Answer
4885.06 mL
Work Step by Step
Let $T_{1} = 75^{o}C + 273 K = 348 K$, $V_{1} = 2500 mL$, $T_{2} = 680 K$, $V_{2} = ?$
Using Charles's Law, we have:
$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$
$\frac{2500 mL}{348 K} = \frac{V_{2}}{680 K}$
$V_{2} = \frac{680 K}{348 K}$ x 2500 mL
= 4885.06 mL