Chapter 12 - Gases and the Kinetic-Molecular Theory - Exercises - Charles's Law: The Volume-Temperature Relationship - Page 440: 26

0.81 L

$V_{1}$ = 0.82 L $T_{1}$ = (26+273)K =299 K $T_{2}$ = (21+273)K =294 K According to Charles's Law, $V_{2}$ = $\frac{V_{1}\times T_{2}}{T_{1}}$ =$\frac{0.82L\times 294K}{299K}$ $\approx$ 0.81 L We can use 0$^{\circ}$C = 273.15 K for more precise results.