Answer :
We can solve the problem by using Charle's law, which states that for a transformation of an ideal gas at constant pressure, the ratio between volume and absolute temperature of the gas remains constant:
[tex] \frac{V}{T}=k [/tex]
where V is the volume of the gas and T its absolute temperature.
The previous law can be rewritten as
[tex] \frac{V_1}{T_1}= \frac{V_2}{T_2} [/tex] (1)
Let's convert the initial and final temperatures of the gas in Kelvin:
[tex]T_1=18C+273=291 K[/tex]
[tex]T_2 = 0C+273 = 273 K[/tex]
and the intiial volume is
[tex]V_1 = 12.5 L[/tex]
so if we re-arrange equation (1) and we use these data, we can find the final volume of the gas:
[tex]V_2=V_1 \frac{T_2}{T_1}=(12.5 L) \frac{273 K}{291 K}=11.7 L [/tex]
[tex] \frac{V}{T}=k [/tex]
where V is the volume of the gas and T its absolute temperature.
The previous law can be rewritten as
[tex] \frac{V_1}{T_1}= \frac{V_2}{T_2} [/tex] (1)
Let's convert the initial and final temperatures of the gas in Kelvin:
[tex]T_1=18C+273=291 K[/tex]
[tex]T_2 = 0C+273 = 273 K[/tex]
and the intiial volume is
[tex]V_1 = 12.5 L[/tex]
so if we re-arrange equation (1) and we use these data, we can find the final volume of the gas:
[tex]V_2=V_1 \frac{T_2}{T_1}=(12.5 L) \frac{273 K}{291 K}=11.7 L [/tex]