## Chemistry 10th Edition

b) $\frac{V_{1}}{T_{1}}$ = $\frac{V_{2}}{T_{2}}$ ( Charles's law) Then, $V_{2}$ =$\frac{31.0L\times311K}{292K}$ =33.0 L c) $V_{2}$ = $\frac{31.0L\times 400K}{292K}$ = 42.5 L d) $V_{2}$ = $\frac{31.0L\times273K}{292K}$ = 29.0 L