1. Ideal gases in a closed container initially have volume V and pressure P. If the final pressure is 4P and the volume is kept constant, what is the ratio of the initial kinetic energy with the final kinetic energy.

__Known :__

Initial pressure (P_{1}) = P

Final pressure (P_{2}) = 4P

Initial volume (V_{1}) = V

Final volume (V_{2}) = V

__Wanted:__ The ratio of the initial kinetic energy with the final kinetic energy (KE_{1} : KE_{2})

__Solution :__

__The relation between pressure (P), volume (V) and kinetic energy (KE) of ideal gases :__

__The ratio of the initial kinetic energy with the final kinetic energy :__

2. What is the average translational kinetic energy of molecules in an ideal gas at 57^{o}C.

__Known :__

Temperature of gas (T) = 57^{o}C + 273 = 330 Kelvin

Boltzmann‘s constant (k) = 1.38 x 10^{-23} Joule/Kelvin

Wanted: The average translational kinetic energy

__Solution :__

The relation between kinetic energy (KE) and the temperature of the gas (T) :

The average translational kinetic energy :

3. A gas at 27^{o}C in a closed container. If the kinetic energy of the gas increases 2 times the initial kinetic energy, thus the final temperature of the gas is…

__Known :__

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

Initial kinetic energy = KE

Final kinetic energy = 4 KE

Wanted:__ The final temperature (T___{2})

__Solution :__

4. An ideal gas is in a closed container, is heated so that the final average velocity of particles of gas increases by 3 times the initial average velocity. If the initial gas temperature is 27^{o}C, then the final temperature of the ideal gas is…

__Known :__

Initial temperature = 27^{o}C + 273 = 300 Kelvin

Initial velocity = v

Final velocity = 2v

__Wanted__ : The final temperature of ideal gas

__Solution :__

The final average velocity = 2 x the initial average velocity

5. Three moles of gas are in a 36 liters volume space. Each gas molecule has a kinetic energy of 5 x 10^{-21} Joule. Universal gas constant = 8.315 J/mole.K and Boltzmann’s constant = 1.38 x 10^{-23} J/K. What is the gas pressure in the container.

__Known :__

Number of moles (n) = 3 moles

Volume = 36 liters = 36 dm^{3 }= 36 x 10^{-3} m^{3}

Boltzmann’s constant (k) = 1.38 x 10^{-23} J/K

Kinetic energy (KE) = 5 x 10^{–21} Joule

Universal gas constant (R) = 8.315 J/mole.K

__Wanted __: Gas pressure (P)

__Solution :__

Calculate the temperature using the equation of kinetic energy of gas.

Calculate the gas pressure using th equation of ideal gas law (in number of moles, n) :

P V = n R T

P (36 x 10^{-3}) = (3)(8.315)(241.5)

P (36 x 10^{-3}) = 6024.22

The gas pressure is 1.67 x 10^{5} Pascal or 1.67 atmospheres.