The entropy change in a reversible isothermal process, when an ideal gas expands to four times its initial volume is
The entropy change in a reversible isothermal process, when an ideal gas expands to four times its initial volume is Correct Answer R log<sub>e</sub> 4
For an isothermal process change in internal energy is equal to zero.So, from first law
$$\eqalign{ & q = - w = - 2.303\,RT\,{\text{lo}}{{\text{g}}_{10}}\left( {\frac{{v2}}{{v1}}} \right) \cr & \Rightarrow q = - RT\,{\log _e}4 \cr & \Rightarrow {\text{We know for reversible isothermal process }}ds = - \frac{{\delta q}}{T} \cr & \Rightarrow {\text{So, }}ds = R\,{\log _e}4. \cr} $$