Challenge Questions

Well, as presented the answer is sufficiently hard to follow that I haven't checked it completely. That said, the inclusion of VB (the volume of vessel B) in the equations is unhelpful, as it is bound to cancel.

So here's my (pathetic) attempt at a less opaque solution:

Expansion of nearly all the gas in A is adiabatic, so we work with P_A^(γ-1).T_A^γ = constant.
As pressure is proportional to number of gas molecules, we can write:
T_A2^γ/T_A1^γ. = M_A1^(γ-1)/M_A2^(γ-1)

Clearly, the proportion of the original thermal energy residing in A is:
E_A2 = E_A1. T_A2/T_A1.M_A2/M_A1
. and substituting and simplifying gives:
E_A2 = E_A1.M_A2^(1/γ/)/M_A1^(1/γ)
The energy in B is:
E_B2 = E_{A1 -} E_A2
and the mass is:

M_B2 = M_A1 - M_A2

Once the gas has settled following closure of the valve, temperature is again proportional to energy per molecule; so the temperature in B is:
T_B2 = T_A1.E_B2/M_B2/(E_B1/M_B1)

Now it's just a matter of substitutions. It's got late, so I'll do that tomorrow (unless someone more capable has corrected my mistakes and/or completed it first).

I realised I made a silly mistake - P is proportional to M.T (not just M)
I'll correct and finish in the UK morning (I will also do a couple of sanity checks to see if I've made any other howlers).

I decided to submit my version of a solution into the main thread...

The initial temperature of 27 (now where is that key?, oh well) degrees C will decrease along with the drop in pressure. The maximum temperature of vessel B will 27 degrees C.

I think this "And the answer is...." is fine. I would give it a GA if that were possible

And I would certainly give wrong answer if that were possible because of using equation for "adiabatic expansion". This expression is only valid in quasistatic, adiabatic processes. Expansion into vacuum ("free expansion") is not quasistatic.

Also the equations describing enthalpy are wrong:
1. You can not calculate the enthalpy of something, only change of enthalpy
2. What happen to T (temperature) in equation with terms p*V??? This equation could be obtained (from incorrect H=m*c*T) only if all the temperatures are the same.

I presented my point of view few times already and I am trying once again here to convince you that presented answer is wrong.

1. Constant enthalpy assumption.
If both vessels have constant volumes gas may not do any work (and no work can be done on gas). Thus under adiabatic conditions, due to the first law of thermodynamics Delta(U) = Q + W, if Q and W are equal to 0 we have constant internal energy system. And it does not matter whether in container B is some gas or it is evacuated.
If the gas is ideal its U remains unchanged (Delta(U)=0), that also means that its temperature has to be constant. Regardles of the volumes A and B, and pressures A and B, if only (at the beginning) temperatures in A and B are the same.

And I really do not know what to write more about this.
You have to take into account, that the equation of adiabatic expansion of an ideal gas, well known from school, has some assumptions, the most important being that the proces is reversible (quasistatic). Of course it is difficult at school to explain the drifference between non- and quasistatic process, so better is not talk about this at all. Thats why most of people (and my students too) try to use Poisson's equation to each adiabatic process, what is wrong.

Going back to enthalpy. Its definition is H=U+pV, and its derivative dH = m*Cp*dT + (-V+T*(dV/dT))dp assuming no chemical reactions and constant m.
In our case (ideal gas) the second term (its dH/dp) is 0, so for an ideal gas dH ( not H!) = m*Cp*dT, giving after integration : Delta(H)=m*Cp*(T2-T1)
Of course we know (sorry, I know, and I am trying to convince you about this) that T2=T1 becasue Delta(U)=0, so Delta(H) is also 0.

Maybe some more detailed explanations you will find here: The Joule Expansion
Free expansion
Free expansion once again, look at the last sentence!

Sorry for this very long post, but I have to fight for the correct answer:-)
At the end some guesses why we can obtain raise of temperature in vessel B in real life:
1. The volumes are not constant - gas is doing some work
2. In a gun (through a piston) we are compressing strongly gas in container laying just behind a bullet. It may become warmer, but only if...
3. If the gas is not ideal.

Greetings for everyone,
Kajot

Hey the deisel works because when the a volume of gas is forced into a smaller space, it gets real hot. Your contention is that when it occupies more space, it does not get colder, because vacuum is special. That's special, too!

You are absoltely right about diesel engine.
But I was talking about ideal gas. Real gases of course get colder during free expansion into vacuum (I mentioned that).

Ideal gas is good approximation only for low pressures, which is not a case in diesel engine.

That means we both are right, I hope so:-)

Ideal gas, despite the lofty name, is a very good approximation of any gas, and is valuable as a simple calculation and a starting point. Thnigs may not go quite the right distance in the predicted direction, but they go in that direction in the vast majorty of situations! Newton's laws of motion are also flawed, Einstein improved them, but they are 100% OK for most rocket science.

I'm sorry, but your insistence that all temperatures remain the same is valid only for the special case where free expansion is permitted. Friction is present in the pipes and valve and so work is done, and the internal energy falls hence the T falls

I think you have very little experience of thermodynamic flow problems. It is true that an absolute enthalpy cannot be given to a gas state but it is common practice to tabulate the enthalpy (and entropy) above some arbitrary datum (usually Tabs = 0). Since we are only ever interested in changes, going from 1 to 2 we have ΔH = H2 - H1 = (H2-Hdatum) - (H1 - Hdatum) = h1 - h2; where h1 and h2 are tabulated values. Look in any Steam Table manual, or Gas properties Tables to see that values are tabulated for Enthalpy. If you insist, you can work with differences only, but it is much more convenient to work as shown Look up Bernoulli's Equation for compressible flow.

You fail completely to explain why the phenomenon occurs in practice.

Hi again,

I was only trying to solve a problem which was presented in the challenge, two vessels, one evacuated, ideal gas. The problem was far from practice, and my solution too.

In most of my posts I tried to show, that using some equations for this process is wrong. The solution presented, incorrect in my opinion is also very far from practice. And there is now word about friction , flow etc. That's why I was also not talking about this (but you are correct, that I am not familiar with problems of gas flow, and I know, this problem is veeery complicated).
Please notice, that I wrote few times, that constant temeperature is only valid for a special, ideal case.

If the question would look like "Why temp in a paintball gun is raising during filling it with gas" my answers would be different

When we introduce friction we probably will have solution like this:
1. gas at the end of a valve has lower temp and pressure than in vessel A, lets call it A'
2. A' is expanding freely to the vessel B. So even if the gas is ideal we have drop of temperature from A to B
3. Because the valve obtained some heat it becomes warmer (the valve, not a gas - the question here was about gas temperature)
4. Because the valve is in contact with gas and vessels it is cooling down, and if no heat is transferred outside of the system (vessels, valve and gas) at the end of heat transport we should have everywhere the same temp, equal to the one at the beginning.
Of course this (point 4) only if we have adiabatically isolated system and ideal gas

In real live none of this conditions is fulfilled. Non ideal gas and no adiabatic system.
I can guess, that cold gas (after expansion to B, no matter if the gas is ideal or more "real") is removed from a gun during a shot. The heat absorbed due to friction by a valve is heating the whole gun. So after many "fill and shoot" processes a gun can become warm or even hot.
And I still have no idea, how it is possible, that gas in B is warmer than in A.

I hope this is better, more real solution...

Regards,
Kajot

I think you are allowing yourself to be diverted from the essentially correct simplicity you identified. The only requirement for your solution is that the orifice is small and the mean free short - both compared with the dimensions of A. No need for randomisation in the orifice (= friction without thermal transfer), no need for any additional assumptions - except that the temperature is only measured after the conditions in B have had time to stabilise.

Yes, the gas gets colder under free expansion - but that is only because the kinetic energy of motion is now away from the original orifice, so you can define a temperature with respect to the average velocity of the stream. The total energy does not reduce.

Consider what happens if enclosure B is large enough that none of the gas hits the walls while the valve is open. You then close the valve and wait for thermal equilibrium. This happens after the gas hits the walls (and other molecules) often enough for the directions to be randomised. All the kinetic energy is thermal motion again - and the gas has warmed up.

I believe we have two extreme conditions under which we model the process.
. One involves no collisions between the particles that make up the gas. In this case I believe that we have already agreed that the gas in B will be warmer due to the higher probability of faster molecules passing through the valve, though we do not have a value for this. (Once the valve is closed the roughness of the surface of the vessel will randomise directions of motion, so we will have a higher temperature in enclosure B - note that the velocity distribution will not be as you would expect for a gas at that temperature in either enclosure - but we can still assign a temperature).
. The other extreme assumes that the mean free path is short compared with the dimensions of the enclosure. I go through the detailed steps of the argument below. I believe the argument is correct; but that only means that it has not yet been properly shot down - so please show the individual holes, and if possible staying at the purely functional level.
(Intermediate conditions will doubtless give intermediate results).

The argument (as I see it)
Suppose we don't consider the details of the expansion into the vacuum, and also ignore initially what happens near the orifice.
We look initially only at what happens to the bulk of the gas in vessel A.
. Does this expand against its own pressure?
. If so, does it conform to the standard adiabatic expansion law?
. In that case, is the temperature of the majority of of the gas in enclosure A (as it expands into enclosure B) given by the standard equations?
[I believe that all the answers are yes - under the given assumption that the mean free path is short compared with the size of the enclosure.]

Now we close the valve.
If we know the proportion of the original mass of gas remaining in enclosure A, we also know its temperature - and hence the total thermal energy remaining in enclosure A. (I'm assuming that the orifice is small enough that there is minimal energy in the bulk movement)
. That means we also know the mass of the gas in enclosure B.
. And, as (for whatever reason) there is no heat or kinetic energy interchange with the enclosure, we must also know the motional energy of the gas in B.
. Initially this motional energy will be distributed between thermal energy and ordered kinetic energy. [It may be that the gas is very cold as it enters enclosure B - but if so it will also be a fast-moving jet (I believe that this depends on the internal geometry of the valve)].
. Now, as you allow things to settle any ordered energy will be converted to heat.
. We know the total motional energy of gas B, therefore we can calculate its temperature.

Regards

Fyz

No collisions between particles of a gas is a pretty good description of absolute zero temperature or pressure (you get to pick).

Did Einstein predicate the existence of fast molecules, as opposed to the usual average C- ones and the special ones on the short bus? I think Boltzman warns us that some of the molecules are far more energetic than others at all temperatures, and we can only measure the average random energy, which is the temperature.

Yes, there are throttling losses as the valve is opened and due to the restriction of the connecting pipe, and surely an excess of gas moves to B before pressure overcomes inertial and sends it back and forth. The change in potential energy between pressure-volume A and A + B will turn to heat. However, I contend that the heating from this throttling friction is not significant, but if you contend elsewise, calculate the exact energy and apply it to the thermal mass to com up with a contributing tempoerature delta. The gas coming from my propane tank is still pretty cool despite the throttling losses of the regulator, but the expansion makes the whole tank so cold it sweats condensation. Admittedly, much of that is from the boiling of the liquid propane, which phase change is more energetic than the cooling of simple lowered pressure.

The presentation of the problem as two containers is there to muddy it up! After all, it is intended to be a test, just a little like real life, so there is excess information and small distracting energy flows to consider and discard! The basic activity is what you would get it you moved a piston out so the cavity above it was doubled in size. The gas inside gets cooler. If it was not modeled as perfectly insulated from the piston, cylinder and head, it would then warm up from the surrounding conductive parts.

When considered in terms of an ideal gas, no collisions represents the limit either of extremely small collision cross section, or of extremely low density.
Paradoxically, perhaps, low temperature and constant pressure would have the effect of increasing the number of collisions, as the following relationships apply:
. Mean free path is proportional to temperature
. . (distance reduces as temperature reduces)
. Velocity is proportional to √(temperature)
. So time between collisions is proportional to mean_free_path/velocity, i.e.
. . Time between collisions is proportional to √(temperature), which means:
. . . Collisions become more numerous as the temperature reduces, and
. . . . At absolute zero (non-QM) particles are in constant collision

However, another property commonly specified for an ideal gas is that the volume modification due to the sizes/interactions of the particles from which it is composed is zero.

This second property is clearly incompatible with a short mean-free-path. This incompatibility of limiting approximations leads to all sorts of confusions, and could well be the basis for the problems that kajot is highlighting.

Coming to another of your points: Maxwell was the first to determine the distribution of particle velocity in a gas at uniform temperature; his conclusion was that the distribution of √(energy/mass) is Gaussian (or normal). This has been confirmed by measurements, both indirectly from the Doppler spectra of light emitted by gases and by direct measurements. BTW, you can measure the velocity of an individual particle - albeit it is hard to see what use you could make of this.

If we allow slide-ruler's constraint that no energy is exchanged with the container, it is irrelevant whether the excess energy is converted to heat by the roughness of a throttling surface or by subsequent multiple collisions.
And the second container is not necessarily purely a distraction - the temperature of air inside a newly inflated tyre is a real issue (though it will not normally be quite as hot as this theory predicts, because it will do work against the atmosphere as the tyre expands). Yes, the temperature inside container A is the same as if the air was expanded in the same proportion by moving a piston - but the temperature of the gas in the second container is also significant.

Returning to the different case of expansion into a vacuum, the expansion away from the orifice means that the kinetic energy of the gas in any region becomes unidirectional, and is then measured as velocity rather than as heat. But the total kinetic energy (including vibrational energy) will still correspond to the values calculated.

Here's what is happening.

There is friction in the connecting pipes and valve .

The gas in vessel A is expanding and doing work against friction. This is done at the expense of its internal energy and so the temperature in A falls. This is a classical adiabatic expansion, and the temperature of A can be calculated by the conventional P^((y-1)/y)/T = constant

The energy of the work done against friction cannot leave the system and becomes heat which is contained in the gas entering B. The flow into B is therefore has increased enthalpy (and entropy) by virtue of this added heat and the temperature in B rises. This section is not adiabatic, and its temperature must be found from the fact that the total system enthalpy remains unchanged

If there were no friction, the flow from A would be free expansion, then of course the temperatures would remain constant at 300K.

But there is friction - despite the fact that no one condidered it!

The effect is apparently well known to paint-ballers, and as I said in an earlier post, it was a problem I had to investigate when filling evacuated vessels with argon gas. Interestingly, the effect impedes filling B with gas. A cols down and holds a greater mass, and B heats up and contains a smaller mass than would be the case if the transfer were isothermal.

The I´Hospital´s rule is well applied in the answer, but does not explain the effect.

I can not see a temperature increase in B unless we used helium or hydrogen that has a very low inversion temperature, and apply the joule Thomson effect that will increase the temperature of expansion.

Ranque-Hilsch vortex tube is a Maxwell Demon device at expense of pressure.

I think that the valve is swirling in such a way that the hot gases are driven to B while the cold fraction is going back to A.

This may be the reason why SlideRuler found, with Argon, that "A cols down and holds a greater mass, and B heats up and contains a smaller mass than would be the case if the transfer were isothermal"

So, the design of air conditioners and refrigerators is flawed, as they have the cold side downstream of the restricting valve and upstream of the compressor. Why do they get cold, if the energy given to the working fluid as it passes through the restricting valve overwhelms the cooling effect of the lower pressure? I'd say it was the phase change in teh evaporator, but the very first air conditioner used air and no phase change.

Let's see some numbers for the simplified case where the valve is opened very slowly, so there is no slosh back of fluids. Apply the extra energy absorbed on the B side to the specific heat of the B gas to get a temperature rise, and calculate the exact temperature of each side.

From what I understand, expanding compressed gas and evaporation absorbs heat, will cool the existing gas and the walls of B, but the expanding gas will not increase its temperature over the ambient temp.

The temperature will rise up to 147ºC, according to paint-ballers. To rise temperature of a gas that much, I see compression, friction, vortex tubes, or a high temperature source.

The temperature of each Vessel derived from the equations given, (for Vessels of equal size) is shown in the graph below. You can consider the valve was opened slowly and then closed at some pressure PB2.

PA1 was taken as 100

That's math. What is the physics or chemical effect that is taking place?

That's mathematics modelling the physics of the given challenge.

There were no chemical effects stipulated. Why try to make thing more difficult? People seem to have had a hard time solving the given challenge without adding complications.

Remember your own quote "Simplicity is the ultimate sophistication"

This is certainly confusing people.

But I think there is a good question hidden in this part of the thread - would a refrigerator of near-conventional design work in principle without utilising latent heat. I think it would*, and the cooling effect would be due to the combination of the velocity of the gas through the evacuated tube and the effective expansion of the gas on the cool side by the pump. But I could be wrong...
*Though it would not be nearly as effective

I'm trying to follow your math. Something here doesn't seem right. You wrote:

Constant mass gives

In terms of the specific heat at constant pressure the enthalpy is given by Therefore, constant enthalpy gives

Adiabatic expansion in vessel A gives

From these three equations we obtain

Comparing the 1st equation to the 2nd equation, if both equations are true then

T_A1 = T_A2 = T_B2

Applying this info to equation 4, and defining the term X where

X = P_B2V_B/P_A1V_A

You get:

(1-X)^1/γ = (1-X)

Which is true only if X = 0 (unless γ = 1, but you already said it was 1.4).

... Meant to say if X = 0 or 1.

The error in your argument lies in the first statement

Comparing the 1st equation to the 2nd equation, if both equations are true then

T_A1 = T_A2 = T_B2 (THIS CONCLUSION IS FALSE)

Consider the simple case solution Pa2=Pb2 and Va = Vb so we have Pa1/Ta1=Pa2/Ta2 + Pb2/Tb2 from the mass; and Pa1=Pa2+Pb2 from the enthalpy

Pa2 = Pb2 = 1 then from the enthalpy Pa1=2

The mass equation becomes 2/Ta1= 1/Ta2 +1/Tb2 and as Ta1=300K we get

2/300 = 1/Ta2 + 1/Tb2 Choose any value for Ta2 and you can find s value for Tb2 which solves. The trick is to know the value for Ta2 which is given by the Third equation leaving 2/300 = 1/246 + 1/384

I think you have one of those math models that use 'familiar' to appear flawless yet 'proves' the impossible. In effect it uses 'enthalpy' to disprove itself and Laws 0 to 3.

In the questions vacuum context; V is initially ∞ in B. On filling, all that changes is the space between the atoms, or initially T can be regarded as - 300 C. Or T fall/rise is a resultant of density fall/rise. Without external inputs, only when density in B = density in A, the T's regain energy density parody. 'Expansion' into B, to 'restoration of density' in B, it is pure adiabatic expansion then recompression – It DEFINES pure enthalpy. Total energy change is ZERO. So at no time can T in B be above T in A, without (prohibited) external inputs.

Add external inputs and the end state resolves under PV but the density in B is less due to T absorption rise. Therefore rise due to external inputs is altering PV null point, or flow cessation occurs ahead of the solution's V independent math.

Excellent empirical 'red herrings' support the solution's math.

Refilling pre-evacuated cylinders; as unavoidable external inputs result in a lower density in B, the cylinder may be 'topped up' by pumping. This compressing to a target density causes T rise due to both extra work and compression of external input T energy acquired during the initial expansion freeze.

Paint ball reservoirs get hot; at minimum there is one atmosphere to recompress, but as 'empty' means 'below the regulator pressure', and many are well above that, recompression produces T rise in the many atmospheres of gas residual (at ambient T ). Similarly residual in SCUBA tanks gets hot when filled from an air bank (at ambient) and the whole hotter when direct filled from a compressor. These herrings play to our experiences. Both 111 C and 147 C are "experience plausible", but only with 'extra work' or 'extra T' - which is 'a totally different topic'

Friction/kinematics? This is work transfer from compressing A and proportional to the mass flow. A typical coefficient in high decimals applies x mass = Zilch T rise. A would need to be infinite in V to not fall in P and T (and velocity). A fall greater than any frictional or kinematics induced rise. The T result would be dimensional per mol, not the volume/mass independent T = y (K) the solution always reduces to.

I'd suggest the crucial logic being obscured is ∞ V and V dependence, but I have to admit I can't bothered to delve into exactly how the math is not right – so congratulations on a great "mind fu*k" and a merry chase. Most enjoyable – and perhaps a suitable "Bernoulli's revenge" on his plagiaristic master.

There is no inconsistency in SlideRuler's models, nor any contradiction of basic laws.
The problem is that, although most of what we were taught at school about gases and gas laws was basically correct*, we don't necessarily understand it fully when we first receive it - i.e. what we take away from that teaching is at most only partially correct (naturally I speak for myself here).
In order to get a correct picture we need to re-examine what we have learned in the light of fundamental principles. If we "can't be bothered", we almost guarantee that we end up maintaining those errors.

For the case of a gas forced into a vacuum, we need to revisit the reason that the temperature is low: it is that most of the thermal energy is converted to ordered kinetic energy. So the temperature will only remain low until that ordered kinetic energy is converted back to heat.
This conversion will happen after just a few collisions, either with other gas molecules, or (at extremely low pressure and/or small cavities) with the walls of the vessel. If the gas is air or nitrogen, the molecules will be travelling at about 1200ft/sec, so this shouldn't take too long.

BTW, the reason that the standard school experiment never shows the rise is that it only looks at the position before equilibrium - typically it looks at the temperature as the gas flows through the tube where the order of the flow is maintained.
Some experiments appear to look at the temperature in a vessel. This can also be low if the vessel is being pumped (because the gas is continually doing work on the pump.

*At least what I was taught was - although when I first revisited it between lessons I thought that my initial incorrect understanding corresponded to the teaching.

1 The gas in A does work accelerating the exiting gas and overcoming friction. This is regardless of what happens to the gas after it has left A

2 This work is at the expense of internal energy of the gas remaining in A, and so Temperature A drops.

3 Since no gas and no energy is allowed to escape, ALL of the energy lost from A is present in B.

4 It doesn't matter if the energy comes form heat of friction, or work done by compressing gas in B in the end, energy lost from A will be found in B.

5 Temperature in B will always be greater than A for no energy loss.

Getting almost serious, I googled up the wiki:

http://en.wikipedia.org/wiki/Ideal_gas_law

and it says:

First, V2 = 2 V1.

Second, nR is constant for this problem (else we have to apply E=Mc^2), so p1V1/T1 = p2V2/T2.

Third, there is also the energies p1V1 + nKT1 = p2V2 + nKT2, where K is some constant that matches the energy units for the gas and it's specific heat. For instance, if P1V1 is ( kilograms / square meter ) * cubic meters, it is in kilogram-meters, and if N is in pounds, the gas has the specific heat of water, and T is in Rankin, nT is in BTUs. Any original energy that becomes heat for any reason will elevate T2, but only so much as is not accounted for by P2V2.

The amount of heating of the gas during the transition is not deterministic, varying due to the viscosity and specific heat characteristics of the gas, the pipes and valve, and how the valve is opened.

Then, I changed my major from Mechanical Engineering to Computer-Electrical, a universe where 1 is one and zero is not.

"The amount of heating of the gas during the transition is not deterministic, varying due to the viscosity and specific heat characteristics of the gas, the pipes and valve, and how the valve is opened."
That is correct so far as it goes. But it considers only the heating, and not the total kinetic energy of the gas. Given that there is no thermal conduction from the gas to other parts of the system, and under a reasonable definition of "pipe" (diameter small compared with the dimensions of vessel A) the total kinetic energy (=heat plus systematic motion) is deterministic.
If you allow time for the draughts to die down after closing the valve, all the kinetic energy will be converted to heat. That means that the final temperature in enclosure B is deterministic.

I agree that transitionally, some of the energy will be kinetic, and that is equally hard to determine at any time. We can only know that all the energy is in there in one form or the other: motion, heat, or pressure. Eventually, the kinetic energy turns to heat and pressure.

My point was that the total energy is independent of the details you cite, and the reason that the thermal energy was dependent on detail was simply that it affected the proportion of the total energy that was initially in the form of heat. However, if we take the challenge to be asking about the settled temperature (hence the closing of the valve), all the energy is then heat, and we can ignore the distributions during the flow.

BTW, I don't believe that gas pressure as such relates to stored energy - though it is clearly a factor in the transfer between thermal and mechanical energy.

When you accelerate a gas (with or without back pressure), you do work on it. As regards changes of energy in vessel B, the only thing that matters is the work that is done expelling new gas from A and its original temperature.

Based on that, and making the assumptions previously stated (mean free path, size of tube, and non-conductivity to enclosure) it becomes obvious that gas entering B has total energy that is (2+N)/N times its thermal energy immediately before leaving A (where N is the number of vibrational degrees of freedom of the gas). Although this energy may initially be arbitrarily distributed between thermal and ordered energy, by the time the gas reaches equilibrium after valve closure it will all be thermal.

So calculating the temperature of the settled gas in B is simply a matter of integration. However, this integration already requires us to calculate the temperature of the gas in A at any stage. Personally, I prefer SlideRuler's method, as it gives the same result without the extra sums.

Well, it's easy to get excited about motion, and acceleration, that's really heady!

But viscosity, Brownian motion and conductivity not being suspended, even in the small connecting pipe and valve, the gas will settle out losing all motion and all temperature variation evens out, give or take the usual Boltzman variation. Now, before the valve was opened, it had PV potential energy of pressure and nKT energy invested in temperature, and the commotion of transfer can be predicted to heat it up. If it had not heated up due to the commotion, since V2 = 2 V1, P1 = 2 P2. But since it did heat, P2 is diminished (energy used to heat it up) and T2 is up. However, if T2 is up, the gas expands, or wants to, so P2 is also increased. The amount of heating (delta E) is variable by viscosity, and perhaps even by how the valve is opened, as turbulance losses may be non-linearly higher at higher velocities (sound barrier, cavitation, energy proportional to mass * velocity squared). The amount of temperature change (delta T per delta E) varies on specific heat. How to calculate even the probable relative size of these contributing components to temperature and pressure leaves me a bit tuckered out.

That is why SlideRuler gave the value for gamma - it uniquely determines the relationship between temperature and energy.

You mean specific heat, like one pound of water goes up 1 degree F for every BTU of energy. OK, necessary to get a) delta temperature from energy, but not sufficient to solve the problem. You need b) how much energy went into heat during the transfer, both to see how much pressure needs to be devoted to creating that energy and to see how much the gas is heated, and c) how much the pressure goes up for this contained gas for the calculated delta in temperature.

Everything, including the pressure law, is based on integral multiples of KT. So you really don't need any more information - and you don't evenneed to know the value of k (Boltzmann's constant). In my day this was basic high-school physics, but I don't expect they do it like that any more.

You will find what you need here in wikipedia. (It's not organised precisely as I would prefer, but that is probably a good thing for other people who don't already have the basics. The main weakness is the lack of definition/detail on"mean free path" - but you can follow up the references once you are happy with the basics).

An engineer's viewpoint

Gamma y is the whole key to this challenge; it is the ratio of the two specific heats (Cp and Cv) and is a number independent of units. In expanding gasses, the expression PVⁿ = K is often applied. It is clear that for isothermal processes n=1. and it can be shown that for adiabatic processes (no heat lost) n=y. Values of y between these values are used to model "some heat is lost"

There is also a fixed relationship Between "work done" 1.e. "energy" and y which is W=R(T2-T1)/(y-1). This means the temperature of vessel A can be calculated using the formula given in the challenge.

Once TA has been determined conservation of mass and energy lead us the the answer for TB. It is not important HOW the Bgas gets heated, it's just important that it MUST become heated by the process.

Ok, I begin to see, the y property is assigned to one side not the both.

The properties that were present throughout the cycle are of equal effect at equivalent cycle points, in both directions, in both A and B, or n = 1. To paraphrase the math outcome T = (y - y = n = 1) K.

From a work point of view;

You can't assign a direct % by V of total energy used in compressing A's contents, because heat lost to ambient is no longer present in the filling of B timeline. The work component is now potential energy. The % by volume transferred to B from A, is expressed as a P/T drop in A due to an increase in total V, or total system energy falls. You are "disbursing" work. Formula right, + sign wrong, "colder" not "hotter"

In practice, when P equalises; Ta will be 'colder' but Tb has 'gained' due to 'relative warmth' of the start injection, plus line friction. Considering friction is miniscule, and Ta has fallen, Tb at cessation of flow will be hotter, with respect to Ta, but not hotter than start Ta.

This is one of the reasons you cannot fully fill a cylinder from another cylinder, short of it being infinite, or significantly above your target pressure. But the uptake from ambient in the expansion phase, or re-compressing of residual charge, is of far greater effect raising Tb so nulling flow.

Congratulations, finally I can understand the temperature increase in B.

I see that you guys really go deep in analyzing processes. I would like to have your comments in my post two phase flow ejector engine, if you don´t mind: http://cr4.globalspec.com/thread/39216/Two-phase-ejector-engine

And a promise is a promise, if I get a grant to build it, I will pay a trip to all active contributors to meet, somewhere nice in the world.

Dear Physicist,

I did say "in practice" and "at cessation of flow", not in theory and at eventual full stasis. Industry (practice) - no time for that.

Anyhow back to theory;

"the second half of that statement contravenes conservation of energy" - a bit harsh.

The concept is the first stuff into B is at A's original system heat and the last in is at A's lower heat. So that starting heat, is re-compressed and recovered and diluted proportionally buy the 'coldest' last portion. Or Tb is at average of Ta's range. But yes, it assumes some significant drop in Ta, between start and end of fill.

I don't see the "contravention" but on with my sins;

"that energy must have gone somewhere" - yes, it's now spread over Va + Vb. Oddly no Laws were harmed in this 'disbursement' of this energy density.

"independent of process"; It can't be as it is connected by a pipe and open valve in the 'fill to cessation of flow' circumstances I predicated the statements upon, but also during any flow occurring.

"so all the energy in vessel B will be converted to heat" - precisely, at first it was 'frozen' in a vacuum, now it's 'heated' by re-compression back to PV conditions in A. Enthalpy, adiabatic, un-harmed.

"hotter than the initial temperature in vessel A"; Can't be as P is equal and V a constant; if T attempts to be higher in B the flow would stop or reverse. Again no Laws harmed.

And as said repeatedly, external absorption the main reason why flow stops 'early' in charge density terms.

And I am conceding "friction", though it's bugger all, because it exists in a pipe with a 'pressure drop' when flow exists. And nit-pickers will pick.

You accuse me of "nit picking"? The situation as I see it is that you are stating that SlideRuler's results are wrong, but without presenting what appears to be an argument that I can follow. Now that is what I call nit-generating.

Anyway, as you appear to be happy that gas spread between container A and container B with contents settled and both containers at lower temperature than originally can contain the same amount of energy as the original condition, I'm giving up on pure logic.

So I'll resort to observations: I have read your statement that the paint-ball tank is not a relevant reference, because it does not start off as a vacuum. So I would ask you - what pressure ratio between initial fill and final fill would you regard as sufficient to convince you that the gas will warm up when pushed into a vacuum?
Then I will give you my example - based on vacuum experience: if you vent an evacuated vessel (typically below 10^-3-torr, or about one-millionth of an atmosphere) to a compressed nitrogen supply (common practice to avoid letting moisture into the chamber), a temperature sensor in the chamber will initially cool while gas is flowing rapidly past, but will then rise above ambient.
(If you vent it to air, the initial cooling will cause condensation; again the temperature will subsequently rise above ambient; but I doubt this would convince on its own, as latent heat confuses the issues.)

Ok, let me break this into 6 and 1/2 parts.

Part 1; re- read the stream; you will notice nit-picking not of your making on 'friction', or the comment is not 'all about you'.

Part 2

"Anyway, as you appear to be happy that gas spread between container A and container B with contents settled and both containers at lower temperature than originally can contain the same amount of energy as the original condition…"

I'm simply applying the Laws, which in this 'theoretical' circumstance say exactly that.

The PV/T concept is of "energy density". Spread it out and it is less dense.

The atom count is unchanged, the total energy is unchanged, but the space is bigger = less dense, hence the 'V' in PV/T effects (the energy effects from impingement of atoms fields upon each other), or T.

Part 2.5

"I'm giving up on pure logic." This tacked onto the end of "Anyway…", infers from your point of view the 'theoretical discussion is at an end – I simply went with the decision. That you seem to wish to continue (post 150) fine – I'm back.

So; Part 3

"you are stating that SlideRuler's results are wrong"

No. Just there are two elements in this. One is an empirical result; the other is a search for explanation. I disagree with the 'explanation path' only.

Serendipitously you then provide an empirical example that 'illustrates' the basis of your 'explanation path' and possibly SlideRuler's too.

To précis a bit for brevity;

"Then I will give you my example - based on vacuum experience: if you vent an evacuated vessel ... to a compressed nitrogen supply … a temperature sensor in the chamber will initially cool while gas is flowing rapidly past, but will then rise above ambient."

Would you agree that a typical "temperature sensor" in a vacuum measures above zero K? Might you accept this is because of 'external inputs' such as radiant heat from the surrounds? If so then, what you and SlideRuler require is a sensor that measures only the heat in the gas, irrespective of "near zero density" and mass flow (kinetically concentrated energy density) This yet to be invented measurement tool, presents a basically a huge metrology problem.

Part 4

Throughout this discussion I have separated the theoretical and the empirical. The former (and question) assumes a magical vessel of ZERO thermal mass/thermal inertia. The latter because it is SO impossibly and irremediably subject to "external inputs", like heat from the vessel during the initial gas expansion, radiation from the environment and, and, and.

Part 5

Perhaps, if you have the time and equipment on hand; repeat your experiment with all at 'gas start T'. Repeat with a liquid nitrogen frozen vessel/pipe. You now have a run with 'controlled inputs' on the expansion phase and a run with 'controlled inputs' on the re-compression phase. Y/N? "Both Sides of the Equation in place"? – average the rises.

I warrant, as soon as I said 'cool the vessel' you 'know instinctively' the rise will be less and the average 'potentially under' "A start T"

Part 6

"(If you vent it to air, the initial cooling will cause condensation; again the temperature will subsequently rise above ambient; but I doubt this would convince on its own, as latent heat confuses the issues.) whereas the 'debate' exist without"

Throughout my comments 'latent heat' is mentioned as enthalpic, as it is, 'in theory'.

In practice the existence of a solid/liquid condensate on a typical un-magic vessel results in a higher transfer rate of heat by conduction from the vessel and additional gains into the vessel from the latent heat absorbed in forming of external condensate. It's not 'confusing', just another external input, giving SlideRuler his empirical vs. theory conundrum. And all you do by attempting to solve it by only using one side of the "process" is jump to erroneous view of 'cause and effect'

I look forward to your dissection of my experiment introducing the "cold half".

You say you are "applying the laws".

The fundamental behind these laws is that the thermal energy of a gas with γ=1.4 is ∑5.k.M.T, where M is the number of molecules at a specific temperature, and k is Boltzmann's constant. It follows (as we have conservation of the energy in the total of the gas) that if you cool one portion of the gas another portion must get hotter (once all kinetic energy has been converted back to heat.

Anything that appears to contradict that has to be misinterpretation of some sort.

Regarding such misinterpretation, I had great difficulty understanding what you meant, so apologies if I'm misinterpreting - but:
. Your comment about energy density makes me suspect that you think that temperature is related to energy density. That is only true if the density of material is constant. In a gas, the energy density (per volume) is irrelevant - what defines the temperature is the energy density per mole (or normalised mass) of the gas - albeit my preference is to express it in terms of the (random) energy per molecule.

As presented the answer is sufficiently hard to follow that I haven't checked it completely.
That said, the inclusion of the vessel volumes and the pressure of the gas in vessel B in the temperature equation is unhelpful, as their contributions can be expressed in terms of the gas missing from A and the pressure in A*.

So here's my attempt at a less opaque "general solution" (you may consider the reduction in opacity to be pathetic, but I'm confident the improved compactness of the solution is worthwhile):

We start with the basic equations:

Expansion of nearly all the gas in A is adiabatic, so we work with
P_A^(1-γ).T_A^γ = constant, or P_A1^(1-γ).T_A1^γ = P_A2^(1-γ).T_A2^γ . . . . (1)
The other equation that we will need to determine the amount of gas in vessel B is the relation between pressure and mass in vessel A:
M_A = C₁.P_A/T_A . . . . . . (2)**
The final basic equation is the energy in a mass of gas -
E = C₂.M.T . . . . . . (3)

This leads to:
E_A2 = E_A1.P_A2/ P_A1

(This result may appear surprising initially, but you can see the inevitability if you consider the mechanical energy available when adiabatically expanding the fixed volume to infinity)
Which gives:

E_B2 = E_A1.(1-P_A2/ P_A1)

Next we need to know the mass of gas in vessel B. (We could easily allow an initial fill at arbitrary temperature in B, but it clutters the expressions, so I won't do it here) –

M_B2 = M_A1 - M_A2 = C₁.P_A1/T_A1 - C₁.P_A2/T_A2
Substituting (1) and simplifying gives -
M_B2 = M_A1.(1 - P_A2^(1/γ)/P_A1^(1/γ))

Now we can calculate the temperature of the gas in vessel B –

T_B2 = E_B2/C₂/M_B2 = T_A1.(1-P_A2/P_A1)/(1 - P_A2^(1/γ)/P_A1^(1/γ))

Now for a couple of sanity checks:

If vessel B is so large that essentially all the gas transfers, the temperature must return to its original value (the gas must settled – so vessel B cannot be truly infinite). For this to happen, P_A2 → 0, and the equation gives T_B2 = T_A1, which is as expected.
If only a tiny amount of gas is transferred, the energy applied by the expelling force means that its temperature will be 1.4.T_A1 (as mentioned in another posting). For this we need the limit as P_A2 → P_A1, which is
T_B2 = T_A1.(1 - (1+δ)/(1 - (1+ δ)^1/γ) = 1.4.T_A1

* Although the temperature in B for a given pressure in vessel A is independent of the size of vessel B, the size of vessel B is important in determining where the equations are valid – i.e. the process can only proceed as defined while P_A ≥ P_B, or:
TA.MA/VA/(TB.MB/VB) ≥ 1

** The constants C₁ , C₂ depend on the (fixed) volume of vessel A and the measurement unit for the gas (as the gas mix is constant this can be actual mass – though moles or number of molecules would be natural for the more general case)

P.S. I decided to cast this so that it shows that the settled temperature of the gases in both chambers is dependent at all times only on the change of pressure in A. If we are only interested in the highest temperature the direct approach using mechanical work done on the gas leaving A is sufficient (given that the temperature and pressure in A are highest at the beginning).

It is true that the answer as given does not give the tedious but simple algebraic manipulations necessary to reduce the three equations to a single expression. This may indeed make it hard to follow for the "faint of heart"

I chose to find the relationship for T_B2 in terms if T_A1 and P_B2 and found:

You found the relationship for T_B2 in terms if T_A1 and P_A2 and found (just before you checked your sanity)

T_B2 = E_B2/C₂/M_B2 = T_A1.(1-P_A2/P_A1)/(1 - P_A2^(1/γ)/P_A1^(1/γ))

Taking my expression, and using the enthalpy equation to replace (P_B2*V_B)/(P_A1*V_A) with (1-(P_A2/P_A1)) we get

T_B2=T_A1*(1-P_A2/P_A1)/(1-(P_A2/P_A1)^(1/y)) which is identical to your expression. It is also more elegant since the V terms disappear

Therefore I think we can both give ourselves a pat on the back for maintaining our sanity.

Note that to get the limit as P_A2--> P_A1 you can still use l'Hopital's rule and the answer is of course still y*T_A1

The "sanity check" was emphatically not on my own sanity - only on the solution.

I fear that you are forgetting that I'm wandering around CR4, complete with its army of white rabbits, Cheshire cats, and people who believe impossible things before breakfast (and in some cases at all times)...

Regards

Fyz

______________________

'I don't want to go among mad people,' Alice remarked. 'Oh, you can't help that,' said the Cat: 'We're all mad here. I'm mad. You're mad.'

A good solution, conveniently ignoring the impossible to quantify conversion of stored pressure energy into heat. One thinks of A Gew Good Men: "You want the truth? You can't handle the truth!" ( . . . with precise numbers and simple formulas!)

A lot of the argument seems to be pitting the "3 second case" (interesting but transient and unstable) against the "infinite time case" (more calculable, but who can wait any more?).

1. When the pressures first reach equality, the gas in B will be hotter, as the gas in A has been expanding and the gas in B has heat from the friction of viscosity and rubbing the walls. This temperature difference means the pressure in B rises to equality before half the gas has been transferred. There's nothing magic about the moment when pressures first equalize; it's just a nice point to take stock of the situation.
2. The initial higher B temperature is probably a bigger factor than the inertia of the flowing gas, so while the flow will continue a little bit due to inertia, it will still stop before half the gas is transferred. There's nothing magic about the moment when the gas flow first stops; it's just a nice point to take stock of the situation. For completeness, let's consider the alternative: How many gasses at any pressure and temperature are heavy and yet have low viscosity? I will allow that if there is a lot of inertia and little viscosity heating, more than half the gas might be transferred before flow stops. There's nothing magic about the moment when half the gas is transferred; it's just a nice point to take stock of the situation.
3. When flow stops, since there is a pressure difference due to the inertia, the flow will reverse.
4. The flow will oscillate back and forth for a bit, which provides some accelerated mixing, but viscosity will damp this out, and almost certainly less gradually than the following process. I suppose that in theory it oscillates forever, but eventually the oscillation motion is lost in the noise of brownian motion. All these movements are converting kinetic energy into heat.
5. Finally, the conduction of heat through the gas in the connecting pipe will very slowly equalize A and B, as B heats A and the remainder of half the gas slowly flows from A to B. The conduction path is long and narrow, the gas is not a great conductor, but eventually it is the predominant activity. This will take a very long time, theoretically forever, but eventually the temperature differences are lost in the noise of normal Boltzman molecular energy variations.

Since heat is (magically by decree) not lost in this problem, the unified final pressure, temperature and heat of the whole will reflect both the expansion and the frictional losses of all the motion, appropriate to the specific heat and viscosity of the gas and the flow losses for this valve and it's opening curve, pipe and containers.

"conveniently ignoring the impossible to quantify conversion of stored pressure energy into heat"
Impossible to quantify?? No, no, no - it's not even significantly uncertain.

Actually, there is one thing significant about the point at which the flow first stops - if there is any backflow B will lose temperature and A will gain (relative to the first occurrence at any given pressures). But first-stop is more difficult than what SlideRuler analysed: he analysed the case where the gas in B mixed very rapidly compared with the flow, and where the flow was slow enough to ignore the effects of momentum when calculating the stop-point - the stop-point was actually equilibrium with thermal insulation between A and B - a very particular (and purely theoretical*) case.

On which last point, it beats me why so many people on these threads take the view that an idealised case is not worth working through properly? It's clearly plenty difficult enough for most of us as a first estimate - and we haven't a hope of making a decent cut at a real situation until we can handle this.

" How many gases at any pressure and temperature are heavy and yet have low viscosity?"
In case this causes confusion (ocnfosuni): we should remember that in general heavy gases have low viscosity (low velocity and short mean-free-paths); also that the viscosity of an ideal gas with constant collision area would be proportional to absolute temperature - but of course collision areas are temperature-dependent, which was first calculated by Sutherland...

*Albeit with surprisingly close correspondence to practical results where the gas-flow is moderate.

Well, I thought going step by step would clarify things!

Sure, ideal cases are worth doing, in case you can disregard some of the effects, but you should know what you are discarding. Like the old saw, forget you assumed at your peril: Ass=U+Me.

I am confused by "the point at which the flow first stops - if there is any backflow". It is a point in time when there first is no flow. Other points in time may have backflow, but not that point. Slow down and deal with each stage before moving to the next.

Yes, as the gasses mix, B cools and A warms. Then, as the heat conducts, B cools and A warms. Finally, A is warming B as fast as B is warming A, give or take random motions, and you have equilibrium.

The problem did not give us the gift of "slow enough to ignore the effects of momentum"; it did give us no heat leakage, and that was pretty much the end of the reality give-away.

So, hydrogen gas is viscous and uranium gas is loose? It must have made it hard to inflate the Hindenburg. Well, the gas rushing down the connecting pipe is not going to suck much extra out behind it, since it is not relatively massive and the pipe is short.

"The problem did not give us the gift of "slow enough to ignore the effects of momentum"; it did give us no heat leakage, and that was pretty much the end of the reality give-away." You are right that the problem did not explicitly state "slow flow" in vessel A (which is what I meant). But, unless otherwise stated, the use of "pipe" implies that the pipe is be small relative to vessel A; so the effects of momentum can be ignored. Similarly, the use of "pressure vessel" implies that the mean-free-paths are short compared with vessel sizes.

Perhaps it is also worth noting that Physicists? solution to which you were replying follows his other postings in which the necessary constraints were made explicit.

The comment about backflow was not intended to apply before first stopping. It was just a note (and was intended to indicate that backflow has a greater equalising effect mixing alone would cause, because the gas that flows back into A has more energy (per mass) than the gas in B).

"So, hydrogen gas is viscous and uranium gas is loose?". Sorry - my statement was technically correct only. It omitted so much that it was effectively wrong. Under basic kinetic theory viscosity is proportional √(m.T)/A (m the molecular mass, T absolute temperature, A the collision cross-section).

As you seem to know, Hydrogen's molecule is large enough that is has lower viscocity than air, but Helium's viscosity is slightly greater than air (as are Argon, Xenon and Radon). If you know where Uranium vapour falls I'd love to hear.

To drop factors as inisgnificant, one should show using an edge of envelope case that the magnitude is demonstrably small, such as 'even if the pipe was a narrow 1/8 inch and a generous ten feet long, and the gas was . . . at 4000 PSI (or whatever the limit of commercially availbalbe vessels would be), the total work done to move the total work done to move the gas is X, which would only change the temperature Y.'

Estimating the probable magnitude of each effect and then calculating the approximate maximum magnitudes in decreasing probable magnitude might make it logical to drop effects that you think less in probable magnitude, but so many scientific and engineering projects have found effects that were discarded in theory to be significant in practice.

I look forward to your posting that defines the limits of accuracy. Shall we say 10% of the temperature change? (You may make of course use "no thermal interaction with the enclosure" as that is in the original challenge).

DGP

the clearest exposition so far; thanks

It is clear as to what we should all have seen at the beginning of this and then let us try and solve the problem in numerical terms.

Sleepy

GA recommended.

I have followed the logic of the discussion with interest but as we got nearer the 'correct' answer I find there are some gaps in my knowledge that I would like filled.

We are told that T_B will be 111°C when V_A=V_B and P_A=P_B.

I've always understood that if the volume doubled, the temperature would stabilise at 27^oC and the pressure would be halved. I accept that the challenge relates to conditions in the first few seconds as the valve is opened and this causes an unbalanced mixture with different temperatures.

Will you therefore please show the formula and the workings to explain how 111^oC has been achieved.

And what would be the associated temperature of vessel A at this particular time.

And subsequently, when VB is not known, what is T_A when T_B is a maximum at 147^oC

The first thing to do is get a clear mental picture of the problem.

Two Vessels, one pressurized with gas; the second containing a vacuum;

A valve between the two is opened and (in the general case) closed before the pressures have equalized between the two vessels. We then have to find the conditions in each vessel (after local and transient effects have ceased).

The key lies in finding the condition in A. Once this is known the condition in B is found from the conservation of mass.

A numerical calculation for the simplified case was posted in 107; the temperatures of both vessels for different amounts of gas in B was published in 117; the steps in solving were given in 138; and the importance of y (gamma) is stressed in 142.

It's all very basic stuff, which depends on work being done by the gas exiting A. The work is done against friction (in the pipe and in the viscosity of the gas ) and we don't have to know where it comes from, simply that is there. The case you quote is the isothermal case of a free expansion (1.e. one doing no work)

Hope this helps SLR

Where does the original problem say the valve is ever closed?

It doesn't. But it was clarified in subsequent postings. (It's almost impossible to cover every eventuality when the word-count is restricted). Also, if you don't allow the gas in B to equilibrate you can't get a unique solution.

If the tube diameter is sufficiently small and B is not too much larger than A (also not stated) there will be time for the gas in B to stabilise anyway.

What do you think is the objective of these challenges?

Last two sentences of the original challenge:-

It is easy to find the Temperature in B for a specific case. For example If and the valve is left open until then will be 111°C; But what if does not equal , and the valve is closed before equals ?

To SlideRuler

Thanks. I'll print the pages and study them tonight.

Cliff

"I've always understood that if the volume doubled, the temperature would stabilise at 27^oC"
That will indeed be the case if/after the gases in the two vessels are been allowed to mix. The way the problem was posed seems to me to imply that the valve is closed before any mixing takes place.

So we can consider that gas escaping from vessel A is driven by the pressure in vessel A. The pressure does work on the combination of the escaping gas and the gas in vessel B. Conversely, the gas in vessel A is doing work and so cools down.

To demonstrate the size of the effect, we can easily make direct calculations of the energy of the escaping gas when vessel B is empty: it is simply the original thermal energy plus the mechanical work done. If the volume (referred to its original pressure) that has escaped is V, the mechanical work done on it will be P_A.V. Once the valve is closed and the gas in B is allowed to settle this energy will be converted to thermal.
From the ideal gas laws:
P.V=1/3.M.v_A(rms)², and (for γ=1.4), and E_A(thermal)=(5/3).M.v_A(rms)²/2.
This leads to a total energy in the gas initially forced into B that is 7/5 times the thermal energy in vessel A, or a temperature of 420K, as given in SlideRuler's solution.

[We could in principle calculate the total energy in vessel B by integrating the imparted energy as the pressure and temperature in vessel A drop. But it is simpler to calculate the temperature in A and using conservation of energy to calculate the temperature in B.]

To Physicist and SlideRuler.

Thanks.

I'll print the pages and read them tonight.

Cliff.

The overall heat in the system cannot change (sentence of the preservation of the energy); if i close the valve in the same moment as i open it so that no gas can flow into vessel B, than gas in vessel A is unchanged with a temperature of 27°C. The Flow of the gas from vessel A to vessel B cools the flowing gas (equivalent to a defrosting spray in the winter what cools down by flowing into the lockmechanic and freezes it deeper). if i double the volume of vessel A than the absolute temperature (Klevin measured) should be half of. No energy leaves the system - also the equation p*V^n/T=const. is still right! This would be a mathematican problem to find the maximum value of the temperature by varying the volume; if the volume expands to infinity the resulting temperaqture will be zero! But with a look of the equation the maximum temperature should be the 27°C or 300K! The system cannot heated up by expanding it (the volume of the system)! This would create energy from nothing! And that is impossible! By all physical laws at the present!

Time to go back to kinetic theory basics:

The temperature of a gas depends only on the energy in each mode of vibration or motion. With gamma = 1.4 there are 5 modes of vibration for each molecule. The volume occupied does not of itself change either the energy of each molecule or (therefore) the temperature.

It is a common miscionception that temperature at constant energy inevitably reduces as a gas expands. This misconception doubtless arises because the most commonly analysed form of constant-energy expansion is when you have a box that is expanding slowly compared with gas velocities - and the reason for the cooling is the gas is expending its energy by doing work on the walls of the box.
But here there is no moving non-gas surface on which work is being done - so all the kinetic energy remains in the gas. If you let the gases mix after the expansion the temperature will be unchanged.
(In your terminology, having a total gas that is cooler after expansion without doing external work would be dissipating energy into nothing).

BTW, freezer spray is a false analogy - the spray is liquid inside the cannister; naturally the main cooling mechanism is latent heat.