I'm trying to comprehend the discussion here about the internal energy of an ideal gas being only dependant to its temperature $dU=nc_vdT$. I have the feeling that there are some assumptions missing.

What I see from the proofs, for example here, is that they assume they can substitute any thermodynamic process with other incremental processes as far as the consequent process has the same start-end. To understand better what I'm talking about you may consider the conservatives forces where the total work done by the force is independent from the path.

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So my question is that if all thermodynamic process with the same start-end are homeomorphic? meaning that this is an intrinsic feature of a thermodynamic system. Or there are certain assumptions to be met? for example has reversibility anything to do here?

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    $\begingroup$ This idea is normally considered in terms of individual quantities, rather than entire systems. In thermodynamics quantities which are path independent are known as functions of state. $\endgroup$ – By Symmetry Feb 6 '18 at 11:54
  • $\begingroup$ So basically the "state" of a thermodynamic system is defined by a subset of all "state variables" and it is independent from the processes leading to that "state"? $\endgroup$ – Foad Feb 6 '18 at 12:12
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    $\begingroup$ Essentially yes $\endgroup$ – By Symmetry Feb 6 '18 at 12:16

A few points that may help…

  1. For any liquid or gas, any property (for example temperature, viscosity) can be expressed in terms of two independent variables. p and V are usually chosen. We say that the particular values of p and V determine the state of the fluid, and that temperature, viscosity etc are functions of state. [We don't have to use p and V as independent variables; we might well choose T and V.] The idea can be extended to solids and more complicated systems, but these usually need more than two independent variables.

  2. It isn't obvious that temperature (determining which way, if at all, heat will flow between two bodies put in contact) is a function of each body's state. Showing that it is, is the point of the Zeroth Law of Thermod.

  3. Likewise, it had to be shown that for any body there is a function of state, U, internal energy, such that $\Delta U = \text{Heat in – Work out}.$ This was done by Joule and others in the nineteenth century. It means that if we go from point A to point B on a pV diagram, $\Delta U$ is independent of route.

  4. So U is a function of state. In general it depends on T and (say) V, but part of the definition of an ideal gas is that U depends only on T. [T itself can be expressed in terms of p and V by means of the so-called equation of state for the gas.]

  • $\begingroup$ to be sure that I have understood correctly : 1. the thermodynamic state of any fluid (i.e. liquid and gas) in equilibrium (set of all imaginable properties) can be determined with a subset of two of all state variables (V, P, T, S, U, H ...) except for the ideal gases where T and U are linearly related 2. Internal energy U being a state variable is an empirical conclusion (Joule's law) $\endgroup$ – Foad Feb 6 '18 at 12:38
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    $\begingroup$ That's it. Well-expressed, if I may say so. Originally the existence of the state variable, $U$, needed establishing empirically because (1) the modern concept of $energy$ was only just emerging, (2) the microscopic structures of systems weren't really known – even atoms were controversial. These days, we're much more confident about substances having a microscopic structure (kinetic theory of gases etc) and this makes it seem obvious that a gas (or whatever) must have an internal energy dependent on its state. $\endgroup$ – Philip Wood Feb 6 '18 at 12:47
  • $\begingroup$ ok. I need some time to understate this and come back. thanks a lot. $\endgroup$ – Foad Feb 6 '18 at 12:57
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    $\begingroup$ A B Pippard: $Classical Thermodynamics$ deals clearly and logically with the fundamentals. Almost entirely on a macroscopic level. No kinetic theory or statistical mechanics. $\endgroup$ – Philip Wood Feb 6 '18 at 13:05
  • $\begingroup$ Engineers regard the relationship between internal energy and temperature of ideal gases as not necessarily being linear. $\endgroup$ – Chet Miller Feb 6 '18 at 16:05

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