Gay-Lussac’s law (constant volume) – problems and solutions

1. Ideal gases initially have pressure P and temperature T. The gas undergoes the isochoric process so that the final pressure becomes 4 times the initial pressure. What is the final temperature of the gas?

Known :

Initial pressure (P1) = P

Final pressure (P2) = 4P

Initial temperature (T1) = T

Wanted: Final temperature (T2)

Solution :

The formula of Gay-Lussac’s law :

Gay-Lussac's law (constant volume) - problems and solutions 1

The final temperature becomes 4 times the initial temperature.

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2. In a closed container, ideal gases initially have a temperature of 27oC. If the final pressure becomes 2 times the initial pressure, what is the final temperature?

Known :

Initial pressure (P1) = P

Final pressure (P2) = 2P

Initial temperature (T1) = 27oC + 273 = 300 K

Wanted: Final temperature (T2)

Solution :

Gay-Lussac's law (constant volume) - problems and solutions 2

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3. A tire is filled to a gauge pressure of 2 atm at 27°C. After a drive, the temperature within the tire rises to 47°C. What is the pressure within the tire now?

Known :

The atmospheric pressure = 1 atm = 1 x 105 Pa

The initial gauge pressure = 2 atm = 2 x 105 Pa

The initial absolute pressure (P1) = 1 atm + 2 atm = 3 atm = 3 x 105 Pa

The initial temperature (T1) = 27oC + 273 = 300 K

The final temperature (T1) = 47oC + 273 = 320 K

Wanted : The final gauge temperature

Solution :

Gay-Lussac's law (constant volume) - problems and solutions 3

The final gauge pressure = final absolute pressure – atmospheric pressure

The final gauge pressure = 3.2 atm – 1 atm

The final gauge pressure = 2.2 atm