1. Interpret the $V$–$T$ graph

   For an ideal gas, $PV = nRT$ with $n=1$ mol.

   In a $V$–$T$ graph:

   • Isochoric process: $V=$ constant $ightarrow$ horizontal line.
   • Isothermal process: $T=$ constant $ightarrow$ vertical line.
   • Isobaric process: from $PV=RT$, at constant $P$, $V = \frac{R}{P}T$ so $V \propto T$, i.e. a straight line through the origin.

2. Identify the four processes from the figure

   From the usual rectangular cyclic diagram in the $V$–$T$ plane:

   • Process I: vertical segment
   • Process II: upper horizontal segment
   • Process III: vertical segment
   • Process IV: lower horizontal segment

   Therefore:

   • Process I and III are isothermal, not isobaric.
   • Process II and IV are isochoric.

   Hence:

   • Statement A: "Process I is an isochoric process" is false.
   • Statement D: "Process I and III are not isobaric" is true.

3. Check heat in process II

   Process II is isochoric, so $W=0$ and therefore $Q = \Delta U$

   For a monatomic ideal gas, $\Delta U = \frac{3}{2}nR\Delta T$

   In process II, temperature increases along the horizontal line, so $\Delta T>0$. Thus, $Q = \frac{3}{2}R\Delta T > 0$

   So the gas absorbs heat in process II.

   Therefore, statement B is true.

4. Check heat in process IV

   Process IV is also isochoric, so again $Q=\Delta U = \frac{3}{2}R\Delta T$

   In process IV, temperature decreases, so $\Delta T<0$. Hence, $Q<0$

   So the gas releases heat in process IV.

   Therefore, statement C is true.

5. Final evaluation of options

   • A: False
   • B: True
   • C: True
   • D: True

Therefore, the correct statements are B, C, D.