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Isothermal Expansion and the 2nd Law

  1. Sep 5, 2013 #1
    The book that I use (Concepts in Thermal Physics by S. and K. Blundell) states the second law in two ways. The way they state the Kelvin version is "no process is possible whose sole result is the complete conversion of heat into work." How does that fit in with the isothermal expansion of an ideal monatomic gas?
    From the first law we have,
    Delta(U)=0=-W(by gas)+Q(added to gas)
    giving
    W(by gas)=Q(added to gas)

    Assuming there is no friction from the piston, this should be correct. However, this seems to suggest that isothermal expansion violates the second law if it is not part of a cycle. How do we reconcile it with the second law?
     
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  3. Sep 5, 2013 #2

    jfizzix

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    For an ideal monatomic gas, the entropy is given by the Sackur Tetrode equation:

    [itex]S(N,V,T) = N k_{B}( \log(\frac{\eta_{q}}{\eta})+\frac{5}{2});[/itex]
    where
    [itex]\eta_{q}\equiv (\frac{m k_{B}T}{2 \pi \hbar^{2}})^{\frac{3}{2}} [/itex] and [itex]\eta \equiv \frac{N}{V}[/itex].

    For an isothermal expansion, [itex]T[/itex] and [itex]N[/itex] are constant. The change in entropy [itex]S_{f}-S_{i}[/itex] is expressed as

    [itex]S_{f}-S_{i}= N k_{B} \log(\frac{V_{f}}{V_{i}})[/itex].

    When the gas expands at constant temperature, the entropy increases.

    Another way of looking at this would be that since heat is flowing to restore thermal equilibrium with the reservoir at temperature [itex]T[/itex], the entropy of the system and reservoir together must increase.
     
  4. Sep 5, 2013 #3

    CAF123

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    I also have this book. One of the key words in that statement is the word 'sole'. In the isothermal expansion, the volume of the gas has changed, and thus the state of the gas at the start and end point has changed.

    Since there is a net change in the state of the gas, the result of the process is that the gas has expanded and work done by gas = heat energy transferred by the first law. Therefore, the result of the process is NOT the sole conversion of heat into work.
     
  5. Sep 5, 2013 #4

    Jano L.

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    The important part of this formulation is the word "sole". This implies that the state of the thermodynamic system should be the same as in the beginning, i.e. it should have the original volume. In your example, the volume of the system has increased, so it does not contradict the formulation.
     
  6. Sep 9, 2013 #5
    Thank you very much
     
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