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Background

Trying to understand how this equation for small heat transfers is obtained [quote is just slightly abbreviated]:

From the first law of thermodynamics and the inner energy as a function of $T$ and $P$ it follows :

$$\delta Q = \left( \frac{\partial U}{\partial T} \right)_{p} dT + \left( \frac{\partial U}{\partial p} \right)_{T} dp + p \left[ \left( \frac{\partial V}{\partial T} \right)_{p} dT + \left( \frac{\partial V}{\partial p} \right)_{T} dp \right] \tag{1}$$

Note that the equation itself is not for constant pressure, otherwise one would not have $dp$ there, but must be of full generality.

I've checked Wikipedia's Internal Energy, Heat Capacity and Fundamental Thermodynamic Relation.

Besides relations like the phase rule and pairs of conjugate variables, I'd expect that since this relation is stated as fully general, it could be fully derived using just the first and second principle of thermodynamics.

Question

So starting from: $\delta Q = dU + \delta W$ or using $TdS$ in place of heat if that's useful, how is the first equation obtained?

I assume this may have to do with Maxwell relations as seen here, but can't fully see how.

It's important that only assumptions of the same generality as the 1st and 2nd Laws are used (preferably those laws themselves only are the assumptions.)

4 Answers4

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I understand the confusion, partly because many presentations of the differentials in thermodynamics lack clarity.

Let me start from the beginning. We write the first principle as $$ \Delta U = Q + W\tag{1} $$ (I prefer using a convention about the sign of work opposite to yours: I consider the work, $W$, done on the system and the heat going into it, $Q$, positive.)

The physical content of the first principle is that the sum of two quantities, $W$ and $Q$, that in principle may not even depend on the thermodynamic variables of the system (that's the case of non-quasistatic processes) turns out to be the difference of a function ($U$) of the thermodynamic variables of the system in the final and initial state.

Under the reasonable assumption that $U$ is a differentiable function, a mathematical theorem tells us that for small differences in the independent variables, the difference $\Delta U$ can be well approximated by the differential $dU$. Moreover, for a quasi-static process with a small volume change, the work can be written as $$ \delta W = -p dV,\tag{2} $$ where p is the internal pressure of the system. Notice that the right-hand side of equation $(2)$ is a differential form but not exact unless the process from the initial to the final state is adiabatic. That's why, in Thermodynamics, it is usual to write the left-hand side as $\delta W$.

Therefore, under the hypothesis of a quasi-static process, from equations (1) and (2), we can write $$ \delta Q = dU - \delta W = dU + p dV \tag{3}. $$ In the last equation, $\delta Q$ is also an inexact differential, being the sum of an exact differential ($dU$) and an inexact differential ($pdV$).

If we use $T$ and $p$ as thermodynamic variables$^{(*)}$

$$ dU = \left( \frac{\partial U}{\partial T}\right)_p dT + \left( \frac{\partial U}{\partial p}\right)_T dp.\tag{4} $$ On the other hand, if also $V$ is considered as a function of the same variables, its differential is $$ dV = \left( \frac{\partial V}{\partial T}\right)_p dT + \left( \frac{\partial V}{\partial p}\right)_T dp.\tag{5} $$

Combining equations ($3$), ($4$), and ($%$), we obtain immediately the equation for $\delta Q$.


$^{(*)}$ As a consequence of the first principle and of the definition of entropy, $dU = TdS -pdV$, showing that the internal energy is a natural function of entropy and volume. However, due to its homogeneity of degree 1 and the minimum principle, we can conclude that it is a convex (almost everywhere strictly convex) function of its natural variables. Such a convexity property ensures the possibility, using the inverse function theorem to invert the definitions of pressure and temperature, $$ \begin{align} p &= - \left( \frac{\partial U}{\partial V} \right)_S \\ T &= \left( \frac{\partial U}{\partial S} \right)_V \end{align} $$ to write the internal energy as a function of $T$ and $p$: $U(S(T,p),V(T,p))$.

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For a single phase material in which the only form of work is by volumetric expansion or compression, $\mathrm dW=P \, \mathrm dV$ and, from the first law, $\mathrm dU=\mathrm dQ+P \, \mathrm dV$ for two closely neighboring thermodynamic equilibrium states. Also, at equilibrium, specific $U$ can be expressed as a function to any two intensive variables, such as $U(T,P), U(S,V), U(T,V)$, etc.

M. A.
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Chet Miller
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2

Trying to understand how [this equation for small heat transfers][1] is obtained [quote is just slightly abbreviated]:

From the first law of thermodynamics and the inner energy as a function of $T$ and $P$ it follows:

$$\delta Q = \left( \frac{\partial U}{\partial T} \right)_{p} dT + \left( \frac{\partial U}{\partial p} \right)_{T} dp + p \left[ \left( \frac{\partial V}{\partial T} \right)_{p} dT + \left( \frac{\partial V}{\partial p} \right)_{T} dp \right] \tag{1}$$

...

Question

So starting from: $\delta Q = dU + \delta W$ or using $TdS$ in place of heat if that's useful, how is the first equation obtained?

As stated in the sentence right above your Eq. (1), the energy is a function of $T$ and $p$: $$ U = U(T,p)\;, $$ which means $$ dU = \left(\frac{\partial U}{\partial T}\right)_p dT + \left(\frac{\partial U}{\partial p}\right)_T dp\;.\tag{A} $$

Given that the volume can also be considered a function of $T$ and $p$ $$ V = V(T,p) $$ we also have: $$ dV = \left(\frac{\partial V}{\partial T}\right)_p dT + \left(\frac{\partial V}{\partial p}\right)_T dp\;.\tag{B} $$

The first law of thermodynamics says: $$ dU = \delta Q - \delta W\;, $$ and we assume we can use $\delta W = pdV$, which gives: $$ \delta Q = dU + pdV\;. \tag{C} $$

Plugging into Eq. (C) with Eq. (A) for $dU$ and Eq. (B) for $dV$ gives: $$ \delta Q = \left[\left(\frac{\partial U}{\partial T}\right)_p dT + \left(\frac{\partial U}{\partial p}\right)_T dp\right] + p\left[\left(\frac{\partial V}{\partial T}\right)_p dT + \left(\frac{\partial V}{\partial p}\right)_T dp\right]\;. \tag{D} $$

hft
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Your expression could be written as

$$ \delta Q = \left [\left({\partial U\over \partial T}\right)_p + p\left({\partial V\over \partial T}\right)_p\right ] dT + \left [\left({\partial U\over \partial p}\right)_T + p\left({\partial V\over \partial p}\right)_T\right ] dp. $$

Now, let's try to take a look in each term alone. By the first law of thermodynamics, we know that

$$ dU = \delta Q - \delta W. $$

Considering $\delta W = p dV$

$$ dU = \delta Q - pdV \Rightarrow \delta Q = dU+pdV. $$

If I say "divide the right-hand expression above by $dT$" someone could get offended, so instead you could just consider that, in first order, the above equality implies

$$ \left ({\partial Q\over \partial X}\right)_Y = \left ({\partial U\over \partial X}\right)_Y + p \left ({\partial V\over \partial X}\right)_Y $$

For $X =p$ and $Y= T$ or vice-versa.

Ruffolo
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