## Chemistry (4th Edition)

c) $203^{\circ}C$
$V_{1}$ = 76.1 L $T_{1}$ = (89.5+273) K = 362.5 K $V_{2}$ = 100.0 L $\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$ (Charles' law) Or $T_{2}$ = $\frac{V_{2}T_{1}}{V_{1}}$ = $\frac{100.0L\times362.5K}{76.1L}$ = 476 K = $(476-273)^{\circ}$C = $203^{\circ}$C