# 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 (P_{1}) = P

Final pressure (P_{2}) = 4P

Initial temperature (T_{1}) = T

Wanted: Final temperature (T_{2})

__Solution :__

__The formula of ____Gay-Lussac’s law :__

The final temperature becomes 4 times the initial temperature.

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

__Known :__

Initial pressure (P_{1}) = P

Final pressure (P_{2}) = 2P

Initial temperature (T_{1}) = 27^{o}C + 273 = 300 K

Wanted: Final temperature (T_{2})

__Solution :__

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 10^{5} Pa

The initial gauge pressure = 2 atm = 2 x 10^{5} Pa

The initial absolute pressure (P_{1}) = 1 atm + 2 atm = 3 atm = 3 x 10^{5} Pa

The initial temperature (T_{1}) = 27^{o}C + 273 = 300 K

The final temperature (T_{1}) = 47^{o}C + 273 = 320 K

__Wanted :__ The final gauge temperature

__Solution :__

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