Phase Transitions: Why is the pressure constant?

AI Thread Summary
During phase transitions, the assumption that pressure (dP) remains constant is based on the fact that the system equilibrates with the surrounding ambient pressure. This allows the relationship dH = TdS to hold true. When phase changes occur in a non-constant pressure environment, the integral of volume times pressure change must be considered. Additionally, temperature is assumed constant during phase transitions because energy input primarily facilitates structural changes rather than increasing molecular motion. Understanding these principles is crucial for accurately analyzing thermodynamic processes.
madshiver
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I have a question regarding phase transitions:

It all textbook I've seen the assumption that dP=0 during a phase transition (so that for example dH=TdS is true during a phase transition). Could anyone explain why this should be true?

I guess I could relate this question to a question I had during intro (Halliday) Physics: Why can we assume that temperature is constant during a phase transition? (Back then I imagined that after a certain point putting in energy to get the molecules to move faster does not really lead to a substantial increase in entropy, so that the rigid structure e.g. of the solid or the liquid has to be broken down, so that subsequent temperature increase would actually lead to a decent entropy increase, at the cost of reduced internal energy.)

And then the final related question: Why can't the pressure and temperature both change by "wandering" along the phase-transition line?
 
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madshiver said:
I have a question regarding phase transitions:

It all textbook I've seen the assumption that dP=0 during a phase transition (so that for example dH=TdS is true during a phase transition). Could anyone explain why this should be true?
For a phase change in the open atmosphere, the pressure of the system is always equal to the ambient pressure of the surroundings ie the atmosphere. If the phase change takes place in an environment in which pressure is not constant, you would have to take the ∫VdP term into account. In such a case ΔH = ∫dH = ∫TdS + ∫VdP

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Thanks :).
 
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