Answer
Please see the work below.
Work Step by Step
(a) We know that
$\Delta U=Q-W$
$\implies W=Q-\Delta U$
$W=\frac{5}{2}nR\Delta T-\frac{3}{2}nR\Delta T$
$\implies W=nR\Delta T$
(b) We know that an adiabatic process occurs without heat transfer; therefore $Q=0$
(c) According to the first law of thermodynamics
$\Delta U=Q-W$
We plug in the known values to obtain:
$\Delta U=0-(-\frac{3}{2}nR\Delta T)$
$\implies \Delta U=\frac{3}{2}nR\Delta T$
(d) We know that at constant volume, $\Delta V=0$
$W=P\Delta V$
$\implies W=0$
(e) We know that
$\Delta U=Q-W$
$\Delta U=\frac{3}{2}nR\Delta T-0$
$\Delta U=\frac{3}{2}nR\Delta T$
(f) We know that
$Q=\Delta U+nRT\ln (\frac{V_f}{V_i})$
$Q=0+nRT\ln (\frac{V_f}{V_i})$
$Q=nRT\ln (\frac{V_f}{V_i})$
(g) We know that in an isothermal process $\Delta T=0$, hence the change in the internal energy $\Delta U=0$.
