I am studying chapter 14 of Burris and Sankappanavar on fully invariant congruences. In Lemma 14.7 is proved that if $\theta$ is a fully invariant congrence on $T(X)$, then for $p = q \in Id(X)$ we have: $$ T(X)/\theta \vDash p=q \Leftrightarrow \langle p,q \rangle \in \theta $$ The proof ends by the following remark: $$ T(X)/\theta \vDash p=q \Leftrightarrow V(T(X)/\theta) \vDash p=q $$ because the same identities are satisfied by a class of algebra's and the variety generated by this class. Upon this notion is concluded that $T(X)/\theta$ is free in $V(T(X)/\theta)$.
But I don't know how to support this claim, maybe because I lack intuition on `being a free algebra'.