Gibbs' paradox and Atmosphere as ideal gas

If you are a weather engineer, there is a wish.

Please read "3.Perpendicular distribution of H2O" before leaving here, if you think that this argument is boring.

"3.Perpendicular distribution of H2O" must be important for you.

It is said that the ingredient of air is as follows.

N2 78%
O2 21%
Ar and etc. 1%
Composition of air is said as follows.
It is applied to a height of 80 km and does not change.
It is as follows supposing there are 10000 particles of air.
N2 is 7800 pieces and, O2 is 2100 pieces, Ar and etc. are 100 pieces.
CO2 is only three pieces.
If atmospheric pressure sets to 1000 hPa, the partial pressure of each ingredient is as follows.
N2:780hPa
O2:210hPa

Ar & etc.:10hPa

Meteorology of Japan says that the small particles of an atomic weight are light.
Furthermore, meteorology of Japan said as follows.
A light gas float and heavy gas are depressed.
You may also agree.  
However, this idea infringes the 2nd law.
The 2nd law said as follows.
Ideal gas does not dissociate freely.
He has not seen the following thing.

Nitrogen and oxygen dissociate automatically.

0. The reason the gas of a small atomic weight is light.
There are three balloons before you.
You stuffed Oxygen into the red balloon and Nitrogen into the blue balloon and air into the pink balloon.

The number of the stuffed particles is the same.
Your answer was as follows.
A Red balloon sinks.
A Blue balloon goes up.

A Pink balloon does not sink and does not go up, either.

Your Answer is required to be the same temperature, the same pressure, and the same volume.

Furthermore, the following two are needed.

It is that only nitrogen is collected.

And it is that nitrogen is intercepted outside with the film of a balloon.

In a pink balloon, the second law does not permit the nitrogen and oxygen are separated automatically.

However, a phenomenon which is dissociation in air happen.

A warm front and the cold front are the example.

2nd law requireｓ explanation.

Air is separated why? 

A weather phenomenon is a very strange phenomenon physically.

1. Gibbs' paradox

A small box is in the room and each is made of thermal insulation. A small box is in the room and each is made of thermal insulation. 

We consider that Nitrogen and Helium are ideal gas.

In the room and a box, the mixed gas of Nitrogen and Helium is contained.

We set up as follows.
1. The temperature and pressure of the room are the same as a box.
2. However, the partial pressure of Nitrogen and Helium is different.
How do you think it becomes, if a lid is opened?
The partial pressure of Nitrogen and Helium in the room and a box becomes the same.
Although it was the same temperature and the same pressure, change occurred.
The 2nd law will say as follows.
He says."When the lid of the box was opened, entropy increased."
This is called Gibbs' paradox.
For example, Nitrogen is filled in one of the boxes of the same size, and one side is made into a vacuum.

真空＝Vacuum

If a hole is made in the wall which separates Nitrogen and a vacuum, free expansion of Nitrogen  will be carried out, and entropy　increases.
 
Temperature is proportional to the mean kinetic energy of particles.
Since kinetic energy does not change, temperature does not change.
Since volume doubled, pressure will be a half.
Entropy is S and presupposes that ΔS increases.
You may have the capability to do troublesome calculation of ΔS.
However, it is not necessary to calculate ΔS here.

It is the same even if it transposes Nitrogen to Helium.
Entropy is S and an increase is ΔS.

Gibbs' paradox said.
When a hole is made in the wall of two boxes containing Nitrogen and helium, entropy increases.
And the increase in entropy is 2ΔS.

The increase in the entropy of C is considered as addition of A and B.
(It will become clear if you study a little statistical mechanics.)

Let's return to the businessman's room.

2.A partial pressure ratio is the same anywhere. 
Suppose that there is a huge room made with thermal insulation.
The room is filled with the mixed gas of Nitrogen and Helium.
If this gas has reached thermodynamic equilibrium in the gravitational field.
This is the same as the next.
This gas has the specific entropy of the same value anywhere in a gravitational field.
Mixed gas is stuffed into a balloon in a height of 700 hPa.

It is not changing entropy although a balloon is compressed.
The energy to compress is potential energy.
As Dry Adiabat Lapse Rate  have discussed, we have two relations at this gas.

A:CpT+mgZ=Const
B: S(Z) =Const    *S (Z) is specific entropy.

A is a principle of conservation of energy.
B expresses a thermodynamic equilibrium.

This means that specific entropy is the same anywhere.
Enthalpy becomes large as a result of compressing by potential energy.

This means that temperature becomes high.

When it gets down to a height of 850 hPa, inner partial pressures and temperature are the same as outside.
Because, if partial pressures or temperature differed from outside, there was no relation of B.

It means that mixed gas had not been a thermal equilibrium situation.

A partial pressure ratio is the same at any height.

Pressure of Z= 0 is set to P (0), and partial pressure of Nitrogen is set to PN (0),He is set to PH(0). They are as follows.
P(0)=PN(0)+PH(0)
Moreover, the pressure P (Z) of arbitrary height is as follows.
We argued by Dry Adiabat Lapse Rate.
Pressure was as follows.

P（Z）＝P（0）EXP－∫（ｍｇ／RT（Z´））ｄZ´　（2－1）

Integration range is Z from 0.
A partial pressure ratio is not based on height, and is constant.
It has the following relations.

P_N（Z）／P（Z）＝P_N（0）／P（0）

P_N（Z）＝P（Z）×P_N（0）／P（0）

　　　　＝P（0）EXP－∫（ｍｇ／RT（Z´））ｄZ´×P_N（0）／P（0）

PN（Z）＝PN（0）EXP－∫（ｍｇ／RT（Z´））ｄZ´　（2－2a）
PH (Z) is as follows similarly.
PH（Z）＝PH（0）EXP－∫（ｍｇ／RT（Z´））ｄZ´  （2－2b）
The physical meanings of (2-2a) and (2-2b) are the following.
The mixed gas of an ideal gas is not separated freely.
That the partial pressure ratio of actual air does not change anywhere is the fact known well.
However, usually it is suggested that (2-2a) and (2-2b) are not realized with a Japanese textbook.
It is suggested by the following explanation.
You raise a balloon in the atmosphere of a helium.
Height of a balloon is HH when pressure of a helium became P/2. 
Moreover, You raise a balloon in the atmosphere of a  nitrogen.
Height of a balloon is Ho when pressure of a nitrogen became P/2. 
At this time, it is as follows.
HH>Ho

A textbook said."A helium floats for this relation."

However, with this conclusion, a partial pressure ratio does not become constant.

Furthermore, the textbook said.
The thing with a constant partial pressure ratio is the following cause.
Air has the force to stir.

・・・・・・？
I criticize a textbook as follows.
You do not consider even the hydrostatic pressure balance of mixed gas.
Hydrogen of mixed gas is subject to the pressure of oxygen and hydrogen.
Hydrogen of mixed gas is subject to the influence of oxygen and hydrogen.
What is the force to stir?
I have been told that the vortex is force to stir.
If air does not have a vortex, he believes truly that Gas is separated.
However, it is the same as the following thing.
If there is no vortex, entropy will decrease.
In this way, he believes radiative cooling and greenhouse effect.
However, he has mistaken.

3. Perpendicular Distribution of H2O

The following graph is a potential temperature emagram of September 8, 2007 00Z Tateno.

 Tateno is an upper-air observation point closest to Tokyo.

00Z will be at 09 am in Japan.

Surface of the earth is already heated with solar radiation, and Potential temperature is the same value up to 750m from the ground.

How the wetness of 750m height do you think?

I surmise that clouds are appearing here from experience.

I explain this moisture as follows.

I judge that specific entropy became the same to 750 m by solar radiation.

If specific entropy is the same, a partial pressure ratio is the same.

Usually, the partial pressure ratio of vapor becomes small with height.

It has the following meanings.

Water vapor was transported to a height of 750m.

The partial pressure of this vapor is as follows by (2-2a).

e（Z）＝e（0）EXP－∫（ｍｇ／RT（Z´））ｄZ´   (3-1)

e(0)/P(0)=e(z)/P(z)=Constant           (3-2)

P is atmospheric pressure and e is water vapor pressure.

Although e and P are reverse, the following graph shows P/e.

There seems to be some regularity to have a partial pressure of water vapor.

If you would like to draw a potential temperature emagram by Excel, please look for "Weather Function Excel VBA". 

Ozenuma’s morning

＊＊＊＊

This is only story.

Much difference from the fact is included.

Short Story　Nuclear power plant No. 1

Earthquake occurrence 14:46 on March 11, 2011.

No. 1 carried out meltdown and melt-through suddenly.

It is the next that it can imagine from the upper figure.

A and D are the pressure of a reactor pressure vessel.

D is 800kPa.

C and F are the pressure of a containment vessel.

F is 800kPa.

They will be D and F at 2:00 on March 12.

I imagine like next.

By 2:00 on March 12, meltdown broke out and the hole opened in the reactor pressure vessel.

It imagines as follows about the circumstances.

You can understand the situation of an accident, if safety valves did not open.

In an interim report, there was no description which operated the safety valve.

When safety valves do not act, the pressure of a pressure vessel goes up in short time.

The isolation condenser for cooling was not helpful.

Possibly it broke.

An isolation condenser of Japan often causes an accident.

＊＊＊＊＊＊＊＊＊＊＊＊

I think that a possibility of having broken is high.

Explanation of an interim report did not have persuasive power.

It said.

”By the sign whose direct-current power supply was lost, the valve closed automatically.”  

However, the closed valve acts by a direct-current power supply.

＊＊＊＊＊＊＊＊＊＊＊＊

The nuclear fuel lost water and melted.

Melt-through occurred by 2:00 on March 12.

The container became breaking.

A fire engine needs to supply water for cooling.

However, in 800kPa, water was not able to be supplied by fire engine.

The "vent" needed and the pressure of the container needed to be decompressed.

However, there was no compressed air for opening a vent valve.

They looked for the compressor.

And the "vent" was acted by the compressor.

Pressure fell and water supply became possible by fire truck.

The upper figure shows the next.

The pressure of the container is not 1 atmosphere =100kPa.

A "vent" is an act which makes pressure of a container same as atmospheric pressure.

Opening a valve was not able to be continued by a compressor.

They hid the problem of compressed air.

In an earthquake, they do not want to leave a broken proof.

They want it as follows.

An accident is not for the earthquake which is for tsunami.

The safety measures of compressed air are not performed.