the pressure that a gas exerts on a sealed container changes from 53.0 atm to 35.2 atm. what is the initial…

the pressure that a gas exerts on a sealed container changes from 53.0 atm to 35.2 atm. what is the initial temperature of the gas if the final temperature is 12.0 °c? t = ? °c
Answer
Explanation:
Step1: Convert final temperature to Kelvin
$T_2=12.0 + 273.15=285.15\ K$
Step2: Apply Gay - Lussac's Law ($\frac{P_1}{T_1}=\frac{P_2}{T_2}$)
$T_1=\frac{P_1T_2}{P_2}$
Step3: Substitute values
$T_1=\frac{53.0\ atm\times285.15\ K}{35.2\ atm}\approx429.9\ K$
Step4: Convert back to Celsius
$T_1 = 429.9 - 273.15=156.75^{\circ}C$
Answer:
$156.75^{\circ}C$