# In a sealed chamber, can a vapor be heated enough, that the pressure increases enough, to condense to liquid?

Assuming a sealed chamber, containing a vapor, where volume is constant:

Given that the ideal-gas law states $$PV = nRT$$, if heat is added to an ideal-gas, the pressure will increase as a function of temperature according to:

$$P = \frac{nR}{V}T$$

So if this constant $$nr/V$$ is great enough, then I'm assuming that water vapor would increase enough for a vapor to enter the liquid phase.

So theoretically you could heat steam into condensing?

Or would there be issues at the boundary due to the reduction in latent heat needed to condense?

So theoretically you could heat steam into condensing?

Try it (graphically)! Choose an amount of water vapor $$n$$ and calculate $$V$$ for a gas-phase $$T_0$$ and $$P_0$$ combination on the phase diagram:

Now draw the line $$P=nRT/V$$ that passes through ($$T_0$$, $$P_0$$). (This will appear as a curve on this log-linear plot.) Can you enter the liquid phase through raising the temperature?

Use your diagram. Pick a point on the vapor-liquid line. The slope is obtained by the Clapeyron equation with vaporization to an ideal gas.

$$\frac{dp}{dT} = \frac{p\ \Delta_{vap}\bar{H}}{RT^2}$$

In this, $$\Delta_{vap}\bar{H}$$ is the molar enthalpy of vaporization. Can we pick a point in the vapor state (steam phase) that can be heated in a closed container and, in doing so, will increase pressure to cause the gas to cross the transition line at that same point?

Pick an arbitrary point at lower temperature and pressure, perhaps even one infinitesimally removed from the vaporization point. In a closed container with an ideal gas, the line that we will follow as we heat the gas has a slope.

$$\frac{dp}{dT} = \frac{R}{\bar{V}} = \frac{p}{T}$$

The two lines are drawn in the diagram below.

By the picture, we want the desired slope for heating in a closed container to be greater than the slope of the line defined by the Clapeyron equation.

$$\frac{p}{T} > \frac{p\ \Delta_{vap}\bar{H}}{RT_{vap}^2}$$

Allow the temperatures to be infinitesimally close so that $$T \approx T_{vap}$$. To cause the steam (modeled as an ideal gas) to condense as it is heated by going back into the liquid region, we must have

$$\Delta_{vap}\bar{H} < RT_{vap}$$

Vaporization is endothermic, so enthalpy is positive. Plot $$\Delta_{vap}\bar{H} / RT_{vap}$$ versus $$T_{vap}$$. To first order, the transformation is impossible at those temperatures where the plot is greater than or equal to one. It can only be possible when the graph crosses below one.

A starting point for the numbers that you need is at this reference.

A quick and dirty calculation at 100 $$^o$$C gives $$\Delta_{vap}\bar{H}/RT_{vap} = 40.5/3.1 = 13 > 1$$. So the process is not possible at that temperature. By 374 $$^o$$C, we still obtain a value of about 7.5. So, the process is not possible at any point along the transition.

A fundamental observation also arises from this analysis. The only way that the proposed process can be induced to happen is when the vaporization enthalpy of the liquid is less than the thermal energy being input at the vaporization temperature. The implication can be carried forward likely to exclude any liquid from being able to spontaneously re-form a liquid when being heated from its ideal gas state.

No. Condensing a vapor requires that you remove the heat of vaporization. If you heat the vapor, you are going "in the wrong direction".

• The pressure is definitely increasing though? Commented Mar 31, 2021 at 22:19
• Yes, and so is the temperature. You will soon get above the critical point as you keep adding heat. And note - if you started with only vapor, you were above the saturation temperature to begin with. Adding more heat and increasing temperature will NOT drive you closer to the saturation temperature. I have a chemical engineering background, and we "played these games" every day as part of our job, so my comment is not just some "off the wall idea". Commented Apr 1, 2021 at 0:25