Quasistatic isothermal expansion

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SUMMARY

The discussion focuses on the relationship between heat input (Q) and entropy change (ΔS) during quasistatic isothermal expansion of a monatomic ideal gas. It establishes that ΔS can be expressed as ΔS = Q/T, where Q is the heat added to the system. The derivation involves integrating the ideal gas law, leading to the conclusion that ΔS = Nk ln(Vf/Vi) when internal energy (U) and particle number (N) are held constant. The discussion also clarifies that this relationship does not hold during free expansion, where no heat is inputted, thus violating the second law of thermodynamics.

PREREQUISITES
  • Understanding of the ideal gas law
  • Familiarity with the concepts of entropy and thermodynamic processes
  • Knowledge of the Sackur-Tetrode equation
  • Basic principles of thermodynamics, particularly the second law
NEXT STEPS
  • Study the derivation of the Sackur-Tetrode equation in detail
  • Explore the implications of the second law of thermodynamics on spontaneous processes
  • Learn about the differences between quasistatic and free expansion processes
  • Investigate the mathematical integration of the ideal gas law for various thermodynamic processes
USEFUL FOR

Students and professionals in physics and engineering, particularly those studying thermodynamics and gas laws, will benefit from this discussion. It is especially relevant for those looking to deepen their understanding of entropy and heat transfer in ideal gases.

jlmac2001
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How would I show that during quastatic isothermal expansion of a monatomic ideal gas, the change in entropy is related to the heat input Q by the simple formula: delta S=Q/T
Show that it is not valid for the free expansion process described.

Answer: Putting heat into a system always increases its entrophy.
deltaS=Nk lnVf/Vi (U,N fixed).

Ireally don't know how to do this. Can someone explain?
 
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A bit late a reply, but I'm currently busy with this stuff as well so reciting it will help me remembering it, here goes:

During quasistatic isothermal expansion the inputted heat Q equals the minus work -W done. You derive this by integrating the ideal gas law over the volume change:
Q = -W = int ( NkT/V *dV ) from Vi to Vf = NkT*ln(Vf/Vi)

Furthermore, the Sackur-Tetrode equation says that delta S = Nk*ln(Vf/Vi) if you hold U and N fixed, which corresponds to a quasistatic isothermal process. (http://en.wikipedia.org/wiki/Sackur%E2%80%93Tetrode_equation" )
So if you derive Q by T you would get delta S.

A free process is still described by the same function, except the inputted heat is zero. Since spontaneous processes have their entropies increased (second law of thermodynamics) delta S must be growing, and so delta S = Q/T can not be valid.
 
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