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The potential energy of a material is its ability to do work. So I would suspect the potential energy to increase during phase changes that increase entropy (solid $\rightarrow$ liquid and liquid $\rightarrow$ gas), and decrease in the opposite direction.

However, what about when the temperature is raised for a solid, up until the melting point - will the potential energy increase? The heat from a 1500 K gold plate could do work on liquid water (evaporate) to a higher extent than a plate at 300 K, so I guess the potential energy has increased. So would the potential energy as a function of temperature, of an arbitrary material, look something like this ($K$ should be $T$)?

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    $\begingroup$ Can you please provide a precise definition of the potential energy of a macroscopic system like the ones you are describing? In what ways is it related to the internal energy? $\endgroup$ – Chet Miller Dec 1 '15 at 23:50
  • $\begingroup$ As was once pointed out, most introductory chemistry textbooks define energy as "the ability to do work," then turn around and say that free energy is "the part of energy in a system that is available to do work." They never note that their two definitions are inherently contradictory: if some of the energy can't do work, then you can't define energy as the ability to do work! Defining the potential energy of a system as "its ability to do work" strikes me as similarly nebulous---is there a more precise definition you could give? $\endgroup$ – chipbuster Dec 8 '15 at 18:11
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Yes, potential energy increases with increasing temperature for at least the following three reasons:

  1. At a higher temperature, more atoms/molecules are in excited electronic states. Higher electronic states correspond to greater potential energy. Potential Energy is -2 times Kinetic Energy. So actually, at higher temperature, when more atoms are in higher electronic states, there is more potential energy and less kinetic energy (just considering electronic energy).

  2. At higher temperature, more molecules are in excited vibrational states. Higher vibrational states correspond to greater potential energy. Half the energy of each vibrational state is kinetic, half potential.

  3. At higher temperature, more molecules are in excited rotational states. While for a rigid rotor, this would not involve increased potential energy, real molecules are not rigid. The rotation of a diatomic molecule about an axis perpendicular to the bond stretches the bond as a centrifugal distortion, which represents an increase in potential energy.

Additionally, for intermolecular forces, to the extent that they involve potentials of the form r^x, the Virial Theorem requires that total energy is divided into potential and kinetic energy in a specific proportion, so total energy can not be increased without increasing potential energy.

Free translational motion is not subject to a potential, nor is rigid rotation.

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  • $\begingroup$ I wanted to ask that we read that temperature is due to random internal kinetic energy. Is temperature also due to the internal potential energy or the total internal energy or only because of kinetic energy $\endgroup$ – Shashaank Mar 5 '17 at 9:54
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Lets talk about a heating curve As heat is constantly added to a substance in the solid phase, the KE increases but the PE remains the same. As the substance reaches its melting point, commonly known as fusion (border between melting and freezing), the KE remains the same while the PE increases. After a while the substance becomes completely a liquid, and heat is still being added constantly and the KE increases while the PE remains the same. The substance now reaches its boiling point, and starts to change into the gas phase, in this process the KE remains the same while the PE increases.

KE = kinetic energy PE = potential energy

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    $\begingroup$ Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the help center. This seems to be much too simple to be true. It would be appreciated if you could add a reason to your statements, i.e. why does the KE increase while the PE stays the same. $\endgroup$ – Martin - マーチン Feb 12 '16 at 5:52

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