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

Answer: Option A

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