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Considering the First law of thermodynamics as an axiom dU=dQ-pdV for any infinitesimal process, we should be able to prove that for any reversible(quasistatic) and cyclic transformation, there is at least one part in the transformation such that dQ<0. So there exists a part in the cycle in which the engine loses heat.
As the efficiency for the cycle is defined as e=1-(Qout\Qin) it follows that e<1. But this is in fact the Kelvin-Planck statement of the Second law for reversible processes! Thus (for quasistatic processes at least) shouldn't this be considered a theorem, rather than an axiom/postulate?

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  • $\begingroup$ I feel like when discussing efficiency you have assumed the second law $\endgroup$ Mar 14 '20 at 4:39
  • $\begingroup$ I also thought about this, but after all efficiency can be defined as the total (useful) mechanical energy produced divided by the heat input e=L/Qin. If now, for example, you consider a cyclic process in p-V coordinates (again, the process must be quasistatic), you can calculate for any point in the transformation dQ=dU+pdV. I am almost certain that there exists (V1,p1) on the graph/transformation such that dQ<0 (thinking of the family of the adiabatic processes makes this quite obvious). So, for prooving the Kelvin-Planck statement I feel like I only used the First Law $\endgroup$
    – smauszmarc
    Mar 14 '20 at 7:16
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The first law is about conservation of energy. The second law is about irreversibility of processes if this involved decreasing entropy in a closed system.

If there are say $N$ degrees of freedom in a system that can be excited, then distributing a given amount of energy $E$ only on say one state (low entropy) of the $N$ or randomly over many of them (high entropy) would both be conserving the energy $E$. So thermodynamically going from a state with high entropy to a state with low entropy concentrating the same amount of energy on few degrees of freedom is not violating the first law. It is however violating the second law in the following sense: going from a high entropy state to such a highly ordered state is extremely unlikely and hence not going to happen in a thermodynamic limit.

Hence the second law of thermodynamics adds new information that cannot be derived from the first one.

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  • $\begingroup$ If the process is cyclic and reversible, the change in entropy is zero. Now, what you have just said is indeed new information (that is called the Second law of thermodynamics stated using entropy), but then again in my particular case, I am only talking about the specific cyclic transformations in which the entropy is constant. So I guess my issue is with the Kelvin-Planck statement (as it should be equivalent with any other formulations of the second law). $\endgroup$
    – smauszmarc
    Mar 13 '20 at 21:42
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    $\begingroup$ A cyclically working machine takes energy in the form of heat and converts it to useful say mechanical energy. This is exactly the case of taking energy from the many degrees of freedom in the heat bath and collapsing all that energy on just one degree of freedom (e.g. of motion). The first law of thermodynamics would not forbid a ship to use a cyclic engine that works by cooling off the ocean, the second law forbids this. $\endgroup$
    – Gianluca
    Mar 13 '20 at 22:50
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you have to prove e < 1 for all processes including irreversible processes to prove second law. your line of reasoning has nothing to say about irreversible changes. this is my first reaction. i too have been thinking about this for a long time. for another perspective, see my book 'The principles of thermodynamics' N.D. Hari Dass. there is a flaw in my argument there proving equivalence of entropy axiom and second law, which has to do precisely with irreversible engines.

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