Change in Entropy for Irreversible Processes

[nayanm]

I was following along in my Thermodynamic textbook and began playing with some definitions. In the following formulation, I somehow managed to prove (obviously incorrectly) that dq = TdS for even irreversible processes. I was hoping someone could point out where in the proof I'm going wrong.

First, consider a both reversible and adiabatic process.

Since dw = -p_{ext}dV for all processes and dq = 0 for adiabatic processes: dU = -p_{ext}dV

We also know for a reversible and adiabatic process, U is a function of V only and not S, so: dU = \frac{∂U}{∂V}dV

Setting the coefficients equal: \frac{∂U}{∂V}=-p_{ext}

This equation involves only state variables and is therefore valid for all process, reversible or irreversible.Next consider a general process (either reversible or irreversible).

As before dw = -p_{ext}dV but now U is a function of S and V, so: dq = dU - dw = (\frac{∂U}{∂S}dS + \frac{∂U}{∂V}dV) - (-p_{ext}dV)

Simplifying: dq = \frac{∂U}{∂S}dS + \frac{∂U}{∂V}dV + p_{ext}dV

Finally, since we know \frac{∂U}{∂V}=-p_{ext}, the last two terms cancel, leaving: dq = \frac{∂U}{∂S}dS = TdS for any processes, reversible or irreversible.

But clearly, this is not true since dq >TdS for irreversible processes.

At which point, then, should I have needed to invoke irreversibility?

Thank you in advance.

 

[UltrafastPED]

See http://web.mit.edu/16.unified/www/FALL/thermodynamics/thermo_7.htm