[froggarana]

I recently bought "thermodynamics and statistical mechanics " by Seddon and Gale and am much perplexed by page 25, 4.1, (a), Isothermal expansion of an ideal gas. Two flasks having equal volumes are connected by a tap. Initially flask A is filled with  gas to pressure P, and flask B is evacuated. When the tap is opened , gas will flow from A to B until the pressures are equal. However ΔU=0 and ΔH=0 . . . . Note that heat flows in from the surroundings, because the gas in A performs work on the gas in B . . . .
Now the only chemistry qualification i have is "O" level which i took at night school nearly forty years ago but i do know a little bit of physics which leads me to believe that the free expansion of an ideal gas is both adiabatic and isothermal. You can see why i'm perplexed ?
In any event , this is not the book i want, can someone please recommend one that also gives the historical background of the development of chemical thermo and statistical mechanics ?   

[Enthalpy]

That story about expansion between bottles is typical from thermo books. Very much disconnected from reality.

They use to assume some sort of "equilibrium" at the end, but if you make the experiment you won't get any equilibrium between the bottles. The bottles get rather the temperature of the surroundings. Letting heat flow only between the bottles would be badly difficult.

And yes, the gas in A pushes on the gas that has already reached B, so in A it gets colder and in B hotter. In addition, the gas gets much speed that converts to heat in B. At one employer, we autoignited electronic equipment when 300bar nitrogen compressed air in a container. The equipment burned.

What thermo books would like to imagine is that, after "some" time, the heat in B flows to A so they reach an "equilibrium" - but perfect isolation from the surroundings please.

[froggarana]

yes but this is an ideal gas , not a real one.

if no work crosses the system boundary and heat enters then delta U can not be zero ?
so the book is wrong ?

[mjc123]

What is between your . ? Two scenarios are envisaged. First, flask B is said to be evacuated. So when the tap is opened, the gas in A expands into a vacuum, and does no work. So ΔT = 0, Q = 0, ΔU = 0, ΔH = 0. Later, it says "the gas in A performs work on the gas in B". So in this case, there is gas in B before opening the tap, and the expansion of A does work on B. If the system is to be isothermal, the heat must be supplied from the surroundings.

[Corribus]

First, flask B is said to be evacuated. So when the tap is opened, the gas in A expands into a vacuum, and does no work.

In an ideal scenario where the entire transfer is instantaneous. However in reality the transfer takes some finite amount of time. Would you agree that as chamber B fills, the remainder of the gas in A has to do work on the gas that has already transferred into B? So in fact the total work isn't 0.

I'd never thought of something like that before... most problems always assume the process is instantaneous.

[froggarana]

thank you Corribus, but,

let us say gas A does work on gas B, work is a form of energy, when the pressure has equalised where and in what form is that energy ? No work has crossed the system boundary, so I'd say it would be heat in the gas

How can the internal energy be constant when heat has crossed the boundary into the system, and no energy has left the system ?
Before i posted my question i checked half a dozen physics and engineering thermo text books and they all have it that the free expansion of an ideal gas is adiabatic and isothermal

[Enthalpy]

You can solve the problem with the available data, just like your book tells. It is indeed a matter of change in U, H and so on.

Whether work has been done from A on B in the meantime doesn't change the final state "at equilibrium", which implies that the gas is at the same temperature in A and B.

By the way, when writing "work on the gas in B", don't imagine some smooth process, with pistons and similar pressures on both sides, turbines... The strong pressure drop in the tube accelerates the gas and nearly nothing exploits this kinetic energy converted from ΔH, which is hence mainly converted to heat by turbulence. The gas in B only receives the work corresponding to its own pressure.

My point is that the book's hypotheses are unrealistic. By putting "equilibrium", the book supposes implicitly that the gas exchanges heat freely between the bottles but not with the surroundings, is this would be extremely hard to achieve. This is what I dislike in the textbook(s).

And that gas cools down in A and heats up in B, I know it experimentally. From 300bar it's brutal. After burning the contents by Diesel effect, there was no way to the book's equilibrium.

[Enthalpy]

Could you tell whether the book gives additional data, like "thermal equilibrium" as the final state?
Other conditions, like insulated bottles and states measured just after the expansion, would give a different state.

[froggarana]

enthalpy, thank you for sharing your real world experience of a real gas, you may find the wiki entry for Z factor interesting as it deals with the extent to which real gasses do, or don't deviate fr the behaviour of ideal gasses

[Enthalpy]

I have complete experimental curves about the Z factor. From NIST mainly, nice people.

Cooling of the source bottle during the expansion, before equilibrium, does not result from the Z factor. It's only because gas in the A bottle provides work to the gas in the tube. That is, the kinetic energy of the gas in the tube equates the change of enthalpy of this gas, but enthalpy does not fully reside there, as opposed to the internal energy. A part (the PV added to E) comes from the gas that surrounds the accelerated element, and here the gas still in A provides work and cools down while the gas already in B receives some work and gets hotter.

An extreme case would be a liquid. During the acceleration provided by a pressure gradient, its volume, temperature and internal energy are constant in a first approximation, and all the kinetic energy results from V×ΔP (put signs as you like) that is provided by the surroundings, nothing by the accelerated liquid.

Cooling or heating in both bottles after the gas reaches equilibrium (does the book want that?) is linked with the Z factor - but again, observing that would be badly difficult in a real experiment.

[froggarana]

i have now bought three physics text books and two engineering thermodynamics books, all agree the free expansion of an ideal gas is adiabatic and isothermal
what i don't understand is that no one here seems to understand the first law of thermoynamics

[mjc123]

If "free" means "unrestricted, unopposed by any pressure" then it implies expansion into an essentially infinite vacuum (or a very large volume where the pressure doesn't rise significantly). In your specified case, once some gas has entered flask B, the expansion of the rest is not "free" because it has to do work on the gas already in B. That's what confused me at first, but Corribus got the point. The scenario is not the idealised one tacitly assumed in the definition of "free expansion of an ideal gas".

[mjc123]

You can't explain what isn't true. There is no violation of the first law. No energy is created or destroyed, or unaccounted for. If part of the gas does work on another part, the work done by A equals the work done on B, so the net work for the whole system is zero. If gas A cools down and B heats up, the heat gained by B balances the heat lost by A. Once the system reaches equilibrium, ΔU, ΔH, ΔT and Q are all zero. If, as your book says, heat flows in from the surroundings (perhaps more likely than waiting for heat to flow from B to A through the tap), then the heat that flows into A to warm it is balanced by the heat that flows out from B to cool it (assuming the system equilibrium temperature is equal to that of the surroundings). The end result - the final equilibrium state - is the same as you would have if the expansion had been adiabatic and isothermal. But in practice the process is more complicated.

[Arkcon]

Hi guys, I've trimmed some of the bickering from this thread.  As it stands now, its not a bad one, and I hope we can all move forward.

froggarana:, I hate to be unwelcoming, but I want to go on record as saying, you arrived with an attitude, and I want to ask you to drop it.   You bought a textbook. Many people do that.  You have O-levels.  Many people have that.  That doesn't make anything about you, your thoughts, or your posts better, or worse, or better because they're worse, or better because they're harder won, or whatever.

Something may be poorly explained across the books and these posts.  We can work with our posts.  You wanna write a letter to the authors of the books, go ahead.  More than that, its out of our scope.

[froggarana]

first law analysis of a closed system is simple arithmetic
if the system contains a certain amount of energy, none is taken out , some is added , then the system now contains more energy
so if ΔU=0,  ΔH=0,  and  ΔW=0 then Q must be zero