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I would like to check whether my analogy is right and ask some related questions.

So, I am studying Thermodynamics right now and there are many calculus techniques are used for the thermodynamic relations between various state functions.

One example is

$ dH = \left(\frac{\partial{H}}{\partial{T}}\right)_{P}dT + \left(\frac{\partial{H}} {\partial{P}}\right)_{T}dP$

It is a total differential of enthalpy when considering $H(P,T)$.

By dividing ${dT}$, a derivative for $H$ can be earned like

$ \left(\frac{\partial{H}}{\partial{T}}\right)_{V} = \left(\frac{\partial{H}}{\partial{T}}\right)_{P} + \left(\frac{\partial{H}}{\partial{P}}\right)_{T}\left(\frac{\partial{P}}{\partial{T}}\right)_{V}$

For me, mathematically this process is clear and understandable.

For some 3-D space with $(H, P, T)$, a well-defined function $H(P,T)$ is a surface and it's total differential $dH$ is a natural change of $H$ earned by an infinitesimal change of $(dT, dP)$ and a gradient of $\left( \left(\frac{\partial{H}}{\partial{T}}\right)_{P}, \left(\frac{\partial{H}}{\partial{P}}\right)_{T}\right)$ at a specific point.

And $\left(\frac{\partial{H}}{\partial{T}}\right)_{V}$ is a directional derivative with a constraint $(V = \textrm{constant})$ therefore satisfying some relationship between $T$ and $P$ expressed by $\left(\frac{\partial{P}}{\partial{T}}\right)_{V}$ (and such direction is along a line also satisfying that constraint.).

So I would like to check that my mathematical analogy is correct.

Also, I really want to know

First, is there any way to explain physically and intuitively for each term of $\left(\frac{\partial{H}}{\partial{T}}\right)_{V}$?

Second, unlike typical geometry dimensions like $(x,y,z)$, definitely, thermodynamic variables like $(P,T)$ are inter-connected, I mean for me they are not independent like $(x, y, z)$. Then why such thermodynamics variables considered mathematically independent? Can I understand this because they are not truly independent so $H(P,T)$ only spans over a limited space (as like a 2-D surface)?

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  • $\begingroup$ The natural dependent variables for $H$ are $S$ and $P$, not $T$ and $P$. The Gibbs free energy is the one which you can naturally express as $G(T,P)$. en.wikipedia.org/wiki/Thermodynamic_potential $\endgroup$ Commented Sep 30, 2020 at 21:49
  • $\begingroup$ I know $H=TdS -VdP$ but is it also possible to express $H$ as a function of $(T,P)$? Anyway, maybe it is inappropriate to ask a physical meaning of each term when using $H(T,P)$. $\endgroup$
    – Youhwan Jo
    Commented Sep 30, 2020 at 21:54
  • $\begingroup$ That's not quite correct; the expression is $dH=TdS-VdP$. Anyway, you can still express $H$ as a function of $T$ and $P$; the fact that you're not using the natural variables only means that the derivatives won't correspond to one of the six usual thermodynamic state variables. $\endgroup$ Commented Sep 30, 2020 at 21:57
  • $\begingroup$ Uh, by the way, that was a typo missing $d$. $\endgroup$
    – Youhwan Jo
    Commented Oct 1, 2020 at 0:06

2 Answers 2

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"Dividing by $dT$" isn't really a well-defined operation, especially when you "divide" $dH$ by $dT$ and end up with $(\partial H/\partial T)_V$. If you want to arrive at this expression, it's best to start with the definition of $H$, instead of $dH$. The enthalpy $H$ is derived from the internal energy $U$ via a Legendre transform:

$$H=U+PV$$

From here, you can differentiate directly:

$$\left(\frac{\partial H}{\partial T}\right)_V=\left(\frac{\partial U}{\partial T}\right)_V+V\left(\frac{\partial P}{\partial T}\right)_V$$

One of these quantities is well-known: $(\partial U/\partial T)_V=C_V$, the heat capacity at constant volume. Likewise, the other derivative corresponds to a physical quantity, namely: $\left(\partial P/\partial T\right)_V=\alpha/\kappa_T$, where $\alpha$ is the coefficient of thermal expansion and $\kappa_T$ is the bulk compressibility at constant temperature. So, we have:

$$\left(\frac{\partial H}{\partial T}\right)_V=C_V+V\frac{\alpha}{\kappa_T}$$

The values of these three coefficients are determined by the specific relation between volume, pressure, and temperature for this system; in other words, they are specified by the equation of state for your system.

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  • $\begingroup$ Thank you for your answer, the expression has a physical meaning. Would you give me more explanation about 'a well-defined operation'? What I did is just deriving a derivative mentioned in my textbook, so is that fundamentally wrong? Comparing $\left(\frac{\partial{H}}{\partial{T}}\right)_{V}$ in your answer and the one in my question, $V = \left(\frac{\partial{H}}{\partial{P}}\right)_{T}$, is that wrong? $\endgroup$
    – Youhwan Jo
    Commented Sep 30, 2020 at 22:57
  • $\begingroup$ Differentiation isn't really "dividing by $dT$", and if your textbook describes it that way, then they aren't being mathematically rigorous. The problem is that merely "dividing by $dT$" does not specify any constraint; the constant-$V$ constraint seems to come from nowhere when you do this. $\endgroup$ Commented Sep 30, 2020 at 22:59
  • $\begingroup$ Actually the textbook said $\left(\frac{\partial{H}}{\partial{T}}\right)_{P}$ is not equal to $\left(\frac{\partial{H}}{\partial{T}}\right)_{V}$ (obviously), and just showing without any detail, they are different as $ \left(\frac{\partial{H}}{\partial{T}}\right)_{V} = \left(\frac{\partial{H}}{\partial{T}}\right)_{P} + \left(\frac{\partial{H}}{\partial{P}}\right)_{T}\left(\frac{\partial{P}}{\partial{T}}\right)_{V}$ and because I failed to understand this equation 'physically' I just tried deriving it and mathematically understand it as the concept of directional derivative. $\endgroup$
    – Youhwan Jo
    Commented Sep 30, 2020 at 23:10
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And ($\frac{∂H}{∂T})_V $is a directional derivative with a constraint (V=constant) therefore satisfying some relationship between T and P expressed by $ (\frac{∂P}{∂T})_V$

Directional derivative is something different.. What you have here is a partial derivative. Consider the gradient vector of $H$ and dot that with a displacement vector of state variable to get the change in $H$ as you move from one point to another on the surface.

$$ dH(T,P) = \nabla H \cdot (dT, dP) = \frac{\partial H}{\partial T} dT + \frac{\partial H}{\partial P} dP$$

Now the above quantity is what is known as the differential or directional derivative(Ref)

(and such direction is along a line also satisfying that constraint.).

Not sure what you mean here.

Second, unlike typical geometry dimensions like (x,y,z), definitely, thermodynamic variables like (P,T) are inter-connected, I mean for me they are not independent like (x,y,z). Then why such thermodynamics variables considered mathematically independent? Can I understand this because they are not truly independent so H(P,T) only spans over a limited space (as like a 2-D surface)?

For a single substance homogenous system, then you can change $P$ and $T$ independently. Basically you can vary two out of three basic state variables at a time whilst the other is held constant. If you are varying pressure with temperature constant that means the volume is changing.

Given all of that, you could even write enthalpy as a function of temperature and volume if you wanted... though those variables are 'unnatural'


More in-depth discussion of differentials:

https://www.youtube.com/watch?v=XGL-vpk-8dU&ab_channel=eigenchris

A post in which I wrote an answer about geometric nature of partials:

Explanation on natural variables concept

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