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In the proof of :

$$dU=C_v dT - \left[T \left( \frac{\partial P}{\partial T} \right)_V - P\right] dV$$

on wikipedia https://en.wikipedia.org/wiki/Internal_energy, they uses the fact that :

$$dS=\left( \frac{\partial S}{\partial T} \right)_V dT + \left( \frac{\partial S}{\partial V} \right)_T dV $$

Then they replace the expression of $dS$ in the thermodynamic relation and they find out the good relationship.

However, the thing that disturbs me is the dependance of the state function on the thermodynamic variables.

I thought that a state function must depend on extensive thermodynamic variables. When we write the expression of $dS$ here we have a $dT$ appearing on the rhs. This is what disturbs me.

So in all generality, how can we know that the expression of $dS$ in terms of its variable is the good one ?

Why can't we write for example :

$$dS=\left( \frac{\partial S}{\partial T} \right)_{V,P} dT + \left( \frac{\partial S}{\partial T} \right)_{V,T} dP + \left( \frac{\partial S}{\partial V} \right)_{T,P} dV $$

My guess about it that I would like to check :

I guess there is an implicit assumption that we know the thermodynamic potentials in our case (gas in a closed system) only depends on two thermodynamic variables.

Then, first thing : we must only have two terms on the rhs of $dS$.

In principle, I could decide to express $S$ in function of the two intensives variables : $(U,V)$. But I can always write $dU=\left(\frac{\partial U}{\partial T} \right)_V dT + \left(\frac{\partial U}{\partial V} \right)_T dV$

Then, I can decide to express $S(T,V)$ instead of $S(U,V)$

Is it correct ?

But then, how do we know that the two independant variables will be $(T,V)$ and not $(P,V)$ for example ? How to know which two independant variable I can chose ? I guess it is linked to the fact I must consider non conjugated variables but I don't really see why.

GiorgioP-DoomsdayClockIsAt-90
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StarBucK
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    using the same letter $S$ to denote the functions $S(T,V)$ and $S(U,V)$ is a misuse of notation and continuous source of misunderstandings. In fact you should think of the entropy as being representable by a function $S_1(T,V)$ and also by a completely different function $S_2(U,V)$, these are as different functions as $a=x^2$ is different from $b=cos(t)$ but for specific values of $x$ and $t$ we can have $a=b$, and so can we have $S_1=S_2$ for some specific values of $T,V$ and $U,V$. – hyportnex Feb 26 '19 at 15:44
  • @hyportnex I agree and I think it is linked to what I say in my "guess" part, when I do the change of variable U->U(T,V). However even with this I would like to know if we can choose any variable to describe our potential. Like would it be valid to write the differential of S in the (P,V) variables ? – StarBucK Feb 26 '19 at 15:52
  • Assuming that the function, say, p=p(U,V) is invertible then you can write that U=U_1(p,V) and that upon substitution into $S_1(U,V)=S_1(U_1(p,V),V)=S_2(p,V)$ to get the entropy as function of p and V. – hyportnex Feb 26 '19 at 15:59
  • Of course it’s valid. It’s just the chain rule. – knzhou Feb 26 '19 at 16:09
  • @hyportnex Allright, I just had a memory in which I thought it wasn't valid to express a thermodynamic function as a function of two conjugates variable (like $(P,V)$ or $(T,S)$). But indeed it doesn't seem to have any problem to do it. And so for the last question : here it is implicitly assumed that we only need 2 variables to describe the system right ? And this is an experimental fact, it is not possible to guess it ? This is why we can focus on any dependance in two variables : $(P,V)$ or $(U,V)$ or $(T,V)$ or whatever if we want to express a thermodynamic potential. – StarBucK Feb 26 '19 at 16:33
  • That only two independent thermodynamic variables are needed to describe a system is an assumption that does not have to be true in general. Water $H_2O$ can be described quite well for most engineering use with two variables, say $T$ and $V$, until you put it into a strong electric field and then you need a third one $E$, etc. – hyportnex Feb 26 '19 at 16:52
  • @hyportnex Yes sure, what I wanted to say is that it is absolutely necessary to know how much thermodynamic variables are necessary to describe our system. In all generality the answer is not obvious at all and come from experimental facts. Once we know it we can express the problem in any of those 2 variables doing an appropriate change of variable. – StarBucK Feb 26 '19 at 16:59
  • you may find this useful: https://physics.stackexchange.com/questions/193677/is-amount-of-entropy-subjective/193863#193863 – hyportnex Feb 26 '19 at 17:01
  • You may find interesting this note, which includes calculation of $dU$ in terms of various combinations of $T$, $P$, $S$, and $V$ (sections "Having fun with it/Expressing anything in terms of anything"). A notable result is that the coefficient of anything other than $dT$ is zero for an ideal gas. – Chemomechanics Feb 26 '19 at 18:32

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