Key Concepts
 An Ideal Gas (perfect gas)is one which obeys Boyle's Law and Charles' Law exactly.
 An Ideal Gas obeys the Ideal Gas Law (General gas equation):
PV = nRT
where
P=pressure
V=volume
n=moles of gas
T=temperature
R = gas constant (dependent on the units of pressure, temperature and volume)
R = 8.314 J K^{1} mol^{1} if
Pressure is in kilopascals(kPa)
Volume is in litres(L)
Temperature is in kelvin(K)
R = 0.0821 L atm K^{1} mol^{1} if
Pressure is in atmospheres(atm)
Volume is in litres(L)
Temperature is in kelvin(K)
 An Ideal Gas is modelled on the Kinetic Theory of Gases which has 4 basic postulates:
 Gases consist of small particles (molecules) which are in continuous random motion
 The volume of the molecules present is negligible compared to the total volume occupied by the gas
 Intermolecular forces are negligible
 Pressure is due to the gas molecules colliding with the walls of the container
 Real Gases deviate from Ideal Gas Behaviour because:
 at low temperatures the gas molecules have less kinetic energy (move around less) so they do attract each other
 at high pressures the gas molecules are forced closer together so that the volume of the gas molecules becomes significant compared to the volume the gas occupies
 Under ordinary conditions, deviations from Ideal Gas behaviour are so slight that they can be neglected
 A gas which deviates from Ideal Gas behaviour is called a nonideal gas.
Ideal Gas Law Calculations
Calculating Volume of Ideal Gas: V = (nRT) ÷ P
What volume is needed to store 0.050 moles of helium gas at 202.6 kPa and 400 K?
PV = nRT
P = 202.6 kPa
n = 0.050 mol
T = 400K V = ? L
R = 8.314 J K^{1} mol^{1}
202.6V = 0.050 x 8.314 x 400
202.6 V = 166.28
V = 166.28 ÷ 202.6
V = 0.821 L (821 mL)
Calculating Pressure of Ideal Gas: P = (nRT) ÷ V
What pressure will be exerted by 20.16 g hydrogen gas in a 7.5 L cylinder at 20^{o}C?
PV = nRT
P = ? kPa
V = 7.5 L
n = mass ÷ molar mass
mass = 20.16 g
molar mass(H_{2}) = 2 x 1.008 = 2.016 g mol^{1}
n = 20.16 ÷ 2.016 = 10 mol
T = 20^{o} = 20 + 273 = 293 K
R = 8.314 J K^{1} mol^{1}
P x 7.5 = 10 x 8.314 x 293
P x 7.5 = 24360.02
P = 24360.02 ÷ 7.5 = 3248 kPa
Calculating moles of gas: n = (PV) ÷ (RT)
A 50 L cylinder is filled with argon gas to a pressure of 10130.0 kPa at 30^{o}C. How many moles of argon gas are in the cylinder?
PV = nRT
P = 10130.0 kPa
V = 50 L
n = ? mol
R = 8.314 J K^{1} mol^{1}
T = 30^{o}C = 30 + 273 = 303 K
10130.0 x 50 = n x 8.314 x 303
506500 = n x 2519.142
n = 506500 ÷ 2519.142 = 201.1 mol
Calculating gas temperature: T = (PV) ÷ (nR)
To what temperature does a 250 mL cylinder containing 0.40 g helium gas need to be cooled in order for the pressure to be 253.25 kPa?
PV = nRT
P = 253.25 kPa
V = 250 mL = 250 ÷ 1000 = 0.250 L
n = mass ÷ molar mass
mass = 0.40 g
molar mass(He) = 4.003 g mol^{1}
n = 0.40 ÷ 4.003 = 0.10 mol
R = 8.314 J K mol^{1}
T = ? K
253.25 x 0.250 = 0.10 x 8.314 x T
63.3125 = 0.8314 x T
T = 63.3125 ÷ 0.8314 = 76.15 K
