Step 1: Apply the first law to the process.
The first law for a closed system reads \( Q = \Delta U + W \).
For an ideal gas, internal energy depends only on temperature, and the process here is isothermal, so \( \Delta U = 0 \).
That makes \( Q = W \) automatically true for any isothermal expansion of an ideal gas, exactly what the question states.
Step 2: Check the second law.
A quasi-static isothermal expansion exchanging heat with a reservoir at the same temperature as the gas is fully reversible, so no entropy is generated and no law is broken.
Nothing in the statement forces any irreversibility or any impossible heat flow.
Final Answer:
\( Q = W \) simply restates the first law for an isothermal ideal gas process, so the process is possible.
\[ \boxed{\text{The process is possible}} \]