The entropy change in a reversible isothermal process, when an ideal gas expands to four times its initial volume is

For an isothermal process change in internal energy is equal to zero.
So, from first law
$$\eqalign{ & q = - w = - 2.303\,RT\,{\text{lo}}{{\text{g}}_{10}}\left( {\frac{{v2}}{{v1}}} \right) \cr & \Rightarrow q = - RT\,{\log _e}4 \cr & \Rightarrow {\text{We know for reversible isothermal process }}ds = - \frac{{\delta q}}{T} \cr & \Rightarrow {\text{So, }}ds = R\,{\log _e}4. \cr} $$