Answer
$$-5.0 \times 10^{3} \mathrm{J}$$
Work Step by Step
Using Table 19-3 and Eq. 19-45, we find
$$\quad \Delta E_{\mathrm{int}}=n\left(\frac{3}{2} R\right) \Delta T $$ $$=(3.0 \mathrm{mol})(8.31 \mathrm{J} / \mathrm{mol} \cdot \mathrm{K})(200 \mathrm{K}-400 \mathrm{K})$$ $$=-5.0 \times 10^{3} \mathrm{J}$$
