How to determine at temperature and pressure to keep Cl2 a gas?

I have chlorine tanks and they will be used industrially. When I use them, they will be exposed to all sorts of varying temperatures. I was wondering if there was any way that when I use them, I could figure out at what pressure I could release the $\ce{Cl2}$, depending on the temperature, so that it remains a gas? 


I already know it only takes $\pu{6.8 atm}$ at $\pu{70^\circ C}$ to liquefy $\ce{Cl2}$ to a volume of $\pu{0.322812 L}$, and its boiling point is $\pu{-33.97 ^\circ C}$ at $\pu{1 atm}$. 




What you need to know is the vapor pressure of $\ce{Cl2}$ as a function of temperature. Just ensure that the release pressure of the chlorine is lower than the vapor pressure, and it will be a vapor. 


I found [one web site](http://www.resistoflex.com/chlorine_graphs.htm) that gives a graph (and a literature reference) of $\ce{Cl2}$ vapor pressure as a function of temperature from -30 °F to 220 °F. The references they give are:



> 
> Reprinted from Chlorine Manual, the Chlorine Institute, 3rd ed. 1959. Data for graph from R. Kapoor and J.J. Martin, Thermodynamic Properties of Chlorine, Eng. Res. Inst. Univ. of Michigan, 1957.
> 
> 
> 


At -30 °F they report $\ce{Cl2}$ vapor pressure is "0" psi (which has to be an approximation; the vapor pressure of all materials is non-zero at any non-zero temperature) and at 220 °F it is 587 psi.


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