Taking Derivatives of the Fundamental Property Relations

[ChemEng]

TL;DR

Help with the application of derivatives to the fundamental property relation in classical thermodynamics.

I am having some trouble following the derivation of the partial derivative of internal energy with respect to volume at constant temperature.

The fundamental property relation is given by:

• dU = TdS - PdV

The text I have shows the result of taking the partial derivative with respect to volume at constant temperature as:

• dU/dV_T = T*dS/dV_T - P

But if I apply the chain rule to the -PdV term, it seems like I would also end up with a term for the partial derivative of pressure we respect to volume at constant temperature:

• -(dP/dV_T)*dV

I am looking for an explanation for why this term doesn't appear.

At a more basic level, I am confused by how the original expression for dU is itself inserted into the partial differential expression.

I know that algebraically, it looks like we just divide both sides by dV, but my understanding is that that is not quite proper use of differentials.

 

[anuttarasammyak]

dU=TdS-pdV
divided by dV with condition T=const.
\frac{dU}{dV}|_T=\frac{TdS}{dV}|_T-\frac{pdV}{dV}|_T
Here if $pdV$ were $d(pV)$,

TL;DR Summary: Help with the application of derivatives to the fundamental property relation in classical thermodynamics.

-(dP/dV_T)*dV

would come from it.

 

[ChemEng]

Reading your post got me googling again, and I came across this. Seems like what you are showing here.

I thought that this type of operation was frowned upon, but I will chalk this up to me being mistaken.

There is also a very in-the-weeds derivation given in Physical Chemistry (Engel and Reid).

 

[Chestermiller]

You need to review your basics on partial differentiation.

 

[PeroK]

TL;DR Summary: Help with the application of derivatives to the fundamental property relation in classical thermodynamics.

I am having some trouble following the derivation of the partial derivative of internal energy with respect to volume at constant temperature.

The fundamental property relation is given by:

• dU = TdS - PdV

The text I have shows the result of taking the partial derivative with respect to volume at constant temperature as:

• dU/dV_T = T*dS/dV_T - P

But if I apply the chain rule to the -PdV term, it seems like I would also end up with a term for the partial derivative of pressure we respect to volume at constant temperature:

• -(dP/dV_T)*dV

I am looking for an explanation for why this term doesn't appear.

At a more basic level, I am confused by how the original expression for dU is itself inserted into the partial differential expression.

I know that algebraically, it looks like we just divide both sides by dV, but my understanding is that that is not quite proper use of differentials.

A term like $dU$ is a "differential". In one sense, it's the limit of a small change. If we have
$$\Delta A = B\Delta C$$Then, for example:
$$\frac{\Delta A}{\Delta t} = B\frac{\Delta C}{\Delta t}$$And taking the limit gives the required result.

 

[ChemEng]

A term like $dU$ is a "differential". In one sense, it's the limit of a small change. If we have
$$\Delta A = B\Delta C$$Then, for example:
$$\frac{\Delta A}{\Delta t} = B\frac{\Delta C}{\Delta t}$$And taking the limit gives the required result.

So this is leading back to my original confusion.

Using the notation this way, why not write (B/dt)*dC or (B*dC)/dt? What permits us to ignore B?

For the first term in the original post, taking T outside the derivative makes sense as we explicitly hold T as constant. But this isn't the case for the PdV term.

 

[PeroK]

So this is leading back to my original confusion.

Using the notation this way, why not write (B/dt)*dC or (B*dC)/dt? What permits us to ignore B?

For the first term in the original post, taking T outside the derivative makes sense as we explicitly hold T as constant. But this isn't the case for the PdV term.

What do you mean by ignoring B? B is some quantity that relates a small change in A to a small change in C.

You could write $B/\Delta t$ but that's no good for taking the limit.

 

[ChemEng]

Maybe "ignore" is the wrong word.

I think the confusion on my end is to understand why we aren't treating the partial of volume as an "operator" acting on each term and then applying the product rule to the PdV term.

$\frac{\partial (dU)}{\partial V}_T = \frac{\partial (TdS - PdV)}{\partial V}_T$
$\frac{\partial (TdS)}{\partial V}_T + \frac{\partial (-PdV)}{\partial V}_T$

From this step, I'm not sure how to proceed to get the answer. I understand constant T means we can take T our of the partial. But what about the PdV term.

 

[anuttarasammyak]

Here if pdV were d(pV),

using the chain rule
pdV \neq d(pV)=pdV+Vdp
d(pV) : differentiaion works on p
p dV : diffentiation does not work on p
g(x)\frac{df(x)}{dx} \neq \frac{d\ g(x)f(x)}{dx}=f'(x)g(x)+f(x)g'(x)

For Enthalpy H=U+PV,
dH=dU+d(PV)=TdS-PdV+PdV+VdP=TdS+VdP

Derivatives of U are division of infinitesimal dU by other infinitesimals dV,dp etc.

 

[ChemEng]

TL;DR Summary: Help with the application of derivatives to the fundamental property relation in classical thermodynamics.

I am having some trouble following the derivation of the partial derivative of internal energy with respect to volume at constant temperature.

The fundamental property relation is given by:

• dU = TdS - PdV

The text I have shows the result of taking the partial derivative with respect to volume at constant temperature as:

• dU/dV_T = T*dS/dV_T - P

But if I apply the chain rule to the -PdV term, it seems like I would also end up with a term for the partial derivative of pressure we respect to volume at constant temperature:

• -(dP/dV_T)*dV

I am looking for an explanation for why this term doesn't appear.

At a more basic level, I am confused by how the original expression for dU is itself inserted into the partial differential expression.

I know that algebraically, it looks like we just divide both sides by dV, but my understanding is that that is not quite proper use of differentials.

Going to thank everyone for the help and suggestions for understanding this.

Ended up coming across a good explanation here for a similar problem.

I'll put what I think is the most straightforward way of getting at my confusion, summarize some of the helpful feedback below, and write out the answer in a way that makes the most sense to me in case someone else comes across this post.

Fundamental Property Relation:


$dU = T dS - P dV$


Total derivative of S(V,T):


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


Substitute total derivative, dS, into the fundamental property relation:


$dU = T \left( \left( \frac{\partial S}{\partial V} \right)_T dV + \left( \frac{\partial S}{\partial T} \right)_V dT \right) - P dV$


Collect terms for dV and dT:


$dU = \left( T \left( \frac{\partial S}{\partial V} \right)_T - P \right) dV + T \left( \frac{\partial S}{\partial T} \right)_V dT$


But we can also express the total derivative of U(T,V):


$dU = \left( \frac{\partial U}{\partial V} \right)_T dV + \left( \frac{\partial U}{\partial T} \right)_V dT$


The key step I was missing is that going from here to an expression for $\left( \frac{\partial U}{\partial V} \right)_T$ is not really about further manipulation. It is recognizing that based on the expression for a total derivative for dU, the term in front of "dV" is in a sense necessarily equal to $\left( \frac{\partial U}{\partial V} \right)_T$.


Thus, we look at the term in front of $\left( \frac{\partial U}{\partial V} \right)_T$ and see it is $T \left( \frac{\partial S}{\partial V} \right)_T - P$.


We don't need further algebraic manipulation. This is no different than steps taken to generate Maxwell Relations.

 

[anuttarasammyak]

I hope the article you find helps you to understand the rule of mathematics behind so that you will get right answers for similar but different problems without help of internet.