Is it possible to deduce the second law of thermodynamics from the first?

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?

• I feel like when discussing efficiency you have assumed the second law Mar 14 '20 at 4:39
• 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 Mar 14 '20 at 7:16

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.