Question

A gas at 20.0°C and 52.1 mL is heated to 93.5°C. What is the new volume?

Answer 1

$\frac{ V_{1} }{ T_{1} }$ = $\frac{ V_{2} }{ T_{2} }$
$V_{1}$ = 0.0521 L
$T_{1}$ = 293.15 K
$T_{2}$ = 366.65 K

Solve for $V_{2}$

$\frac{0.0521 L}{293.15 K}$ = $\frac { V_{2} }{366.65 K}$

$V_{2}$ = 0.06516 L or 65.2 mL