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# 2.6: The Entropy

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$$\newcommand\Vsigma ParseError: invalid DekiScript (click for details) Callstack: at (Template:MathJaxArovas), /content/body/div/p[1]/span[26], line 1, column 1 at template() at (Under_Construction/Arovas_Texts/Book:_Thermodynamics_and_Statistical_Mechanics_(Arovas)/02:_Thermodynamics/2.06:_The_Entropy), /content/body/p/span, line 1, column 23 $$
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$$\newcommand\Vtau ParseError: invalid DekiScript (click for details) Callstack: at (Template:MathJaxArovas), /content/body/div/p[1]/span[28], line 1, column 1 at template() at (Under_Construction/Arovas_Texts/Book:_Thermodynamics_and_Statistical_Mechanics_(Arovas)/02:_Thermodynamics/2.06:_The_Entropy), /content/body/p/span, line 1, column 23 $$
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$$\newcommand\Vphi ParseError: invalid DekiScript (click for details) Callstack: at (Template:MathJaxArovas), /content/body/div/p[1]/span[30], line 1, column 1 at template() at (Under_Construction/Arovas_Texts/Book:_Thermodynamics_and_Statistical_Mechanics_(Arovas)/02:_Thermodynamics/2.06:_The_Entropy), /content/body/p/span, line 1, column 23 $$
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## Entropy and heat

The Second Law guarantees us that an engine operating between two heat baths at temperatures $$T\ns_1$$ and $$T\ns_2$$ must satisfy ${Q\ns_1\over T\ns_1} + {Q\ns_2 \over T\ns_2} \le 0\ ,$ with the equality holding for reversible processes. This is a restatement of Equation \ref{qteqn}, after writing $$Q\ns_1=-\CQ\ns_1$$ for the heat transferred to the engine from reservoir #1. Consider now an arbitrary curve in the $$p-V$$ plane. We can describe such a curve, to arbitrary accuracy, as a combination of Carnot cycles, as shown in Fig. [mcarnot]. Each little Carnot cycle consists of two adiabats and two isotherms. We then conclude $\sum_i {Q\ns_i\over T\ns_i} \longrightarrow \oint\limits_\CC\!{\dbar Q\over T}\le 0\ ,$ with equality holding if all the cycles are reversible. Rudolf Clausius, in 1865, realized that one could then define a new state function, which he called the entropy, $$S$$, that depended only on the initial and final states of a reversible process: $dS={\dbar Q\over T} \quad\Longrightarrow\quad S\subB-S\subA=\int\limits_{\RA}^{\RB} \!{\dbar Q\over T}\ . \label{dseqn}$ Since $$Q$$ is extensive, so is $$S$$; the units of entropy are $$[S]=\RJ/\RK$$.

## The Third Law of Thermodynamics

Equation [dseqn] determines the entropy up to a constant. By choosing a standard state $$\RUps$$, we can define $$S\ns_\RUps=0$$, and then by taking $$\RA=\RUps$$ in the above equation, we can define the absolute entropy $$S$$ for any state. However, it turns out that this seemingly arbitrary constant $$S\ns_\RUps$$ in the entropy does have consequences, for example in the theory of gaseous equilibrium. The proper definition of entropy, from the point of view of statistical mechanics, will lead us to understand how the zero temperature entropy of a system is related to its quantum mechanical ground state degeneracy. Walther Nernst, in 1906, articulated a principle which is sometimes called the Third Law of Thermodynamics,

Again, this is not quite correct, and quantum mechanics tells us that $$S(T=0)=\kB\ln g$$, where $$g$$ is the ground state degeneracy. Nernst’s law holds when $$g=1$$.

We can combine the First and Second laws to write $dE + \dbar W=\dbar Q \le T\,dS\ ,$ where the equality holds for reversible processes.

## Entropy changes in cyclic processes

For a cyclic process, whether reversible or not, the change in entropy around a cycle is zero: $$\RDelta S\ns_\ssr{CYC}=0$$. This is because the entropy $$S$$ is a state function, with a unique value for every equilibrium state. A cyclical process returns to the same equilibrium state, hence $$S$$ must return as well to its corresponding value from the previous cycle.

Consider now a general engine, as in Fig. [engref]. Let us compute the total entropy change in the entire Universe over one cycle. We have $(\RDelta S)\ns_\ssr{TOTAL}=(\RDelta S)\ns_\ssr{ENGINE} + (\RDelta S)\ns_\ssr{HOT} +(\RDelta S)\ns_\ssr{COLD}\ ,$ written as a sum over entropy changes of the engine itself, the hot reservoir, and the cold reservoir10. Clearly $$(\RDelta S)\ns_\ssr{ENGINE}=0$$. The changes in the reservoir entropies are \begin{aligned} (\RDelta S)\ns_\ssr{HOT}&=\!\!\int\limits_{T=T\ns_2}\!\!\!{\dbar Q_\ssr{HOT}\over T} = -{Q\ns_2\over T\ns_2}\ <\ 0\\ (\RDelta S)\ns_\ssr{COLD}&=\!\!\int\limits_{T=T\ns_1}\!\!\!{\dbar Q_\ssr{COLD}\over T} ={\CQ\ns_1\over T\ns_1}=-{Q\ns_1\over T\ns_1}\ >\ 0\ ,\end{aligned} because the hot reservoir loses heat $$Q\ns_2>0$$ to the engine, and the cold reservoir gains heat $$\CQ\ns_1=-Q\ns_1>0$$ from the engine. Therefore, $(\RDelta S)\ns_\ssr{TOTAL}=-\bigg({Q\ns_1\over T\ns_1} + {Q\ns_2\over T\ns_2}\bigg) \ge 0\ .$ Thus, for a reversible cycle, the net change in the total entropy of the engine plus reservoirs is zero. For an irreversible cycle, there is an increase in total entropy, due to spontaneous processes.

## Gibbs-Duhem relation

Recall Equation [dwork]: $\dbar W=-\sum_j y\ns_j\,dX\ns_j -\sum_a \mu\ns_a\,dN\ns_a\ .$ For reversible systems, we can therefore write $dE=T\,dS + \sum_j y\ns_j\,dX\ns_j +\sum_a \mu\ns_a\,dN\ns_a\ . \label{dErev}$ This says that the energy $$E$$ is a function of the entropy $$S$$, the generalized displacements $$\{X\ns_j\}$$, and the particle numbers $$\{N\ns_a\}$$: $E=E\big(S,\{X\ns_j\},\{N\ns_a\}\big)\ .$ Furthermore, we have $T=\pabc{E}{S}{\{X\ns_j,N\ns_a\}}\quad,\quad y\ns_j=\pabc{E}{X\ns_j}{S,\{X\ns_{i(\ne j)},N\ns_a\}}\quad,\quad \mu\ns_a=\pabc{E}{N\ns_a}{S,\{X\ns_j,N\ns_{b(\ne a)}\}}$ Since $$E$$ and all its arguments are extensive, we have $\lambda E=E\big(\lambda S,\{\lambda X\ns_j\},\{\lambda N\ns_a\}\big)\ .$ We now differentiate the LHS and RHS above with respect to $$\lambda$$, setting $$\lambda=1$$ afterward. The result is $\begin{split} E&=S\,{\pz E\over \pz S} + \sum_j X\ns_j\,{\pz E\over\pz X\ns_j} + \sum_a N\ns_a\, {\pz E\over\pz N\ns_a}\\ &=TS+\sum_j y\ns_j\,X\ns_j + \sum_a \mu\ns_a\,N\ns_a\ .\label{ETS} \end{split}$ Mathematically astute readers will recognize this result as an example of Euler’s theorem for homogeneous functions. Taking the differential of Equation [ETS], and then subtracting Equation [dErev], we obtain $S\,dT+\sum_j X\ns_j\,dy\ns_j + \sum_a N\ns_a\,d\mu\ns_a=0\ . \label{GDR}$ This is called the Gibbs-Duhem relation. It says that there is one equation of state which may be written in terms of all the intensive quantities alone. For example, for a single component system, we must have $$p=p(T,\mu)$$, which follows from $S\,dT -V\,dp + N\,d\mu=0\ .\label{GDRa}$

## Entropy for an ideal gas

For an ideal gas, we have $$E=\half f N\kT$$, and $\begin{split} dS&={1\over T}\,dE + {p\over T}\,dV - {\mu\over T}\,dN\\ &=\half f N\kB\,{dT\over T} + {p\over T}\,dV +\bigg(\half f\kB- {\mu\over T}\bigg)\,dN\ . \end{split}$ Invoking the ideal gas equation of state $$pV=N\kT$$, we have $dS\big|\nd_N=\half f N\kB \,d\ln T + N\kB d\ln V\ .$ Integrating, we obtain $S(T,V,N)=\half f N\kB \,\ln T + N\kB\,\ln V + \varphi(N)\ ,$ where $$\varphi(N)$$ is an arbitrary function. Extensivity of $$S$$ places restrictions on $$\varphi(N)$$, so that the most general case is $S(T,V,N)=\half f N\kB \,\ln T + N\kB\,\ln \bigg({V\over N}\bigg) +N a\ ,$ where $$a$$ is a constant. Equivalently, we could write $S(E,V,N)=\half f N\kB \,\ln\!\bigg({E\over N}\bigg)+ N\kB\,\ln \bigg({V\over N}\bigg) +N b\ , \label{SEVN}$ where $$b=a-\half f\kB\ln(\half f\kB)$$ is another constant. When we study statistical mechanics, we will find that for the monatomic ideal gas the entropy is $S(T,V,N)=N\kB\Bigg[\frac{5}{2} + \ln\!\bigg({V\over N\lambda_T^3}\bigg)\Bigg]\ ,$ where $$\lambda\ns_T=\sqrt{2\pi\hbar^2/m\kT}$$ is the thermal wavelength, which involved Planck’s constant. Let’s now contrast two illustrative cases.

• Adiabatic free expansion – Suppose the volume freely expands from $$V\ns_\Ri$$ to $$V\ns_\Rf=r\,V\ns_\Ri$$, with $$r>1$$. Such an expansion can be effected by a removal of a partition between two chambers that are otherwise thermally insulated (see Fig. [AFE]). We have already seen how this process entails $\RDelta E = Q = W = 0\ .$ But the entropy changes! According to Equation [SEVN], we have $\RDelta S = S\ns_\Rf-S\ns_\Ri=N\kB\ln r\ . \label{AFEdS}$
• Reversible adiabatic expansion – If the gas expands quasistatically and reversibly, then $$S=S(E,V,N)$$ holds everywhere along the thermodynamic path. We then have, assuming $$dN=0$$, $\begin{split} 0=dS &=\half f N\kB\,{dE\over E} + N\kB\,{dV\over V}\\ &=N\kB \,d\ln\big(V E^{f/2}\big)\ . \end{split}$ Integrating, we find ${E\over E\ns_0}= \bigg({V\ns_0\over V}\bigg)^{\!2/f}\ .$ Thus, $E\ns_\Rf=r^{-2/f}\,E\ns_\Ri\quad\Longleftrightarrow\quad T\ns_\Rf=r^{-2/f}\,T\ns_\Ri\ .$

## Example system

Consider a model thermodynamic system for which $E(S,V,N)={aS^3\over NV}\ ,$ where $$a$$ is a constant. We have $dE=T\,dS-p\,dV + \mu\,dN\ ,$ and therefore \begin{aligned} T&=\pabc{E}{S}{V,N}={3aS^2\over NV}\label{EXA}\\ p&=-\pabc{E}{V}{S,N}={aS^3\over NV^2}\bvph\label{EXB}\\ \mu&=\pabc{E}{N}{S,V}=-{aS^3\over N^2 V}\ .\label{EXC}\end{aligned} Choosing any two of these equations, we can eliminate $$S$$, which is inconvenient for experimental purposes. This yields three equations of state, ${T^3\over p^2} = 27a\,{V\over N}\qquad,\qquad {T^3\over\mu^2}=27a\,{N\over V}\qquad,\qquad {p\over\mu}=-{N\over V}\ , \label{TEOS}$ only two of which are independent.

What about $$C\ns_V$$ and $$C\ns_p$$? To find $$C\ns_V$$, we recast Equation [EXA] as $S=\bigg({NVT\over 3a}\bigg)^{\!1/2}\ .$ We then have $C\ns_V=T\pabc{S}{T}{V,N}={1\over 2}\,\bigg({NVT\over 3a}\bigg)^{\!1/2}={N\over 18a}\, {T^2\over p}\ ,$ where the last equality on the RHS follows upon invoking the first of the equations of state in Equation [TEOS]. To find $$C\ns_p$$, we eliminate $$V$$ from eqns. [EXA] and [EXB], obtaining $$T^2/p=9aS/N$$. From this we obtain $C\ns_p=T\,\pabc{S}{T}{p,N}={2N\over 9a}\,{T^2\over p}\ .$ Thus, $$C\ns_p/C\ns_V=4$$.

We can derive still more. To find the isothermal compressibility $$\kappa\ns_T= -{1\over V}\spabc{V}{p}{T,N}$$ , use the first of the equations of state in Equation [TEOS]. To derive the adiabatic compressibility $$\kappa\ns_S=-{1\over V}\spabc{V}{p}{S,N}$$ , use Equation [EXB], and then eliminate the inconvenient variable $$S$$.

Suppose we use this system as the working substance for a Carnot engine. Let’s compute the work done and the engine efficiency. To do this, it is helpful to eliminate $$S$$ in the expression for the energy, and to rewrite the equation of state: $E=pV=\sqrt ParseError: EOF expected (click for details) Callstack: at (Under_Construction/Arovas_Texts/Book:_Thermodynamics_and_Statistical_Mechanics_(Arovas)/02:_Thermodynamics/2.06:_The_Entropy), /content/body/div[6]/p[4]/span[1], line 1, column 2  \>V^{1/2}\,T^{3/2}\qquad , \qquad p=\sqrt ParseError: EOF expected (click for details) Callstack: at (Under_Construction/Arovas_Texts/Book:_Thermodynamics_and_Statistical_Mechanics_(Arovas)/02:_Thermodynamics/2.06:_The_Entropy), /content/body/div[6]/p[4]/span[2], line 1, column 2  \>{T^{3/2}\over V^{1/2}}\ .$ We assume $$dN=0$$ throughout. We now see that for isotherms, $dT=0\ :\ {E\over \sqrt{V}}={constant}$ Furthermore, since $\dbar W\big|\nd_T=\sqrt ParseError: EOF expected (click for details) Callstack: at (Under_Construction/Arovas_Texts/Book:_Thermodynamics_and_Statistical_Mechanics_(Arovas)/02:_Thermodynamics/2.06:_The_Entropy), /content/body/div[6]/p[4]/span[3], line 1, column 2  \>T^{3/2}\>{dV\over V^{1/2}}=2\,dE\big|\nd_T\ ,$ we conclude that $dT=0\ :\ W\ns_{\Ri\Rf}=2(E\ns_\Rf-E\ns_\Ri) \quad,\quad Q\ns_{\Ri\Rf}=E\ns_\Rf-E\ns_\Ri+W\ns_{\Ri\Rf}=3(E\ns_\Rf-E\ns_\Ri)\ .$ For adiabats, Equation [EXA] says $$d(TV)=0$$, and therefore $\qquad\qquad\dbar Q=0\ :\ TV={constant}\quad,\quad {E\over T}={constant}\quad,\quad EV={constant}$ as well as $$W\ns_{\Ri\Rf}=E\ns_\Ri-E\ns_\Rf$$. We can use these relations to derive the following: $E\subB=\sqrt{V\subB\over V\subA}\,E\subA\quad,\quad E\subC={T\ns_1\over T\ns_2}\sqrt{V\subB\over V\subA}\,E\subA\quad,\quad E\subD={T\ns_1\over T\ns_2}\,E\subA\ .$ Now we can write \begin{aligned} W\subAB&=2(E\subB-E\subA)=2\Bigg(\sqrt{V\subB\over V\subA}-1\Bigg)E\subA\\ W\subBC&=(E\subB-E\subC)=\sqrt{V\subB\over V\subA}\Bigg(1-{T\ns_1\over T\ns_2}\Bigg)E\subA\\ W\subCD&=2(E\subD-E\subC)=2{T\ns_1\over T\ns_2}\Bigg(1-\sqrt{V\subB\over V\subA}\Bigg)E\subA\\ W\subDA&=(E\subD-E\subA)=\Bigg({T\ns_1\over T\ns_2}-1\Bigg)E\subA\end{aligned} Adding up all the work, we obtain $\begin{split} W&=W\ns_\ssr{AB}+W\ns_\ssr{BC}+W\ns_\ssr{CD}+W\ns_\ssr{DA}\vph\\ &=3\Bigg(\sqrt{V\subB\over V\subA}-1\Bigg)\Bigg(1-{T\ns_1\over T\ns_2}\Bigg)E\subA\ . \end{split}$ Since $Q\subAB=3(E\subB-E\subA)=\frac{3}{2}W\subAB=3\Bigg(\sqrt{V\subB\over V\subA}-1\Bigg)E\subA\ ,$ we find once again $\eta={W\over Q\subAB}=1-{T\ns_1\over T\ns_2} \ .$

## Measuring the entropy of a substance

If we can measure the heat capacity $$C\ns_V(T)$$ or $$C\ns_p(T)$$ of a substance as a function of temperature down to the lowest temperatures, then we can measure the entropy. At constant pressure, for example, we have $$T\,dS=C\ns_p\,dT$$, hence $S(p,T)=S(p,T=0)+\int\limits_0^T\!\!dT'\,{C\ns_p(T')\over T'}\ .$ The zero temperature entropy is $$S(p,T=0)=\kB\ln g$$ where $$g$$ is the quantum ground state degeneracy at pressure $$p$$. In all but highly unusual cases, $$g=1$$ and $$S(p,T=0)=0$$.