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

A. R log_e 4

B. R log₁₀ 4

C. C_v log₁₀ 4

D. C_v log_e 4

Answer: Option A

Solution(By Examveda Team)

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