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
The stress applied to the wire is (in this theoretical example) the exact same.
There is the same force applied through the whole length of the wire (neglecting small effects due to loading mechanism and weight); and the wire in each case has the same diameter/area, so the stress is also the same in every part of both wires.
Because of the known stress-strain relationships, we know that in the linear-elastic range, stress is directly proportional to strain. Strain is a measure of relative elongation compared to a total length. So strain basically measures the ratio of elongation compared to original unstained length. For the simple strain considered here, we can define it as $$\epsilon = \frac {\Delta L}{L}$$
Because strain is a measure of ratio, and the stresses are the same, value of $\epsilon$ will be the same, therefore the change in length ($\Delta L$) will vary in proportion to the total length ($L$).
You can also consider this like springs in series. (A good question on that here) Basically, if you put several springs with the same stiffness together end to end, then apply a force to all of them (like by hanging a weight off them for example), the springs in series will have a lot more motion than a single spring of the same stiffness. This is because each element experiences the same load, and stretches the same amount. The more total elements you have, the more total stretch when identical loads are applied.