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

$$I_{\rm A}\approx \mbox{6.022 1 10$\POW9,{23}$}$$

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,

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

$$\left(\frac{\partial Z}{\partial T}\right)_{V{\rm constant... ...ntom' E}^{\rm S}_q e^{-{\vphantom' E}^{\rm S}_q/{k_{\rm B}}T}.$$

(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

$$\fbox{$\displaystyle E = {k_{\rm B}}T^2 \left(\frac{\partial \ln Z}{\partial T}\right)_{V{\rm constant}} $} %$$ (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

$$-P{ \rm d}{V}.$$

(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:

$$\fbox{$\displaystyle P = - \sum_{{\rm all}\;q}P_q \frac{{\rm d}{\vphantom' E}^{\rm S}_q}{{\rm d}V} $} %$$ (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:

$$\fbox{$\displaystyle P = k_{\rm B}T \left(\frac{\partial \ln Z}{\partial V}\right)_{T{\rm constant}} $} %$$ (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 $\vphantom{0}\raisebox{1.5pt}{$-$}$$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

$$\fbox{$\displaystyle H = E + P V $} %$$ (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:

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

In words, the in­ter­nal en­ergy of the sys­tem changes by the amount $\delta{Q}$ of heat added plus the amount $\vphantom{0}\raisebox{1.5pt}{$-$}$$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,

$${\rm d}e = \delta q - P{ \rm d}v,$$

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:

$$\fbox{$\displaystyle C_v \equiv \left(\frac{\partial e}{\partial T}\right)_v $} %$$ (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

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