## 2019年7月26日 (金)

### エントロピー　Entropy

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

The internal energy (E int) only depends on the temperature (T) but neither on either pressure nor volume.

E int = nCv T

On the P-V diagram, any gas on the same temperature line (the same red line) has the same internal energy.

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

ΔE int = Q - W,

Because ΔE int = 0, the entire heat (external energy: Q) given to the system goes into volume change (Work), not to the pressure.

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

0 = Q - W

T (temperature) = constant Each red line represents the gas status with the same temperature (T) as well as the same internal energy (E int).

When a certain amount of heat (external energy:Q) is applied to the system with no change in the temperature (T), the amount of heat (Q) is equal to the work (W) by which the volumeof the system did to the external environment.

There is no heat exchange (Q = 0), but the internal energy (E int) changes as well as the temperature (T) does.

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

ΔE int = Q - W

Because Q = 0, the work (W) is expressed by the following area under the curve and is purely a result of or resulting in the internal energy (E int).

ΔE int = 0 - W

S (entropy) = constant (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 (ΔE int).  H = E int + W. )

Another isothermal 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 = Q - W ] part and [ΔE int = 0 - W ] part.

net W = Q in - Q out This area (Q in - Q out) has the same size as the following area on the TS diagram. Compare the PV diagram and the TS diagram.

Isothermal curves. If you straighten the curves, you get this.  If you straighten the curves, you get this. On the TS diagram, both isothermal and adiabatic curves are straightened.  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 (heat: Q) 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 (Q) given to the system. And the area described above has the same size as the following area. So this is equal to the heat (Q) given. 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 (W) can be represented as the following yellow area under the PV curve.

However on this diagram, we can not see how much amount of heat (Q) is given or lost.   The net amount of heat (Q) absorbed to the system could be represented as the area under the PV curve of isothermal change, as the heat (Q) is equal to Work (W: area under the curve) in the isothermal change.

0 = Q - W 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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