I'm going to combine two criteria to find the wall thickness (the only unknown here).
The first criterion is based on yielding of the wall material prior to to failure as a result of crack formation and propagation, so the inspection team can observe the plastic deformation and the pressure within the tank be released before total failure of the tank. So materials with large critical crack length are suitable, what you can do, is, rank the materials with higher fracture toughness. how ? See further.
The second criterion is Leak before break. The cracks may grow and penetrate throughout the wall thickness, but, they never propagate rapidly, again inspector sees the leaking of liquid, and the catastrophic failure can be prevented.
Here, i assume the condition of plain stress, and i assume the pressure vessel is in one whole piece, and i neglect what ever shape, the nozzles introduce on the tank. So a simple cylindrical vessel with spherical heads and no stress risers. The volume of the tank and the pressure are known.
The stress cause by pressure inside are: $$\begin{bmatrix}
\sigma_{rrp} & 0 & 0 &\\
0 & \sigma_{\theta\theta p} & 0 & \\
0 & 0 & 0 &
\end{bmatrix}
= \frac{pr}{2t}
\begin{bmatrix}
1 & 0 & 0 &\\
0 & 2 & 0 & \\
0 & 0 & 0 &
\end{bmatrix}$$
You can find the $\sigma_{rr p}$ and $\sigma_{\theta\theta p}$ by solving the ODE (you can find it by writing the equilibrium equations for free body diagram ) or simply using the Lamé equations.
The thermal effect, is also important, you didn't specify how many degrees are outside, so i can't jump into details now, but it will introduces new stresses $\sigma_{rrt}$ and $\sigma_{\theta\theta t}$. The subscript t stands for thermal.
Now we are dealing with a complex stress condition, i choose to work with von Mises criterium.
I call the result of von Misses yield criterium $\sigma_v$.
Let's use our criteria to find the wall thickness.
The first design criterion (Yielding before failure):
I borrow this simple equation from fracture mechanics: $$K_{Ic} = Y\sigma \sqrt{\pi a} \qquad (1)$$
$K_{Ic}$ is called plain strain fracture toughness. $Y$ is a dimensionless parameter $Y \cong 1.1$, $a$ is the half of the crack length, and $\pi$ is the mathematical constant.
Before we use the expression above, i want to make a small modification, i replace the $\sigma$ with $\sigma_y$ (yield strength) and divid it by $N$ safety factor.
Now solving for $a$:$$a_c = \frac{N^2}{\pi Y^2} (\frac{K_{Ic}}{\sigma_y})^2$$. Now we find the critical crack length, we can find the material using the ratio $\frac{K_{Ic}}{\sigma{y}}$. (it's a material properties). How to find $a_c$? That's not something you can exactly specify the value at this moment, but we should estimate it, it's practically the wall thickness.
After choosing a material, we use the $\sigma_v$ ( it contains the wall thickness), and replace $a$ in equation (1) with $t$ the wall thickness and $\sigma$ with $\sigma_v$. and solve it for the wall thickness. The wall thickness, that we have so far found should be equal or slightly bigger than that we already estimated. If not got back there, improve it and repeat the last step until you get it right.
This is the naivest, and the most simplified version of design the pressure vessel, based on the information provided in the question.