2.  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

3.  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)

4.  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

5.  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

6.  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

7.  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.

9.  What volume is needed to store
0.050 moles of helium gas at
202.6kPa and 400K?

10.  P = 202.6 kPa
 n = 0.050 mol
 T = 400K
 V = ? L
 R = 8.314 J K-
1 mol-1
 202.6(V) = 0.050
x 8.314 x 400
 202.6(V)= 166.28
 V = 166.28/202.6
 V = 0.821 L
(821mL)

11.  What pressure will be exerted by
20.16g hydrogen gas in a 7.5L
cylinder at 20oC?

12.  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
 P x 7.5 = 10 x 8.314 x
293
 P x 7.5 = 24360.02
 P = 24360.02 ÷ 7.5
 P = 3248kPa

13.  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?

14.  P = 10130.0kPa
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

15.  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?

16.  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
 253.25 x 0.250 = 0.10
x 8.314 x T
63.3125 = 0.8314 x T
 T = 63.3125 ÷ 0.8314
 T = 76.15K

17. Problem
A hydrogen gas thermometer is found to
have a volume of 100.0 cm3 when
placed in an ice-water bath at 0°C.
When the same thermometer is
immersed in boiling liquid chlorine, the
volume of hydrogen at the same
pressure is found to be 87.2 cm3. What is
the temperature of the boiling point of
chlorine?

18. Solution
For hydrogen, PV = nRT, where P is pressure,
V is volume, n is number of moles, R is the
gas constant, and T is temperature.
Initially:
 P1 = P, V1 = 100 cm3, n1 = n, T1 = 0 + 273 =
273 K
 PV1 = nRT1
Finally:
 P2 = P, V2 = 87.2 cm3, n2 = n, T2 = ?
 PV2 = nRT2

19. Note that P, n, and R are the same.
Therefore, the equations may be
rewritten:
P/nR = T1/V1 = T2/V2
and T2 = V2T1/V1
Plugging in the values we know:
 T2 = 87.2 cm3 x 273 K / 100.0 cm3
 T2 = 238 K
Answer
238 K (which could also be written
as -35°C)

20.  http://www.ausetute.com.au/idealgas.h
tml
 http://chemistry.about.com/od/chemistr
yglossary/a/idealgaslawdef.htm