Is the mean distance to a body's surface times its area always proportional to its volume? This question came up because the ideal gas law says that for, say, a gas enclosed in a bottle the pressure is:
$ P \propto \frac{N \cdot T}{V}$
Where P: Pressure, N: Number of Gas particles in the bottle, T: Temperature and V: Volume of the bottle.
I am only investigating the dependence on volume:
The Pressure does not depend on the shape of the volume.
I find it intuitive that the pressure would be inversely proportional to the bottle's surface area. I also find it plausible that the pressure would be inversely proportional to the average distance of a point inside the volume to the volume's surface, as the particles inside the volume will take more time to impinge on the surface (and cause a momentum transfer that causes pressure) the farther the particles are away from the bottle's surface.
The average distance to the surface should be understood as the average over all points inside the volume and all solid angles subtending each point to the surface, and not averaged over the surface itself. This is because the average distance needed here should be a sum of the contributing distances weighed by the likelihood this distance will be traversed by a particle: This likelihood is identical for all directions. If one were to calculate the average by weighing the distances proportionally to the (infinitesimal) surface areas that are connected to the point whose average distance is being calculated, one would weigh those directions more that make a very small or large angle with the surface, as the solid angle cones intersect a larger piece of the surface there. 
I checked the conjecture for a sphere, and it seems to work, but I do not see how to extend this result to all "well behaved" volumes.
Also, the proportionality factor between (Average Distance multiplied by Surface Area) and Volume would have to be identical for all body shapes....
 A: So if $P$ is in a solid, $M$, and the boundary of $M$ is $S$, then you're talking about the min of $d(P,s)$, where $s$ is any point of $S$. In that case, the conjecture is almost certainly wrong. Consider, on the unit sphere, drilling a "well" from the north pole towards the south pole, but not quite reaching it. If the well is very narrow, then the volume is essentially unchanged, as is the surface area. But all points in the "middle" of the sphere (distance < $\frac{1}{2}$ from the origin) are now "closer" to the surface than they were, by about an average of $\frac{1}{4}$.
I would add that the conjecture is also false if for each interior point, instead of minimum distance to the surface, an average is taken. For that, it suffices to consider very long, thin cylinders ($l \gg r$). For any point, the average distance to the surface is at least $l/4$ and no more than $l+2r$ (in fact, a much stricter bound can be put on that, but there is no need here). The average for all points is thus also within these bounds. But the ratio of volume to surface area can be made arbitrarily close to $(\pi r^2l)/(2\pi rl) = r/2$, so the two quantities cannot possibly be proportional to each other for cylinders of sufficiently different "aspect ratios" $r/l$.
A: The first conjecture is not quite right, but it can be modified into a second conjecture which is correct.
Conjecture 1:
The average distance-to-surface (over points in the interior and solid angles from them) times the surface area is proportional to the volume.
We can disprove this by contradiction.  Consider two blobs $B_1$ and $B_2$ connected by a short (negligible) thread to form the full volume $B_3$.  If the conjecture were true, we would have:
 \begin{align}
 V_1 & = k S_1 \hat d_1 \tag{1} \\
 V_2 & = k S_2 \hat d_2 \tag{2} \\
 V_3 & = k S_3 \hat d_3 \tag{3}
 \end{align}
By construction, we know that:
 \begin{align}
 V_1 + V_2 & = V_3 \tag{4} \\
 S_1 + S_2 & = S_3 \tag{5} \\
 \frac{V_1}{V_3} \hat d_1 + \frac{V_2}{V_3} \hat d_2
  & = \hat d_3 \tag{6}
 \end{align}
and then we can use (5) and (6) to rewrite (3) as
 \begin{align}
 V_3
  &= k (S_1 + S_2)
   \left(
    \frac{V_1}{V_3} \hat d_1 + \frac{V_2}{V_3} \hat d_2
   \right)
    \tag{7}
 \\
 V_3^2
  &= k (S_1 + S_2)
   \left(
    V_1 \hat d_1 + V_2 \hat d_2
   \right)
    \tag{8}
 \\
 V_3^2
  &=  k S_1 V_1 \hat d_1
    + k S_1 V_2 \hat d_2
    + k S_2 V_1 \hat d_1
    + k S_2 V_2 \hat d_2
    \tag{9}
 \\
 V_3^2
  &=  V_1^2
    + k S_1 V_2 \hat d_2
    + k S_2 V_1 \hat d_1
    + V_2^2
    \tag{10}
 \\
 V_3^2 - V_1^2 - V_2^2
  &= k (  S_1 V_2 \hat d_2
        + S_2 V_1 \hat d_1 )
    \tag{11}
 \\
 \left( \frac{V_3^2 - V_1^2 - V_2^2}{k} \right)
  &=      S_1 V_2 \hat d_2
        + S_2 V_1 \hat d_1
    \tag{12}
 \end{align}
where we get from (9) to (10) by using (1) and (2).
Now we see that we are in a strange situation.
If we squish the blobs around without changing their volumes,
then the left side of (12) is constant,
while the right side has strange, physically meaningless cross-terms.
(These arose because the conjecture is multiplying quantities that essentially combine additively over independent subparts.)
In particular,
if we use a low-distance, high-surface volume (e.g. a wide wrinkled crêpe or a very long wrinkled thread) for $B_2$, then we can hold $V_2$ constant while making $\hat d_2$ arbitrarily small, with $S_2$ arbitrarily large.  If we leave $B_1$ alone while we do this, this will make the right hand side of (12) arbitrarily large, contradicting the intransigence of the left hand side.$\tag*{$\blacksquare$}$
However, your original intuition is correct.
But particles hitting the surface many times due to a short distance between bounces contribute a lot to the pressure while particles that rarely hit the surface due to long distances are almost irrelevant,
so instead of the average distance,
we should use the average inverse total distance,
where the total distance is the full distance traveled between bounces
 (i.e. the sum of the two distances in opposite directions, the width).
Conjecture 2:
The harmonic mean (over points in the interior and solid angles from them) of the width (at that point, in that direction), times the surface area, is equal to 4 times the volume.
\begin{align}
\frac{1}{\mbox{harmonic mean }}\ 
=\ & \frac{1}{V} \int_{p \in B} \frac{1}{2\pi} \int_\phi \frac{1}{\mbox{width (through $p$ to surfaces in direction $\pm\phi$)}}\;\;
 dB d\Phi
\\=\ & \frac{1}{2\pi V}
 \int_\phi \int_{p \in B} \frac{1}{\mbox{width}}\ 
 dB d\Phi
\\=\ & \frac{1}{2\pi V \cdot 2 \epsilon}
 \int_\phi \int_{p \in B} \left\{{
 \begin{array}{ll}
  1 & \mbox{if $p$ is within $\epsilon$ of a surface
   in direction $\pm \phi$} \\
  0 & \mbox{otherwise}
 \end{array}}\right\}
 dB d\Phi
\\=\ & \frac{1}{2\pi V \cdot 2 \epsilon}
 \int_{p \ \in \  surface(B)} \int_{\phi\ pointing\ out\ of\ B}
  \epsilon \sin (\mbox{angle from $\phi$ to surface})
 \ dS
\\=\ & \frac{1}{2\pi V \cdot 2 \epsilon}
 \int_{p \ \in \  surface(B)}
  2 \pi \cdot \frac{1}{2} \epsilon
 \ dS
\\=\ & \frac{1}{2\pi V \cdot 2 \epsilon}
  \cdot S \cdot \pi \epsilon
\\=\ & \frac{S}{4V}
 \tag*{$\blacksquare$}
\end{align}
