Answer
See explanation.
Work Step by Step
(a) $f(v) = 4\pi \frac{m}{2 \pi kT} v^2 e^{-mv^2/2kT}$
The velocity distribution depends only on T, so in an isothermal process it does not change.
(b) $\Delta S = k ln \frac{w_2}{w_1} $
Where $ 3V \space is \space w_2 = (3)N w_1$ So
$\Delta S = k ln (3)^N $
$\Delta S =N k ln (3) $
$\Delta S =nN_A k ln (3) $
$\Delta S =nR ln (3) $
$\Delta S =(2.00 mol)(8.3145 J/mol⋅K)ln(3) $
$\Delta S = +18.3 J/K$
(c) $Q = nRT ln (\frac{V_2}{V_1}) $
$\Delta S = Q/T$
So $\Delta S = nR ln (\frac{V_2}{V_1})$
$\Delta S = (2.00 mol)(8.3145 J/mol ⋅K) ln (\frac{3V_1}{V_1})$
$\Delta S = +18.3 J/K$ The result obtained in (b) are the same
