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So, I have been studying thermodynamics for a while and some mathematical steps involving partial derivatives have now started to hurt my head. First of all, I understand that whenever we take a $PVT$ system, there are only two independent variables for any process to occur. Using those two variables (choice is ours), we can construct many state functions like $U,F,H,S$ and $G$, which help us in analyzing the system according to our needs. So, I was trying to prove the following relation: $$C_P-C_V=T \Big(\dfrac{\partial P}{\partial T}\Big)_V \Big( \dfrac{\partial V}{\partial T} \Big)_P$$ I tried and got a solution on the following lines:
Consider $S=S(T,V)$, thus we have, $$dS=\Big(\dfrac{\partial S}{\partial T} \Big)_V dT +\Big(\dfrac{\partial S}{\partial V} \Big)_TdV$$ Now, if we consider the process to occur at a constant pressure, we have $$\Big(\dfrac{\partial S}{\partial T} \Big)_P=\Big(\dfrac{\partial S}{\partial T} \Big)_V +\Big(\dfrac{\partial S}{\partial V} \Big)_T \Big (\dfrac{\partial V}{\partial T} \Big)_P$$ Multiplying both sides by $T$, using the definitions of $C_P$ and $C_V$ and then using Maxwell's second relation $$\Big(\dfrac{\partial S}{\partial V} \Big)_T =\Big(\dfrac{\partial P}{\partial T} \Big)_V,$$ we can arrive at the desired relation. The second last equation is, however something that I am not being able to convince myself of. When we impose the restraint of constant pressure, why don't the quantities $(\partial S/\partial T)_V$ and $(\partial S/ \partial V)_T$ somehow change? In other words, why does it remain unaffected even though we have applied a restraint? Is this a sensible question to ask? Can someone elaborate that step mathematically as well as give me some physical intuition into what exactly is going on?

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  • $\begingroup$ I can't fully understand your question too. why do you mean by saying "When we impose the restraint of constant pressure, why don't the quantities (∂S/∂T)V and (∂S/∂V)T somehow change?" in my opinion, the second last equation means that if we fix the pressure and change the temperature, the rate of entropy change with respect to temperature change equals to the rate of entropy change w.r.t temperature change at constant pressure plus an additional term. $\endgroup$
    – FaDA
    Oct 20, 2020 at 7:04
  • $\begingroup$ @SarthakGirdhar It is the definition of partial derivative: $\frac{\partial f(x,y)}{\partial x}=\lim_{\epsilon\rightarrow 0}\frac{f(x+\epsilon,y)-f(x,y)}{\epsilon}$. "Changing the temperature keeping the volume constant will obviously change the pressure", yes and therefore $\left(\frac{\partial p}{\partial T}\right)_V \neq 0$ $\endgroup$
    – Umaxo
    Oct 20, 2020 at 8:19
  • $\begingroup$ @Sarthak I might have made a mistake, i thought it was like indicating that a general function S(T,V,P) is not used but a subspace S(T,V) (if one has a subscript P), but it might be the other way around (that it indicates it IS a parameter). The stuff about partial derivatives holding every parameter except the one in the derivative direction constant should still be true. I will remove my erroneous comment. $\endgroup$
    – Emil
    Oct 20, 2020 at 21:32

2 Answers 2

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Clarifying such an issue goes in parallel between Math and Physics.

An explicit rewording of the key relation helps a lot.

$$dS=\Big(\dfrac{\partial S}{\partial T} \Big)_V dT +\Big(\dfrac{\partial S}{\partial V} \Big)_TdV ~~~~~~~~~~~~~~~~~~~[1]$$ is the general expression of the differential of a function $S(T,V)$. What does it mean?

It says that if one knows the local quantities $\Big(\dfrac{\partial S}{\partial T} \Big)_V$ and $\Big(\dfrac{\partial S}{\partial V} \Big)_T$ at a given point in the $T,V$ plane, it is possible to obtain the changes of $S$ in the neighborhood of that point, correct at the first order in the deviation from the reference point. In other words, if one chooses a point $(T',V')$ different from $(T,V)$, $\Big(\dfrac{\partial S}{\partial T} \Big)_V (T'-T) +\Big(\dfrac{\partial S}{\partial V} \Big)_T(V'-V)$ differs from $S(T',V')-S(T,V)$ by terms going to zero faster than $T'-T$ and $V'-V$ when $T' \rightarrow T$ and $V' \rightarrow V$.

The key point is that the partial derivatives have to considered fixed at one point and $dT=(T'-T)$ and $dV=(V'-V)$ express the deviation from that point.

It should now be clear that the partial derivatives in the original form for $dS$ do not depend on the way one gets the deviation from $(T,V)$ in practice. That information goes into $dT$ and $dV$ only. Therefore, if those deviations follow a constant pressure path, one can re-express them in terms of partial derivatives with respect to the pressure. But this applies only to the $dT$ and $dV$ terms.


Elaboration following the OP comment (but also after Umaxo's mathematical answer).

Since the differential $[1]$ coincides, within first order terms in $(T'-T)$ and $(V'-V)$ with the variation of $\Delta S$ of entropy between the two states, one can use that expression to evaluate partial derivatives of $S$, when keeping another thermodynamic variable constant. For the partial derivative with respect to temperature at constant pressure, we have just to recall that keeping constant $P$ induces a dependence of the volume on temperature ($V=V(P,T)$). This is the important step both from the mathematical and physical point of view.

Therefore, we have to evaluate the limit $$ \lim_{T' \rightarrow T} \frac{\Delta S}{(T'-T)} $$ where $$ \Delta S = S(T',V(T',P))-S(T,V(T,P)) = d S + \mathcal{o}(T'-T) $$ which immediately gives $$ \Big(\dfrac{\partial S}{\partial T} \Big)_P=\Big(\dfrac{\partial S}{\partial T} \Big)_V +\Big(\dfrac{\partial S}{\partial V} \Big)_T \Big (\dfrac{\partial V}{\partial T} \Big)_P$$.

Once again, a notation which takes trace of which quantity is function of what, and a clear understanding of the functional restrictions induced by processes where a thermodynamic quantity is kept fixed are pivotal to manage thermodinamic formulae without doubts.

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  • $\begingroup$ I can understand quite a bit from what you have written. And I think you have answered my question. But can you elaborate a bit more on this? It would be great. $\endgroup$ Oct 20, 2020 at 7:28
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I think @GiorgioP answer quite nicely explains how to understand partial derivatives role in the total derivative. I will show you, how is the formula $$\Big(\dfrac{\partial S}{\partial T} \Big)_P=\Big(\dfrac{\partial S}{\partial T} \Big)_V +\Big(\dfrac{\partial S}{\partial V} \Big)_T \Big (\dfrac{\partial V}{\partial T} \Big)_P$$ derived.

Let say you have function of 2 variables $f(x,y)$. The partial derivative by $x$ while keeping $y$ constant is the change of functional value as you are moving along the curve $y=const.:$ $$\left(\frac{\partial f(x,y)}{\partial x}\right)_y=\lim_{\epsilon\rightarrow 0}\frac{f(x+\epsilon,y)-f(x,y)}{\epsilon}.$$

Now suppose you have some equation $\phi(x,y,z)=0,$ where $\phi(x,y,z)$ is some function of the variables $x$,$y$ and $z$. You can use this equation to get $y$ as function of $x$ and $z$ and transforming function $f(x,y)$ into a funciton of $f(x,z):$ $$f(x,z)=f(x,y(x,z)).$$ You can then compute the change in the function along the trajectory that keeps $z$ constant instead of $y$, i.e. partial derivative of $f$ by $x$ keeping $z$ constant:

$$\left(\frac{\partial f(x,z)}{\partial x}\right)_z=\lim_{\epsilon\rightarrow 0}\frac{f(x+\epsilon,y(x+\epsilon,z))-f(x,y(x,z))}{\epsilon}=\lim_{\epsilon\rightarrow 0}\frac{f(x+\epsilon,y(x+\epsilon,z))-f(x,y(x+\epsilon,z))+f(x,y(x+\epsilon,z))-f(x,y(x,z))}{\epsilon}=\lim_{\epsilon\rightarrow 0}\frac{f(x+\epsilon,y(x+\epsilon,z))-f(x,y(x+\epsilon,z))}{\epsilon}+\lim_{\epsilon\rightarrow 0}\frac{f(x,y(x+\epsilon,z))-f(x,y(x,z))}{\epsilon}.$$

You can see, that in the first term, the $y$ is not changing - it is of the same value $y(x+\epsilon,z)$ in both terms. Therefore the first limit is just the definition of partial derivative when keeping $y$ constant $(\partial f/\partial x)_y$.

In the second term, it is now $x$ that is kept constant in $f$, and $z$ that is kept constant in $y$. This is basically derivative of one variable function $g(y(x))$, that is created from $f(x,y(x,z))$ by supplying the constant values $$g(y(x))=f(\text{value of x},y(x,\text{value of z})).$$ The derivative is thus $$\frac{dg(y(x))}{dx}=\frac{dg(y)}{dy}\frac{dy}{dx}=\frac{df(\text{value of x},y)}{dy}\frac{dy(x,\text{value of z})}{dx}=\left(\frac{\partial f}{\partial y}\right)_x\left(\frac{\partial y}{\partial x}\right)_z.$$

The result is therefore: $$\left(\frac{\partial f(x,z)}{\partial x}\right)_y=\left(\frac{\partial f(x,y)}{\partial x}\right)_y+\left(\frac{\partial f(x,y)}{\partial y}\right)_x\left(\frac{\partial y(x,z)}{\partial x}\right)_z$$

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  • $\begingroup$ Great answer. Thanks! $\endgroup$ Oct 20, 2020 at 9:48

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