Question

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?

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.

Answer 1

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

For example, we know that `dG` is an exact differential, whose Maxwell relation is:

`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 an inexact differential and `d` indicates an exact differential.

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, `deltaq_(rev)` is an inexact differential (assuming ideal gases). We can then make it an exact differential by multiplying through by the integrating factor `1/T`:

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

We recognize this as the entropy, `dS = (deltaq_(rev))/T`, and expect this to yield an exact differential.

Right now, `P` is dependent on temperature, so let's rewrite this for an ideal gas, which has `P/T = (nR)/V`:

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

This now does yield `0 = 0`, which makes this an exact differential, and shows that entropy is a state function, as it should be.