# Ideal gas law – problems and solutions

1. Ideal gases in a closed container initially have volume V and temperature T. The final temperature is 5/4T and the final pressure is 2P. What is the final volume of the gas?

__Known :__

Initial volume (V_{1}) = V

Initial temperature (T_{1}) = T

Final temperature (T_{2}) = 5/4 T

Initial pressure (P_{1}) = P

Final pressure (P_{2}) = 2P

__Wanted:__ Final volume (V_{2})

__Solution :__

2. Determine the volume of 2.00 moles of gases (ideal gas) at STP. STP = Standard Temperature and Pressure.

__Known :__

Moles of gas (n) = 2 moles

Standard temperature (T) = 0 ^{o}C = 0 + 273 = 273 Kelvin

Standard pressure (P) = 1 atm = 1.013 x 10^{5} Pa

Universal gas constant (R) = 8.315 Joule/mole.Kelvin

__Wanted __: Volume of gases (V)

__Solution :__

**Equation of Ideal gas law **__(in the number of moles, n)__

Volume 2 moles of gases is 44.8 liters.

Volume 1 mol of gases is 45.4 liters / 2 = 22.4 liters.

Volume 1 mol of any gases is 22.4 liters.

3. 4 liters of oxygen gas has a temperature of 27°C and pressure of 2 atm (1 atm = 10^{5} Pa) in a closed container. Universal gas constant (R) = 8.314 J.mole^{−1}.K^{−1} and Avogadro’s number (N_{A}) = 6.02 x 10^{23} molecules/mole. What are the molecules of oxygen gases in the container?

__Known :__

Volume of gases (V) = 4 liters = 4 dm^{3} = 4 x 10

^{-3}m

^{3}

Temperature of gases (T) = 27^{o}C = 27 + 273 = 300 Kelvin

Pressure of gases (P) = 2 atm = 2 x 10^{5 }Pa

Universal gas constant (R) = 8.314 J.mole^{−1}.K^{−1}

Avogadro’s number (N_{A}) = 6.02 x 10^{23}

__Wanted __: What is the molecules of oxygen gases in the container (N)

__Solution :__

In 1 mole oxygen gases, there are 1.93 x 10

^{23}oxygen molecules.

4. A container containing a neon gas (Ne, atomic mass = 20 u) at standard temperature and pressure (STP) has a volume of 2 m^{3}. Determine the mass of the neon gas!

__Known :__

Atomic mass of neon = 20 gram/mole = 0,02 kg/mole

Standard temperature (T) = 0^{o}C = 273 Kelvin

Standard pressure (P) = 1 atm = 1.013 x 10^{5} Pascal

Volume (V) = 2 m^{3}

__Wanted ____:__ mass (m) of neon gas

__Solution :__

At standard temperature and pressure (STP), 1 mole of any gases, include neon gas, have volume 22.4 liters = 22.4 dm^{3} = 0.0448 m^{3}.

In the volume of 2 m^{3}, there are 44.6 moles of neon gas.

Relative atomic mass of neon gas is 20 gram/mole.

This means that in 1 mole there are 20 grams or 0.02 kg of neon gas. Because in 1 mol there are 0.02 kg of neon gas then in 44.6 mole there are 44.6 moles x 0.02 kg/mole = 0.892 kg = 892 gram of neon gases.