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 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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