Option D) 180K is the correct answer.
Hence, the temperature of the gas before the pressure changed is 180K
Gay-Lussac's law states that "the pressure exerted by a given mass of gas at constant volume varies directly with its absolute temperature".
It is expressed as;
[tex]\frac{P_1}{T_1} = \frac{P_2}{T_2}[/tex]
Given the data in the question;
Initial pressure of gas; [tex]P_1 = 0.5atm[/tex]
Final pressure of gas; [tex]p_2 = 2.5atm[/tex]
Final temperature; [tex]T_2 = 900K[/tex]
Initial temperature before pressure change; [tex]T_1 = \ ?[/tex]
We substitute our given values into the expression above.
[tex]\frac{P_1}{T_1} = \frac{P_2}{T_2} \\\\\frac{0.5atm}{T_1} = \frac{2.5atm}{900K}\\\\T_1 = \frac{0.5atm\ *\ 900K}{2.5atm}\\ \\T_1 = \frac{450K}{2.5}\\\\T_1 = 180K[/tex]
Option D) 180K is the correct answer.
Hence, the temperature of the gas before the pressure changed is 180K.
Learn more about Gay-Lussac's law here :https://brainly.com/question/2683502
