Exact (Perfect) and Inexact (Imperfect) Differentials

2. Exact (Perfect) and Inexact (Imperfect) Differentials

If the sum Mdx+Ndy (where M = M (x,y) and N = N (x,y)) can be written as d(sum), it is an exact differential which can be expressed in the form

dZ = M(x,y) dx + N(x,y) dy. (17) If the sum cannot be written in the form d(sum), it is more properly expressed as δZ = M(x, y) dx + N(x, y) dy,

(18) For instance, the expression xdy + ydx is an exact differential, since it can be written as d(xy),

i.e., dZ = xdy + ydx = d (xy), where Z = xy + C. A plot of the function Z versus x and y is a surface. The difference Z 2 -Z 1 =x 2 y 2 -x 1 y 1 depends only on the points (x 1 ,y 1 ), (x 2 ,y 2 ) and not on the path connecting them. But the expression x 2 dy + ydx cannot written as d(xy) and hence the sum x 2 dy + ydx denoted as

δZ = x 2 dy + ydx.

Exact differentials may also be defined through simple integration. Consider a differ- ential expression that is equal to 9x 2 y 2 dx + 6x 3 ydy, the integration of which can be problem- atic. It is possible to specify a particular path (Say path AC in Figure 9 ), e.g., by first keeping x constant while integrating the expression with respect to y, we obtain Z = 3 x 3 2 y (for a mo- ment let us ignore the integration constant). On the other hand we can keep y constant and integrate with respect to x and obtain Z = 3 x 3 y 2 , which is same as before. Only exact differ- entials yield such identical integrals. Hence, the the sum 9x 2 2 y 2 dx + 6x 3 3 ydy is an exact differ- ential. Consider 6x ydx + 6x ydy. If we adopt a similar procedure we get different results (along constant x, Z = 3x 3 y 2 and along constant y, Z = 2 x 3 y). Instead of integrating the alge- braic expressions, we can also integrate the differentials between two given points. For an ex- act differential of the form dZ = Mdx + Ndy the integrated value between any two finite points

(x 1 ,y 1 ) and (x 2 ,y 2 ) is path independent. If the integration is path dependent, an inexact differen- tial (of the form δZ = Mdx + Ndy) is involved.

Consider the exact differential dZ = xdy + ydx. The term xdy represents the elemental area bounded by the y–axis (GKJH) in Figure 10 and ydx is the corresponding area bounded by the x–axis (KLMJ). The total area xdy + ydx is to be evaluated while moving from point

F(x 1 ,y 1 ) to point C(x 2 ,y 2 ). An arbitrary path

“a” can be described joining the points F and

C. Integrating along “a” from F to C, the integrated area ∫(xdy) + ∫(ydx) is a sum of the areas EFaCD + AFaCB. Using another path

“b” results in exactly the same area, since regardless of the path that traversed to con- nect F and C, the integrated value is the

same, i.e., Z 2 –Z 1 = (x 2 y 2 –x 1 y 1 ). Therefore,

∫(xdy + ydx) is path independent.

c. Example 3 Is the function

(A) an exact or inexact differential? Prove or

9x 2 y 2 dx + 6x 3 ydy

disprove by adopting path integration ( Figure 9 ).

Solution Figure 9: Illustration of an exact differential

The difference Z A –Z B can be determined

using path integration.

Figure 10: Illustration of an exact differential, dz = xdy + ydx.

by moving along paths ACB or ADB, as illustrated in Figure 9 . Consider the path ACB along AC for which x = 1. Integrating the relation while keeping x constant,

(A) Along the path CB, y is held constant (y = 4). Integrating the relation at constant y, Z

(B) Z B –Z A = Eq. (A)+Eq. (B) = 381.

= 336, i.e.,

(C) The integration can also be performed along path ADB, i.e., along AD, keeping y at a

constant value of 1. Using the relation

(D) Similarly, x is constant (x = 2) along DB, so that Z –Z

From Eqs. (D) and (E), Z B –Z A = 381.

The integral is the same for paths ACB and ADB. Thus the differential is an exact differential. Remarks

If the integration is performed along the path ACB and continued from B to A along BDA, the cyclic integral dz

dz ∫ = 381 – 381 = 0. =∫

dz

+∫ BDA

ACB

The difference (Z B –Z A ) is independent of the path selected to reach point B from point A, since Eq. (A) is an exact differential. The function Z is a point function, since it only depends upon the selected coordinates.

In the context of Figure 9 , the value of Z B -Z A via path C will be the same as via path

D. Thus if we take a cyclic process from A-B via path C and then from B-A via path

D, there will be no net change in Z , i.e ∫ dZ = 0.

d. Example 4

Determine if 6x 2 y 2 dx + 6x 3 ydy is an exact or inexact differential.

Solution Consider the path ACB along which Z

(6x y / 3) 3 2 24 14 , , = 224 so that (Z B –Z A ) = 269.

Likewise, following the path ADB Z

= 14 and Z B –Z D = (3x y ) 21 , = 360 so that (Z B –Z A ) = 374.

The value of (Z B –Z A ) along the path ADB does not equal that along path ACB. Conse- quently, the expression for Z is not a property, since it is path dependent, and is, therefore, an inexact differential.

Remark If the integration is first performed along the path ACB and continued from B back to A along BDA, the integrated value is 269 (ACB)–374 (BDA) = –105. If the integration is first performed along the path ACB and continued from B back to A along BDA, the inte- grated value is ∫ δZ ≠ 0 since 269 (ACB)-374 (BDA) = -105. In general, the cyclic in-

tegral of an inexact differential is nonzero.

a. Mathematical Criteria for an Exact Differential i.

Two Variables (x and y) The path integration procedure helps determine whether a differential is exact or in-

exact. However, the mathematical criterion that will now be discussed avoids lengthy path integration and saves time. Example 2 shows that a point function of the form P = P(T,v) may

be written as dP = M(T,v) dv + N(T,v) dT,

(19) and it possesses the property ∂M/∂T = ∂N/∂v,

(20) i.e., ∂ 2 P/ ∂v∂T = ∂ 2 P/ ∂T∂v. Substituting for x = v, y = T, and Z = P, dZ = ( ∂Z/∂x) y dx + ( ∂Z/∂y) x dy = M(x,y) dx + N(x,y) dy.

(21) The function M is called the conjugate of x, and N is the conjugate of y. The necessary and

sufficient condition for Z to be a point function is given by Eq. 20, namely,

(22) This is another criterion describing an exact differential, and it is also referred to as the condi-

2 ∂ 2 Z/ ∂y ∂x = ∂ Z/ ∂x∂y or (∂M/∂y)

x =( ∂N/∂x) y .

tion of integrability. A differential expression of the form M(x,y) dx + N(x,y) dy is said to be in the linear differential (or Pfaffian) form. A differential expression derived from a point function or a scalar function, such as P = P(T,v), in the Pfaffian form satisfies the criterion for being an exact differential.

e. Example 5 Consider the expression –(Ry/x 2 )dx + (R/x) dy, where R is a constant. Is this an exact dif- ferential? If so, integrate and determine “Z”. Solution M = –R y/x 2 , and N = R/x. Therefore, (

∂M/∂y) 2 x = –R/x 2 , and ( ∂N/∂x) y = –R/x , i.e., the criterion for being an exact differential is satisfied by the expression. Therefore,

dZ = –Ry/x 2 dx + R/x dy, i.e.,

(A) (A)

eg: P = RT/v

Figure 11. Z-x-y or P-T-v surface.

∂Z/∂y = R/x = N, and ∂Z/∂x = –Ry/x 2 = M. (B) To determine Z, we can integrate along constant x to obtain Z = Ry/x + f(x).

(C) Upon differentiating Eq. (C) with respect to x ∂Z/∂x = –Ry/x 2 + f´(x).

(D) However, from Eqs. (A) and B, (–Ry/x 2 ) = M so that f´(x) = 0, i.e., f(x) is a constant, and Z = Ry/x + C.

(E) Remarks

Assume that C = 0 and R = 8. Once x and y are specified (say, respectively with val- ues of 2 and 4) in Eq. (E), the value of Z is fixed (= 16) irrespective of the path along which the point (x = 2, y = 4) is reached. In this case Z is called a point function. If C = 0, y = T, x = v, and Z = P, then using Eq. (E) the point function that is obtained is of the form

P = RT/v, which is the familiar ideal gas equation of state. A plot of P with respect to T is presented in

Figure 12 while a plot of P versus both T and v describes a surface ( Figure 11 ) Starting a process at point (T 1 , v 1 ) (i.e., point A, of Figure 12 ), the pressure P 2 can be determined at a point (T 2 ,v 2 ) (i.e., Point B of the figure) using either of the paths ACB or ADB. From the preceding discussion we note that dz ∫ = 0 if Z is a point function. There-

fore, if Z denotes the temperature T, then ∫ dT =0 . However, if Z denotes the heat transfer Q which is not a point function, ∫ δQ = 0. Point functions such as T = T(P,v) = Pv/R can be fore, if Z denotes the temperature T, then ∫ dT =0 . However, if Z denotes the heat transfer Q which is not a point function, ∫ δQ = 0. Point functions such as T = T(P,v) = Pv/R can be

ii. Three or More Variables The exactness criteria can be generalized to systems involving more than two vari-

ables. Consider that Z is described by three variables x 1 ,x 2 , and x 3 , i.e.,

(23) The total differential of Z is

Z = Z(x 1 ,x 2 ,x 3 ).

(24) Since dZ is exact

dZ = ∂Z/∂x 1 dx 1 + ∂Z/∂x 2 dx 2 +( ∂Z/∂x 3 ) dx 3 .

(25) We now have three conditions in terms of all three variables. Generalizing these expressions

∂Z/∂x 1 = ∂Z/∂x 2 , ∂Z/∂x 2 = ∂Z/∂x 3 , and ∂Z/∂x 3 = ∂Z/∂x 1 .

for k variables when

(26) The total differential may be written in the form dZ k =∑ ( ∂∂ Z / x dx

Z = Z(x 1 ,x 2 ,x 3 ,…,x k ).

and by analogy the criteria describing an exact differential for this case are ∂ ∂ Z

(28) ∂ x j ∂ x i

( ), j ≠1 .

When Z is a function of two variables alone, i.e., Z(x 1 ,x 2 ), one criterion describes an exact differential. If more than two variables are involved, i.e., Z (x 1 ,x 2 ,x 3 ,…,x k ) it is possible to write the following equations in terms of x 1 , namely, 2 2 ∂ 2 / ∂x 1 ∂x 2 = ∂ / ∂x 2 ∂x 1 , ∂ / ∂x 1 ∂x 3 = ∂ 2 / ∂x

3 ∂x 1 ,..., etc., and generate (k–1) equations for the k variables. Likewise, in terms of

, ..., etc. However, ∂ 2 ∂x 2 ∂x 1 ∂ ∂x 1 ∂x 2 ∂ ∂x 2 ∂x 3 ∂ ∂x 3 ∂x 2 ∂ /

∂x 2

1 ∂x 2 = ∂ / ∂x 2 ∂x 1 , which appears in both equations so that only ((k–1)–1) equations can be generated in terms of x 2 .

Similarly, ((k–1)–2) criteria can be generated for x 3 , and so on resulting in ((k–1) + ((k–1)–1) + ((k – 1) – 2) + ((k – 1) – 3)+....) criteria. Simplifying, k(k–1) – (1 + 2 + 3 + ... + k–1) = k(k–1) – (1/2)(k–1)k = k(k–1)/2. Therefore, the number of criteria describing an exact differ-

ential of Z(x 1 ,x 2 ,x 3 ,…,x k ) are k(k–1)/2. When a point function involves more than two vari- ables i.e. Z = Z(x 1 ,x 2 ,x 3 ,…,x k ) a hypersurface is produced, e.g., the plot of P = N 1 R T/V +

N 2 R T/V +.. = P (T,V, N 1 ,N 2 ,N 3 ,...).