Let's say I have a cylindrical piston containing saturated liquid ammonia that is fitted with an electrical heater and a paddle wheel for stirring at an initial pressure and an initial temperature. When the gas expands, some of it evaporates to form a two phase, liquid-vapor solution inside the system. By the way, this process occurs at a constant pressure.

My real question is, what happens to the temperature as the pressure is constant and the volume increases, also causing the specific volume to increase. I would assume the temperature increases, but when I look at the steam table, there is only one temperature corresponding to the pressure. For example, if the initial temperature is 20C, the initial pressure is 8.57bar. If that pressure is constant so the final pressure is 8.57bar, the steam table still says the temperature is 20C.

Why is this? Because I was just looking at the ideal gas equation and it essentially says temperature increases as volume increases. But can I use this intuition with the ideal gas law when the system is 2 phase?


1 Answer 1


If you only heat the cylinder, the ammonia boils, during which time

  • The whole apparatus remains at saturation temperature,
  • The vapor phase expands simply because more ammonia is being added to it, and
  • Some of the heat-energy goes into pushing up the piston to accommodate more gas.

After all the liquid has boiled off, the vapor heats up and expands as suggested by the ideal gas law.

If you pull on the piston, the pressure drops and more liquid boils to restore the vapor pressure. But this takes energy out of the system and lowers its temperature, which makes a new equilibrium pressure and temperature lower than what you started with. If you pull on the piston after all the liquid has boiled, the vapor cools down further.

  • $\begingroup$ Oh right, temperature does not increase during boiling even though heat is added but the substance does fall into the two phase liquid-vapor reagion, right? And does all this happen even when $P < P_{c}$? $\endgroup$ Commented Nov 24, 2011 at 13:01
  • 1
    $\begingroup$ Yes, it's in the two-phase region because there are two phases in equilibrium. That means each phase has its own equation of state – or its own unique solution to a single gas-liquid equation of state. It doesn't matter whether this happens above or below $P_c$ – but it cannot happen above $T_c$, because then a liquid cannot form. $\endgroup$
    – Chel
    Commented Nov 24, 2011 at 13:35

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