Some Comments on Partial Derivatives in Thermodynamics E. W. Anacker, Steven E. Anacker', and Willlam J. SwarLz Montana State University, Bozeman, MT 59717 In any serious study of thermodynamics, one is certain to encounter the following relationships2 in some form or another (Appendix 1provides examples):
For the differentialprocess in which P and Tare the variables,
then, since.. . [eq 41 refers to a process at constant enthalpy,dH = 0, and
For this reason, writers of texts on physical chemistry and thermodynamics often provide derivations. Although appearing to be straightforward and unambiguous, many of these treatments contain a subtle inconsistency. Chemistry ipstructors should use alternatives free of this objection and in harmony with principles taught by our colleagues in mathematics. For purposes of illustration, we paraphrase a common development of eq 3 and then, because it possesses the same flaw, quote part of the discussion of the Joule-Thomson effect from a thermodynamics text. First the derivation:
The quantity (JTIJP)His called the Joule-Thomson coefficient and denoted by p. Thus
If z in the derivation and H in the quotation are constants, the differentials dz and d H are indeed zero. But, so also are (dzlax),, (azlay),, ( a H l a P ) ~and , (aH1aT)p. Once this is established, subsequent manipulations are meaningless. I t ispossible, of course, for H to he avariable (as such, (8Hl ~ P ) and T (aHlaT).~need not vanish) and dlf to be zero, simultaneously. In this situation, (aHIaP)@ = -(aHl aT),dT, and we can write
From the functional relation z = z(x,y) we can write the total differential of z as
The condition of constant z is obtained by setting dz = 0. Thus,
On rearranging terms we get
Recognizing that dx,ldy, = (Jrlay),, using eq 2 to replace (JzlJy), by ll(aylJz), and multiplying through by (JylJz),, we obtain eq 3. And now the quotation:
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Brackets on the left side of the equation emphasize the fact that at this point dTldP is a ratio of differentials and not a derivative. Further progress now hinges upon showing that . path to eq 6 can be tortuous [ d T l d P ] d ~ = ~( a T l d P ) ~This (see Appendix 2), and one is advised to seek a more direct route. Although certainly not the only way of deriving relations 1,2, and 3, the following method is pedagogically attractive, especially when classroom time is at a premium (see Appepdix 3 for alternative methods). It is based on the well-known chain rule for partial differentiation:
Here it is understood that for new variables r and s, 2 = f(x,y), x = x(r,s), y = y(r,s), and, therefore, z = g(r,s). The rule remains valid when r (or s) is taken as x, y, or z. Partial derivatives such as (azlaz), and (azlar), have vaiues of 1and 0, respectively, and should not be interpreted as implying that 2 is a function of s or r as well as z. Obviously, r f s. When r is taken to be w and s to bey, eq 7 becomes
08541.
For the sake of clarity, we limit the number of independent variables tp two in our discussion. The relationships are not so restricted. Thus, if n, represents the additional independent variables n,, n,, n,. . . . eq 1 becomes
.
Equations 2 and 3 are generalized in similar fashion
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Journal of Chemical Education
Since (aylaw), = 0,eq 1follows immediately. With r = z and s = x, eq 7 becomes
Since (a~laz),= 0 and (azlaz), = 1,this reduces t o
which is eq 2. If we set r = y and s = z , we get from eq 7
We have utilized the fact that (azlay), = 0 and that (ay/ay), = 1. Combining eqs 8 and 9, we obtain
Multiplication through by (ayldz), produces eq 3. One may use the chain rule directly to get eq 6. With z = s = H, x = r = P , and y = T, eq 7 appears as
(E) = (E) (E) + (E) JP T aP
JP
d~
H
P
(9 JP
Finding relationship between the partial derivative (JTlaO~andthe differentials dTmd dP.
Example 2 show that ( a s ~ a=~( )a ~ v/a~)~/(a~/a~)~.
According to eq 3
H
(S) dT
Simple rearrangement gives the desired result. One may prefer the following approach instead. With x = H,y=P,andz=T,eq3is
P
0(E) JP s dS
= -1
T
We combine this with eq 10 and get
(E)(E) ( X ) = -1 JP
T
dT
H
JH
P
Thus
Example 3 Show that ( a s / a n p = Cp/T.
Equation 2 permits us to write
(E) = JP
T
(9(E) JP
H
JT
Cp i~a designation for (JH/JT)p. We can relate Cp and (JS/JT)p with the help of eq 1. Thus
P
Substitution of fi for ( ~ T / ~ P and ) H C, for (dHlaT)p gives eq 6.
+
According to dH = TdS VdP, which is another combined form of the first and second laws, T = (E)
as
P
We invoke eq 2 and get
Appendix 1
Hence We recall one of the combined forms of the first and second laws applicable to a closed system for which volume expansion work is the only possibility:
d G = VW-SdT The system's free energy, volume, pressure, entropy, and tempernlure aredmot~dhy G,1'. P,S,and T, respectively. Smce VdP - SdT is an exact differential,
We note that by combining eqs 11 and 12 we obtain
cp= T(JV/JTIP (aT/JP)s which has been used to determine heat capacities of liquids and gases at constant pressure.
According to eq 2, Part A Show that [dT/dP]dH=o= (dTIaP)H.
Therefore,
Proof: Let M in the figure designate the paint (PO,To,Ha) on surfaceF(P,T,H) = 0 (surface is not pictured). Draw line a throughMand tangent to the trace of the surface in the plane P = Po. Slope of a is (dH/dT)p. Volume 64
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Draw line 8through M and tangent to the trace of the surface in the plane T = TO.Slope of 8is (JH1aP)T. Draw line y through M and tangent to the trace of the surface in the plane H = Ho. Slope of y is ( ~ T I J P ) H . Since a,8, and y all lie in the tangent plane (see Parts B and C for comment), any two of the tangent lines may he used to define it. We select n and 8. Let ME be an arbitrary value for dT.The line segment M B is in plane P = Po and is parallel to the T axis. The magnitude of the partial differential (aHlJTlpdT is BC (also AG since GM = ME). Now choose d P (in plane T = To and parallel to P axis) so that the corresponding partial differential ( a H I a P ) d P (i.e., DE) is equal in magnitude but opposite in sign to (aH1JT)pdT. Thus dH = (JHl 0.Since GElies in ~ l a n e H = Ho and since dTlodT+ IdH/aP>RdP= . 1C and 1)Earc equal and parallrl to theHaxis, ADlies i t , a plnneuf wn-tnnt H. 111) i, pnrnllel to (:Land therrfore has the snme slupe. namely, [dTldP]d~+ Because AD and y lie in the tangent plane defined by or and B and also lie in planes of constant H, they are parallel and have the same slope. Hence, [dT/dPJd~.o = (JT/JP)H.
..~
~
= anBHi
+ aH&j - a+&
Since y lies in the plane defined by a and 8,it is perpendicular t o e X
B Therefore,
We rearrange to get
~
YT -=..yp
(BHIPP) (a$=~)
Consequently
Part B T h e result obtained i n P a r t A rests o n the assumption that tangents a,6, a n d y a r e coplanar. W h a t is t h e proof? Let S be the surface F(P,T,H) = 0 and C be any smooth curve lying on S and passing through M. If we let C have parametric representation
Appendlx 3.
Other Ways of Gettlng Eqs 1, 2, a n d 3
Method A Suppose z = z(x,y). Then
differentiation of F(P(t),T(t),H(t))and evaluation at M (denoted by subscript M) gives
We solve for dy and get ((azlJy), f 0).
and dPldt, respectively. FT, FH.T', Here Fp and F' are (~FIJP)T,H and H' are defined in an analogous manner. The left side of the equation is the dot product of the vectors V F = (Fp)ui + ( F d ~ j ( F H ) ~and k P h i V7',j HMk,where i, j, and k are unit vectors in the P, T, and H direct~ons,respectively. VFis normal to S at M. The second vector has the direction of the tangent line to C a t M. Since the dot product is 0, the two vectors are perpendicular. What we have found for C will also hold for all other smooth curves lying on S and passing through M, that is, their tangent lines at M are all perpendicular to VF. Hence, they are coplanar. Since the traces of S in the planes P = Po, T = To,and H = Ho all lie an S, their tangents-a, 8, y-lie in the tangent plane and are therefore copla-
For y = y(z,x)
+
+
+
Since r and x may be varied independently in eqs 14 and 15 without destroying the relationships, we must have equality between the coefficients of dz and between the eoeffcients of dx, that is,
nar.
Part C It is interestine to note that we can derive eq 5 from the fact that A . .?, and > arr cuplanar. Treating the ulngenrs as wctors, we let or, z,~,nndur, be the cornpunrntv o i a in rhr P, T, and Hdirrctionsand define the mrnpmcntr of B a n d 7 in a similar farhion. Thus
Equation 16 rearranges directly to eq 2. Sinee (Jylax), = l/(Jxlay), and (JylJz), = ll(azlJy),, we can rewrite eq 17 as
This transforms immediately to eq 3. Now suppose x = x(w,y). Then r =YP~+YT~
The slopes of a,8, and 7 at M are, respectively,
We substitute this into eq 13 and get
For z = z ( w , ~ )
The coefficients of dw in eqs 18 and 19 must he equal. Thus We note that a p = PT = YH = 0. a X 8 is a vector normal to the plane generated by u and 8. Its components are easily found.
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Journal of Chemical Education
which is eq 1.
MethodB Application of Chain Rule to an Implicit Function Consider the function F(x,y,z) = 0 . W e apply the chain rule for partial differentiation: F
(E),+
+ Fy
F
(E),+ (g)y ($)>+ Fz($Iy + ($Iy + Fz
Fy
= F,
= F,
F*
=0
F,
= 0 (21)
F,
($Iz
F
($Ir ($Ix(g), ($Iz
Fx ($)x
+ F,
($)*+ F, ($)% = F,
=0
(22)
+ Fa = 0
(23)
($)= + F, (z),F, + F, ($)= =0
(24)
+ Fy
+ F,
+ F,
(20)
+ Fz
= Fy
=
where F, = (JFlax),,, F, = (JFldy),, and F, = ( a F l a ~ ) ,Elimina,~ tion of F, between eqs 20 and 21 gives
which is eq 1. From eqs 23 and 24 we find that
Hence eq 2. Equations 2 1 , 2 2 , and 23 may be written
Multiplication of these three equations together gives eq 3, that is,
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August 1987
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