# Question about entropy=0 in an irreversible process

1. Feb 22, 2012

### Ruko15

We learned about calculating entropy in my physics 2 class, and i came up with an example that my professor couldn't answer, so I was hoping someone here could shed some light on this for me.

We learned that change in entropy for an irreversible process be represented by changes in entropy of reversible process as long as both reversible and irreversible paths have the same initial and final states.

so here are my two states for 1 mole of a monatomic gas.
P0=1kPa
V0=1m3
T0=$\frac{500000}{4157}$K≈120.279K

P=500Pa
V=23/5m3≈1.516m3
T=$\frac{906987}{9950}$K≈91.154K

These two states lie on a reversible adiabatic path.
Calculating the entropy using (using fractions, instead of approximate values)
ΔS=nCvln$(\frac{T}{T0})$+nRln$(\frac{V}{V0})$
ΔS=$(1)(\frac{3}{2})(8.314)ln(\frac{≈120.279}{≈91.154})+(1)(8.314)ln(\frac{≈1.516}{1})$
ΔS=0

Would this not imply then that any real path between these two states is reversible, or that it is possible for an ireversable path to have a ΔS=0? Both of which we learned were not possible.

Is this in the correct place or should it be in Homework & Coursework Questions?

Last edited: Feb 22, 2012
2. Feb 22, 2012

### eis2718

For an irreversible process (not path!) ending up in that same state the entropy of something else will increase.

For example, if we allow the gas to free expand to the final volume (an irreversible process), its final temperature will be be the same as the initial temperature. To reach the desired state we could place it in contact with a temperature reservoir at 91.154K. The heat transferred will increase the entropy of that reservoir.

Any irreversible process will end up with ΔS>0, when you include all of the interacting parts.