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I am confused about the relation between the internal energy and temperature of ideal gases and single-phase materials. The following two statements from two distinct sources are mainly the reasons:

In single-phase, materials such as solids, liquids, and gases, the internal energy of a system depends primarily on the temperature.

quoted from Introduction to Thermal and Fluids Engineering by Deborah A. Kaminski and Micheal K. Jensen

... in an isothermal process, the internal energy of an ideal gas is constant. ... Note that this is true only for ideal gases; the internal energy depends on pressure as well as on temperature for liquids, solids, and real gases.

quoted from Isothermal Process, Wikipedia

Does the second article not suggest that they are single-phase when referring to liquids, solids and real gases? And also, why does the internal energy of single-phase materials depend primarily on temperature?

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For a single phase closed system (solid, liquid, or gas), the differential change in internal energy dU as a function of dT and dV is given by the following general equation (neglecting electric and magnetic fields): $$dU=C_vdT-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]dV$$ Are you familiar with this equation? For an ideal gas, the term in brackets is zero, and for an incompressible solid or liquid, dV is zero. So, for such ideal materials, dU is a function only of dT. And many solids, liquids, and gases approach these ideal situations.

Can you write out the expression for dV in terms of dP and dT? When you combine this with the above equations, you get the relationship for dU in terms of dT and dP.

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  • $\begingroup$ I understand how an ideal gas and incompressible solids and liquids the equation reduces down to show that the internal energy is only a function of temperature, though I am not really familiar with the equation. I will examine it to digest it thoroughly . In the end then, can we say that the property of solids, liquids and real gases that may make their internal energy only depend on temperature is not the fact that they are single-phase rather it is the fact that they are incompressible? $\endgroup$ – Elruz Rahimli Sep 30 '17 at 5:15
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    $\begingroup$ The derivation of the equation I presented can be found in virtually all thermodynamics textbooks. It is based on a combination of the first- and second laws of thermodynamics. Regarding your question: solids and liquids are typically nearly incompressible, but real gases are not. For real gases, it is the typically small value of the term in brackets (at low pressures) that causes their internal energy to be primarily a function of temperature. $\endgroup$ – Chet Miller Sep 30 '17 at 12:15
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The 1st law of thermodynamics states that the change of the internal energy is $dU=\delta Q + \delta W$. that is the sum of the absorbed heat and work, so if you want to change the internal energy you have to transfer either heat or work. When combined with the 2nd law you may write this as the Gibbs equation spelling out the various interactions explicitly: $dU = TdS -pdV + \mu dN - edE - mdH + etc.,$
For and ideal gas we assume only thermal and mechanical interaction, ie., $dU=TdS-pdV$, and assume that $U=f(T)$. This is essentially a definition but derived from experience with dilute gasses irrespective of the interaction it participates in. This has nothing to do with the process, instead it is in the nature of the material system, ie., the ideal gas.

Practical liquids and solids are essentially incompressible with everyday pressures, $dV=0$, therefore the mechanical interaction term is missing from the Gibbs' equation, and what remains is the thermal interaction and everything else, if any: $dU = TdS + \mu dN - edE - mdH + etc.,$ But if the material is electrically/magnetically neutral and the process is such that $dN=0$, then all you have left is $dU=TdS$.

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