## Chemistry 10th Edition

$V_{1}$ = 256 mL $P_{1}$ = 2.75 atm $T_{1}$ = (16+273) K = 289 K $P_{2}$ = 1.00 atm $T_{2}$ = (100+273) K = 373 K Using combined gas law, $V_{2}$ = $\frac{P_{1}\times V_{1}\times T_{2}}{T_{1}\times P_{2}}$ = $\frac{2.75atm\times256 mL\times373K}{289K\times1.00atm}$ $\approx$ 909. mL