Chemistry Why can we differentiate this entropy total derivative with repect to Temperature?

[zenterix]

Homework Statement

I am going through a thermodynamics course and there is a section about natural variables.

Relevant Equations

In that section, there is the following derivation

Ignoring chemical potential for now, the natural variables of $U$ are $S$ and $V$. Thus

$$dU=\left (\frac{\partial U}{\partial S}\right )_VdS+\left (\frac{\partial U}{\partial V}\right )_SdV=TdS-pdV\tag{1}$$

which we can rewrite for $dS$ as

$$dS=\frac{dU}{dT}+\frac{pdV}{T}\tag{2}$$

We wish to determine how the entropy depends on temperature.

We can differentiate the expression for $dS$ with respect to temperature at constant volume to obtain $\left (\frac{\partial S}{\partial T}\right )_V$.

Apply the derivative to obtain

$$\left (\frac{\partial S}{\partial T}\right )_V=\frac{1}{T}\left (\frac{\partial U}{\partial T}\right )_V+\frac{p}{T}\left (\frac{\partial V}{\partial T}\right )_V$$

$$=\frac{1}{T}\left (\frac{\partial U}{\partial T}\right )_V=\frac{C_V}{T}$$

Mathematically, what is happening exactly?

In other words, why/how does $dS$ become $\left (\frac{\partial S}{\partial T}\right )_V$?

Here is my current understanding.

$dS$ is the total derivative function for $S(T,V)$.

$dS$ is a linear function of $dT$ and $dV$ and this linear function is a function of $T$ and $V$.

That is, at each $T$ and $V$ we have a different function $dS$.

If we differentiate $dS$ relative to $T$ we are determining the rate of change of $dS$ relative to $T$ at fixed $V$.

Notation-wise, it seems we would have $\frac{\partial (dS)}{\partial T}$, which is a notation I have never seen before.

If this is correct so far, then it seems that $\frac{\partial (dS)}{\partial T}$ is written as just $\left (\frac{\partial S}{\partial T}\right )_V$.

But this latter step isn't very clear to me.

So my question is essentially how to understand what is going on from a mathematical perspective in this derivation.

 

[DrDu]

No, don't differentiate dS, rather use the differential of S with variables T and V, $dS(T,V)=\left(\partial S/\partial T\right)_V dT + \left (\partial S /\partial V\right)_T dV$ and a similar expression for dU(T,V).

 

[ChemEng]

I am working through quite a bit of Thermodynamics myself lately, and the impression I get is that the Thermo books aren't doing a great job with explaining the "calculus" of how these things work. As luck would have it, I asked a similar question today.

There is probably more than one way to go about this, but here is a method following "Introduction to Chemical Engineering Thermodynamics" (Elliot and Lira). I think all the steps below are using calculus as you are expecting to find it.

dU = TdS - PdV (Fundamental Property relation).

dU = (dU/dT_V)*dT + (dU/dT_V)*dV (the righthand side are partials with respect to T (constant V) and V (constant T), respectively.). Here we are just taking an exact differential of U with respect to T and V.

To find dS in relation to dT, we hold V constant (dV = 0). From the first equation, this let's us relate dU to dS.

dU = TdS and dU = (dU/dT_V)*dT at constant V. We can therefore set these expressions equal to each other.

TdS = (dU/dT_V)*dT. But dU/dT_V is Cv by definition.

dS = (Cv*dT)/T at constant V.

And lastly, we can also right:
dS = (dS/dT_V)*dT + (dS/dV_T)*dV (again these are partials). And we then hold volume constant, dV = 0.
dS = (dS/dT_V)*dT

We can now equate our expressions for dS.

(dS/dT_V)*dT = (Cv/T)*dT with both partials at constant V. This expression allows us to write the following.
(dS/dT_V) = (Cv/T)

For completeness, I will also provide a link to the following which is more in line with guidance on explanation for the mathematical manipulation given in your question: Divide-Through Rule.

 

[Chestermiller]

In my judgment, the thernodynamics books do a fine job of representing the differential changes in thermodynamic functions in terms of partial derivatives, fully consistent with how partial differentiation is done in mathematics books. Perhaps you need to review the subject of partial differential equations in your math books first before judging the thermodynamics books as "not doing a great job."

 

[ChemEng]

That's fair.

 

[DrDu]

In my judgment, the thernodynamics books do a fine job of representing the differential changes in thermodynamic functions in terms of partial derivatives, fully consistent with how partial differentiation is done in mathematics books. "

When I was taught partial derivatives, my prof pointed out to us that there actually is an important difference between thermodynamic notation and mathematical notation: Specifically, we designate for example internal energy always with the letter "U" irrespective whether we assume it to depend on V and S or on p and T, while, for a mathematician U(V,S) And U(p,T) are clearly different functions. While this makes sense physically, it leads to the need to allways specify these variables, i.e. we have to write $\partial U/\partial S| _V$ , i.e., the variable which is held constant, has always to be made explicit.