Answer:
[tex]156.1^{\circ}C[/tex]
Explanation:
We can solve the problem by using Gay-Lussac's law, which states that:
"For a fixed mass of ideal gas kept at constant volume, the pressure of the gas is directly proportional to its absolute temperature"
Mathematically:
[tex]\frac{p}{T}=const.[/tex]
where
p is the pressure
T is the absolute temperature
The equation can be also rewritten as
[tex]\frac{p_1}{T_1}=\frac{p_2}{T_2}[/tex]
where, in this problem:
[tex]p_1 = 53.0 Bar[/tex] is the initial pressure
[tex]p_2=35.2 Bar[/tex] is the final pressure
[tex]T_2=12.0^{\circ}+273=285 K[/tex] is the final temperature
Solving for T1, we find the initial temperature:
[tex]T_1=\frac{p_1 T_2}{p_2}=\frac{(53.0)(285)}{35.2}=429.1 K[/tex]
Converted into Celsius degrees,
[tex]T_2=429.1-273=156.1^{\circ}C[/tex]
