# Closed form approximation for the equilibrium temperature of matter in a container if the pressure is not constant?

I know that many closed form expressions exists for finding the final equilibrium temperature of a gas/liquid/solid mixture if the pressure is held constant, or if the solid/liquid does not change phase, but what about a simple case of a single solid being heated through a perfectly rigid and insulating container?

Let's say we have a perfectly rigid and insulating container with volume $$V$$ that contains a block of ice (amount defined in moles $$n$$) chilled to near absolute zero, then we pull a near-perfect vacuum. We then magically apply heat to the block of ice through the container (heat $$Q$$ is defined in Joules). If the energy applied is high enough, the block of ice would be instantly vaporized, but as the pressure increases, some of the gas will liquefy (or even solidify), following the phase diagram of water.

Given enough time for this container's insides to reach equilibrium, what would be the final temperature or the change of temperature? Finding the partial pressure is not important for me but might be needed to find the temperature?

I guess probably that an exact solution does not exist, but any approximation would be adequate to simulate this phenomenon in a computer. I've tried calculating this using gas equations, but they all either assume that pressure or temperature is constant.

If a solution can be found for the problem above, what about a slightly more complicated version where we put many different materials inside of that box, where each material is separated from others with a flexible membrane that only allows heat to pass through (the materials never mix). What would be the change of temperature after adding some heat?

• The volume of ice is << than V, right? Commented Feb 11, 2023 at 16:02
• @ChetMiller Yes, we can assume the amount of matter inside of the volume never gets compressed to extreme pressures. I'm looking at relatively "normal" temperature and pressure ranges (eg. 50-5000K and 0-5000kPa) Commented Feb 11, 2023 at 20:56

This is a constant-volume process. The heat is equal to $$\Delta U$$, so at the final state we know $$U$$ and $$V$$: $$V = \frac{V^\text{tank}}{n}, \quad U = U_\text{initial}+Q$$ In principle these two variables fix the final state. What we need is a method to calculate $$U$$ at the very low temperature of the initial state. But if instead the initial state is above 0$$^\circ$$C, we can do it using the steam tables.