Explain the isothermal and free expansion of an ideal gas.

(N/A) For isothermal $(T = \text{constant})$ expansion of an ideal gas into a vacuum,the external pressure $(p_{ex})$ is $0$. Since work done $w = -p_{ex} \Delta V$,it follows that $w = 0$.
According to the first law of thermodynamics,$\Delta U = q + w$. For an ideal gas,internal energy $(U)$ is a function of temperature only. Thus,for an isothermal process,$\Delta U = 0$. Consequently,$q = -w = 0$.
For other isothermal processes:
$1$. For isothermal irreversible change: $q = -w = p_{ex}(V_f - V_i)$.
$2$. For isothermal reversible change: $q = -w = nRT \ln \frac{V_f}{V_i} = 2.303 nRT \log \frac{V_f}{V_i}$.
$3$. For adiabatic change: $q = 0$,therefore $\Delta U = w_{ad}$.