Explicit and Implicit Functions and Total Differentiation

1. Explicit and Implicit Functions and Total Differentiation

If P is a known function of T and v, the explicit function for P is P = P(v,T),

(14) and its total differential may be written in the form ∂ P 

dP =

 (15) ∂ v  T

dv

 ∂ T  v

dT .

Consider the P, T, v relation

(16) where a and b are constants. Equation (16) is explicit with respect to P, since it is an explicit

P = RT/(v–b) – a/v 2 ,

function of T and v. On the other hand, v cannot be explicitly solved in terms of P and T, and, hence, it is an implicit function of those variables. The total differential is useful in situations that require the differential of an implicit function, as illustrated below.

b. Example 2 If state equation is expressed in the form

(A) find an expression for ( ∂v/∂T) P , and for the isobaric thermal expansion coefficient β P =

P = RT/(v–b) – a/v 2 ,

(1/v) ( ∂v/∂T) P . Solution

For given values of T and v, and the known parameters a and b, values of P are unique (P is also referred to as a point function of T and v). Using total differentiation

dP = ( ∂P/∂v) T dv + ( ∂P/∂T) v dT. (B) From Eq. (A)

( ∂P/∂v) T = –RT/(v – b) 2 + 2a/v 3 , and

(C)

( ∂P/∂T) v = R/(v – b) (D) Substituting Eqs.(C) and (D) in Eq. (B) we obtain

(E) We may use Eq. (E) to determine ( ∂v/∂T) P or ( ∂v/∂P) T . At constant pressure, Eq. (E)

dP = (–RT/(v – b) 2 + 2a/v 3 ) T dv + (R/(v – b)) v dT.

yields

(F) so that (

0 = (–RT/(v – b) 2 + 2a/v 3 ) T dv + (R/(v – b)) v dT,

2 ∂v 3 P / ∂T P )=( ∂v/∂T) P = –(R/(v – b))/(–RT/(v – b) + 2a/v ), (G) and the isobaric compressibility

= 1/v ( = –R/(v(–RT/(v–b) + 2a(v–b)/v β 3 P ∂v/∂T) P )). (H) Remarks

It is simple to obtain ( ∂P/∂T) v or ( ∂P/∂v) T from Eq. (A). It is difficult, however, to obtain values of ( ∂v/∂T) P or ( ∂v/∂P) T from that relation. Therefore, the total differentiation is em- ployed.

Note that Eqs. (C) and (D) imply that for a given state equation: ( ∂P/∂T) v = M(T,v), and

(I) ( ∂P/∂v) T = N(T,v), and

(J) Since,

dP = M(T,v) dv + N(T,v) dT, (K) Differentiating Eq. (C) with respect to T,

(L) Likewise, differentiating Eq. (D) with respect to v,

∂/∂T (∂P/∂v) = (∂M/∂T) v = –R/(v – b) 2 .

(M) From Eqs. (L) and (M) we observe that

∂/∂v (∂P/∂T) = (∂N/∂v) T = –R/(v – b) 2 .

2 ∂M/∂T = ∂N/∂v or ∂ 2 P/ ∂T∂v = ∂ P/ ∂v∂T. (N) Eq. (N) illustrates that the order of differentiation does not alter the result. The equation

applies to all state equations or, more generally, to all point functions (see next section for more details). From Eq. (B), at a specified pressure

( ∂P/∂v) T dv + ( ∂P/∂T) v dT = M(T,v) dv + N(T,v) dT = 0. Therefore, ( ∂v/∂T) P = –M(T,v)/N(T,v) = –( ∂P/∂T) v /( ∂P/∂v) T .

(O) Eq. (O) can be rewritten in the form

( ∂v/∂T) P ( ∂T/∂P) v ( ∂P/∂v) T = –1, (P) ( ∂v/∂T) P ( ∂T/∂P) v ( ∂P/∂v) T = –1, (P)