### エントロピー Entropy

**The First Law of Thermodynamics**

When heat (Q) is applied, it (thermal energy) goes into the internal energy (U) and/or work (W).

ΔQ = ΔU + ΔW

ΔQ: Heat flows into a system

ΔW: Work done by the system

ΔU: Increased internal energy

The change in temperature (ΔT) is determined by the internal energy (ΔU).

ΔU = mcv ΔT

This equation is always true for any process.

m: mass

cv: the specific heat capacity of a gas at constant volume, cv = 3/2 R for mono-atomic gases, 5/2 R for di-atomic gases, 3 R for more complex gases

The change in temperature (ΔT) represents heat (ΔQ), and

ΔQ = mcv ΔT, for isovolumic process (volume is constant: ΔV =0), or

ΔQ = mcp ΔT, for isobaric process (pressure is constant: ΔP = 0).

m: mass

cv: the specific heat capacity of a gas at constant volume, cv = 3/2 R for mono-atomic gases, 5/2 R for di-atomic gases, 3 R for more complex gases

cp: the specific heat capacity of a gas at constant pressure, cp = 5/2 R for mono-atomic gases, 7/2 R for di-atomic gases, 4 R for more complex gases

R: universal gas constant (8314 J/kmol or 1.98 cal/mol)

Therefore, based on the first law,

mcv ΔT = ΔU, in isovolumetric process (ΔW = 0), and

mcp ΔT = ΔU + P ΔV = mcv ΔT + P ΔV, in isobaric process.

Solution technique.

1. Calculate ΔQ by using mc ΔT (unless the process is isothermal change)

2. Calculate ΔW

3. Claculate ΔU by ΔQ - ΔW

**Isothermal change**

In the ideal gas, when heat (ΔQ) is applied while the temperature (T) is kept constant (ΔT = 0), the pressure decreases and the volume expands.

Do not use

ΔQ = mcv ΔT

or

ΔQ = mcp ΔT

as it always ends up ΔQ = 0, when ΔT = 0. Wrong. Please note that the equation ΔQ = mcv ΔT or ΔQ = mcp ΔT is only for isovolumetric or isobaric change.

Instead, use

ΔU = mcv ΔT

This equation is true universally (not only for isovolumic change).

Therefore ΔU = 0, when ΔT = 0.

Any gas on the same temperature (T: the same red line) has the same internal energy (U) on the P-V diagram.

When there is no change in the temeperature (ΔT = 0), the energy within the system remains also constant (ΔU = 0).

Therefore, from the 1st law of thermodynamics,

ΔQ = ΔU + ΔW

ΔQ = 0 + ΔW

ΔQ = ΔW

When ΔU = 0, the entire heat given to the system (ΔQ: external energy) goes into volume change (ΔW: Work out), not to the pressure in the isothermal procss.

The work done (ΔW) is expressed by the following area under the curve and amounts purely to a result of or results in the external energy (ΔQ: the amount of heat applied).

In this process, T (temperature) = constant

The area, the work done (ΔW) from condition (P1, V1, T is constant) to (P2, V2, T is constant) is given

ΔW = ∫PdV

= ∫nRT(1/V)dV (∵PV = nRT)

= nRT∫(1/V)dV

= nRT ln(V2/V1)

= P1V1 ln(V2/V1)

Please note, in this equation P is not constant and is a function of V, however T is constant.

(When P is constant, just use ΔW = P ΔV)

Each red line represents the gas status with the same temperature (T) as well as the same internal energy (U).

When a certain amount of heat (ΔQ: external energy) is applied to the system with no change in the temperature (ΔT = 0), the amount of heat (ΔT = 0) is equal to the work (ΔW) by which the volume of the system did to the external environment.

**Adiabatic change (Isentropic change).**

There is no heat applied (ΔQ = 0), but the internal energy changes as well as the temperature does.

From the 1st law of thermodynamics,

ΔQ = ΔU + ΔW

0 = ΔU + ΔW

0 = mcvΔT + ΔW

The work done by this system (ΔW) is expressed by the following area under the curve and any work done by this system (ΔW) is done purely at the expense of the internal energy (ΔU) because ΔQ = 0.

In this process, S (entropy) = constant

In this system, the change in the condition (P1, V1, T1) to (P2, V2, T2) is given by

P_{1}V_{1}^{γ} = P_{2}V_{2}^{γ}

or

T_{1}V_{1}^{γ-1} = T_{2}V_{2}^{γ-1}

where γ is the specific gas ratio (1.67 for mono-atomic gases, 1.40 for di-atomic gases).

γ = cp/cv = R/M

cv: the specific heat capacity of a gas at constant volume, cv = 3/2 R for mono-atomic gases, 5/2 R for di-atomic gases, 3 R for more complex gases

cp: the specific heat capacity of a gas at constant pressure, cp = 5/2 R for mono-atomic gases, 7/2 R for di-atomic gases, 4 R for more complex gases

R: universal gas constant (8314 J/kmol or 1.98 cal/mol)

M: molecular mass of the gas

ΔQ = ΔU + ΔW

0 = ΔU + ΔW

0 = mcvΔT + ΔW

0 = mcvΔT + ΔW

0 = mcvΔT + ∫PdV

0 = mcvΔT + ∫(k/V^{γ})dV (∵PV^{γ} =k, in adiabatic change)

0 = mcvΔT + kcvV^{(1-cp/cv)}/(cv-cp)

0 = mcvΔT - kcvV^{(1-γ)}/R

0 = mΔT - kV^{(1-γ)}/R^{}

0 = mΔT - PV^{γ}V^{(1-γ)}/R

0 = mΔT - PV/R

PV = mRΔT

What remains constant in this sytem is called "Entropy".

(Enthalpy (H): On the PV diagram, Enthalpy (H) is the same as the heat (Q) in the isothermal change. Also Enthalpy (H) is the same as the internal energy change (ΔU). H = U + W. )

Another isothermal change.

Another adiabatic change.

The total energy exchange (net Work) is expressed in the following diagrasm.

The net energy exchanged (net Work).

And this called the Carnot cycle. This cycle is composed of [0 = ΔQ - ΔW ] part and [ΔU = 0 - ΔW ] part.

The net work (ΔW) = Q in - Q out

This area (Q in - Q out) has the same size as the following area on the TS diagram.

But, what is S?

S is Entropy.

In isothermal prosess,

ΔQ = T ΔS

Compare the PV diagram and the TS diagram.

Isothermal curves.

If you straighten the curves, you get this.

Adiabatic curves.

If you straighten the curves, you get this.

On the TS diagram, both isothermal and adiabatic curves are straightened, but perpendicularly.

And this is a 3D version of PV curve with Temperature as a height.

If you look down it from the top, it looks like this.

And the area (Q in - Q out) can be converted as follows:

The same logic can be applied to the following area.

Therefore, these area have logically the same value to each other.

Think of the following area.

This area expresses the external energy (ΔQ) applied to or extracted from the system while the temperature is kept constant.

This area has the same area as the following blue area, believe it or not. So this are is equal to the heat (ΔQ) given to the system.

And the area described above has the same size as the following area. So this is equal to the heat (ΔQ) given.

S is Entropy.

The area under the curve matches heat given into the system (Q).

In isothermal prosess, where T is constant,

ΔQ = T ΔS

In adiabatic process, where S is constant (ΔS = 0),

ΔQ (the area under the curve) = 0

The same logic can be applied to the following area.

## ||

These are the same theoretically.

Suppoese the status from A to B contains both isothermal ans well as adiabatic change.

The net work-out (ΔW) can be represented as the following yellow area under the PV curve.

ΔW = ∫PdV

However, it is very difficult to express P as a function of V.

It is almost impossible to descrive P as a function of V.

The net amount of heat (ΔQ) absorbed to the system could be represented as the area under the PV curve of isothermal change, as the heat (ΔQ) is equal to Work (W: area under the curve) in the isothermal change.

ΔQ = 0 + ΔW

But there are too much overlapping in the PV diagram. You can avoid this problem with TS diagram.

On TS diagram, there is no overlapping and you can see the exact area.

On TS diagram, the horizontal change is only accounted for the isothermal change.

Therefore, the yellow area under the curve on this diagram purely represents the net amount of heat (Q) absorbed.

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