During a non-flow thermodynamic process (1-2) executed by a perfect gas, the heat interaction is equal to the work interaction (Q_1-2 = W_1-2) when the process is

The correct option is Isothermal.

To understand why this is correct, we can analyze the process using the First Law of Thermodynamics for a non-flow (closed system) process. The first law is expressed mathematically as:

$$Q_{1-2}-W_{1-2}=\Delta U$$

where:
$Q_{1-2}$ is the heat interaction,
$W_{1-2}$ is the work interaction, and
$\Delta U$ is the change in internal energy of the system during the process from state 1 to state 2.

For the heat interaction to be equal to the work interaction ($Q_{1-2}=W_{1-2}$), the change in internal energy must be zero:

$$\Delta U=0$$

For a perfect gas (or ideal gas), the internal energy $U$ is a function of temperature only (Joule's law). The change in internal energy is given by:

$$\Delta U=m{C}_v(T_2-T_1)$$

where $m$ is the mass, $C_v$ is the specific heat at constant volume, and $T$ represents the temperature.
Therefore, for $\Delta U$ to be zero, the temperature must remain constant throughout the process:

$$T_1=T_2$$

A thermodynamic process that occurs at a constant temperature is defined as an isothermal process. Hence, during an isothermal process for a perfect gas, all the heat added to the system is converted entirely into work output, making $Q_{1-2}=W_{1-2}$.