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I am currently taking the freshman thermodynamics course and have asked several times for explanations of the relationships which can, and which cannot, and why they cannot. But I still don't really understand it.

So far we have covered the following state quantities: $T$, $p$, $V$, $u$, $c_v$ and $h$ (temperature, pressure, volume, internal energy, specific heat and enthalpy). The professor (if I remember correctly) indicated that for a single compressible system, consisting of a pure substance, the state of a system can be described by: v and T ($p = p(v, T)$), v and P ($T = T(v, P)$) but not by $T$ and $p$ (see the added picture below). I have quite a hard time seeing why you cannot describe the state of a system with the state quantities $T$ and $p$.

If possible, based on this could someone also explain how to reason out where $u$ and $h$ depend on then? Because those depend on $T$, $p$ and $V$, right? The lecture slides state that $c_v$ can be described by $T$ and $p$ ($c_v = c_v(T, p))$

Some things make perfect sense, but for things like this I get very confused.

statement in the presentation

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    $\begingroup$ You can have two containers having different volumes, but both can have the same $p$ and $T$. $u$ and $h$ are different beasts. They are not parameters; they are functions of parameters. Beyond that the story is more involved than appropriate for a comment. $\endgroup$
    – garyp
    Jun 8, 2023 at 12:24
  • $\begingroup$ @garyp so the important thing is to understand what the possible combinations of paramaters are I assume? As you indicate yourself, not every possible combination gives a unique state of the system, right? $\endgroup$ Jun 8, 2023 at 16:36

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For a simple compressible system you can relate $T$ to $p$ and $V$ via the equation of state. It can be quite helpful to plot the result on a 3D graph, with axes $p$, $V$, $T$. Here is a typical example:

enter image description here

For most points on this surface (i.e. for most states of the system) you can pick any two out of $p,V,T$ and deduce the third. However there are some sets of states which all have the same $p,T$ but differing $V$. They lie along a line parallel to the $V$ axis on the diagram, and they occur during a phase transition.

You can also have two different possible temperatures at the same $p,V$ for a given mass (an example occurs in water near the freezing point), but this is rare.

It follows that specifying $p,T$ does not always fix the system to one possible state, and nor does specifying $p,V$. However the second is a special case (i.e. $p$ and $V$ are usually sufficient) and I don't know of an example where $T$ and $V$ are not sufficient, but in principle it could happen I think.

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  • $\begingroup$ So if I understand correctly, you're saying: when you look at this $p-V-T$ surface, there are perspectives where the values at a combination of state variables are not unique? And therefore, for example, knowing just $p$ and $T$ would not be sufficient to "capture" the state of a subtance because there can be multiple values for $v$ with this combination as an example? $\endgroup$ Jun 8, 2023 at 16:16
  • $\begingroup$ It's rather confusing, but if I interpreted it correctly how would you know which combinations of state quantities are bound to give unique combinations? $\endgroup$ Jun 8, 2023 at 16:21
  • $\begingroup$ Your first comment is correct. Your second comment is also basically right: there's no way to say in general how such surfaces might go, so there is no general rule. $\endgroup$ Jun 8, 2023 at 16:39
  • $\begingroup$ I think we are only looking at water, so should I then assume that combinations of $P$ and $V$, and $T$ and $V$ yield unique combinations and thus the state of water the water at that combination is fixed? $\endgroup$ Jun 8, 2023 at 16:43
  • $\begingroup$ @Masterrun80 Actually, chemical potential and fugacity are functions of $T$ and $P$ in the mathematical sense, i.e., they have a unique value at all $T$ and $P$. The are also continuous, although their derivatives are discontinuous at the phase boundary. So, it is not true that no property $\phi$ can be expressed as a function of $T$ and $P$. I mean, if we are going to be nitpicking we should nitpick all the way :) $\endgroup$
    – Themis
    Jun 9, 2023 at 22:15
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You must have misunderstood your professor. we can use any two variables among $(T,P,V)$ as the independent variables to describe the thermodynamic state. The combination $(T,P)$ is especially useful because temperature and pressure can be measured and controlled in experiment very easily.

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  • $\begingroup$ I added an image from the presentation to my question! $\endgroup$ Jun 8, 2023 at 16:22
  • $\begingroup$ @Masterrun80 What is $\phi$? $\endgroup$
    – Themis
    Jun 8, 2023 at 18:19
  • $\begingroup$ A state quantity. He states that each state is either a function of volume and temperature, or volume and pressure. I just watched the lecture again, and he does not seem to notice the combination $p$ and $T$ in it, only that its graph is the phase diagram. $\endgroup$ Jun 8, 2023 at 18:30
  • $\begingroup$ @Masterrun80 This doesn't make sense. Take the ideal gas law, for example, $PV=RT$, and write it as $V = RT/P$: there we have $V$ as a function of $P$ and $T$. Your professor cannot possibly be making this mistake, he/she must mean something else. Can you share the complete slide? $\endgroup$
    – Themis
    Jun 8, 2023 at 19:22
  • $\begingroup$ @Masterrun80 OK, in retrospect it sounds like your teacher means that we can have multiple phases at same $T$ and $P$, therefore, $T$ and $P$ alone does not identify the true state. While strictly true, this is a weird way to put it, certainly not how I would teach it. Ignore it and move on. $\endgroup$
    – Themis
    Jun 8, 2023 at 19:31
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You either misunderstood your professor, or your professor is wrong. For a single-phase pure component, specification of any two intensive, independent properties is sufficient to fix all the rest. Pressure $p$ and temperature $T$ are intensive properties.

For example, the ideal gas equation can be written as $Pv=RT$ where $R$ is the specific gas constant, and $v$ is the specific volume. $P$ and $T$ determine $v$.

Specific volume $v$, specific internal energy $u$, and specific enthalpy $h$ are also intensive properties.

Hope this helps.

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  • $\begingroup$ As you point out, $v$ is specific volume. Equations of state are often written with the extensive variable $V$. $\endgroup$
    – garyp
    Jun 8, 2023 at 13:26
  • $\begingroup$ Your post brings to mind this: physicists typically deal with fixed volume (certainly in introductory expositions), where $u$ is the useful energy function. For chemists, it's typically pressure that is fixed, and there $h$ is useful. Maybe that will help the OP. $\endgroup$
    – garyp
    Jun 8, 2023 at 13:31
  • $\begingroup$ @garyp Yes, but when $V$ is used the equation includes either mass $m$ or moles $n$ which, together with $V$ give you specific quantities $\endgroup$
    – Bob D
    Jun 8, 2023 at 13:35
  • $\begingroup$ I guess I misinterpreted his answer then. I have attached an image of what he added to the presentation slides. $\endgroup$ Jun 8, 2023 at 16:09
  • $\begingroup$ @Masterrun80 What is "maar"? $\endgroup$
    – Bob D
    Jun 8, 2023 at 17:12

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