ε1 + p1v1 - p2v2 = ε2,

or

ε1 + p1v1 = ε2 + p2v2.

According to the symbols chosen by Gibbs, χ1 = χ2.

As χ1 is determined by T1 and p1, and χ2 by T2 and p2, we obtain, if we take T1 and p2 as being constant,

( δχ1 ) dp1 = (δχ2 ) dT2.
δp1 T1δT1 p2

If T2 is to have a minimum value, we have

( δχ1 ) = 0 or (δχ1 ) = 0.
δp1 T1δv1 T1

From this follows

( δε1 ) + [δ(p1v1) ] = 0.
δv1 T1δv1 T1