Question

A neon gas at 1 atm has a volume of 12.0 L and a temperature of 44°C. Find the new volume of the gas after the temperature has been increased to 85°C.

Answer 1

$\frac {V_1} {T_1}=\frac{V_2}{T_2}$

$V_2=\frac {12.0L×358K}{317K}$

V2=13.55L