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Thermodynamics

Key Point: learn how to describe macroscopic thermal properties of many-body systems. Compared with thermotics, we intend to present a more general and comprehensive theory, with more rigorous and advanced methods.

Thermodynamics is an empirical theory for macroscopic systems, typically based on experiments or observations, rather than microscopic details of the system. Historically, it arose before the atomic nature of matter was understood. The idea that heat is a form of energy was suggested by Count Rumford (1798) and Davy (1799), stated explicitly by R. J. Mayer (1842), and gained acceptance after Joule’s experiments (1843–1849). The theory was formulated consistently by Clausius and Lord Kelvin around 1850, and greatly developed by J. W. Gibbs (1876–1878). Kinetic theory was developed by Clausius, Maxwell, and Boltzmann, and statistical mechanics grew out of Boltzmann’s work (1872) and Gibbs’s work (1902).

Thermal equilibrium and temperature

macroscopic system: A system consists of many “microscopic” particles or elements (e.g. gases, liquids, solids, plasmas, granular matter, celestial matter, human beings, companies).

“NOTE

“microscopic”: is relative to “macroscopic”; it need not be in the quantum-mechanical sense.

“many”: $1\ \mathrm{cm}^3$ gas consists of about $10^{19}$ molecules, which is theoretically set to be infinite.

Thermodynamics is an empirical theory for macroscopic systems, typically based on experiments or observations, rather than microscopic details of the system.

Why study thermodynamics independently of statistical mechanics?

• It is the history.
• Thermodynamics shows a certain beauty and simplifies the theoretical framework. It is based on some “universal” laws, independent of the details of many-body systems.
• One learns not only logical methods, but also various quantitative methods for describing thermal properties of many-body systems. These methods are also useful in statistical mechanics.

Why not directly tackle the system with mechanics or quantum mechanics?

• Historically, one did not know what “heat” is, and had little knowledge of the microscopic structure of macroscopic systems.
• It is too difficult to solve the equations of motion.
• Compared with few-body systems, many-body systems exhibit complexity. For example, why does water suddenly become gas at $100^\circ\mathrm{C}$, although the microscopic equations of motion remain the same? The butterfly effect is another example: a small deviation in initial conditions may lead to a large change in the dynamic evolution of nonlinear complex systems.

Therefore, one introduces macroscopic parameters, or observables, to describe macroscopic properties.

• Mechanical parameters: pressure $P$, volume $V$.
• Chemical parameters: density $n$, chemical potential $\mu$.
• Electromagnetic parameters: electric field $\vec{E}$, magnetic field $\vec{B}$.
• Thermal parameters: temperature $T$, heat, entropy $S$.

$T$ and $S$ are exclusively from thermodynamics.

A macroscopic state is described by “a complete set” of parameters.

• The number of parameters in the complete set is fixed for a system.
• All other parameters can be expressed as functions of the complete set of parameters.

If the complete set is $x_1,x_2,\dots,x_M$, then a state $i$ can be represented by a point $(x_1^i,\dots,x_M^i)$ in parameter space. When the state changes, the point moves and forms a “surface” in parameter space. The number of parameters in the complete set is fixed, e.g. for a dilute gas it is two. This is an assumption based on experimental facts.

Classifications of macroscopic states:

• Equilibrium state: parameters do not depend on time $t$.
• Non-equilibrium state: some parameters evolve with time $t$, or more complicated, e.g. local parameters may be needed.

If not specified, we study equilibrium states in this course.

Classifications of systems:

• An isolated system: nothing is exchanged with the environment.
• A thermally isolated system: no “heat” is exchanged with the environment.

Classifications of parameters:

• Extensive parameters: proportional to $V$.
• Intensive parameters: independent of $V$.

“NOTE

Why is it meaningful to define extensive and intensive parameters? In the thermodynamic limit, a standard parameter is either extensive or intensive. But in a singular state, it can be more complicated, e.g. there may exist a non-integer power of $V$. For a system not too far from equilibrium, we may divide it into many subsystems, and extensive parameters can be added up.

Thermal equilibrium and temperature

Pressure, volume, density, etc. can be defined with knowledge from other subjects. So let us concentrate on thermal observables, e.g. temperature first.

In daily life, we feel “hot” or “cold”. What is it? How to describe it quantitatively?

Observation: Initially, two systems are differently hot or cold. After they thermally touch each other for a long time, they become equally hot or cold. Then we say they are in thermal equilibrium.

The zeroth law: If systems $A$ and $B$ are each in thermal equilibrium with a third system $C$, then $A$ and $B$ are in thermal equilibrium with each other.

Introduce a parameter $T$, the temperature, to describe how hot or cold the system is, and assume that the temperatures of two systems in thermal equilibrium are the same. This assumption is very important; otherwise we cannot compare temperatures of different systems, and $T$ is useless. The zeroth law defines the temperature.

$T$ is a property of the macroscopic state. This is clear from the definition, which does not depend on the history of how the system comes to thermal equilibrium. If $(x_1,\dots,x_M)$ is a complete set, then $T$ should be a function of them:

This is called the equation of state.

“NOTE

E.g.

• The Ideal gas

$$\begin{aligned} PV = NkT \end{aligned}$$

• Real gases

$$\begin{aligned} PV &= A + BP + CP^2 + \cdots\\ &= A + \frac{B'}{V} + \frac{C'}{V^2} + \cdots \end{aligned}$$

Temperature scales

To complete the definition of temperature, we need to fix its scale. We may proceed as follows:

1. Choose a specific system and a proper property that monotonously depends on temperature.
2. Fix the relation between this property and $T$.

Practically, one may proceed in two steps: i) assume the relation to be linear, quadratic, or a special form; ii) fix the unit including the origin. The former is somewhat theoretical, while the latter relies on experimental phenomena for convenience.

E.g. mercury and its volume; a linear relation between $V$ and $T$. To fix the unit and the origin, set $T=0$ for the transition point of water from liquid to solid at standard atmospheric pressure, and $T=100$ for that from gas to liquid.

Mathematically,

This way of defining temperature is not so elegant theoretically, because it depends on the system and the property chosen. It is not “universal”, and it is difficult to measure accurately over the whole temperature range.

A better choice is to choose the ideal gas as the specific system. Experimentally, this is set up by Boyle’s law: for a given ideal gas at a fixed temperature,

and the constant depends on the temperature. Fix $V$, set a linear relation between $T_V$ and $P$, and take the transition point of water from liquid to solid at standard atmospheric pressure as $T=0$, and that from gas to liquid as $T=100$. Thus

Exercise: Prove $T_V = T_P$.

“NOTE

Hint: assume $PV = a_V T_V + b_V$ and $PV = a_P T_P + b_P$, and prove $a_V=a_P$, $b_V=b_P$.

Proof: For fixed $V=V_i$,

$$P V_i = a_V 100\frac{P-P_i}{P_f-P_i} + b_V$$

Comparing the coefficient of $P$ and the constant,

$$a_V = \frac{P_f-P_i}{100}V_i,\qquad b_V = P_i V_i$$

For fixed $P=P_i$,

$$P_i V = a_P 100\frac{V-V_i}{V_f-V_i} + b_P$$

therefore

$$a_P = \frac{V_f-V_i}{100}P_i,\qquad b_P = P_i V_i$$

At the boiling point of water, by Boyle’s law, $P_f V_i = V_f P_i$, therefore $a_P=a_V$ and $b_P=b_V$. Hence $T_V=T_P$.

• The ideal-gas temperature scale

• In fact, for any single-valued and monotonous function $F(T)$, setting $PV=F(T)$ defines a temperature scale.

The first law

Thermodynamics transformation

A change of state is a thermodynamic transformation. The transformation can be realized only by external conditions, which are typically macroscopic. Here we only consider equilibrium states.

quasi-static: If the external condition changes so slowly that at any moment the system is approximately in equilibrium, the transformation is called quasi-static. In other words, the relaxation at each time $t$ from the non-equilibrium state to the equilibrium one is much faster than the change of the external condition.

For example, for an ideal gas, every point on the $P$-$V$ space represents an equilibrium state, and a quasi-static transformation is a continuous path on the $P$-$V$ diagram.

If not specified, we concern ourselves only with quasi-static transformations in this course.

The concept of work is taken over from mechanics. For a system described by $(P,V)$, the work done by the system in an infinitesimal transformation is

For example, $dW = P \cdot S \cdot dx = P \dd V$. For a finite process,

Important: Work is not fixed by the initial and final states, but depends on the specific process between them. Therefore, the infinitesimal work is denoted by $\dbar W$ rather than $\dd W$.

Now comes another thermal concept, “heat”. “Heat” is what is absorbed by a system if its temperature increases while no work is done.

$C$ is called the heat capacity. It depends on the detailed nature of the system, and is also process-dependent. In general, $C$ cannot be simply written as $C(P,V)$; an additional direction has to be specified at each point $(P,V)$, such as $C_V(P,V)$ and $C_P(P,V)$.

Experimentally, we may fix the unit of $Q$ with a specific system, e.g. by increasing the temperature of water by one degree.

The first law

For an arbitrary (not only quasi-static) transformation given the initial and final states, let

• $\Delta Q$ denote the heat absorbed by the system;
• $\Delta W$ denote the work done by the system.

A minus sign of $\Delta Q$ or $\Delta W$ represents the opposite direction. The first law states that the quantity $\Delta U$ defined by

is the same for all possible transformations, i.e. it only depends on the initial and final states.

This immediately defines a state function $U$, called the internal energy. Of course, it is defined only up to a constant. Experimentally, it is found that $U$ is extensive. $U$ being a state function means that for an infinitesimal transformation,

is exact. That is, $\dd U$ is a differential.

For example, if

then $\dd U$ being exact leads to

“NOTE

Question: how to prove that $\dd U$ being exact is equivalent to $U$ being a state function?

Actually, it is also equivalent to that any loop integral is zero, or any integral between two states is path independent.

• If the number of independent parameters is one,

• If the number of independent parameters is two,

What is the physical meaning of the first law? It states that heat is also a kind of energy, the total energy is conserved, and it is feasible to convert mechanical work into heat or vice versa.

Then, what is heat?

Heat is the energy transferred through microscopic irregular motion of molecules, particles, or elements. “Irregular” here is from the macroscopic level. Of course, the microscopic motion is still governed by Newton’s law or the Heisenberg equation, etc. Temperature measures how irregular the microscopic motion is. This will be understood in statistical physics.

$U$ excludes the global mechanical energy. $U$ includes the energy of irregular motion, interaction energy between molecules, energy inside a molecule, and energy in a field.

Different from temperature $T$, which is rather universal, $\Delta U$ is only defined for an individual system, but one may compare it with that of another system through work.

Some applications

1. Exercises:

These are called the heat equations. It can be deduced from the last two equations that

$H = U + PV$ is called enthalpy.

2. Free expansion of an ideal gas.

In the experiment shown in Fig. 1.4, the finding is that the temperature of the water bath is unchanged, $T_1=T_2$. For the ideal gas:

Since $\Delta W=0$, $\Delta T=0$,

Thus $\Delta Q=0$ (since $\Delta T=0$), $\Delta U=0$ (from the first law).

We may take $T$ and $V$ as independent variables, $U=U(T,V)$. Therefore, $U=U(T)$ (since $\Delta V$ is non-zero).

1
\begin{tikzpicture}[>=stealth, thick, font=\small]
2
  \draw (0,0) rectangle (2,2);
3
  \draw (2,0) rectangle (4,2);
4
  \draw[dashed] (2,0) -- (2,2);
5
  \node at (1,1) {Gas};
6
  \node at (3,1) {Vacuum};
7
  \draw[->] (2.1,1) -- (2.9,1);
8
  \node at (2,2.35) {Before: $V_1$};
9

10
  \draw (6,0) rectangle (10,2);
11
  \node at (8,1) {Gas};
12
  \node at (8,2.35) {After: $V_2>V_1$};
13
\end{tikzpicture}

3. Internal energy of an ideal gas.

Since $U$ depends only on $T$,

Assuming $C_V$ to be independent of $T$, we obtain

4. $C_P-C_V$ for an ideal gas.

The enthalpy of an ideal gas is a function of $T$ only:

Hence

Although the heat capacity of an ideal gas is constant along a certain path, it is generally path dependent.

The second law

Reversible and irreversible transformations

The direction of processes is the main concern of the second law.

Observation: All naturally occurring processes proceed in one direction only. They never, of their own accord, proceed in the opposite direction. For example:

• A pipe of water never flows from the low-pressure end to the high-pressure end.
• The water of a river never flows in the opposite direction.
• A cup of hot coffee never becomes hotter without external interactions.
• The molecules of a drop of ink in a glass of water diffuse uniformly throughout the whole volume. They never recluster into a drop-shaped clump.

A reversible transformation is a transformation in which the system retraces its history in time when the external condition retraces its history in time.

Important: If only the system retraces its history, but the environment does not, the transformation is not reversible.

Examples of irreversible transformations: All naturally occurring processes are irreversible. These are processes in which systems relax from non-equilibrium states to equilibrium states; they never go in the opposite direction.

• A process or part of a process contains a macroscopically finite relaxation from a non-equilibrium state to an equilibrium state.

Examples of reversible transformations:

• Systems $A$ and $B$ are in thermal equilibrium. Increase the temperature of $B$ by $dT$; then a certain amount of heat $dQ$ is transferred to $A$. Then cool $B$ by $dT$; the same $dQ$ is transferred from $A$ back to $B$.

Questions:

• How to increase the temperature of $B$ by $dT$?
• If $dT$ is finite, is it still reversible?
• How to construct a reversible transformation such that one may increase the temperatures of $B$ and $A$ by a finite $\Delta T$?

Here we note that even for a reversible transformation, in some sense, an “infinitesimal” trace may be left after both the system and environment retrace their histories. This belongs to a kind of common approximation in thermodynamics.

To increase the temperature of $B$ by $dT$, for example, one may let $B$ contact a heat reservoir with temperature $T+dT$. To decrease the temperature of $B$ back to $T$, $B$ should contact a heat reservoir with temperature $T$. This infinitesimal trace can be further diminished by dividing $dT$ into many $ddT$, and increasing the temperature step by step. Then the trace becomes “a higher-order infinitesimal”.

This example tells that for a reversible transformation, the change of the environment at a time instant must be infinitesimal. More rigorously, $dF/dt$ should be finite, with $F$ being the external force, although standard thermodynamics does not directly consult time $t$.

Therefore, for example, the transformation is not reversible if one increases the temperature $dT$ by “fast pressing” system $B$.

A heat reservoir is a system so large that the gain or loss of any finite amount of heat does not change its temperature.

In expansion, the system absorbs heat from the reservoir and transfers energy to the spring. This process can be reversed, unlike free expansion.

• Reversible transformations have to be quasi-static, while in a certain sense, the reverse is not necessarily true. But such examples are nonstandard. We may usually think that quasi-static and reversible are equivalent. In our concept in this course, “any smooth path in the parameter space is a reversible transformation”!

For example, a gas that freely expands into successive infinitesimal volume elements undergoes a quasi-static transformation in the sense that it is “slow”, but this transformation is not reversible. Naturally, this is an irreversible transformation, since it is a relaxation to the equilibrium state, and the gas loses the capacity of doing work.

Although such a process looks slow, it is not a standard quasi-static transformation. Essentially, it is an artificial division of an irreversible transformation into many “infinitesimal” pieces. In each piece, there exists a “rapid” change of pressure.

A similar irreversible transformation is that heat from a high-temperature reservoir is directly transferred to a low-temperature reservoir with a finite temperature difference by successive infinitesimal amounts. In such a transformation, heat loses its capacity for doing work.

A reversible transformation is necessarily a quasi-static transformation, but the simple concept “slow” is not sufficient to ensure reversibility. The change of the environment must be infinitesimal, contain no non-equilibrium relaxation, and keep the capacity for doing work.

Finally, dissipative interactions such as friction, which are beyond the standard thermodynamic formalism, naturally violate reversibility.

Discussion: Why are there macroscopic irreversible transformations?

Newton’s equations are generally time-reversal symmetric. Therefore, the microscopic mechanical motion of particles is reversible. Why is the macroscopic process irreversible?

Indeed, microscopically, if we start from a microscopic configuration in the final macroscopic state with $V_2$, and reverse the motion of all particles, the particles will come back to the microscopic state in the initial state with $V_1$. But this probability is very small.

The probability for a particle to stay in $V_1$ is $V_1/V_2$. Since the number $N$ of particles is a big number, $(V_1/V_2)^N$ is infinitesimally small. In simple words, the macroscopic initial and final conditions violate the reversibility of Newton’s law in time. After the volume is changed to $V_2$, the fraction of microscopic states purely in $V_1$ becomes infinitesimally small.

Question: When the particles described by Newton’s equations retrace their history in time, could the external environment retrace its history in time?

The second law

• Kelvin statement: There exists no thermodynamic transformation whose sole effect is to extract a quantity of heat from a given heat reservoir and to convert it entirely into work.

Otherwise, we would not have any energy crisis.

• Clausius statement: There exists no thermodynamic transformation whose sole effect is to transfer a quantity of heat from a colder reservoir to a hotter reservoir.

That is why we need refrigerators and air conditioners.

For example, when an ideal gas is expanded reversibly and isothermally, a quantity of heat absorbed from the reservoir is entirely converted to work. However, this process is allowed by the second law because the volume of the gas is changed, i.e. the word “sole” is violated.

Adiabatic expansions do not violate the second law, since the work is not converted from heat, but from the change of state.

From the view of a relaxation from a non-equilibrium state to an equilibrium state, the Clausius statement is obvious. One may also consider that the process of converting work into heat is a relaxation from an ordered state to a disordered state; the Kelvin statement is also natural.

Kelvin and Clausius statements are equivalent.

\begin{tikzpicture}[>=stealth, font=\small]
\node (T1) at (0,0) {$T_1$};
\node[draw, circle] (M) at (0,-1.5) {$M$};
\node (Q) at (0.5, -0.75) {$Q$};
\node (W) at (1.5,-1.5) {$W$};
\node (T2) at (0,-3) {$T_2$};
\draw (-1,0.2) -- (-1,-0.2) -- (1,-0.2) -- (1,0.2);
\draw[->] (T1) -- (M);
\draw[->] (M) -- (W);
\draw (-1,-3.2) -- (-1,-2.8) -- (1,-2.8) -- (1,-3.2);
\node (Kelvin) at (0,-3.5) {Kelvin};

\node (T1b) at (6,0) {$T_1$};
\node (Qb) at (6.5, -1.5) {$Q$};
\node (T2b) at (6,-3) {$T_2$};
\draw (5,0.2) -- (5,-0.2) -- (7,-0.2) -- (7,0.2);
\draw (5,-3.2) -- (5,-2.8) -- (7,-2.8) -- (7,-3.2);
\draw[->] (T2b) -- (T1b);
\node (Clausius) at (6,-3.5) {Clausius};
\end{tikzpicture}

Proof of equivalence:

• $K$ false $\Rightarrow$ $C$ false. If $K$ is false, we may extract heat from a reservoir at temperature $T_1$ and convert it entirely into work, with no other effect. We can convert this work into heat and deliver it to a reservoir at $T_2>T_1$. Then the net result is the transfer of an amount of heat from a colder reservoir to a hotter reservoir. Hence $C$ is false.
• $C$ false $\Rightarrow$ $K$ false. Suppose $C$ is false; we may transfer heat $Q_2$ from a reservoir at temperature $T_1$ to a reservoir at $T_2>T_1$. Then we extract $Q_2$ from the reservoir at $T_2$, convert part of it to work, and reject certain heat to the reservoir at $T_1$. The net result is that an amount of heat is extracted from the reservoir at $T_1$ and converted entirely into work, with no other effect. Hence $K$ is false.

Carnot’s theorem

The Carnot engine

An engine is a machine which converts heat into work. An engine which does everything in a reversible way is called a Carnot engine.

For example, for $T_2>T_1$:

• $a \rightarrow b$: isothermal, heat $Q_2$ is absorbed by the system.
• $b \rightarrow c$: adiabatic.
• $c \rightarrow d$: isothermal, heat $Q_1$ is rejected by the system.
• $d \rightarrow a$: adiabatic.

This is a reversible engine, but it can be generalized to an irreversible one. Work is done by the system in every step, positively from $a\to b$ and $b\to c$, and negatively from $c\to d$ and $d\to a$. For a car engine, $T_1$ is the temperature of the environment, and $T_2$ is kept by burning gasoline.

In one cycle, $\Delta U=0$. The total work done by the system, according to the first law, is

The efficiency of the engine is

We may prove that if $W>0$, then $Q_1>0$, $Q_2>0$.

Proof: If $Q_1=0$, it violates the Kelvin statement. Suppose $Q_1<0$; this means the engine absorbs heat $Q_2$ from $T_2$ and $-Q_1$ from $T_1$, and converts the total amount $Q_2-Q_1$ into work. Then we convert the work $Q_2-Q_1$ to heat and deliver it to $T_2$. Finally the net result is to transfer an amount of heat $-Q_1$ from $T_1$ to $T_2$. By the Clausius statement, this is impossible. Therefore $Q_1>0$. From $W=Q_2-Q_1>0$, it follows $Q_2>0$.

Exercise: If $W<0$ and $Q_1<0$, then $Q_2<0$. This is a refrigerator, but it is trivial. Better: if $Q_1<0$, then $W<0$, $Q_2<0$.

For a refrigerator, the system is compressed from $c\to b$ and $b\to a$, releasing heat to $T_2$, and expanding $a\to d$ and $d\to c$, absorbing heat from $T_1$.

If there are only two heat reservoirs, the form of the Carnot engine is unique.

If the system is an ideal gas, for a Carnot engine,

where $T$ is the ideal-gas temperature scale.

Proof: We first derive the equation for an adiabatic process. Assuming $PV=NkT$, $U=C_V T+\mathrm{constant}$, and $C_V$ is a constant,

Since $PV=NkT$,

Eliminating $\dd T$,

therefore, this leads to the so-called adiabatic equation. Note that such a simple adiabatic equation is possible only for the ideal gas. Alternatively,

Therefore, for adiabatic processes $b\to c$ and $d\to a$, the relations hold.

1
\begin{tikzpicture}[>=stealth, x=1.3cm, y=1.1cm, font=\small]
2
  % axes
3
  \draw[->] (0,0) -- (5.5,0) node[below] {$V$};
4
  \draw[->] (0,0) -- (0,6.2) node[left]  {$P$};
5
  \node[below left] at (0,0) {$O$};
6

7
  % parameters
8
  \def\CTh{8.25}
9
  \def\CTc{7.01}
10
  \def\Ktwo{12.80}
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  \def\Kfour{9.70}
12
  \def\gam{1.4}
13

14
  % 1->2 high-T isothermal expansion
15
  \draw[thick, red] plot[domain=1.5:3.0, samples=100] (\x, {\CTh/\x});
16
  \draw[->, red, thick] (2.30, 3.587) -- (2.40, 3.437);
17

18
  % 2->3 adiabatic expansion
19
  \draw[thick, blue] plot[domain=3.0:4.5, samples=100] (\x, {\Ktwo/pow(\x,\gam)});
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  \draw[->, blue, thick] (3.65, 2.090) -- (3.75, 2.012);
21

22
  % 3->4 low-T isothermal compression
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  \draw[thick, red] plot[domain=4.5:2.25, samples=100] (\x, {\CTc/\x});
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  \draw[->, red, thick] (3.40, 2.062) -- (3.30, 2.124);
25

26
  % 4->1 adiabatic compression
27
  \draw[thick, blue] plot[domain=2.25:1.5, samples=100] (\x, {\Kfour/pow(\x,\gam)});
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  \draw[->, blue, thick] (1.95, 3.807) -- (1.85, 4.100);
29

30
  % four points
31
  \fill (1.5,5.5)   circle (1.5pt) node[above left,  xshift=-3pt, yshift=2pt]  {$a$};
32
  \fill (3.0,2.75)  circle (1.5pt) node[above right, xshift=3pt,  yshift=2pt]  {$b$};
33
  \fill (4.5,1.56)  circle (1.5pt) node[below right, xshift=3pt,  yshift=-2pt] {$c$};
34
  \fill (2.25,3.12) circle (1.5pt) node[below left,  xshift=-3pt, yshift=-2pt] {$d$};
35

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  % labels
37
  \node[red]  at (2.75,3.95) {$T_2$ , in $Q_2$};
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  \node[red]  at (3.40,1.55) {$T_1$ , out $Q_1$};
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  \node[blue] at (4.20,2.45) {$Q=0$};
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  \node[blue] at (1.10,4.55) {$Q=0$};
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\end{tikzpicture}

For $a\to b$ isothermal,

For $c\to d$ isothermal,

Finally,

Carnot’s theorem

No engine operating between two given temperatures is more efficient than a Carnot engine.

Proof: Construct a Carnot engine $C$ and an arbitrary engine $X$ as shown in Fig. 1.12.

Let

where $N,N'$ are integers. This equality will be valid by making $N,N'$ sufficiently large.

It is important to keep in mind that the Carnot engine can be operated in reverse, i.e. the signs of $W,Q_1,Q_2$ can be simultaneously changed.

\begin{tikzpicture}[>=stealth, thick, font=\large\sffamily]
    % two reservoir lines
    \draw (-1, 3) -- (7, 3) node[right, xshift=10pt] {$T_2$};
    \draw (-1, 0) -- (7, 0) node[right, xshift=10pt] {$T_1 < T_2$};

    % engine C
    \draw (1, 1.5) circle (0.6);
    \node at (1, 1.5) {C};

    % engine X
    \draw (5, 1.5) circle (0.6);
    \node at (5, 1.5) {X};

    % arrows for C
    \draw[->] (1, 3) -- (1, 2.1) node[midway, left] {$Q_2$};
    \draw[->] (1, 0.9) -- (1, 0) node[midway, left] {$Q_1$};
    \draw[->] (1.6, 1.5) -- (2.4, 1.5) node[midway, above] {$W$};

    % arrows for X
    \draw[->] (5, 3) -- (5, 2.1) node[midway, left] {$Q'_2$};
    \draw[->] (5, 0.9) -- (5, 0) node[midway, left] {$Q'_1$};
    \draw[->] (5.6, 1.5) -- (6.4, 1.5) node[midway, above] {$W'$};
\end{tikzpicture}

Now operate the $C$ engine $N$ cycles in reverse, and the $X$ engine $N'$ cycles,

On the other hand, from the first law and $(\Delta U)_{\mathrm{total}}=0$,

If $W_{\mathrm{total}}>0$, it violates the Kelvin statement. Therefore

In other words,

From the ratio equation,

i.e.

Since $X$ can be a Carnot engine, we have

Corollary: All Carnot engines between two given temperatures have the same efficiency.

In the proof of Carnot’s theorem, the key point is to choose one of $W_{\mathrm{total}}$, $(Q_2)_{\mathrm{total}}$, and $(Q_1)_{\mathrm{total}}$ to be zero, then apply the first and second laws.

An alternative way to prove Carnot’s theorem: Operate the $C$ engine $N$ cycles in reverse and the $X$ engine $N'$ cycles, and let

From the second law,

Hence,

therefore,

Note that assuming $W_{\mathrm{total}}=W'-W=0$, one may also prove Carnot’s theorem. However, this is not sufficiently general since the efficiency may depend on the work $W$.

With Carnot’s theorem at hand, one can deduce that the efficiency of Carnot engines is independent of $W$, since the $X$ engine can also be a Carnot engine, and both $W'$ and $W$ are arbitrary. But such a conclusion does not exist for an arbitrary engine, for example, the engine of a car.

Absolute scale of temperature

Carnot’s theorem tells something “absolute” related to temperatures, and it may be used for defining a better temperature scale.

If the efficiency of a Carnot engine is $\eta$, we define the absolute scale of temperature by

If $\eta=1$, it violates the second law, therefore $0\le \eta<1$. Then we conclude $T_1>0$ and $T_2>0$, i.e. all temperatures are positive. Of course, it is also possible to set $T_1<0$ and $T_2<0$, but this only means we add a minus sign to $T$.

Note that in this definition, $\eta$ does not fix the unit, since $1-\eta=T_1/T_2=\alpha T_1/(\alpha T_2)$.

To define the unit, we note that

hence,

Similarly, we fix $W=Q_{n+1}-Q_n$ independent of $n$, and have

i.e. both $W$ and the ratio $T_n/Q_n$ are independent of $n$.

Choosing $W$ such that

That is the unit of the absolute scale of temperature.

The above argument not only defines the unit of the absolute scale of temperature, but also completely fixes the absolute scale. Here we assume that $1\ \mathrm{K}$ is empirically known. If it is not, we may define the unit $T_{n+1}-T_n$ by fixing $W$ to be a certain amount.

Practically, setting the freezing temperature of water to be $T_1$, and the boiling one to be $T_2$, and $T_2-T_1=100\ \mathrm{K}$, one may extract $T_1=273.15\ \mathrm{K}$ from experiments. On the other hand, the triple point of water is at $T=273.16\ \mathrm{K}$, i.e. $0.01^\circ\mathrm{C}$. The experimental measurement of the triple point is more reliable and accurate. In some sense, it may be used to define the unit $1\ \mathrm{K}$, but such a definition is possible because of the above theoretical argument.

• The absolute scale of temperature is independent of the specific properties of any substance.
• No Carnot engine exists with a reservoir of $T_1=0$, otherwise $\eta=1$. Since $T_2$ is a higher temperature, it is naturally not possible that $T_2=0$. Thus $T=0$ is the greatest lower bound, and is called the absolute zero. But this has not yet meant that $T$ cannot be zero at all.
• The absolute scale is identical with the ideal-gas scale $PV=NkT$.

Entropy

The fact that some processes are irreversible already indicates that the initial and final states are characteristically different. There should be a state function which describes this property.

Clausius theorem

A cyclic transformation is a transformation whose final state is identical with the initial state. A reversible cyclic transformation is a cyclic transformation which is reversible.

In any cyclic transformation through which the temperature is defined,

where the integral extends over one cycle of the transformation, $\dbar Q$ is the heat absorbed by the system. The equality holds if the cyclic transformation is reversible.

Proof: For an engine with two reservoirs, it is already known that

i.e.

$Q_1$ is the heat rejected by the system, i.e. $-Q_1$ is the heat “absorbed” by the system. For two adiabatic processes, $\dbar Q=0$ and $T$ is defined in every step.

For a general reversible cyclic transformation $C$, it can be divided by isothermal lines and adiabatic lines into many infinitesimal Carnot transformations. Equivalently, it can be divided by adiabatic lines or isothermal lines into infinitesimal Carnot strips.

Since all transformations are reversible, if we add up all the infinitesimal cyclic transformations, the resulting transformation will be the original transformation $C$.

For each Carnot cyclic transformation,

therefore,

$Q_j<0$ means exothermic. Here $j$ sums over all Carnot cyclic transformations. Since all processes inside $C$ will not contribute,

If $C$ is irreversible, it is straightforward to prove

“NOTE

Hint: independent of reversible or irreversible, the integration over an infinitesimal transformation can be approximated by that over an isothermal one and an adiabatic one.

Corollary: For a reversible transformation,

is independent of the path, and depends only on the initial and final states.

This allows us to define a state function such that

or

In mathematical terminology, $1/T$ is an integrating factor. $S$ is called the entropy.

Properties of $S$:

a) For an arbitrary transformation,

The equality holds for a reversible transformation. This property can be simply understood from the Clausius theorem by constructing a reversible transformation from $B$ to $A$, and an arbitrary transformation from $A$ to $B$.

b) If $\dbar Q=0$,

For a thermally isolated system, $S$ never decreases. The equality holds for a reversible transformation.

For irreversible transformations, $S$ always increases is equivalent to the second law.

c) $S$ is extensive. It is clear from the definition since $\dbar Q$ is extensive.

Calculations of $S$

• By definition, if the thermodynamic transformation is reversible.
• Constructing a reversible transformation, if the thermodynamic transformation is irreversible.
• With $T\dd S = \dd U + \dbar W$ for a reversible transformation. This equation is a combination of the first and second laws, sometimes called the fundamental equation of thermodynamics.
• $S$ is extensive.

“NOTE

Reversible isothermal expansion of an ideal gas

Since the gas is ideal, $U=U(T)$, hence $\Delta U=0$.

For the gas,

$$\begin{align} (\Delta S)_{\mathrm{gas}} &= \int \frac{\dbar Q}{T} = \frac{\Delta Q}{T}\\ &= \frac{W}{T}\\ W = \int P\cdot \dd V &= NkT \int \frac{1}{V}\dd V\\ &= NkT \ln\frac{V_2}{V_1}\\ (\Delta S)_{\mathrm{gas}} &= Nk \ln\frac{V_2}{V_1} \end{align}$$

The reservoir supplies the amount of heat $-\Delta Q$,

$$\begin{align} (\Delta S)_{\mathrm{reservoir}} = -Nk \ln\frac{V_2}{V_1} \end{align}$$

The change in entropy of the whole system is zero. This is natural since the whole system is adiabatic, and the process is reversible. The work is stored in the spring, which can be used for reversing the process.

“NOTE

Free expansion of an ideal gas

This process is a little puzzling:

$$\begin{align} \Delta W = 0,\quad \Delta T=0,\quad \Delta Q=0,\quad \Delta U=0 \end{align}$$

Why or how is $S$ changed? Because it is not a reversible transformation.

But $S$ is a state function, therefore

$$\begin{align} (\Delta S)_{\mathrm{gas}} = Nk \ln\frac{V_2}{V_1} \end{align}$$

the same as in a). Although the gas is actually adiabatic, the entropy increases.

But in this case,

$$\begin{align} (\Delta S)_{\mathrm{reservoir}} = 0 \end{align}$$

since nothing is changed in the reservoir.

In comparison with a), an amount of useful energy $W=T(\Delta S)_{\mathrm{gas}}$ is just “lost”. An irreversible transformation is “wasteful”. Therefore, entropy may be viewed as a measure of the unavailability of useful energy in that state.

c) For a general reversible transformation,

Since $\dd S$ is exact,

hence,

and it leads to

therefore,

This is the equation of entropy. If $C_V$ and the equation of state are known, one can derive the entropy.

Compared with the equation of heat, the equation of entropy is more simplified because $\dd S$ is exact.

For an ideal gas,

i.e. $U=U(T)$, consistent with the experiment.

Exercises: Derive the entropy of an ideal gas. Prove

Hint: $(\partial x/\partial y)_w$ implies that there are only two independent variables.

The second law for reversible transformations

For reversible transformations, the entropy being a state function is equivalent to the second law.

a) Adiabatic lines never intersect each other.

An adiabatic line is just an isentropic line, since

If two adiabatic lines intersect, $S$ is obviously not a state function.

b) An isothermal line and an adiabatic line intersect at most only once.

Since

hence,

therefore $S$ is a monotonously increasing function of $V$ along an isothermal line. Note that $(\partial P/\partial T)_V>0$ is an experimental fact. Thus the isothermal line and the adiabatic line cannot intersect twice.

As a direct result, it is not possible that $S_1=S_2$. Otherwise, such an adiabatic line may intersect an isothermal line twice.