User talk:Paul Wormer/scratchbook

The second law of thermodynamics, as formulated in the middle of the 19th century by William Thomson (Lord Kelvin) and Rudolf Clausius, states that it is impossible to gain mechanical energy by letting heat flow from a cold to a warm object. The law states, on the contrary, that mechanical energy (work) is needed to transport heat from a low- to a high-temperature heat bath.

If the second law would be invalid, there would be no energy crisis. For example, it would be possible—as already pointed out by Lord Kelvin—to fuel ships by energy extracted from sea water. After all, the oceans contain immense amounts of internal energy. If it would be possible to extract a small portion of this energy—whereby a slight cooling of the sea water would occur—and to use this energy to move the ship (a form of work), then the seas could be sailed without any net consumption of energy. It would not violate the first law of thermodynamics, because the the ship's rotating propellers do heat the water and in total the energy of the supersystem "ship-plus-ocean" would be conserved, in agreement with the first law. Unfortunately, it is not possible, because a ship is warmer than the sea water that it moves in (or at least the ship is not colder) and hence no work can be extracted from the water by the ship.

Entropy

Clausius was able to give a mathematical expression of the second law. In order to be able do that, he needed the concept of entropy. Following his footsteps entropy will be introduced in this subsection.

The state of a thermodynamical system is characterized by a number of (dependent) variables, such as pressure p, temperature T, amount of substance, volume V, etc. In general a system has a number of energy contacts with its surroundings. For instance, the prototype thermodynamical system, a gas-filled cylinder with piston, can perform work DW = pdV on its surroundings, where dV stands for a small increment of the volume V of the cylinder, p is the pressure inside the cylinder and DW stands for a small amount of work. This small amount is indicated by D, and not by d, because DW is not necessarily a differential of a function. However, when we divide by p the quantity DW/p becomes equal to the differential of the state function V. State functions are local, they dependent on the actual values of the parameters, and not on the path along which the state was reached. Mathematically this means that integration from point 1 to point 2 along path I is equal to integration along another path II

$V_{2}=V_{1}+{\int \limits _{1}\limits ^{2}}_{{\!\!}^{(I)}}dV=V_{1}+{\int \limits _{1}\limits ^{2}}_{{\!\!}^{(II)}}dV\;\Longrightarrow \;{\int \limits _{1}\limits ^{2}}_{{\!\!}^{(I)}}{\frac {DW}{p}}={\int \limits _{1}\limits ^{2}}_{{\!\!}^{(II)}}{\frac {DW}{p}}$

The amount of work (divided by p) performed along path I is equal to the amount of work (divided by p) along path II, which proves that the fraction DW/p is a state variable.

Absorption of a small amount of heat DQ is another energy contact of the system with its surroundings. In a completely analogous manner, the following result can be shown for DQ (divided by T) absorbed by the system along two different paths:

${\int \limits _{1}\limits ^{2}}_{{\!\!}^{(I)}}{\frac {DQ}{T}}={\int \limits _{1}\limits ^{2}}_{{\!\!}^{(II)}}{\frac {DQ}{T}}$

(1)

Hence the quantity dS defined by

$dS\;{\stackrel {\mathrm {def} }{=}}\;{\frac {DQ}{T}}$

is the differential of a state variable S, the entropy of the system. Before proving equation (1) from the second law, it is emphasized that this definition of entropy only fixes entropy differences:

$S_{2}-S_{1}\equiv \int _{1}^{2}dS=\int _{1}^{2}{\frac {DQ}{T}}$

Note further that entropy has the dimension energy per degree temperature (joule per degree kelvin) and recalling the first law of thermodynamics (the differential dU of the internal energy satisfies dU = DQ + DW), it follows that

$dU=TdS-pdV.\,$

(For convenience sake only a single work term was considered here, namely DW = pdV). The internal energy is an extensive quantity, that is, when the system is halved, U is halved too. The temperature T is an intensive property, independent of the size of the system. The entropy S, then, is also extensive. In that sense the entropy resembles the volume of the system. An important difference between V and S is that the former is a state variable with a concrete meaning, whereas the latter is introduced by analogy and therefore is not easily visualized.

Proof that entropy is a state variable

After equation (1) has been proven, the entropy S has been shown to be a state variable. The standard proof, as given now, is physical and by means of the construct of Carnot cycles and is derived from the Clausius/Kelvin formulation of the second law given in the introduction.

PD Image
Fig. 1 Carnot engine (E) moves to or from condensor (C) heat from or to infinite heat reservoir (R).

In figure 1 a two finite heat baths C ("condensors") of constant volume with variable temperature T are shown. They are connected to an infinite heat reservoir R through reversible Carnot engines E. Because R is infinite its temperature T₀ is constant, addition or extraction of heat does not change T₀. It is assumed that always T ≥ T₀. The Carnot engines cycle and per cycle either generate work DW_out when heat is transported from high temperature to low temperature (path II), or need work DW_in when heat is transported from low to high temperature (path I).

The definition of thermodynamical temperature is such that for path II,

${\frac {DW_{\mathrm {out} }}{DQ}}={\frac {T-T_{0}}{T}},$

while for path I it holds by definition that

${\frac {DW_{\mathrm {in} }}{DQ_{0}}}={\frac {T-T_{0}}{T_{0}}}.$

The first law of thermodynamics states that both I and II, respectively, it holds that

$DW_{\mathrm {in} }=DQ-DQ_{0}\quad {\hbox{and}}\quad DW_{\mathrm {out} }=DQ-DQ_{0}$

For I,

${\begin{aligned}{\frac {DW_{\mathrm {in} }}{DQ_{0}}}&={\frac {DQ-DQ_{0}}{DQ_{0}}}={\frac {DQ}{DQ_{0}}}-1\\&={\frac {T-T_{0}}{T_{0}}}={\frac {T}{T_{0}}}-1\\&\Longrightarrow DQ_{0}=T_{0}\left({\frac {DQ}{T}}\right)\end{aligned}}$

For II,

${\begin{aligned}{\frac {DW_{\mathrm {out} }}{DQ}}&={\frac {DQ-DQ_{0}}{DQ}}=1-{\frac {DQ_{0}}{DQ}}\\&={\frac {T-T_{0}}{T}}=1-{\frac {T_{0}}{T}}\\&\Longrightarrow DQ_{0}=T_{0}\left({\frac {DQ}{T}}\right)\end{aligned}}$