Question 1

What are the following quantities functions of: volume, pressure, temperature and mass? For I am very confused when I should be using $d$ or $\partial$ for my derivatives in thermodynamics.

For example, If I take the total derivative of internal energy [$U(S,V)$] as a function of temperature, I get the following:

$\frac{d}{dT}$U(S,V) = $ \frac{ \partial U}{ \partial T} + (\frac{ \partial U}{ \partial S}) \frac{ds}{dT} + (\frac{ \partial U}{ \partial V}) \frac{dV}{dT} $

If I multiply both sides by $dT$ to get my desired infinitesimal, I get:

$dU(S,V) = \frac{ \partial U}{ \partial t} dT + (\frac{ \partial U}{ \partial S}) ds + (\frac{ \partial U}{ \partial V})dV$

However, my textbook uses the following notation in some questions and I do not think $\partial t$ and $dt$ cancel each other out, but I am not sure

$\frac{d}{dT}$U(S,V) = $ \frac{ \partial U}{ \partial T} + (\frac{ \partial U}{ \partial S}) \frac{ \partial s}{ \partial T} + (\frac{ \partial U}{ \partial V}) \frac{ \partial V}{ \partial T} $

Question 2

When you turn to infinitesimal form, which variable are we taking the derivative with respect to? I always assumed time, but my previous question is casting doubles. For example:

$\frac{d}{dt}$U(S,V) = $ \frac{ \partial U}{ \partial t} + (\frac{ \partial U}{ \partial S}) \frac{ds}{dt} + (\frac{ \partial U}{ \partial V}) \frac{dV}{dt} $

Some authors "cheat" and do the derivative calculations without the numerator. For example:

$F + \Delta F = U + \Delta U - (P + \Delta P)(V + \Delta V)$

How is this connected to a normal calculus (i.e. total/partial derivatives)?


I understand kinds types of derivatives (partial/total), but do know know which type thermodynamics uses or when

Previous Research

Partial derivatives vs total derivatives in thermodynamics


3 Answers 3


Thermodynamics is a minefield for issues like this.

The internal energy, as you say, is a function of two variables - $S$ and $V$. We can define its partial derivatives with respect to $S$ and $V$ as follows:

$$\left(\frac{\partial U}{\partial S}\right)_V = \lim_{h\rightarrow 0} \frac{U(S+h,V)-U(S,V)}{h}$$

$$\left(\frac{\partial U}{\partial V}\right)_S = \lim_{h\rightarrow 0} \frac{U(S,V+h)-U(S,V)}{h}$$

But now you want to talk about $\frac{\partial U}{\partial T}$, where $T \equiv \left(\frac{\partial U}{\partial S}\right)_V$, and it's not immediately clear what that means, since $U$ is not even a function of $T$.

What we mean is the following: we change $U$ by changing both $S$ and $V$ at the same time:

$$dU = U\big(S+dS,V+dV\big) = \left(\frac{\partial U}{\partial S}\right)_V dS + \left(\frac{\partial U}{\partial V}\right)_S dV $$

Because $T\equiv \left(\frac{\partial U}{\partial S}\right)_V$ is also a function of $S$ and $V$, by changing both $S$ and $V$ at the same time, we also end up changing $T$ :

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

and so we define $\frac{\partial U}{\partial T}$ to be the ratio of these two changes:

$$\frac{\partial U}{\partial T} = \frac{\left(\frac{\partial U}{\partial S}\right)_V dS + \left(\frac{\partial U}{\partial V}\right)_S dV}{ \left(\frac{\partial T}{\partial S}\right)_V dS + \left(\frac{\partial T}{\partial V}\right)_SdV}$$

Now, this quantity is not well-defined because we need to specify precisely how we change $S$ and $V$. For example, we could ask about $\left(\frac{\partial U}{\partial T}\right)_V$ - the rate of change of the internal energy with respect to $T$ when we hold $V$ constant. In this case $dV=0$ and we would find

$$\left(\frac{\partial U}{\partial T}\right)_V = \left(\frac{\partial U}{\partial S}\right)_V \big/\left(\frac{\partial T}{\partial S}\right)_V = T \left(\frac{\partial S}{\partial T}\right)_V \equiv c_V$$

where $\left(\frac{\partial S}{\partial T}\right)_V \equiv 1\big/\left(\frac{\partial T}{\partial S}\right)_V$ and we note the right hand side as being simply the definition of the specific heat at constant volume $c_V$.

If you choose some other way to change $S$ and $V$, the result will be different. For example, we could choose not to keep $V$ constant, but rather the pressure $P \equiv -\left(\frac{\partial U}{\partial V}\right)_S$.

  • $\begingroup$ Can you explain more about $π‘‘π‘ˆ=π‘ˆ(𝑆+𝑑𝑆,𝑉+𝑑𝑉)=(βˆ‚π‘ˆβˆ‚π‘†)𝑉𝑑𝑆+(βˆ‚π‘ˆβˆ‚π‘‰)𝑆𝑑𝑉$ or explain what the technique is called? I looked in my calculus book and didn't find it... It looks like a gradient dotted with a vector ($\Del U \circ dL$), where $dL = dx , dy , dz$. It works with potential energy $U(x,y,z ,t)$. However, it does not work for free energy (F= U -TS) $\endgroup$
    – Edward
    May 6, 2020 at 20:11
  • $\begingroup$ @Edward If $U$ is a function of $S$ and $V$, then a small change $dS$ and a small change $dV$ will cause a small change $dU = \left(\frac{\partial U}{\partial S}\right)_VdS + \left(\frac{\partial U}{\partial V}\right)_S dV$. This is just linearization (i.e. the total differential ) applied to a function of several variables. $\endgroup$
    – J. Murray
    May 6, 2020 at 20:15
  • $\begingroup$ Thank you! The second you said that a light went off! $\endgroup$
    – Edward
    May 6, 2020 at 20:21

If U=U(S,V), then, mathematically, you should be writing $$dU=\left(\frac{\partial U}{\partial S}\right)_VdS+\left(\frac{\partial U}{\partial V}\right)_SdV=TdS-PdV$$Treating S as a function of V and T, you also have $$dS=\left(\frac{\partial S}{\partial T}\right)_VdT+\left(\frac{\partial S}{\partial V}\right)_TdV=\frac{C_v}{T}dT+\left(\frac{\partial S}{\partial V}\right)_TdV$$So, combining these two equations, you have $$dU=C_vdT-\left[P-T\left(\frac{\partial S}{\partial V}\right)_T\right]dV$$


Your first equation represents the differential of U with respect to T, as shown here: https://en.wikipedia.org/wiki/Differential_of_a_function.

This function represents a linearization about a specific point in the multi-variable equation, and I seriously doubt that it is correct to multiply that equation by dT in order to eliminate it.


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