# Why can we differentiate this entropy total derivative with repect to Temperature?

• Chemistry
• zenterix
zenterix
Homework Statement
I am going through a thermodynamics course and there is a section about natural variables.
Relevant Equations
In that section, there is the following derivation
Ignoring chemical potential for now, the natural variables of ##U## are ##S## and ##V##. Thus

$$dU=\left (\frac{\partial U}{\partial S}\right )_VdS+\left (\frac{\partial U}{\partial V}\right )_SdV=TdS-pdV\tag{1}$$

which we can rewrite for ##dS## as

$$dS=\frac{dU}{dT}+\frac{pdV}{T}\tag{2}$$

We wish to determine how the entropy depends on temperature.

We can differentiate the expression for ##dS## with respect to temperature at constant volume to obtain ##\left (\frac{\partial S}{\partial T}\right )_V##.

Apply the derivative to obtain

$$\left (\frac{\partial S}{\partial T}\right )_V=\frac{1}{T}\left (\frac{\partial U}{\partial T}\right )_V+\frac{p}{T}\left (\frac{\partial V}{\partial T}\right )_V$$

$$=\frac{1}{T}\left (\frac{\partial U}{\partial T}\right )_V=\frac{C_V}{T}$$

Mathematically, what is happening exactly?

In other words, why/how does ##dS## become ##\left (\frac{\partial S}{\partial T}\right )_V##?

Here is my current understanding.

##dS## is the total derivative function for ##S(T,V)##.

##dS## is a linear function of ##dT## and ##dV## and this linear function is a function of ##T## and ##V##.

That is, at each ##T## and ##V## we have a different function ##dS##.

If we differentiate ##dS## relative to ##T## we are determining the rate of change of ##dS## relative to ##T## at fixed ##V##.

Notation-wise, it seems we would have ##\frac{\partial (dS)}{\partial T}##, which is a notation I have never seen before.

If this is correct so far, then it seems that ##\frac{\partial (dS)}{\partial T}## is written as just ##\left (\frac{\partial S}{\partial T}\right )_V##.

But this latter step isn't very clear to me.

So my question is essentially how to understand what is going on from a mathematical perspective in this derivation.

No, don't differentiate dS, rather use the differential of S with variables T and V, ## dS(T,V)=\left(\partial S/\partial T\right)_V dT + \left (\partial S /\partial V\right)_T dV## and a similar expression for dU(T,V).

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