Isochoric process

Constant $V$, so $\Delta V = 0$, so $W = -P\Delta V = 0$, and the first law collapses to

Every joule transferred by heating goes straight into internal energy, and for an ideal monatomic gas straight into temperature through $U = \frac{3}{2}nRT$. Heat a gas inside a rigid sealed container and its pressure climbs while its volume cannot respond.

On a pressure-volume diagram the path is a vertical line. A vertical line spans no width on the volume axis, so there is no area beneath it, which is the graphical form of $W = 0$.

A vocabulary point that costs marks on a stem. The AP Physics 2 required course content calls this case isovolumetric, at 9.4.B.3, where the special cases are listed as constant volume (isovolumetric), constant temperature (isothermal) and constant pressure (isobaric), plus adiabatic. "Isochoric" is the commoner textbook word, and in the course description it turns up only inside an optional sample instructional activity. They name the same process. Recognize both.

Do not read zero work as zero energy transfer. Only the work term vanishes. $Q$ can be large in either direction, and the temperature moves with it.

The mirror image is the adiabatic case, where the other term vanishes instead: $Q = 0$ and $\Delta U = W$.