Suppose 4.0 mol of an ideal gas undergoes a reversible isothermal expansion from volume V₁ to volume V₂ where $V_2=2.0V_1$  at temperature $T=400K$ . Find the entropy change of the gas . (Take $\ln 2=0.693$ )

Here , $n=4$  mol, $R=8.314$  J/mol K

$T=400K$ , $V_2=2.0V_1$

Work done by an ideal gas during isothermal expansion is

$W=nRT\ln \left(\frac{V_2}{V_1}\right)$ ,

For an isothermal expansion, $Q=W=nRT\ln \left(\frac{V_2}{V_1}\right)$   …….(i)

For reversible isothermal process, the change in entropy is

$\Delta S=\frac{Q}{T}$         

  $\Delta S=\frac{nRT\ln \left(\frac{V_2}{V_1}\right)}{T}$       (using (i))

 $=nR\ln \left(\frac{V_2}{V_1}\right)$

Substituting the given value, we get

$\Delta S=4\times 8.314\times \ln 2=4\times 8.314\times 0.693=23.0$  J/K