Answer
0.47 L
Work Step by Step
Let $T_{1} = 18^{o}C + 273 K = 291 K$, $V_{1} = 0.5 L$, $T_{2} = 0^{o}C + 273 = 273 K$, $V_{2} = ?$
Using Charles's Law, we have:
$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$
$\frac{0.5 L}{291 K} = \frac{V_{2}}{273 K}$
$V_{2} = \frac{273 K}{291 K}$ x 0.5 L
= 0.47 L
