Let the volume of the gas at 0°C = V ml
Thus, we have:
V1 = V ml,
V2 = 2V ml
TI = 0 + 273
= 273 K
T2 =?
By applying Charles Law,
\(\frac{V_1}{T_1}=\frac{V_2}{T_2}\)
Substituting the corresponding values, we have
\(\frac{V}{273}=\frac{2V}{T_2}\)
T2 = \(\frac{2V\times273}{V}\) = 546 K
T2 = 546 – 273 = 273oC
