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.(adsbygoogle = window.adsbygoogle || []).push({});

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.

P_{0}=1kPa

V_{0}=1m^{3}

T_{0}=[itex]\frac{500000}{4157}[/itex]K≈120.279K

P=500Pa

V=2^{3/5}m^{3}≈1.516m^{3}

T=[itex]\frac{906987}{9950}[/itex]K≈91.154K

These two states lie on a reversible adiabatic path.

Calculating the entropy using (using fractions, instead of approximate values)

ΔS=nC_{v}ln[itex](\frac{T}{T0})[/itex]+nRln[itex](\frac{V}{V0})[/itex]

ΔS=[itex](1)(\frac{3}{2})(8.314)ln(\frac{≈120.279}{≈91.154})+(1)(8.314)ln(\frac{≈1.516}{1})[/itex]

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?

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Question about entropy=0 in an irreversible process

**Physics Forums | Science Articles, Homework Help, Discussion**