For now, let's ignore the expansion of the container due to the heating and just focus on the stress in the wall of the pressure vessel. I will also examine the case of a thin-walled, spherical vessel, but the same procedure may be applied for other geometries.
Compute Stress State
First, you must compute the pressure in the walls. To do this, imagine a cutting plane through any diameter of the vessel. Now, balance forces on one half of the vessel. In one direction, you have the force of the fluid pressure, which totals to $F = \pi R^2 P$. In the other direction, the only balancing force is the stress in the walls acting over the exposed cross-section. The force here equals $F = 2\pi R t \sigma_{wall}$, where $t$ is the thickness of the wall. By setting these forces equal, we can compute that the stress in the wall is
$$
\sigma_{wall} = \frac{P R}{2t}.
$$
Now, due to the symmetry of the problem, and making the (valid) simplification that of zero through-thickness stress, we can write the full stress state of any point in the wall:
$$
\mathbf{\sigma} = \sigma_{wall}
\left(
\mathbf{e}_{\theta}
\otimes
\mathbf{e}_{\theta}
+
\mathbf{e}_{\phi}
\otimes
\mathbf{e}_{\phi}
\right)
+
0\;
(\mathbf{e}_{r}
\otimes
\mathbf{e}_{r})
$$
Check Yield Criterion
To compute the point where yield happens, you must compute the Mises equivalent tensile stress. There are a variety of equivalent definitions, but all will lead you to the conclusion that in order to prevent plastic deformation, $\sigma_{wall} < \sigma_y$, where $\sigma_y$ is the material's uniaxial tensile yield stress. Thus, the maximum pressure the vessel can contain before yield is
$$P < \frac{2\,t\,\sigma_{y}}{R}$$
Other Considerations
You still have to consider fracture, but that's a separate discussion that is more suited to an engineering class. (If this is what you're looking for, I can go there...) With this comes the "leak before break" criterion, which puts an upper bound (counter-intuitive, but it checks out) on the safe thickness of the vessel walls.
Another point to keep in mind is that the pressure in the tank will change with temperature. Be sure to account for this in your analysis with $PV = nRT$ or some other appropriate state equation.
As others have mentioned in the comments on your question, there are very strict codes that put factors of safety on essentially every aspect of the tank design. This would be the place to start if you're actually going to build something.
Best Answer
In physics elasticity is defined as the ability of a material to resist a distorting influence and return to its original size and shape when the distorting influence is removed (source Wikipedia).
Young’s Modulus is the ratio of applied stress to resulting strain in the linear elastic region of behavior. Therefore, they greater Young’s modulus the stiffer a material is, that is, the greater the materials ability to resist a distorting influence (applied stress). For this reason, given the above general definition of elasticity, Young’s modulus is also called the modulus of elasticity.
Due to the highly overall non linear behavior of rubber I believe Young’s modulus for rubber is usually quoted forces small loads. The values for rubber are much lower than steel meaning that rubber is much less able to resist a distorting influence than steel as would obviously be expected.
Given the above definitions of elasticity and Young’s modulus we would conclude that the “elasticity” of rubber is less than steel. It is admittedly counter intuitive because on the one hand we think of rubber bands as being highly elastic in the sense that they’re easy to stretch. But on the other hand if they’re easier to stretch that means they’re less able to resist distortion, which is consistent with the physics definition of elasticity and Young’s modulus.
Hope this helps.