For English speakers!

For the benefit of English speakers, I’ve hired stunt doubles of Geo and Pleven to reenact the last two posts in English according to Google translate (never mind that sometimes it says the opposite of what was really said):

Google Translate

Bulgarian to English translation

Full of smoke
Oh – the town is still full of smoke! There’s a huge fire somewhere. Outside there is fog of smoke, and covers the town like a blanket. The smell is very strong, and I think it might be a dangerous person to be outside. Here is the weirdest place I’ve lived.Smoke should not bother me, but I think only the theorem of Duhamel …
Often I have the same problem!
(next post)
Well what can I do?
Surely there will be many errors will, however, began to speak Bulgarian.
Because my posts are so terrible that if only the Bulgarians understand them (and so do many people) will feel much better!
Great new ways to torture a crocodile.

June 8, 2011 at 1:07 am Leave a comment

Пълно е с дим

Ох – градчето  пак е пълно с дим! Има огромен пожар някъде. Навън има мъгла от дим, и покрива градчето като одеяло. Миризмата е много силна, и мисля че може би е опасно човек да бъде навън. Тука е най-странното място на което съм живяла.

Димът трябва да ме притеснява, обаче мога да мисля само за теоремата на духамел…

Често имам същия проблем!

June 7, 2011 at 2:52 am Leave a comment

Ами какво да правя?

Със сигурност ще има много грешки, обаче ще започна да говоря на български.


Защото моите постове са толкова ужасни, че ако само българите ги разбират (и затова не много хора), ще почувствам много по добре!

Чудесно, нови начини за изтезание на един крокодил.

June 6, 2011 at 1:02 am Leave a comment

Sanity control

Hello, my name is Shnoorko. I work behind the scenes to keep Pleven and Geo from going too crazy. For example, I make sure that Geo eats right and that Pleven doesn’t get too grouchy. I just wanted to introduce myself, because I’ll be stopping by sometimes, and I didn’t want to startle you!

November 12, 2010 at 4:17 am Leave a comment

Isotherms and adiabats, part III

Because parts I and II were just so wildly popular, here comes the third installment of Isotherms and adiabats.

Geo, may I offer a little piece of advice? First of all, you might try not posting about obscure topics from an obscure corner of the internet. Second, if you insist on doing this thing, you should use titles that draw people in, such as:


No mercy? What the hell does that even mean? Where the hell was I? Oh, yeah, I was busy showing that isotherms and adiabats must have a certain structure, independent of the substance under consideration.

Claim ~A. Adiabats can never cross each other (even on a bad day).

Proof. Assume the opposite of ~A, that is A, and pick an isotherm that crosses both of the crossing adiabats, as shown in the figure below.

Consider a process that follows the arrows in the diagram. The total work done by the system is the area between the curves. What is the total heat absorbed after a full cycle, starting, say, at the upper left corner and ending there? The only heat involved must be along the top curve, because the other two are adiabats. At the end of the cycle, the change in the internal energy is zero: U(end)-U(start) = 0. But by the first law, U(end)-U(start) = w + q, so -w = q is the work done by the system in one cycle. It is equal to the heat added along the isotherm! But there is no lower isotherm along which part of q can be transferred someplace else; hence, heat has been taken from a constant temperature reservoir and converted entirely into work, without affecting the thermodynamic state of any third body (the first two bodies being the reservoir and the working substance). This state of affairs is contrary to statement T (from Isotherms and adiabats, part I). Because statement T is true, statement A must be false: ~A. QED.

November 11, 2010 at 9:39 pm Leave a comment

Isotherms and adiabats, part II

For the next claim about relationships between isotherms and adiabats, we’ll need to appeal to a general fact about isotherms, and a general fact about temperature. For the former, let objects A and B be separated by a barrier that blocks mass flow but allows the flow of heat.

Statement I. If objects A and B are brought into contact until thermal equilibrium is reached, there are a variety of pressures P(A) and volumes V(A) of object A that are consistent with it’s being in thermal equilibrium with object B, and vice versa. The curve (P(A),V(A)) in P-V space thus corresponds to a single temperature, T(A), of object A.

Isotherms often look something like this:

Statement R. Temperatures are accurately represented by real numbers.

Claim. An adiabat cannot touch an isotherm tangentially – it must cross it.

Proof. Statements I and R together imply that P-V space is infinitely dense with isotherms. If an adiabat were to touch an isotherm tangentially, and then veer away, it would thus have to cross a different isotherm more than once (see figure below)!

This we have already shown to be impossible in the previous post (part I). Hence, the adiabat must not only touch, but cross the isotherm. QED.

November 9, 2010 at 3:12 am Leave a comment

Isotherms and adiabats, part I

In most thermodynamics textbooks that I’ve seen, it’s shown that isotherms (processes of constant temperature) and adiabats (processes with no heat exchange) must take a particular form when the working substance is an ideal gas. That form is as shown on the dry erase board in the CRAZY THERMODYNAMICS WOMAN picture. But did you know that the isotherms and adiabats look like that for ANY working substance; not just ideal gases?

Um, yes. Yes I did know that (Лъжа… Аз съм крокодил, и хич не ми пука!).

Yes, it’s true! How do you go about showing it? You start by assuming the following two statements, both inductive generalizations based on a vast amount of experience:

Statement J (after Joule, who did the first experiments providing evidence for this statement). Work done on an adiabatically isolated system increases the temperature of that system, and the required amount of work to get a given temperature change is the same, no matter how that work is carried out (for the proof below, we won’t need this second part of statement J).

Statement T (after Thompson). It is impossible to take heat from an isothermal reservoir and convert it completely into work, without affecting the thermodynamic state of any other body.

Corollary of T. Distinct thermodynamic states have different internal energies.

Proof. Suppose that states A and B are different, but that U(A) = U(B). By the first law, U(B)-U(A) = w + q, where w is work done on the system and q is heat transferred to the system. Hence, 0 = w + q or -w = q. The work done BY the system is equal to the heat transferred to the system. If we assume that the process does not appreciably affect the temperature of the environment (which is always the case if we consider the environment to be big enough), then heat from a constant T reservoir has been converted entirely into work, without affecting the state of any system, contradicting statement T. QED.

If statement T were false, it would be possible to extract heat from some not particularly hot place in the ground and power things with it for a very, very long time. My backyard would become a great energy supply! Unfortunately, to use heat from a reservoir, you have to dump a fraction of that heat into a cooler reservoir, and the maximum amount of work that you can get is determined by the ratio of the temperatures of the hot and cold reservoirs (find a thermodynamics book, and read the sections on Sadi Carnot and reversible heat engines). Now for the part that isn’t as easy to find.

Claim 1. An adiabat cannot cross an isotherm more than once. We can symbolize this statement as ~C.

Proof. Suppose that Claim 1 is false (that is, assume ~~C = C). Then there will be distinct points A and B, both lying on an adiabat, and such that the temperature at A is the same as the temperature at B. Consider a process that starts at A, follows the adiabat, and ends at B. Because A and B are distinct states, the internal energy of the system at A is different from that at B (by the corollary to statement T above). By the first law of thermodynamics, U(B)-U(A) = w + q, and so the work done by the system is

-w = q – [U(B)-U(A)] = U(A)-U(B),

the last equality following because q = 0 along an adiabat. Now either U(A)-U(B) < 0 or U(A)-U(B) > 0. Consider the first case. In this case, -w is also negative, meaning that work was done ON the system in going from A to B. But by statement J, the temperature at B would have to be higher than that at A, which it isn’t, because A and B lie on an isotherm! So it is impossible that U(A)-U(B) < 0. Consider the second case, U(A)-U(B) > 0. Then -w is also positive, and work was done BY the system in going from A to B. Now note that, because q = 0, no other system has to be thermodynamically affected. We can just use -w to lift a weight or something. Because q =0 and -w > 0, this weight-lifting feat is equivalent to converting heat from a constant T reservoir completely into work, which is impossible by statement T. In summary, we have

C -> (~T v ~J)

T & J



And so the proof is complete.

November 7, 2010 at 4:40 pm Leave a comment

Older Posts Newer Posts

June 2018
« Feb    

Blog Stats

  • 1,190 hits