For an isothermal reversible compression of an ideal gas?

Answer: Option 3

Actually the internal energy ($$U$$) of a substance is a function of $$\eqalign{ & dU = CvdT - \left[ {P + T\left( {\frac{{\left( {\frac{{\partial V}}{{\partial T}}} \right)p}}{{\left( {\frac{{\partial V}}{{\partial P}}} \right)T}}} \right)dV} \right] \cr & {\text{For an ideal gas, }}PV = RT \cr & {\text{So, }}\left( {\frac{{\partial V}}{{\partial T}}} \right)p = \frac{R}{P}\,\,{\text{and }}\left( {\frac{{\partial V}}{{\partial P}}} \right)T = - \frac{{RT}}{{{P^2}}} \cr & {\text{Hence, }}dU = CvdT \cr} $$ Similarly we can prove $$dH = {C_p}dT$$ So, enthalpy and internal energy are solely functions of temperature in case of ideal gas hence $$\Delta E = \Delta H = 0$$