Thermodyanamics

[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.)