The Ideal Gas Law
Section 4.4
Pg. 172-175
IDEAL GAS LAW

Before getting too far into this law, it is important to understand
the difference between an ideal gas and a real gas....


IDEAL GAS – does not really exist, it is hypothetical

Follows all gas laws perfectly under all conditions

Does not condense when cooled

Assumes that the particles have no volume and are not attracted to each
other
REAL GAS – does not follow gas laws exactly, it deviates at low
temperatures and high pressures

Condenses to liquid or sometimes solid when cooled or under pressure

Particles are attracted to each other and have volume

Behaves like an ideal gas at higher temperatures and lower pressures
Ideal vs. Real Gases
(a) In an ideal gas, the molecules collide like perfectly hard spheres
and rebound very quickly after collision.
(b) In a real gas, the molecules are “soft” (can be deformed) and
intermolecular attractions are important. The process of collision
takes a slightly longer time, as a result.
In 1873, Johannes van der Waals
hypothesized the existence of attractions
between gas molecules to explain
deviations from the ideal law.
The general forces of attraction, called van
der Walls forces, include dipole-dipole and
London forces.
We will be
dealing with
gases as if they
were ideal
Johannes van der Waals
(1837-1923)
IDEAL GAS LAW

Describes the interrelationship of pressure, temperature, volume and amount
(moles) of matter; the four variables that define a gas system

REMEMBER:
 Boyle’s
Law: Volume is inversely proportional (α) to pressure V α 1/P
 Charles’ Law: Volume is directly proportional to temperature
VαT
 Avogadro’s Theory: Volume is directly proportional to chemical amount (mol) V

From all of these comparisons to volume:
V α
1
P
x
T
x
n
=
VαTn
P
V=RTn
P
R is a constant called the universal gas constant
(allows us to change from α to =)
V = nRT
P
PV = nRT (ideal gas law)
αn
R = universal gas constant

Depends on STP or SATP, atm or kPa

Units: L • kPa/mol • K
value = 8.314 L • kPa/mol • K

Units: L • atm/mol • K
value = 0.0821 L • atm/mol • K
Make sure you look at the unit for pressure to decide which R value to use

Any idea how we came up with the number??

You substitute SATP or STP conditions for one mole into the ideal gas law
and solve for R
R = PV
nT
= (101.325 kPa)(22.414L)
(1.0 mol)(273.15K)
= 8.314 L • kPa/mol • K
.
Using the Ideal Gas Law

When solving for the ideal gas law, start by listing your variables. If three
are known of the four, you can solve for the last one.
Example One: What mass of neon gas should be introduced into an
evacuated 0.88L tube to produce a pressure of 90 kPa at 30°C?
P = 90 kPa
V = 0.88L
T = 30°C  303K
R = 8.314 L • kPa/mol • K
m=?
n =?
PV = nRT
n = PV
RT
n =
(90kPa)(0.88L)
(8.314 L•kPa/mol•K)(303K)
n = 0.0314 mol x 20.18 g = 0.63 g
1 mol
Using the Ideal Gas Law – Practice
2.
A rigid steel cylinder with a volume of 2.00 L is filled with nitrogen gas
to a final pressure of 20.0 atm at 27°C. How many moles of N2 gas
does the cylinder contain?
P = 20.0 atm
PV = nRT
V = 2.00 L
n = PV/RT
T = 27 °C = 300 K
R = 0.0821 L•atm/mol•K
n=?
n=
(20.0 atm)(2.00L)
(0.0821 L•atm/mol•K) (300K)
n=
1.624 mol
n=
1.62 mol
Using the Ideal Gas Law – Practice
3.
Predict the volume occupied by 0.78 g of hydrogen at 22°C and 125 kPa
P = 125 kPa
V=?
T = 22 °C = 295 K
R = 8.314 L•kPa/mol•K
m = 0.78g
n=?
n H2 = 0.78 g x 1 mol = 0.386 mol
2.02g
PV = nRT 
V = nRT
P
V = (0.386 mol)(8.314L•kPa/mol•K)(295K)
125 kPa
V = 7.573 L
V = 7.6 L
Homework

Pg. 174 #3-5

Pg .176 # 1,2,6,9,11