A student says: 'In an isothermal expansion of an ideal gas the temperature does not change, so no…

Step 1. The mistake. A steady temperature means $\Delta U = 0$, not $Q = 0$. The gas does work as it expands, so heat must flow in to replace that energy: $Q = W$.

Step 2. Heat. $Q = 0.50\ \text{mol} \times 8.31\ \text{J mol}^{-1}\text{K}^{-1} \times 400\ \text{K} \times \ln 3 = 1826\ \text{J}$.