A party balloon holds 1.5 mol of helium, a monatomic gas, at 280 K…

Step 1. Internal energy. $U = \tfrac32 \times 1.5\ \text{mol} \times 8.31\ \text{J mol}^{-1}\text{K}^{-1} \times 280\ \text{K} = 5235\ \text{J}$.

Step 2. The rise. $\Delta U = \tfrac32 \times 1.5\ \text{mol} \times 8.31\ \text{J mol}^{-1}\text{K}^{-1} \times 100\ \text{K} = 1870\ \text{J}$.

Step 3. The path. $U$ depends only on $T$, so any way of warming from $280\ \text{K}$ to $380\ \text{K}$ gives the same $1870\ \text{J}$.