User talk:Paul Wormer/scratchbook1: Difference between revisions

From Citizendium
Jump to navigation Jump to search
imported>Paul Wormer
imported>Paul Wormer
No edit summary
Line 1: Line 1:
==Rotations in <math>\mathbb{R}^3</math> ==
==Entropy==
Consider a real 3&times;3 matrix '''R''' with columns
Clausius was able to give a mathematical expression of the second law of thermodynamics. To that end he needed a totally new thermodynamic concept, one that had no mechanical analogy and that had no intuitive meaning like  temperature. He called the new thermodynamic property  [[entropy]] from the classical Greek έν + τροπη (tropè = change, en = at). Following in his footsteps entropy will be introduced in this section.
'''r'''<sub>1</sub>, '''r'''<sub>2</sub>,  '''r'''<sub>3</sub>,
 
i.e.,
The state of a [[thermodynamic system]] (a point in state space) is characterized by a number of variables, such as [[pressure]] ''p'', [[temperature]] ''T'', amount of substance ''n'', volume ''V'', etc.  Any thermodynamic parameter  can be seen as a function of an arbitrary independent set of other thermodynamic variables, hence the terms  "property", "parameter""variable" and "function" are used interchangeably. The number of ''independent'' thermodynamic variables of a system is equal to the number of energy contacts  of the system with its surroundings.
 
An example of a reversible (quasi-static) energy contact is offered by the prototype thermodynamical system, a gas-filled cylinder with piston. Such a cylinder can perform work on its surroundings,  
:<math>
:<math>
\mathbf{R} = \left(\mathbf{r}_1,\, \mathbf{r}_2,\, \mathbf{r}_3 \right)
DW = pdV, \quad dV > 0,  
</math>.
</math>  
The matrix '''R''' is ''orthogonal''  if
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. Work by expansion is a form of energy contact between the cylinder and its surroundings. This process can be reverted, the volume of the cylinder can be decreased, the gas is compressed and the surroundings perform work  ''DW'' = ''pdV''  ''on'' the cylinder. 
 
The small amount of work is indicated by ''D'', and not by ''d'', because ''DW'' is not necessarily a differential of a  function.  However, when we divide ''DW'' by ''p'' the quantity ''DW''/''p'' becomes obviously equal to the differential ''dV'' of the differentiable state function ''V''. State functions depend only on the actual values of the thermodynamic parameters (they are local), and ''not'' on the path along which the state was reached (the history of the state). Mathematically this means that integration from point 1 to point 2 along path I  in state space is equal to integration along a different path II,
:<math>
:<math>
  \mathbf{r}_i \cdot \mathbf{r}_j = \delta_{ij}, \quad i,j = 1,2,3 .
V_2 - V_1 = {\int\limits_1\limits^2}_{{\!\!}^{(I)}} dV
= {\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}
</math>
</math>
The matrix '''R''' is a ''proper rotation matrix'', if it is
The amount of work (divided by ''p'') performed along path I is equal to the amount of work (divided by ''p'')  along path II. This condition is necessary and sufficient that  ''DW''/''p'' is a differentiable state function. So, although ''DW'' is not a differential, the quotient ''DW''/''p'' is one.
orthogonal ''and'' if '''r'''<sub>1</sub>,  '''r'''<sub>2</sub>,
 
'''r'''<sub>3</sub> form a right-handed set, i.e.,
Reversible absorption of a small amount of heat ''DQ'' is another energy contact of a system with its surroundings;  ''DQ'' is again not a differential of a certain function.  In a completely analogous manner to ''DW''/''p'',  the following result  can be shown for the heat ''DQ'' (divided by ''T'')  absorbed by the system along two different paths (along both paths the absorption is reversible):
:<math>
 
\mathbf{r}_i \times \mathbf{r}_j = \sum_{k=1}^3 \, \varepsilon_{ijk}
<div style="text-align: right;" >  
\mathbf{r}_k .
<div style="float: left; margin-left: 35px;" >
<math>{\int\limits_1\limits^2}_{{\!\!}^{(I)}}\frac{DQ}{T} = {\int\limits_1\limits^2}_{{\!\!}^{(II)}} \frac{DQ}{T} .
</math>
</math>
Here the symbol &times; indicates a
</div>
[[cross product]] and <math>\varepsilon_{ijk}</math> is the
<span id="(1)" style="margin-right: 200px; vertical-align: -40px; ">(1)</span>
antisymmetric [[Levi-Civita symbol]],
</div>
<br><br>
Hence the quantity ''dS'' defined by
:<math>
:<math>
\begin{align}
dS \;\stackrel{\mathrm{def}}{=}\; \frac{DQ}{T}
\varepsilon_{123} =&\; \varepsilon_{312} = \varepsilon_{231} = 1 \\
\varepsilon_{213} =&\; \varepsilon_{321} = \varepsilon_{132} = -1
\end{align}
</math>
</math>
and <math>\varepsilon_{ijk} = 0</math> if two or more indices are equal.
is the differential of a state variable ''S'', the ''entropy'' of the system. In a later subsection  equation (1) will be proved from the Clausius/Kelvin principle. Observe that this definition of entropy  only fixes entropy differences:
 
The matrix '''R''' is an ''improper rotation matrix'' if
its column vectors form a left-handed set, i.e.,
:<math>
:<math>
\mathbf{r}_i \times \mathbf{r}_j = - \sum_{k=1}^3 \, \varepsilon_{ijk}
S_2-S_1 \equiv \int_1^2 dS = \int_1^2 \frac{DQ}{T}  
\mathbf{r}_k \; .
</math>
</math>
The last two equations can be condensed into one equation
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'' &minus; ''DW''), it follows that
:<math>
:<math>
\mathbf{r}_i \times \mathbf{r}_j = \det(\mathbf{R}) \sum_{k=1}^3 \;
dU = TdS - pdV.\,
\varepsilon_{ijk} \mathbf{r}_k
</math>
</math>
by virtue of the the fact that
(For convenience sake  only a single work term was considered here, namely ''DW'' = ''pdV'', work done ''by'' the system).
the determinant of a proper rotation matrix is 1 and of an improper
The internal energy is an extensive quantity, that is, when the system is doubled, ''U'' is doubled too. The temperature ''T'' is an intensive property, independent of the size of the system. The entropy ''S'', then, is an extensive property. In that sense the entropy resembles the volume of the system.
rotation &minus;1. This can be proved as follows:
 
The determinant of a  3&times;3  matrix with column vectors  '''a''',
An important difference between ''V'' and ''S'' is that  the former is a state function with a well-defined mechanical  meaning, whereas entropy is introduced by analogy and is not easily visualized. Indeed, as is shown in the next subsection, it requires a fairly elaborate reasoning to prove that ''S'' is a state function, i.e., equation [[#(1)|(1)]] to hold.
'''b''',  and '''c''' can be written as [[scalar triple product#Triple product as determinant|scalar triple product]]
 
:<math>
 
\det\left(\mathbf{a},\,\mathbf{b},\, \mathbf{c}\right) =
 
\mathbf{a} \cdot (\mathbf{b}\times\mathbf{c})
 
</math>.
===Proof that entropy is a state function===
It was just shown that for a proper rotation
When equation [[#(1)|(1)]] has been proven, the entropy ''S''  is shown to be a state function. The standard proof, as given now, is physical, by means of [[Carnot cycle]]s, and is based on the Clausius/Kelvin formulation of the second law given in the introduction.
the columns of '''R''' are orthonormal and satisfy,
{{Image|Entropy.png|right|350px|Fig. 1.  ''T'' > ''T''<sub>0</sub>. (I): Carnot engine E moves  heat from  heat reservoir R to "condensor" C and needs input of work DW<sub>in</sub>. (II): E generates work DW<sub>out</sub> from the heat flow from C to R. }} An alternative, more mathematical proof, postulates the existence of a state variable ''S'' with certain properties and derives the existence of [[thermodynamical temperature]] and the second law from these properties.
:<math>
 
\mathbf{r}_1 \cdot (\mathbf{r}_2 \times \mathbf{r}_3 ) = \mathbf{r}_1 \cdot\left(\sum_{k=1}^3 \,
In figure 1 a finite heat bath C ("condensor")<ref>Because of a certain similarity of C with the condensor of a steam engine C is referred as "condensor". The quotes are used to remind us that nothing condenses, unlike the steam engine where steam condenses to water</ref> of constant volume and variable temperature ''T'' is shown. It is connected to an infinite heat reservoir R through a reversible Carnot engine E. Because R is infinite its temperature ''T''<sub>0</sub> is constant, addition or extraction of heat does not change ''T''<sub>0</sub>.  It is assumed that always ''T'' &ge; ''T''<sub>0</sub>.  One may think of the system E-plus-C as a ship and the heat reservoir R as the sea. The following argument then deals with an attempt of extracting energy from the sea in order to move the ship, i.e., with an attempt to let E perform net outgoing work in a cyclic (i.e., along a closed path in the state space of C) process.
\varepsilon_{23k} \,
 
  \mathbf{r}_k \right) = \varepsilon_{231} =  1 .
A Carnot engine performs reversible cycles (in the state space of E, not be confused with cycles in the state space of C) and per cycle either generates work ''DW''<sub>out</sub> when heat is transported from high temperature to low temperature (II), or needs work ''DW''<sub>in</sub> when heat is transported from low to high temperature (I), in accordance with the Clausius/Kelvin formulation of the second law.
</math>
Likewise the determinant is &minus;1 for an improper rotation.


====Theorem====
The definition of [[thermodynamical temperature]] (a positive quantity) is such that for  II,
A proper rotation matrix '''R''' can be
factorized thus
:<math>
:<math>
\mathbf{R} = \mathbf{R}_z (\omega_3 ) \; \mathbf{R}_y (\omega_2 ) \; \mathbf{R}_x (\omega_1 )
\frac{DW_\mathrm{out}}{DQ} = \frac{T-T_0}{T},
</math>
</math>
which is referred to as the ''Euler z-y-x parametrization'',
while for  I
or also as
:<math>
:<math>
\mathbf{R} = \mathbf{R}_z (\alpha) \; \mathbf{R}_y (\beta ) \; \mathbf{R}_z (\gamma ) \quad
\frac{DW_\mathrm{in}}{DQ_0} = \frac{T-T_0}{T_0}.
</math>
</math>
the ''Euler z-y-z parametrization''.


Here  the matrices representing rotations around the ''z'', ''y'', and ''x'' axis, respectively, over arbitrary angle &phi;, are
The first law of thermodynamics states for  I and II, respectively,
:<math>
:<math>
\mathbf{R}_z (\varphi ) \equiv
-DW_\mathrm{in} -DQ_0 + DQ=0\quad\hbox{and}\quad DW_\mathrm{out} + DQ_0-DQ=0
\begin{pmatrix}
\cos \varphi & -\sin \varphi & 0 \\
\sin \varphi &  \cos \varphi & 0 \\
          0 &  0          & 1 \\
\end{pmatrix}, \quad
\mathbf{R}_y (\varphi ) \equiv
\begin{pmatrix}
\cos \varphi & 0 & \sin \varphi \\
      0    & 1 &        0    \\
-\sin \varphi& 0 & \cos \varphi \\
\end{pmatrix}, \quad
\mathbf{R}_x (\varphi ) \equiv
\begin{pmatrix}
1 & 0 & 0 \\
0 & \cos \varphi & -\sin \varphi \\
0 & \sin \varphi & \cos \varphi \\
\end{pmatrix} .
</math>
</math>
====Proof====
{{Image|Cycle entropy.png|right|150px|Fig. 1. Two paths in the state space of the "condensor" C.}}
First the Euler ''z-y-x''-parametrization will be proved by describing an
For  I,
algorithm for the factorization of '''R'''.
Consider to that end the matrix product
:<math>
:<math>
\mathbf{R}_z (\omega_3 ) \, \mathbf{R}_y (\omega_2 ) =
\begin{align}
\begin{pmatrix}
\frac{DW_\mathrm{in}}{DQ_0} &= \frac{DQ- DQ_0}{DQ_0} = \frac{DQ}{DQ_0} -1 \\
\cos \omega_3 \cos \omega_2 & -\sin \omega_3  & \cos \omega_3 \sin \omega_2 \\
&=\frac{T-T_0}{T_0} = \frac{T}{T_0} - 1 \;
\sin \omega_3 \cos \omega_2 & \cos \omega_3 & \sin \omega_3 \sin \omega_2 \\
\Longrightarrow DQ_0 = T_0 \left(\frac{DQ}{T}\right)
            -\sin \omega_2 &              0 & \cos \omega_2 \\
\end{align}
\end{pmatrix} \equiv
          (\mathbf{a}_1 , \mathbf{a}_2 , \mathbf{a}_3 ) .
</math>
</math>
The columns of the matrix product are for ease of reference designated by '''a'''<sub>1</sub>,  '''a'''<sub>2</sub>, and '''a'''<sub>3</sub>. 
 
Note that the multiplication by
For II we find the same result,
'''R'''<sub>''x''</sub>(&omega;<sub>1</sub>) on the right
does not affect the first column, so that '''a'''<sub>1</sub> =
'''r'''<sub>1</sub> (the first column of the matrix to be factorized).
Solve <math>\omega_2\;</math> and <math>\omega_3\;</math> from the first column of
'''R''',
:<math>
\mathbf{a}_1 =
\begin{pmatrix}
  \cos \omega_3 \; \cos \omega_2 \\
  \sin \omega_3 \; \cos \omega_2 \\
                  -\sin \omega_2 \\
\end{pmatrix} =
\begin{pmatrix}
  R_{11} \\
  R_{21} \\
  R_{31} \\
\end{pmatrix} \equiv \mathbf{r}_1 .
</math>
This is possible. First solve <math>\omega_2\;</math> for <math> -\pi/2 \leq \omega_2
\leq \pi/2</math> from
:<math>
\sin \omega_2 = - R_{31}. \,
</math>
Then solve <math>\omega_3\;</math> for <math>0 \leq \omega_3 \leq 2 \pi</math> from the two equations:
:<math>
:<math>
\begin{align}
\begin{align}
\cos \omega_3 =& {R_{11} \over \cos \omega_2} \\
\frac{DW_\mathrm{out}}{DQ} &= \frac{DQ- DQ_0}{DQ} = 1- \frac{DQ_0}{DQ} \\
\sin \omega_3 =& {R_{21} \over \cos \omega_2} .
&=\frac{T-T_0}{T} =  1- \frac{T_0}{T}  
\;\Longrightarrow DQ_0 = T_0 \left(\frac{DQ}{T}\right)
\end{align}
\end{align}
</math>
</math>
Knowledge of <math>\omega_2\;</math> and <math>\omega_3\;</math>  determines the vectors '''a'''<sub>2</sub> and '''a'''<sub>3</sub>.
In figure 2 the state diagram of the "condensor" C is shown. Along path I the Carnot engine needs input of work to transport heat from the colder reservoir R to the hotter C and the absorption of heat by C raises its temperature and pressure. Integration of ''DW''<sub>in</sub> = ''DQ'' &minus; ''DQ''<sub>0</sub> (that is, summation over many cycles of the engine E) along path I gives
 
Since '''a'''<sub>1</sub>, '''a'''<sub>2</sub> and '''a'''<sub>3</sub> are the columns of a
proper rotation matrix  they form an orthonormal right-handed system. The plane spanned by '''a'''<sub>2</sub> and '''a'''<sub>3</sub> is orthogonal to <math> \mathbf{a}_1 \equiv \mathbf{r}_1</math> and hence the plane contains <math>\mathbf{r}_2</math> and
<math>\mathbf{r}_3</math>. Thus the latter two vectors are a linear combination of the first two,
:<math>
:<math>
( \mathbf{r}_2 , \mathbf{r}_3 ) = (\mathbf{a}_2 , \mathbf{a}_3 )
W_\mathrm{in} = Q_\mathrm{in} - T_0 {\int\limits_1\limits^2}_{{\!\!}^{(I)}} \frac{DQ}{T} \quad\hbox{with}\quad  Q_\mathrm{in} \equiv  {\int\limits_1\limits^2}_{{\!\!}^{(I)}} DQ.  
\begin{pmatrix}
    \cos \omega_1 & -\sin \omega_1 \\
    \sin \omega_1 & \cos \omega_1 \\
\end{pmatrix} .
</math>
</math>
Since <math>\mathbf{r}_2,\; \mathbf{a}_2,\; \mathbf{a}_3</math> are
Along path II the Carnot engine delivers work while transporting heat from C to R. Integration of ''DW''<sub>out</sub> = ''DQ'' &minus; ''DQ''<sub>0</sub> along path II gives
known unit vectors we can compute
:<math>
:<math>
\begin{align}
W_\mathrm{out} = Q_\mathrm{out} - T_0 {\int\limits_2\limits^1}_{{\!\!}^{(II)}} \frac{DQ}{T}
\mathbf{a}_2 \cdot \mathbf{r}_2 =& \cos \omega_1 \\
\quad\hbox{with}\quad  Q_\mathrm{out} \equiv  {\int\limits_2\limits^1}_{{\!\!}^{(II)}} DQ
\mathbf{a}_3 \cdot \mathbf{r}_2 =& \sin \omega_1.
\end{align}
</math>
</math>
These equations give <math>\omega_1\;</math> with <math> 0 \leq \omega_1 \leq 2 \pi</math>.


Augment the 2&times;2 matrix  to the 3&times;3 matrix <math>\mathbf{R}_x(\omega_1)</math>, then
Assume now that the  amount of heat ''Q''<sub>out</sub> extracted (along path II) from C and the heat ''Q''<sub>in</sub> delivered (along I) to C are the same in absolute value. In other words,  after having gone along a closed path in the state diagram of figure 2, the condensor C has not gained or lost heat. That is,
:<math>
:<math>
\begin{align}
Q_\mathrm{in} + Q_\mathrm{out} = 0, \,
\mathbf{R} \equiv ( \mathbf{r}_1 , \mathbf{r}_2 , \mathbf{r}_3 ) = (
\mathbf{r}_1 , \mathbf{a}_2 , \mathbf{a}_3 )
\mathbf{R}_x (\omega_1 )
= (\mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3)\mathbf{R}_x (\omega_1 )
= \mathbf{R}_z (\omega_3 ) \, \mathbf{R}_y (\omega_2 ) \, \mathbf{R}_x (\omega_1 ) .
\end{align}
</math>
</math>
This concludes the proof of the ''z-y-x'' parametrization.
then
 
The Euler ''z-y-z'' parametrization is obtained by a small modification of the previous proof. Solve
<math>\omega_2\;</math> and
<math>\omega_3\;</math> from <math>\mathbf{r}_3 = \mathbf{a}_3 </math> (the rightmost multiplication by '''R'''<sub>''z''</sub>(&omega;<sub>1</sub>) does not affect '''r'''<sub>3</sub>)
and then consider
:<math>
:<math>
( \mathbf{r}_1, \; \mathbf{r}_2 ) = (\mathbf{a}_1, \; \mathbf{a}_2 )
W_\mathrm{in} + W_\mathrm{out} = - T_0 {\int\limits_1\limits^2}_{{\!\!}^{(I)}} \frac{DQ}{T}  
\begin{pmatrix}
- T_0 {\int\limits_2\limits^1}_{{\!\!}^{(II)}} \frac{DQ}{T}.
\cos \omega_1 & -\sin \omega_1 \\
\sin \omega_1 & \cos \omega_1 \\
\end{pmatrix}
</math>
or,
<math>
\mathbf{a}_1 \cdot \mathbf{r}_1 = \cos \omega_1 \; , \quad \mathbf{a}_2 \cdot
\mathbf{r}_1 = \sin
\omega_1 .
</math>
</math>
The equation for '''R''' can be written as
If the total net work ''W''<sub>in</sub> + ''W''<sub>out</sub>  is positive (outgoing), this work is done by  heat obtained from R, which is not possible because of the Clausius/Kelvin principle.  If the total net work ''W''<sub>in</sub> + ''W''<sub>out</sub> is negative, then by inverting all reversible processes, i.e., by going down path I and going up along II, the net work changes sign and becomes positive (outgoing). Again the Clausius/Kelvin principle is violated. The conclusion is that the net work is zero and that
:<math>
:<math>
( \mathbf{r}_1 , \mathbf{r}_2 , \mathbf{r}_3 ) =
T_0 {\int\limits_1\limits^2}_{{\!\!}^{(I)}} \frac{DQ}{T} +
( \mathbf{a}_1 , \mathbf{a}_2 , \mathbf{r}_3 ) \, \mathbf{R}_z (\omega_1 ) = \mathbf{R}_z (\omega_3 ) \,
T_0 {\int\limits_2\limits^1}_{{\!\!}^{(II)}} \frac{DQ}{T} = 0
\mathbf{R}_y (\omega_2 ) \, \mathbf{R}_z (\omega_1 ) \; ,
\;\Longrightarrow\{\int\limits_1\limits^2}_{{\!\!}^{(I)}} \frac{DQ}{T} =  {\int\limits_1\limits^2}_{{\!\!}^{(II)}} \frac{DQ}{T}.
</math>
</math>
which proves the Euler ''z-y-z'' parametrization. It is common in this parametrization to write
From this independence of path it is concluded that
:<math>
:<math>
\omega_3 = \alpha,\quad \omega_2 = \beta, \quad \omega_1 = \gamma.
dS \equiv \frac{DQ}{T}
</math>
</math>
is a state (local) variable.

Revision as of 08:19, 30 October 2009

Entropy

Clausius was able to give a mathematical expression of the second law of thermodynamics. To that end he needed a totally new thermodynamic concept, one that had no mechanical analogy and that had no intuitive meaning like temperature. He called the new thermodynamic property entropy from the classical Greek έν + τροπη (tropè = change, en = at). Following in his footsteps entropy will be introduced in this section.

The state of a thermodynamic system (a point in state space) is characterized by a number of variables, such as pressure p, temperature T, amount of substance n, volume V, etc. Any thermodynamic parameter can be seen as a function of an arbitrary independent set of other thermodynamic variables, hence the terms "property", "parameter", "variable" and "function" are used interchangeably. The number of independent thermodynamic variables of a system is equal to the number of energy contacts of the system with its surroundings.

An example of a reversible (quasi-static) energy contact is offered by the prototype thermodynamical system, a gas-filled cylinder with piston. Such a cylinder can perform work 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. Work by expansion is a form of energy contact between the cylinder and its surroundings. This process can be reverted, the volume of the cylinder can be decreased, the gas is compressed and the surroundings perform work DW = pdV on the cylinder.

The small amount of work is indicated by D, and not by d, because DW is not necessarily a differential of a function. However, when we divide DW by p the quantity DW/p becomes obviously equal to the differential dV of the differentiable state function V. State functions depend only on the actual values of the thermodynamic parameters (they are local), and not on the path along which the state was reached (the history of the state). Mathematically this means that integration from point 1 to point 2 along path I in state space is equal to integration along a different path II,

The amount of work (divided by p) performed along path I is equal to the amount of work (divided by p) along path II. This condition is necessary and sufficient that DW/p is a differentiable state function. So, although DW is not a differential, the quotient DW/p is one.

Reversible absorption of a small amount of heat DQ is another energy contact of a system with its surroundings; DQ is again not a differential of a certain function. In a completely analogous manner to DW/p, the following result can be shown for the heat DQ (divided by T) absorbed by the system along two different paths (along both paths the absorption is reversible):

(1)



Hence the quantity dS defined by

is the differential of a state variable S, the entropy of the system. In a later subsection equation (1) will be proved from the Clausius/Kelvin principle. Observe that this definition of entropy only fixes entropy differences:

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 = DQDW), it follows that

(For convenience sake only a single work term was considered here, namely DW = pdV, work done by the system). The internal energy is an extensive quantity, that is, when the system is doubled, U is doubled too. The temperature T is an intensive property, independent of the size of the system. The entropy S, then, is an extensive property. 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 function with a well-defined mechanical meaning, whereas entropy is introduced by analogy and is not easily visualized. Indeed, as is shown in the next subsection, it requires a fairly elaborate reasoning to prove that S is a state function, i.e., equation (1) to hold.



Proof that entropy is a state function

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

PD Image
Fig. 1. T > T0. (I): Carnot engine E moves heat from heat reservoir R to "condensor" C and needs input of work DWin. (II): E generates work DWout from the heat flow from C to R.

An alternative, more mathematical proof, postulates the existence of a state variable S with certain properties and derives the existence of thermodynamical temperature and the second law from these properties.

In figure 1 a finite heat bath C ("condensor")[1] of constant volume and variable temperature T is shown. It is connected to an infinite heat reservoir R through a reversible Carnot engine E. Because R is infinite its temperature T0 is constant, addition or extraction of heat does not change T0. It is assumed that always TT0. One may think of the system E-plus-C as a ship and the heat reservoir R as the sea. The following argument then deals with an attempt of extracting energy from the sea in order to move the ship, i.e., with an attempt to let E perform net outgoing work in a cyclic (i.e., along a closed path in the state space of C) process.

A Carnot engine performs reversible cycles (in the state space of E, not be confused with cycles in the state space of C) and per cycle either generates work DWout when heat is transported from high temperature to low temperature (II), or needs work DWin when heat is transported from low to high temperature (I), in accordance with the Clausius/Kelvin formulation of the second law.

The definition of thermodynamical temperature (a positive quantity) is such that for II,

while for I

The first law of thermodynamics states for I and II, respectively,

PD Image
Fig. 1. Two paths in the state space of the "condensor" C.

For I,

For II we find the same result,

In figure 2 the state diagram of the "condensor" C is shown. Along path I the Carnot engine needs input of work to transport heat from the colder reservoir R to the hotter C and the absorption of heat by C raises its temperature and pressure. Integration of DWin = DQDQ0 (that is, summation over many cycles of the engine E) along path I gives

Along path II the Carnot engine delivers work while transporting heat from C to R. Integration of DWout = DQDQ0 along path II gives

Assume now that the amount of heat Qout extracted (along path II) from C and the heat Qin delivered (along I) to C are the same in absolute value. In other words, after having gone along a closed path in the state diagram of figure 2, the condensor C has not gained or lost heat. That is,

then

If the total net work Win + Wout is positive (outgoing), this work is done by heat obtained from R, which is not possible because of the Clausius/Kelvin principle. If the total net work Win + Wout is negative, then by inverting all reversible processes, i.e., by going down path I and going up along II, the net work changes sign and becomes positive (outgoing). Again the Clausius/Kelvin principle is violated. The conclusion is that the net work is zero and that

From this independence of path it is concluded that

is a state (local) variable.

  1. Because of a certain similarity of C with the condensor of a steam engine C is referred as "condensor". The quotes are used to remind us that nothing condenses, unlike the steam engine where steam condenses to water