First law of thermodynamics, q=w for a reversible isothermal process

[jarman007]

according to first law of thermodynamics,q=w for a reversible isothermal process.this means all the energy absorbed is being used to to do the work ,but according to second law of thermodynamics,there cannot be 100% effeciency.please tell where i am going wrong

 

[Mapes]

More precisely, the first law says that the sum of the heat input ($q_\mathrm{in}$) and the work input ($w_\mathrm{in}$) equals the sum of the heat output ($q_\mathrm{out}$) and the work output ($w_\mathrm{out}$). The second law says that in a heat engine (where $w_\mathrm{in}=0$ and $w_\mathrm{out}/q_\mathrm{in}$ is a measure of efficiency), there must be some $q_\mathrm{out}$,* so $w_\mathrm{out}/q_\mathrm{in}<0$. Does this answer your question?

*(To remove the entropy brought in by the heat input; entropy can't decrease and isn't transferred by work.)