# How do you prove that đq is not an exact differential for an isobaric process using #đq = dU - đw# where the only work is PV work?

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I understand that heat is a path function and this:

Where dP is an exact differential because #(del/(delV)((delP)/(delT))_V)=(del/(delT)((delP)/(delV))_T)#

That is very easily calculable, however I'm not sure how to set up the equation to prove that đq is an inexact differential.

I understand that heat is a path function and this:

Where dP is an exact differential because

That is very easily calculable, however I'm not sure how to set up the equation to prove that đq is an inexact differential.

##### 1 Answer

An **exact differential** would have that the cross-derivatives are equal.

For example, we know that

#dG = -SdT + VdP = ((delG)/(delT))_PdT + ((delG)/(delP))_TdP#

so we would have that

#(del)/(delP)[((delG)/(delT))_P]_T = (del)/(delT)[((delG)/(delP))_T]_P#

or

#-((delS)/(delP))_T = ((delV)/(delT))_P# .

For the first law of thermodynamics, we have:

#dU = deltaq + deltaw# where

#delta# indicates aninexactdifferential and#d# indicates anexactdifferential.

A reversible process with PV-only work in general would have that:

#dU = deltaq_(rev) - PdV#

By definition, the **constant-volume heat capacity** is

#C_V = ((delU)/(delT))_V# ,

so

#C_VdT + PdV = deltaq_(rev)#

If we took the cross-derivatives at this point:

#((del C_V)/(delV))_T stackrel(?" ")(=) ((delP)/(delT))_V#

The constant-volume heat capacity is only a function of temperature for ideal gases, so:

#0 ne ((delP)/(delT))_V#

So, at this point, **inexact differential** (assuming ideal gases). We can then make it an exact differential by multiplying through by the integrating factor

#C_V/TdT + P/TdV = (deltaq_(rev))/T#

We recognize this as the entropy,

Right now,

#((del(C_V//T))/(delV))_T stackrel(?" ")(=) ((del(nR//V))/(delT))_V#

This now does yield