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?
Best Answer
If you only heat the cylinder, the ammonia boils, during which time
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.