Air at 1 atm was closed into a rigid container. It was then heated to a temperature of 911
K. The pressure inside the container was measured to be 2.7 atm. What was the initial
temperature of the gas (in Kelvin)? Round answers to the ones place.

Answer:

Final temperature, T2 = 2460 Kelvin.

Explanation:

Given the following data;

Initial pressure = 1atm

Final pressure = 2.7atm

Initial temperature = 911K

To find the final 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 T2 as the subject formula, we have;

[tex] T_{2}= \frac{P_{2}}{P_{1}} * T_{1}[/tex]

Substituting into the equation, we have;

[tex] T_{2}= \frac{2.7}{1} * 911[/tex]

[tex] T_{2}= 2.7 * 911 [/tex]

Final temperature, T2 = 2459.7 ≈ 2460 Kelvin.

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