Given a regular tiling of the hyperbolic plane, with p-sided polygons and q polygons meeting at each vertex, is the Gaussian curvature of the plane determined by (p,q)?
Intuitively it seems like having five pentagons meeting at each point would need greater (negative) curvature than four pentagons. But I'm just guessing? (And have never formally studied hyperbolic geometry.)
This depends on the edge length of the polygons.
Usually the convention is that you use a hyperbolic plane with curvature $-1$ unless indicated otherwise. In such a plane, you can observe that small figures look almost Euclidean, but the larger a figure is, the more pronounced the difference in geometry becomes. Actually the angle deficit can be directly used as a measure of area. So for five pentagons meeting at each corner, you need a smaller interior angle, which means a larger angle deficit and thus a larger area than for for pentagons meeting at each corner.
But you could go about this the other way. You could fix the edge length of your pentagon, and then observe that the curvature has to be different in the two situations. Because essentially all lengths in hyperbolic geometry are relative to the curvature of the plane, which is a Gaussian curvature and thus a square of lengths. If you quadruple the curvature, you double all the lengths. Fixing the edge length of the polygon, you would observe an increasing angle deficit the more negative your curvature becomes, so you can fit in more of them around each corner, just as your intuition told you.
The formula $\text{surplus angle}=\text{curvature}\times\text{area}$ holds in general. For hyperbolic geometry you have negative curvature and an angle deficit which is a negative surplus. For elliptic geometry surplus and curvature are positive, and for Euclidean both are zero. To fit more pentagons around a corner, you need more deficit and therefore have to increase either the absolute value of the curvature or the area of each pentagon. If you have a $p$-sided polygon, the Euclidean interior angle is $\frac{p-2}p\pi$. If you fit $q$ polygons around each corner the hyperbolic interior angle is $\frac2q\pi$. So the surplus angle is $\left(\frac2q-\frac{p-2}p\right)\pi$. For four pentagons this is $-0.1\pi$, for five pentagons it is $-0.2\pi$ so the value of the curvature would have to be twice as high in the latter case if you fix the area, or the area would have to be twice as large (corresponding to a $\sqrt2$ increase in edge length) if you fix the curvature.