11.7 The Basic Thermodynamic Variables

This section introduces the most important basic players in thermodynamics.

The primary thermodynamic property introduced so far is the temperature. Recall that temperature is a measure of the hotness of the substance, a measure of how eager it is to dump energy onto other systems. Temperature is called an “intensive variable;“ it is the same for two systems that differ only in size.

The total number of particles $I$ or the total volume of their box $V$ are not intensive variables; they are “extensive variables,“ variables that increase in value proportional to the system size. Often, however, you are only interested in the properties of your substance, not the amount. In that case, intensive variables can be created by taking ratios of the extensive ones; in particular, $I$$\raisebox{.5pt}{$/$}$$V$ is an intensive variable called the “particle density.” It is the number of particles per unit volume. If you restrict your attention to only one half of your box with particles, the particle density is still the same, with half the particles in half the volume.

Note that under equilibrium conditions, it suffices to know the temperature and particle density to fully fix the state that a given system is in. More generally, the rule is that:

Two intensive variables must be known to fully determine the intensive properties of a simple substance in thermal equilibrium.
(To be precise, in a two-phase equilibrium like a liquid-vapor mixture, pressure and temperature are related, and would not be sufficient to determine something like net specific volume. They do still suffice to determine the specific volumes of the liquid and vapor parts individually, in any case.) If the amount of substance is also desired, knowledge of at least one extensive variable is required, making three variables that must be known in total.

Since the number of particles will have very large values, for macroscopic work the particle density is often not very convenient, and somewhat differently defined, but completely equivalent variables are used. The most common are the (mass) “density” $\rho$, found by multiplying the particle density with the single-particle mass $m$, $\rho$ $\vphantom0\raisebox{1.5pt}{$\equiv$}$ $mI$$\raisebox{.5pt}{$/$}$$V$, or its reciprocal, the “specific volume” $v$ $\vphantom0\raisebox{1.5pt}{$\equiv$}$ $V$$\raisebox{.5pt}{$/$}$$mI$. The density is the system mass per unit system volume, and the specific volume is the system volume per unit system mass.

Alternatively, to keep the values for the number of particles in check, they may be expressed in “moles,” multiples of Avogadro’s number

\begin{displaymath}
I_{\rm A}=6.022{,}1\;10^{23}
\end{displaymath}

That produces the “molar density” $\bar\rho$ $\vphantom0\raisebox{1.5pt}{$\equiv$}$ $I$$\raisebox{.5pt}{$/$}$$I_{\rm A}{}V$ and “molar specific volume” $\bar{v}$ $\vphantom0\raisebox{1.5pt}{$\equiv$}$ $VI_{\rm A}$$\raisebox{.5pt}{$/$}$$I$. In thermodynamic textbooks, the use of kilo mol (kmol) instead of mol has become quite standard (but then, so has the use of kilo Newton instead of Newton.) The conversion factor between molar and nonmolar specific quantities is called the “molar mass” $M$; it is applied according to its units of kg/kmol. Note that thermo books for engineers may misname $M$ to be the “molecular mass”. The numerical value of the molar mass is roughly the total number of protons and neutrons in the nuclei of a single molecule; in fact, the weird number of particles given by Avogadro’s number was chosen to achieve this.

So what else is there? Well, there is the energy of the system. In view of the uncertainty in energy, the appropriate system energy is defined as the expectation value,

\begin{displaymath}
\fbox{$\displaystyle
E = \sum_{{\rm all\ }q}P_q {\vphantom' E}^{\rm S}_q
$} %
\end{displaymath} (11.6)

where $P_q$ is the canonical probability of (11.4), (11.5). Quantity $E$ is called the “internal energy.” In engineering thermodynamics books, it is usually indicated by $U$, but this is physics. The intensive equivalent $e$ is found by dividing by the system mass; $e$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E$$\raisebox{.5pt}{$/$}$$mI$. Note the convention of indicating extensive variables by a capital and their intensive value per unit mass with the corresponding lower case letter. A specific quantity on a molar basis is lower case with a bar above it.

As a demonstration of the importance of the partition function mentioned in the previous section, if the partition function (11.5) is differentiated with respect to temperature, you get

\begin{displaymath}
\left(\frac{\partial Z}{\partial T}\right)_{V{\rm\ constan...
...tom' E}^{\rm S}_q e^{-{\vphantom' E}^{\rm S}_q/{k_{\rm B}}T}.
\end{displaymath}

(The volume of the system should be held constant in order that the energy eigenfunctions do not change.) Dividing both sides by $Z$ turns the derivative in the left hand side into that of the logarithm of $Z$, and the sum in the right hand side into the internal energy $E$, and you get
\begin{displaymath}
\fbox{$\displaystyle
E = {k_{\rm B}}T^2
\left(\frac{\partial \ln Z}{\partial T}\right)_{V{\rm\ constant}}
$} %
\end{displaymath} (11.7)

Next there is the “pressure” $P$, being the force with which the substance pushes on the surfaces of the box it is in per unit surface area. To identify $P$ quantum mechanically, first consider a system in a single energy eigenfunction ${\vphantom' E}^{\rm S}_q$ for certain. If the volume of the box is slightly changed, there will be a corresponding slight change in the energy eigenfunction ${\vphantom' E}^{\rm S}_q$, (the boundary conditions of the Hamiltonian eigenvalue problem will change), and in particular its energy will slightly change. Energy conservation requires that the change in energy ${{\rm d}}{\vphantom' E}^{\rm S}_q$is offset by the work done by the containing walls on the substance. Now the work done by the wall pressure on the substance equals

\begin{displaymath}
-P{\,\rm d}{V}.
\end{displaymath}

(The force is pressure times area and is normal to the area; the work is force times displacement in the direction of the force; combining the two, area times displacement normal to that area gives change in volume. The minus sign is because the displacement must be inwards for the pressure force on the substance to do positive work.) So for the system in a single eigenstate, the pressure equals $P$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-{{\rm d}}{\vphantom' E}^{\rm S}_q$$\raisebox{.5pt}{$/$}$${{\rm d}}V$. For a real system with uncertainty in energy, the pressure is defined as the expectation value:
\begin{displaymath}
\fbox{$\displaystyle
P = - \sum_{{\rm all}\;q}P_q \frac{{\rm d}{\vphantom' E}^{\rm S}_q}{{\rm d}V}
$} %
\end{displaymath} (11.8)

It may be verified by simple substitution that this, too may be obtained from the partition function, now by differentiating with respect to volume keeping temperature constant:

\begin{displaymath}
\fbox{$\displaystyle
P = k_{\rm B}T
\left(\frac{\partial \ln Z}{\partial V}\right)_{T{\rm\ constant}}
$} %
\end{displaymath} (11.9)

While the final quantum mechanical definition of the pressure is quite sound, it should be pointed out that the original definition in terms of force was very artificial. And not just because force is a poor quantum variable. Even if a system in a single eigenfunction could be created, the walls of the system would have to be idealized to assume that the energy change equals the work $\vphantom0\raisebox{1.5pt}{$-$}$$P{\,\rm d}{}V$. For example, if the walls of the box would consist of molecules that were hotter than the particles inside, the walls too would add energy to the system, and take it out of its single energy eigenstate to boot. And even macroscopically, for pressure times area to be the force requires that the system is in thermal equilibrium. It would not be true for a system evolving in a violent way.

Often a particular combination of the variables defined above is very convenient; the“enthalpy” $H$ is defined as

\begin{displaymath}
\fbox{$\displaystyle
H = E + P V
$} %
\end{displaymath} (11.10)

Enthalpy is not a fundamentally new variable, just a combination of existing ones.

Assuming that the system evolves while staying at least approximately in thermal equilibrium, the “first law of thermodynamics” can be stated macroscopically as follows:

\begin{displaymath}
\fbox{$\displaystyle
{\rm d}E = \delta Q - P{\,\rm d}V
$} %
\end{displaymath} (11.11)

In words, the internal energy of the system changes by the amount $\delta{Q}$ of heat added plus the amount $\vphantom0\raisebox{1.5pt}{$-$}$$P{\,\rm d}{V}$ of work done on the system. It is just energy conservation expressed in thermodynamic terms. (And it assumes that other forms of energy than internal energy and work done while expanding can be ignored.)

Note the use of a straight ${\rm d}$ for the changes in internal energy $E$ and volume $V$, but a $\delta$ for the heat energy added. It reflects that ${\rm d}{E}$ and ${\rm d}{V}$ are changes in properties of the system, but $\delta{Q}$ is not; $\delta{Q}$ is a small amount of energy exchanged between systems, not a property of any system. Also note that while popularly you might talk about the heat within a system, it is standard in thermodynamics to refer to the thermal energy within a system as internal energy, and reserve the term “heat” for exchanged thermal energy.

Just two more variables. The “specific heat at constant volume” $C_v$ is defined as the heat that must be added to the substance for each degree temperature change, per unit mass and keeping the volume constant. In terms of the first law on a unit mass basis,

\begin{displaymath}
{\rm d}e = \delta q - P{\,\rm d}v,
\end{displaymath}

it means that $C_v$ is defined as $\delta{q}$$\raisebox{.5pt}{$/$}$${\rm d}{T}$ when ${\rm d}{v}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. So $C_v$ is the derivative of the specific internal energy $e$ with respect to temperature. To be specific, since specifying $e$ normally requires two intensive variables, $C_v$ is the partial derivative of $e$ keeping specific volume constant:
\begin{displaymath}
\fbox{$\displaystyle
C_v \equiv \left(\frac{\partial e}{\partial T}\right)_v
$} %
\end{displaymath} (11.12)

Note that in thermodynamics the quantity being held constant while taking the partial derivative is shown as a subscript to parentheses enclosing the derivative. You did not see that in calculus, but that is because in mathematics, they tend to choose a couple of independent variables and stick with them. In thermodynamics, two independent variables are needed, (assuming the amount of substance is a given), but the choice of which two changes all the time. Therefore, listing what is held constant in the derivatives is crucial.

The specific heat at constant pressure $C_p$ is defined similarly as $C_v$, except that pressure, instead of volume, is being held constant. According to the first law above, the heat added is now $de+P{\,\rm d}{v}$ and that is the change in enthalpy $h$ $\vphantom0\raisebox{1.5pt}{$=$}$ $e+Pv$. There is the first practical application of the enthalpy already! It follows that

\begin{displaymath}
\fbox{$\displaystyle
C_p \equiv \left(\frac{\partial h}{\partial T}\right)_P
$} %
\end{displaymath} (11.13)