A sample of 2.00 mol of an ideal gas at 300 K expands from 10.0 L to 40.0 L, while its temperature…

(a) Reversible.

(b) One step. The final pressure is $p_2 = \dfrac{nRT}{V_2} = \dfrac{4988.4}{40.0 \times 10^{-3}}\ \text{Pa} = 1.2471 \times 10^{5}\ \text{Pa}$ (about $1.247\ \text{bar}$). Then

(c) Into a vacuum. Nothing pushes back: $p_{ext} = 0$, so $w = 0$.

(d) Heat. $\Delta U = 0$, so $q = -w = +6915\ \text{J}$: the gas takes in from its surroundings all the energy it gives out as work.

(e) Entropy. $\Delta S = \dfrac{q_{rev}}{T} = \dfrac{6915}{300}\ \text{J K}^{-1} = 23.05\ \text{J K}^{-1}$. Entropy is a state function, so the same start and end give the same $\Delta S$ whichever way the gas goes.