011 (UTC)

Mercury is more dense than lead?
Is this correct? So if I have equal volumes of lead and mercury, the mercury is heavier, correct? ScienceApe (talk) 03:28, 22 October 2011 (UTC)


 * It seems to be - see Density:
 * Lead: 11340 Kg/m3
 * Mercury: 13546 Kg/m3
 * If you are looking for something dense try Osmium: 22570 Kg/m3


 * AndyTheGrump (talk) 03:37, 22 October 2011 (UTC)


 * Or Neutronium: 8.4 × 1016 − 1 × 1018 kg/m3 :-) 67.169.177.176 (talk) 04:53, 22 October 2011 (UTC)
 * Mercury is more dense 13.54 to 11.34 grams per cubic centimeter. Dualus (talk) 08:26, 22 October 2011 (UTC)


 * Shall I say it a third time for good measure? Mercury has a density of 13.546 g cm-3, lead has a density of 11.340 g cm-3. Plasmic Physics (talk) 09:11, 22 October 2011 (UTC)
 * But that was in cgs rather than mks. Now, what is it in debens per cubic cubit? Clarityfiend (talk) 22:02, 22 October 2011 (UTC)
 * I would prefer the densities in slugs per cubic furlong. Edison (talk) 05:10, 23 October 2011 (UTC)
 * Of all of these strange units for density, I find that the cgs units are simplest to relate to real life (though ounces per cubic inch would also make good sense for me). Perhaps this is just because of the easy comparison with water (see Specific gravity).     D b f i r s   09:21, 23 October 2011 (UTC)
 * Perhaps more useful as a side note is that gold is considerably more dense that either of them. 19.30 g/cubic centimetre. Grandiose (me, talk, contribs) 09:19, 22 October 2011 (UTC)


 * You can try floating lead in mercury, but do not try sinking gold rings in mercury as it will form an amalgam on the surface damaging your gold object's appearance. Graeme Bartlett (talk) 02:37, 23 October 2011 (UTC)

Archery
Why would a right-handed person use the bow with the left-hand? Is the dominant hand really suppose to draw the string while the recessive hand holds and aims the bow? From what I've seen, you use the bow with your recessive hand while you use the sword with your dominant hand. But the way I see it, you're comfortable with using your strong hand to do both tasks of aiming the bow and wielding the sword. --Arima (talk) 08:58, 22 October 2011 (UTC)
 * (Assuming a right hander for this example) The bow is held with the left hand for a couple reasons which are fairly closely linked. The left arm only basically needs to remain locked and steady holding the bow out at arms length.  The right hand meanwhile will be doing most of the aiming.  It also needs more strength than the left because it's pulling the string back while also holding it steady.  Dismas |(talk) 09:04, 22 October 2011 (UTC)
 * In target archery, you also need to be able to aim with your dominant eye. So if you are right-eye dominant, you need to draw the arrow to the right side of your face with your right hand. Left-eye dominant archers need to draw with the left hand. A test for eye dominance is here. Usually right-handed people are right-eye dominant and vice versa. however, some are right-handed and left-eyed or vice-versa; this is known as cross dominance. Target archers usually go for eye dominance over left or right handedness. This is less of an issue with field archery where both eyes are kept open to improve the field of vision. I would imagine that if you are shooting a traditional longbow or a compound bow with a very heavy draw weight, then you would want to go with the strongest hand as Dismas suggests. I am a GNAS youth instructor. Alansplodge (talk) 12:16, 22 October 2011 (UTC)

More redox questions
Few questions. When a redox reaction occurs they usually say that energy is given off as heat. Is this heat produced directly from the reaction or is it merely from the photons emitted by the electrons that are then absorbed by the surrounding environment?

Is it fair to say that energy is being emitted from the substance that is being oxidized (I know it's specifically coming from the electrons that move from a high energy state to a lower energy state but I want to know if it's fair to say that the energy is coming from the substance being oxidized) and that the oxidizer is one of the catalysts? ScienceApe (talk) 15:30, 22 October 2011 (UTC)


 * OK, let me tackle your questions in the order given. (1) The heat is produced directly from the electrons moving to a lower energy state (see molecular orbital theory and Gibbs free energy), and is released as photons in the infrared (and sometimes also visible or even ultraviolet) spectrum that may or may not be absorbed by the surrounding environment. (2) The heat is released during the transition from the activated complex to the final product, so it is incorrect to say that the energy is being emitted from the substance being oxidized -- a better way to think about it is that the heat comes from both reagents, or from the product as it forms. (3) The oxidizer is NOT a catalyst, because it's used up in the reaction -- therefore, it's a reagent. 67.169.177.176 (talk) 22:01, 22 October 2011 (UTC)
 * 1) In order for heat to be produced the photons have to be absorbed by the environment right? Otherwise they just carry the energy away from the immediate vicinity correct? 2) The energy comes from the electrons moving from a high energy state to a low energy state, and that's it correct? ScienceApe (talk) 22:11, 22 October 2011 (UTC)
 * (1) Not necessarily. (2) Correct, and true for all other exothermic reactions as well. 67.169.177.176 (talk) 22:15, 22 October 2011 (UTC)
 * I'm not sure I understand. If the photons are not absorbed by the environment, how can heat be produced? Are you saying that the electrons are producing heat directly in addition to photons? ScienceApe (talk) 22:43, 22 October 2011 (UTC)


 * Don't forget about kinetic energy, energy can either be given of as a photon, or as kinetic energy of the particle. Plasmic Physics (talk) 23:48, 22 October 2011 (UTC)
 * But what is kinetic energy being imparted to in a redox reaction? I thought all that happens is that the electrons move from a higher energy state to a lower energy state and give off a photon. ScienceApe (talk) 00:08, 23 October 2011 (UTC)
 * The product molecule(s), of course. 67.169.177.176 (talk) 00:11, 23 October 2011 (UTC)
 * So lets say you react Hydrogen with Oxygen. The product is a water molecule that has some kinetic energy imparted to it, as well as some ultraviolet photons? The water molecule didn't have to absorb any photons in order to have that kinetic energy imparted to it, the KE was imparted to it from the redox reaction itself? ScienceApe (talk) 00:28, 23 October 2011 (UTC)
 * Exactly. 67.169.177.176 (talk) 00:56, 23 October 2011 (UTC)


 * Also don't forget that electromagnetic radiation (made of photons) is a form of heat. Dauto (talk) 00:07, 23 October 2011 (UTC)
 * Err it is? I thought it only becomes heat when matter absorbs the photons and heat is produced. ScienceApe (talk) 00:10, 23 October 2011 (UTC)
 * Sure. Read for instance Heat transfer. Dauto (talk) 06:29, 23 October 2011 (UTC)


 * Perhaps surprisingly, there's some imprecision as to what exactly the word "heat" means. See Heat and Heat.  Red Act (talk) 23:19, 23 October 2011 (UTC)


 * It is an either or situation, one molecule either emits the energy as an individual photon, or it transforms it into kinetic energy. Both are a result of a loss of momentum of an electron, as it shifts to a lower energy state. The increase in kinetic energy is distributed over the whole molecule, this in turn leads to an increase in temperature of the reaction, hence an exothermic reaction results. Plasmic Physics (talk) 00:45, 23 October 2011 (UTC)

Measurement uncertainty
If two measurements have uncertainties of 0.01 and 0.1, what's the uncertainty of their product? --75.33.218.167 (talk) 15:41, 22 October 2011 (UTC)


 * See error propagation. Your query appears to be the third case in the table of examples.  Dragons flight (talk) 16:03, 22 October 2011 (UTC)
 * So what's the uncertainty? Those equations don't make sense. --75.33.218.167 (talk) 17:17, 22 October 2011 (UTC)


 * You haven't provided enough information in order to find the uncertainty. If you have numbers $$A \plusmn \sigma_A \text{ and } B \plusmn \sigma_B$$, then the error propagation estimate of their product is:


 * $$(A \times B) \plusmn \sqrt{ ( A \times \sigma_B )^2 + ( B \times \sigma_A )^2 }$$


 * Provided that we can assume that the errors arise from a normally distributed process and are uncorrelated with each other. Dragons flight (talk) 17:37, 22 October 2011 (UTC)
 * Assuming the OP is working at a lower level than the above, it is generally considered a good estimate if the percentage uncertainties are added. Say both those uncertainties - which look like absolute uncertainties - are on measurements of 5. 0.01/5 = 0.002 = 0.2%; 0.1/5 = 0.02 = 2%. The total uncertainity it 2.2%, which you should probably then convert back to an absolute number. In my example, it would be 25 +/- 2.2% = 25 +/- 0.11. This is only the same as the absolute uncertainty sum because they were both on measurements of 5, which they won't be in practice. Grandiose (me, talk, contribs) 17:42, 22 October 2011 (UTC)
 * To the OP: Notice what Grandiose is saying is equivalent to the product rule for differentiation, because sufficiently small uncertainties can be approximated by differentials. 24.92.85.35 (talk) 03:30, 23 October 2011 (UTC)

Plutogenic
The word plutogenic has been used in medical conversation recently. I have not found a definition for it. I believe it means the opposite of or being a better or reasonable approach to healthcare compared to treating patients in a pathogenic format.

Thanks for any help,

Willy Holden — Preceding unsigned comment added by 12.145.227.121 (talk) 16:36, 22 October 2011 (UTC)


 * Added a heading. &#x2013; b_jonas 17:04, 22 October 2011 (UTC)


 * I guess it's supposed to mean "wealth generating" (from the same Greek root as Plutocracy); but it's not in any dictionary I can see. But I don't really think that helps you, as (without a definition or context) it's not clear for whom this wealth is being generated. Someone could as easily say "This kind of healthcare system is plutogenic" (meaning that healthy people are able to work and produce wealth) or "This procedure is unnecessary and downright plutogenic" (meaning the physicians performing it are doing so only to enrich themselves). -- Finlay McWalterჷTalk 17:23, 22 October 2011 (UTC)


 * Are you sure the word wasn't "pleurogenic", which means "beginning in the pleura"? That word would sound somewhat similar to "plutogenic", and would be a word used in a medical conversation.  Red Act (talk) 21:34, 22 October 2011 (UTC)


 * This page claims that 'salutogenic' is the opposite of 'pathogenic'. Warning: link may contain quackery. --Heron (talk) 09:59, 23 October 2011 (UTC)

malleability and ductility
Todays featured pictured referes to these properties (see ductility). which article says gold is both M & D whilst lead is only M but not D. Is there any material which is D but not M? -- SGBailey (talk) 17:07, 22 October 2011 (UTC)
 * Ductility is the ability to deform under tensile stress while malleability is the ability to deform under compression stress? If that's the case, then Carbon nanotubes might fit the bill. I know they have strong tensile strength but poor compression strength, but I don't know if they can plastically deform under tensile stress. ScienceApe (talk) 20:49, 22 October 2011 (UTC)
 * The article says that they can. 67.169.177.176 (talk) 22:08, 22 October 2011 (UTC)
 * Ok so carbon nanotubes are ductile but not malleable then. :) ScienceApe (talk) 22:40, 22 October 2011 (UTC)
 * Wouldn't rubber, as in a rubber band, not be prone to compression distortion nearly as much as it is to tensile distortion, or am I just mixing in an entirely different bird?  DRosenbach  ( Talk 00:16, 24 October 2011 (UTC)

Minerals and radiactivity
Several Wikipedia articles on radioactive mineral specimens state that they should not be kept with minerals that are damaged by radioactivity, but there are no lists or charactristics of minerals that might be damaged by radioactivity of other specimens. Is there a list of minerals that are damaged by radioactivity? If such a list is very long, are there characteristics of minerals that make them likely to be damaged by housing with radioactive specimens? Or, as a last resort, is there a listing of minerals that are not damaged by radioactivity?76.26.140.45 (talk) 21:04, 22 October 2011 (UTC)
 * It somewhat depends on what type of radiation; alpha radiation, the stuff given off by U238, can't even penetrate paper, so I doubt it would damage anything. Other types, though, such as beta, would probably damage the lattice of any crystal such as quartz, zircon, flourite, etc. --  T H F S W  (T · C · E) 21:28, 22 October 2011 (UTC)
 * What is flourite? I've heard of fluorite, but not flourite.  Is it some kind of mineral that produces rock flour? 67.169.177.176 (talk) 22:05, 22 October 2011 (UTC)
 * See http://www.rockandmineral.org/light%20faded%20minerals-%20don%20kauffman%200110.pdf for some minerals damaged by ultraviolet or visible light. I would expect minerals like mercury or silver minerals would be seriously damaged by radiation. Graeme Bartlett (talk) 02:31, 23 October 2011 (UTC)
 * Diamonds are actually treated using radiation to enhance their color, as described in this article: Diamond_enhancement. Under undesired or uncontrolled conditions this would no doubt be considered "damage". Not that this answers the question, but it's a clue.Vespine (talk) 23:17, 23 October 2011 (UTC)

Faster than light travel, causality, and time dilation
FTL travel violates causality correct? But is it fair to say that causality is violated because time dilation occurs when you approach the speed of light? So lets say in an alternate universe where time dilation simply does not occur, there should be no problems with causality and FTL travel correct? ScienceApe (talk) 22:51, 22 October 2011 (UTC)


 * Yes, FTL travel does violate causality, but it is not violated because of time dilation. Plasmic Physics (talk) 23:42, 22 October 2011 (UTC)


 * Time dilation is not the cause for the acausal nature of faster than light (FTL) motion but it is related to it. In a universe where Absolute time and space exist there would be no speed limit and there would be no time dilation either. But notice that in such universe light might very well be instantaneous and FTL motion would still violate causality because in order to be faster than instantaneous you must move backwards in time. Dauto (talk) 23:57, 22 October 2011 (UTC)


 * (EC) I think you'd have to define your hypothetical alternate universe more thoroughly than just it being one in which time dilation does not occur, in order for your question to be answered meaningfully. Of all possible hypothetical universes in which time dilation does not occur, I think there would be some such universes in which FTL travel raises no causality issues, other such universes in which FTL travel would raise causality issues, and others in which the question would be unanswerable because FTL and/or causality wouldn't be well-defined.  Red Act (talk) 00:09, 23 October 2011 (UTC)

Here's the webpage I read that explained to me why ftl travel violates causality. It seemed to me that if time dilation wasn't a factor, then causality wouldn't have been violated. http://sheol.org/throopw/tachyon-pistols.html ScienceApe (talk) 00:07, 23 October 2011 (UTC)
 * If you read it carefully you will see that the important point is the relativity of simultaneity. Dauto (talk) 00:57, 23 October 2011 (UTC)
 * See also tachyon. ~ AH1 (discuss!) 15:03, 24 October 2011 (UTC)

Effect of blood sugar on weight loss
There's so much literature about diabetes it's hard to go through it all. I'm trying to understand why people who develop prediabetes or type II diabetes don't easily or automatically lose weight once the blood sugar level increases. My impression was that low blood sugar was a major reason why people eat, so I'd think it would make a huge difference. (Actually, weight loss is potentially a symptom of type II diabetes, but so is weight gain, or neither may happen...) Along these lines...:


 * Is it useful for a normal person on a reducing diet to monitor fasting blood glucose to decide how much to cut back on food from day to day (so as to prevent uncomfortably low blood sugar)?


 * Does untreated type II diabetes help at least some people to lose weight? Are some of these people lost from the statistics because they don't remain diabetics, or fail the original blood glucose criterion on the second test due to weight loss?


 * Does deliberate poor management of diabetes by an insulin-dependent person (keeping it at, say, a consistent 125 or 150 mg/dL) make it easier to go on a drastic diet to lose weight?

Wnt (talk) 23:53, 22 October 2011 (UTC)
 * A diabetic whose blood sugar is high and out of control, or an undiagnosed diabetic, may have weight loss as a symptom, along with frequent urination and deteriorating vision. These are classic signs, noted centuries ago. There are health costs, such as permanent vision impairment, poor infection fighting, or kidney failure, such that high blood sugar=weight loss is not a sensible weight loss strategy. Edison (talk) 05:06, 23 October 2011 (UTC)
 * Well, but either blood sugar has to be very high or the high levels need to go on for a long time in order for the more serious effects to manifest. If dieting is actually done while the levels are at a moderately high level, the outcome might be more control over blood sugar in the long term.  So it's not clear to me that this wouldn't be worth it - if it works.  But I'm still not all that clear about that.  Basically, I'm wondering whether type II diabetes can be thought of as some kind of last-ditch homeostatic mechanism to control body weight. Wnt (talk) 15:01, 23 October 2011 (UTC)