Several forcing axioms imply $2^{\aleph_0 }= \aleph_2$. What about $2^{\aleph_1}$?

On the one hand, it seems intuitive that $2^{\aleph_1 }> 2^{\aleph_0}$, because $\aleph_1 > \aleph_0$. However, I also know that, like many things involving the continnum function, that's actually independent of ZFC. So, since these forcing axioms seem to resolve what $2^{\aleph_0}$ is, do they also resolve $2^{\aleph_1}$? If so, is it also $\aleph_2$, or is it something bigger?

Martin's axiom, $\mathsf{MA}$, already implies that $2^\kappa=2^{\aleph_0}$ for every $\kappa<2^{\aleph_0}$. In particular, this means that (for example) $\mathsf{PFA}$ implies $2^{\aleph_1}=\aleph_2$.