The process occurs at constant volume, so we can use Gay-Lussac law, which states that for a gas transformation that occurs at constant volume, the ratio between the gas pressure and its temperature is constant:
[tex] \frac{p}{T}= const. [/tex]
which can also be rewritten as
[tex] \frac{p_1}{T_1}= \frac{p_2}{T_2} [/tex]
where all the temperatures must be expressed in Kelvin.
The initial temperature is [tex]T_1 = 20.0 ^{\circ} C= 293 K[/tex]
Therefore if we use the pressures given by the problem, we can calculate the final temperature of the gas:
[tex]T_2 = T_1 \frac{p_2}{p_1} = (293 K) \frac{1.00 atm}{3.00 atm} =97.7 K[/tex]
Which can be converted in Celsius degrees:
[tex]T_2 = 97.7 K - 273 = -175.3^{\circ}C[/tex]
