Work In A Polytropic Cycle

Because the volume is held constant, the process is isochoric, so the gas does no work and the First Law reduces to $$\begin{gathered} Q=\Delta \\ cup=n{C}_V\Delta T \end{gathered}$$. For a monatomic ideal gas $C_V=\frac{3}{2}R$, since it has three translational degrees of freedom each contributing $\frac{1}{2}R$. Substituting $n=1$, $C_V=\frac{3}{2}(8.31)=12.47{\text{ J mol}}^{-1}{\text{K}}^{-1}$, and $\Delta T=600-300=300\text{ K}$ gives $Q=1\times 12.47\times 300\approx 3740\text{ J}$. The value 6232 J wrongly uses $C_P=\frac{5}{2}R$, appropriate for constant pressure, not constant volume. The value 2494 J uses $R$ alone without the factor $\frac{3}{2}$. The value 1247 J halves the correct $C_V$ contribution. Since no work is done, all supplied heat raises internal energy, so $Q$ equals $\Delta U$ exactly. Had the same temperature rise been carried out at constant pressure instead, more heat would be required because part of it would go into expansion work, illustrating why $C_P$ exceeds $C_V$ by $R$. As a magnitude check, doubling the temperature of one mole of monatomic gas should add a few kilojoules of internal energy, and 3740 J sits squarely in that expected range.