This is the image which I have.
When I was a freshman at college, and studied entropy.
However, I was not able to understand.
Words did not become a question easily.
I think that there are people with the same question.
And I even forget having had a question.
The next is a clear thing.
I do not understand entropy.
The internal energy of n mol ideal gas is expressed as follows.
U=nCvT
Experiment person put n mole ideal gas into the box.
He took care that the size of a box did not change.
He found that energy required to warm 1 ℃ was nCv.
This is known as experimental
fact.
The
theorist explained as follows.
It will be set to nCvT if n mole
kinetic energy is all added.
Temperature was set to 0k when kinetic energy was 0.
This is the absolute temperature.
Internal energy is likely to have
such a meaning.
Energy is not given by the wall although particles were perfectly elastic collision.
A box is changed into a piston and compressed.
Particles said.
"Piston’s side is approaching“
Particles get energy from the
piston’s side.
nCvdT =-PdV (1-1)
dV is dV = Sdx.
S is the piston’s area.
dx is the distance.
I think P does not change even if you press only dx.
(1-1) is perhaps the following meanings.
When you press the piston only dx, the temperature was raised only dT.
I think so・・・.
I expressed as follows: U internal energy of an ideal gas is increased only dU.
dU = nCvdT =-PdV
I may not know.
What is the necessity to think of P does not change?
Cv is a proportionality constant that does not change the volume.
dV is small, the volume is changed.
I would be using the Cv?
Do not know if I will always
think of them.
However, I decided to be convinced that it is right.
(1-1). may be called quasistatic process.
In Japan, it is a freshman at
college and learns.
That
I learned in first grade college is not always easy.I was explained as follows.
Quasistatic process takes infinite time.
I thought as follows.
May I be based on such a thing?
Moreover, there are also the following talks.
Quasistatic process does not need time so much.
2．Entropy was not changing.
ｎCvｄT＝－PｄV （１－１）
This
is the energy conservation law (1-1).We have asked the definition of potential temperature equation from this equation.
Japanese meteorologist said.
"A potential temperature is defined from the law of conservation of energy nCvdT =－PdV"
I feel this lacking something.
Consider
the ideal gas (T2, P2, V2) and (T1, P1, V1).Two gases are very similar.
I want the difference between the two gas was like following.
nCv (T1－T2) =－ (P1－P2) (V1－V2)
If this equation holds, the
entropy of (T2, P2, V2) is the same as that (T1, P1, V1).
Someone
said I think it reached the definition of entropy.“On quasistatic process, Entropy does not change.”
(It may be my mistake.)
If nCvdT =－PdV,
entropy is not changing.
As
good as saying that entropy is nCvdT =－PdV..
We have described the entropy in
entropy.
However, nCvdT =-PdV What is the special law of conservation of energy.
Even if it heats the inside of a box with a heater, a principle of conservation of energy is realized (δQ).
It is realized even if it turns a fan in a box (δW).
The principle of conservation of
energy is fulfilled also by the following formula.
dU = nCvdT = δQ + δW
3. Dry Aiabatic lapse rate.
We find the dry adiabatic lapse rate.
Please see the Dry Adiabatic Lapse Rate for more information.
If premised on Hydrostatic
equilibrium, there is the following relation.
dP (Z) / P (Z) = －
(mg / RT (Z)) dz (3-1)
I
do the following with ease.dP / P = - (mg / RT) dz
Consider the ideal gas of n mole gravitational field.
The state equation is as follows.
P (Z) V (Z) = nRT (Z)
It is as follows if this is differentiated from Z.
dP / P + dV / V = dT / T
The left hand side of (3-1) is now modified as follows.
dP / P = dT / T － dV / V
= (dT / T)-PdV / PV
= dT / T-PdV / nRT
Here, We use the (1-1) nCvdT =－PdV
.
ｄP/P ＝（ｄT/T）＋CvｄT/（RT） (3－2)
(3-1) was the following.
ｄP/P ＝－（ｍｇ/RT）ｄｚ (3－1)
It is as follows from (3-1) and
(3-2).
（ｄT/T）＋CvｄT/（RT）＝－（ｍｇ/RT）ｄｚ
We multiplied by RT to both
sides.
RｄT＋CvｄT＝－ｍｇｄZ (3－3)
It
is Cp = R + Cv from Cp-Cv = R equation of Meyer.
(3－3) is as follows.
CpｄT＝－ｍｇｄZ
We divide both sides by Cpdz.
ｄT/ｄZ＝－ｍｇ/Cｐ (3－4)
We will be integral as follows.
CpT＋ｍｇZ＝Const (3－5)
Let's define a θ as follows.
Cpθ＝CpT＋ｍｇZ＝Const (3－6)
θ is the value does not change by
the Z height.
I asked for the definition of potential temperature equation from the equation of Hydrostatic equilibrium
Hydrostatic equilibrium was that of the following.
dP (Z) / P (Z) = － (mg / RT (Z)) dz (3-1)
We integrate both sides.
The integration range of the left
side is P (Z) from P0.
∫ｄP/P＝lnP(Z)－lnP0
From
(3-4), the right hand side is the Cp / RdT / T.
The integration range is from T0
to T (Z).
∫（Cp/RT(Z)）ｄｚ＝（Cp/R）（lnT(Z)－lnT0)）
Because it is left-hand side =
right side, it looks like this.
lnP(Z)－lnP0 ＝（Cp/R）（lnT(Z)－lnT0)）
lnP(Z)－（Cp/R）lnT(Z)＝lnP0－（Cp/R）lnT0 (4－2)
Even if Z is any value, lnP(Z)－（Cp/R）lnT(Z) is the same value as
lnP0－（Cp/R）lnT0 which is an initial
value.
lnP(Z)－（Cp/R）lnT(Z)＝lnP0－（Cp/R）lnT0＝Const (4－3)
An amount that does not change is
the entropy of the (4－3)
.(4－3) is as follows.
ln（P(Z)/P0）＝（Cｐ/R）ln（T(Z)/T0） (4－4)
P0 is set to 1000 and T0 is set
to θ.
ln（P(Z)/1000）＝（Cｐ/R）ln（T(Z)/θ） (4－5)
This definitional equation has a serious problem.
Please confirm the serious
problem by Definition
of potential temperature.
5．Enthalpy
nCvdT
=-PdV (1-1)I have studied (1-1) in first grade college.
I do not know what that is (1-1) today.
However, it was found that the energy conservation law is entropy does not change (1-1).
The ideal gas, there is a law of conservation of energy that does not change entropy another.
It is the law of conservation of
energy using the enthalpy.
It is as follows if the equation
of state of ideal gas is differentiated from Z.
VdP＋PdV＝nRdT
－PｄV＝VｄP－nRdT
(3－1) is as follows.
nCvdT=VｄP－nRdT
nCvdT+nRdT=VdP
Mayer's expression of relations R=Cp-Cv is
used.
nCpdT=VdP (5－1)
They are the Ideal gas law, Mayer's equation, and a principle of conservation of energy (1-1) which does not change entropy.
(5－1) can be said to be the principle of conservation of energy which does not change entropy.
We can have an image which pushes a piston by (1-1).
However, we cannot have even an
image by (5-1).
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