Isochoric Process

As described in NCERT Class 11, Chapter 12 (Thermodynamics), the work done by a gas during a volume change is $\Delta W=\int PdV$. In an isochoric (constant volume) process the volume is fixed, so $dV=0$ and therefore $\Delta W=0$, no matter how much pressure or temperature rises. A rigid steel container enforces exactly this condition. By the First Law, $$\begin{gathered} \Delta Q=\Delta \\ cup+\Delta W=\Delta \\ cup+0 \end{gathered}$$, so all 400 J of supplied heat goes entirely into raising the internal energy of the gas, increasing its temperature. The option that work equals 400 J is wrong because that would require the heat to be converted to mechanical work, which is impossible without volume change. The option of 200 J is an arbitrary fraction with no physical basis here. The option that it cannot be determined is wrong because the constant-volume condition alone fixes the work at zero independent of temperature. A check on units and logic: $PdV$ has units of energy, and with $dV=0$ the integral is identically zero, confirming the result.