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During a phase change, internal energy of a system will change because energy must be added or subtracted (depending on "direction" of change) from the system in the form of latent heat.

However, the temperature of the system will stay constant (assuming the change occurs at constant pressure) because the latent heat is used only to "break" the attractions between particles, and does not contribute to the average kinetic energy.

I've learned that for an ideal gas, internal energy is solely a function of temperature (because we ignore the potential energy from attractions). However, because during the phase change temperature will remain constant but the internal energy will not, how can this be so? Doesn't that mean that there are two U values for a single T value?

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    $\begingroup$ Ideal gases do not experience a phase change. $\endgroup$ – David White Oct 4 '19 at 3:57
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I've learned that for an ideal gas, internal energy is solely a function of temperature (because we ignore the potential energy from attractions).

That is correct.

The molecules of an ideal gas are considered to be point masses with no intermolecular (van der Waal) forces. Consequently, the point masses of an ideal gas do not interact with each other. As such it does not take energy for the molecules of an ideal gas to separate from each other, nor is energy released when the molecules come closer together, as long as the molecules remain far apart so that the gas continues to exhibit ideal gas behavior. This means that the internal energy of an ideal gas consists only of kinetic energy (no potential energy) and that any energy added to or withdrawn from the gas results only in a change in its internal kinetic energy and a corresponding change in temperature, according to

$$\Delta U=mC_{v}\Delta T=\Delta KE$$

Where $C_{v}$ is the specific heat of the gas at constant volume. This applies to any process, not just a constant volume process (you can look up the proof).

However, because during the phase change temperature will remain constant but the internal energy will not, how can this be so? Doesn't that mean that there are two U values for a single T value?

Since, by definition, the molecules of an ideal gas are so far apart that there are no intermolecular forces and no potential energy, an ideal gas will not undergo a phase change.

But a gas will exhibit ideal gas behavior only as long as the gas pressure is low enough and/or temperature is high enough so that the molecules remain far apart. If the pressure becomes too high and/or the temperature too low, the molecules become closer together and the gas no longer behaves like an ideal gas as there will be intermolecular (van der Waal) forces between them. The gas now has an internal potential energy component of internal energy of the gas is due to the intermolecular forces, along with its kinetic energy component, so that

$$U=KE+PE$$

Now a phase change becomes possible. At that point, yes, the temperature will remain constant (if the phase change occurs at constant pressure), but it does not mean there will be "two $U$ values for a single $T$ value". It means there will be two components of $U$, a kinetic energy component and a potential energy component, where the temperature is due to the KE component.

During the phase change, ideally only the internal potential energy component of $U$ changes, that is

$$\Delta U=\Delta PE$$

and ideally there is no temperature change,

$$\Delta KE=0$$

Hope this helps.

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  • $\begingroup$ Thank you for another great explanation Bob! Great conceptual explanation, instead of just math. $\endgroup$ – F16Falcon Oct 4 '19 at 20:17
  • $\begingroup$ @F16Falcon Glad it was helpful. $\endgroup$ – Bob D Oct 4 '19 at 22:18

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