Demonstrate that the straight line 1−5 corresponding to the isothermal-isobaric phase transition cuts the Van der Waals isotherm so that areas I and II are equal (Fig. 2.5).
2.204. Let us apply Eq. (2.4e) to the reversible isothermic cycle 1−2−3−4−5−3−1:T∮dS=∮dU+∮pdV Since the first two integrals are equal to zero, ∮pdV=0 as well. The latter equality is possible only when areas I and II are equal. Note that this reasoning is inapplicable to the cycle 1−2−3−1, for example. It is irreversible since it involves the irreversible transition at point 3 from a single-phase to a diphase state.