Ideal
Gas Law
- 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 non-ideal 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 202.6kPa and 400K?
PV = nRT
P = 202.6 kPa
n = 0.050 mol
T = 400K
V = ? L
R = 8.314 J K-1 mol-1
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 (821mL)
202.6 V = 166.28
V = 166.28 ÷ 202.6
V = 0.821 L (821mL)
Calculating
Pressure of Ideal Gas: P = (nRT) ÷ V
What pressure will be exerted by
20.16g hydrogen gas in a 7.5L cylinder at 20oC?
PV = nRT
P = ? kPa
V = 7.5L
n = mass ÷ MM
mass = 20.16g
MM(H2) = 2 x 1.008 = 2.016g/mol
n = 20.16 ÷ 2.016 = 10mol
T = 20o = 20 + 273 = 293K
R = 8.314 J K-1 mol-1
V = 7.5L
n = mass ÷ MM
mass = 20.16g
MM(H2) = 2 x 1.008 = 2.016g/mol
n = 20.16 ÷ 2.016 = 10mol
T = 20o = 20 + 273 = 293K
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 = 3248kPa
P x 7.5 = 24360.02
P = 24360.02 ÷ 7.5 = 3248kPa
Calculating
moles of gas: n = (PV) ÷ (RT)
A 50L cylinder is filled with argon
gas to a pressure of 10130.0kPa at 30oC. How many moles of argon gas
are in the cylinder?
PV = nRT
P = 10130.0kPa
V = 50L
n = ? mol
R = 8.314 J K-1 mol-1
T = 30oC = 30 + 273 = 303K
V = 50L
n = ? mol
R = 8.314 J K-1 mol-1
T = 30oC = 30 + 273 = 303K
10130.0 x 50 = n x 8.314 x 303
506500 = n x 2519.142
n = 506500 ÷ 2519.142 = 201.1mol
506500 = n x 2519.142
n = 506500 ÷ 2519.142 = 201.1mol
Calculating
gas temperature: T = (PV) ÷ (nR)
To what temperature does a 250mL
cylinder containing 0.40g helium gas need to be cooled in order for the pressure
to be 253.25kPa?
PV = nRT
P = 253.25kPa
V = 250mL = 250 ÷ 1000 = 0.250L
n = mass ÷ MM
mass = 0.40g
MM(He) = 4.003g/mol
n = 0.40 ÷ 4.003 = 0.10mol
R = 8.314 J K mol-1
T = ? K
V = 250mL = 250 ÷ 1000 = 0.250L
n = mass ÷ MM
mass = 0.40g
MM(He) = 4.003g/mol
n = 0.40 ÷ 4.003 = 0.10mol
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 = 76.15K
63.3125 = 0.8314 x T
T = 63.3125 ÷ 0.8314 = 76.15K
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