# How they derived this eqn?

In the lecture notes
http://webraft.its.unimelb.edu.au/640322/pub/notes/lectures_common/lecture3.pdf [Broken]

How did they derive the equation on the line above "L.H.S bracket is not a function of V thus"?

I can see that (d/dv)(ds)= "that equation except with the 'equals sign' replaced by '+' "

(dU/dV)=0 because it is an ideal gas.

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## Answers and Replies

The expression for dS has three terms so we can write it as

dS = AdT + BdV + CdV

The condition for dS being an exact differential is

d(A)/dV = d(B+C)/dT ( partial differentiation)

But dB/dT is zero because B is held at fixed T, so

d(A)/dV = d(C)/dT

which is the equation you asked about.

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I don't understand why the condition on dS in order for it to be an exact differential. The reasoning for the exact differential on the notes I don't follow either.

I'm not clear what's troubling you.

dS has to be an exact differential for physical reasons. It just means there are no other sources of variation in S but these.

The little equation in brackets ( 'dz = ...') is a theorem, which says that if a total differential is the sum of two sources, then the sources have a special relationship. Just accept it for now. It's used to get the next line.

I'm not clear what's troubling you.

dS has to be an exact differential for physical reasons. It just means there are no other sources of variation in S but these.

The little equation in brackets ( 'dz = ...') is a theorem, which says that if a total differential is the sum of two sources, then the sources have a special relationship. Just accept it for now. It's used to get the next line.

I see now but I seem to be more interested to see why the exact differential demands this form. You want dS to be an exact differential because S is a function of two variables and is a state function which behaves just like a perfect mathematical function of two variables.

That's how I see it, but talk to other people about this, it might be important.