Answer:
[tex] T_{1}= 94.75 K[/tex]
Explanation:
Given the following data;
Final pressure, P2 = 760 Torr
Initial pressure, P1 = 320 Torr
Final temperature, T2 = 225K
To find the initial temperature, we would use Gay Lussac's law;
Gay Lussac law states that when the volume of an ideal gas is kept constant, the pressure of the gas is directly proportional to the absolute temperature of the gas.
Mathematically, Gay Lussac's law is given by;
[tex] PT = K[/tex]
[tex] \frac{P1}{T1} = \frac{P_{2}}{T_{2}}[/tex]
Making T1 as the subject formula, we have;
[tex] T_{1}= \frac{P_{1}}{P_{2}} * T_{2}[/tex]
Substituting into the equation, we have;
[tex] T_{1}= \frac{320}{760} * 225 [/tex]
[tex] T_{1}= 0.4211 * 225 [/tex]
[tex] T_{1}= 94.75 K[/tex]
Therefore, the initial temperature of the gas is 94.75 Kelvin.
