Various Confusing Issues on Applications of D + f = C + 2

Consider a pure liquid $A$ in contact with the air. The degrees of freedom can be determined in several equilvalent ways.

A

Consider the system composed of two components, the pure liquid $A$ and air and restrict that the total pressure is 1 atm.
$(D + f = C + 2) \rightarrow (D+f = C + 1)$.
Therefore, $D = 2-2 +1 = 1$.

B

Considering that the system consists of three components: $A$, O$_2$, N$_2$ and has two additional restrictions: 1) $\Sigma P =1$ ${P_{\mbox{O}_2}}/{P_{\mbox{N}_2}}=$ constant, then $(D + f = C + 2) \rightarrow (D+f = C + 0)$.
$D=3-2+0=1$ as before.

C

Disregard the air: $C=1$. $f=2$ and therefore $D=1$. The liquid has an equilibrium vapor pressure which is a function of temperature. One can pick either the vapor pressure or the temperature independently, but not both.