Adiabatic-Lines-Intersection-Example
Solved Example:
(a) Prove that no two reversible adiabatic lines intersect each other on graphical plane.
(b) Prove that reversible adiabatic lines are steeper than reversible isothermal lines n P
– V plane.
Solution:
(a) Let us assume that two adiabatic curves intersect, as shown in figure at point ‘a’ and an isotherm intersect these curves at ‘b’ and ‘c’. A cycle is thus formed for which the heat is supplied during isothermal process and no heat is rejected because other two processes are adiabatic.
⇒ The cycle represents a PMM 2, Since a PMM-2 is not possible to be designed our assumptions are wrong. Hence, two adiabatic lines should never intersect on P – V plane.

(b) The rev. adiabatic and rev. isothermal curves are represented on P – V plane as shown in Figure (b). They are given by PV y = C and PV = constant. The slope of the curve is thus given by (dP/dV).
Now, d/dV (P.V) = dC / dV
⇒ P + V dP/dV = 0 …(1)
⇒ d/dV (PV g ) = dC / dV
⇒ P.y . v y–1 + v y . dP/dV = 0 …(2)
(dP/dV)
Isothermal
= – P/V …(3)
–ve shows q > 90 o
Similarly (dP/dV) adiabatic
= –y.P/V …(4)
–ve sign shown θ > 90 o
From equations (3) and (4)
(dP/dV) adiabatic
> (dP/dV)
Isothermal

Hence rev. adiabatic curve is more steeper than rev. Isothermal curve.