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.

Step-by-step Explanation:

For a non-flow (closed system) thermodynamic process, the First Law of Thermodynamics is stated as:

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

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

According to the given condition, the heat interaction is equal to the work interaction:

$Q_{1-2}=W_{1-2}$

Substituting this condition into the First Law of Thermodynamics equation yields:

$\Delta U=0$

For a perfect (ideal) gas, internal energy is a function of temperature only. Therefore, the change in internal energy ($\Delta U$) is expressed as:

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

Since $\Delta U=0$, and the mass ($m$) and specific heat capacity at constant volume ($C_v$) are non-zero constant values, we must have:

$T_2-T_1=0$

⇒ $T_2=T_1$

This indicates that the temperature of the perfect gas remains constant throughout the process. A process that occurs at a constant temperature is called an isothermal process.