Solution

In the general case, the change in entropy of an ideal gas when transitioning from state $1$ to state $2$ is described by the formula:
$$\Delta S = \frac{m}{M} C_V \ln\left(\frac{T_2}{T_1}\right) + \frac{m}{M} R \ln\left(\frac{V_2}{V_1}\right)$$
where $\frac{m}{M}$ is the amount of substance (moles), $C_V$ is the molar heat capacity at constant volume, and $R$ is the universal gas constant.

Let us consider both cases:

b) In an isothermal process
According to the given conditions, the process is isothermal, hence the temperature is constant ($T_1 = T_2$). The first term in the entropy equation (which depends on temperature) becomes zero.
The change in entropy is determined solely by the change in volume from $V_1 = V$ to $V_2 = 2V$:
$$\Delta S_b = \frac{m}{M} R \ln\left(\frac{2V}{V}\right) = \frac{m}{M} R \ln 2$$

a) Upon expansion in a vacuum
The expansion of a gas into a vacuum is an adiabatic process with no work done ($Q = 0$, $A = 0$). According to the first law of thermodynamics, the change in internal energy is $\Delta U = Q - A = 0$.
Since the internal energy of an ideal gas depends only on its temperature, a constant internal energy implies that the gas temperature remains unchanged ($T_1 = T_2$).

Thus, the initial and final states of the gas upon expansion in a vacuum are exactly the same as in the isothermal expansion. Since entropy is a state function, its change does not depend on the path the system took to get from the initial state to the final one.
Therefore, the entropy increment in a vacuum will be exactly the same:
$$\Delta S_a = \Delta S_b = \frac{m}{M} R \ln 2$$