CHE 1000 e-LEARNING

GASES
 Lecture 1- Gases

• Matter exists in three states namely , solids, liquids and

gases.
• The difference between gases and the other two states
of matter is that gas particles have very large distances
between them.
• A pure gas may be made up of individual atoms (e.g. a
noble gas like neon), elemental molecules made from
one type of atom (e.g. oxygen), or compound
molecules made from a variety of atoms (e.g. carbon
dioxide).

2
 Characteristics of gases
Gases:
i. are highly compressible
You can pump a bicycle tyre or a ball etc
ii. are thermally expandable
When heated, gases expand- hot air balloons get filled this way
iii. have low viscosity
Compared to liquids gases have no friction and flow easily
iv. have relatively low density
Compared to solids and liquids, gases are very light
v. infinitely miscible
Gases mix in all proportions at all concentrations
Gases exert pressure on the walls of the container. This
pressure can be measured using a barometer or a
manometer
3
BAROMETER

4
UNITS OF GAS PRESSURE

5
 Gas Laws
• Gases obey the following physical laws
• Boyle’s Law
• Charles's Law
• Avogadro’s Law
• Gay-Lussac’s Law and
• Ideal gas Law
• Combined Law

Remember BCA or ABC to memorize names of these laws

and relationships between variables. The others are
combinations of the three.

6
 Boyles Law states that:

• the volume of a fixed amount of gas held at a constant

temperature varies inversely with the pressure.
1 1
• Mathematically, V∝ or P=
P V
• Introducing a proportionality constant k, we have an equation
k 1
• V= =k hence PV=k where P=Pressure, V=Volume and
P P
k=proportionality constant
• Plotting volume against pressure yields a graph which
decreases exponentially as shown
Practice and
memorize these
Summary of Boyle’s Law P1V1=P2V2
mathematical
expressions
7
 Plot of volume against pressure

An increase in pressure reduces the volume or vice versa

8
Example
The following data was obtained at constant temperature. Plot volume
against pressure and also volume against 1/pressure
Volume (L) 117.5 87.2 70.7 58.8 44.2 35.3 29.1
Pressure (atm) 12.0 16.0 20.0 24.0 32.0 40.0 48.0
Pressure x Volume 14.0 x 14.0 14.0 x 14.0 x 14.0 14.0 x 14.0
(atm x L) 102 x102 102 102 x 102 102 x 102
Solution Solution
A plot of volume against A plot of volume against
pressure yields a curve with 1/pressure yields a straight line
decreasing gradient while with positive gradient
150 Volume (L) 150
Volume (L)

100 100

50 50

0 0
0 20 40 60 0.00 0.02 0.04 0.06 0.08 0.10
Pressure (atm) 1/Pressure (1/atm) 9
 Example
1. A gas has volume of 255 mL at 725 torr. What volume will it occupy
at 365 torr?

Solution
First, understand what we want! It’s the volume. What we have is
pressure 1, volume 1 and pressure2.What is missing is volume 2
P1 V1 = P2 V2
=725 torr x 255 mL = 365 torr x V2 .Make V2 the subject and solve
V2 = 506.5 mL
2. A diver blows a 0.75-L air bubble 10 m under water. As it rises to the
surface, the pressure goes from 2.25 atm to 1.03 atm. What will be the
volume of air in the bubble at the surface?
Solution
P1 V1 = P2 V2
=2.25 atm x 0.75 L = 1.03 atm x V2 .Make V2 the subject and solve
V = 1.64 L
10
 Charles’s Law states that:
• the volume of a given amount of gas is directly proportional to its
Kelvin temperature at constant pressure. In other words, the
volume of a fixed mass of gas at constant pressure increases linearly
with the increase in temperature.
• Mathematically, vµT or V = b T T = Temperature in K, V = volume b
= proportionality constant

A plot of various
gases at constant
pressure yields a
straight line. Slopes of
these lines INDICATE
different number of
moles for each gas.
11
A very interesting feature of these plots is that the volumes of all the
gases extrapolate to zero at the same temperature, -273OC. On the
Kelvin temperature scale this point is defined as 0 K, which leads to the
following relationship between the Kelvin and Celsius scales: K= OC +273

Example
1. A gas has volume of 3.86 L at temperature of 45 OC. What will be volume
at 80 OC?

Solution
We have V1=3.86 L , T1=45 OC and T2=80 OC. We need V2.

REMEMBER TO CHANGE TEMPERATURE TO KELVINS ALWAYS

V1 V2 V2 3.86 L×353 K
= = 3.86 L = = V2 = = 4.28 L
T1 T2 318 K 353 K 318 K

12
Example
2. A helium balloon in a closed car occupies a volume of 2.32 L at 40.0°C. If
the car is parked on a hot day and the temperature inside rises to 75.0°C,
what is the new volume of the balloon, assuming the pressure remains
constant?

Solution
Again let us convert Celsius temperature to Kelvin temperature

40.0°C= 273 +40.0 = 313 K and 75.0°C= 273 + 75=348 K

We have V1=2.32 L , T1=313 K and T2=348 K. We need V2.

We replace the values and make V2 the subject of the formula and solve
for V2
V1 V2 V2 2.32 L×348 K
= = 2.32 L = = V2 = = 2.58 L
T1 T2 313 K 348 K 313 K
13
 Avogadro’s Law
Avogadro’s Law or principle states that equal volumes of gases at the
same temperature and pressure contains equal number of particles
(moles).
Mathematically, V µ n or V= an where V = volume, n = number of
moles, a = proportionality constant
• Recall that 1 mol contains 6.02 × 1 023 particles.
• The molar volume of a gas is the volume that 1 mol occupies at
0.00°C (273 K) and 1.00 atm pressure. The conditions of 0.00°C
and 1.00 atm are known as standard temperature and pressure
(STP).
• Avogadro showed experimentally that 1 mol of any gas occupies a
volume of 22.4 L at STP.

14
Example
The main component of natural gas used for home heating and
cooking is methane (CH4 ). Calculate the volume that 2.00 kg of
methane gas will occupy at STP.
Solution
What do we know ? Pressure= 1atm, temperature 273 K (remember
STP), mass= 2.00 kg. What we are looking for is volume.

Lets revise calculations of molar mass by determinig molar mass of

methane from first principles
12.01 amu 1.01 amu
Mr =1 C atom +4 H atoms
1 C atom 1 H atom

Cancel out common units

Mr =12.01 amu + 4.04 amu= 16.05 amu = 16.05 g/mol

Next we will determine the number of moles of methane gas from mass
and molar mass
15
 mass in grams Please, don’t
number of moles n =
molar mass in g/mol
> forget this
formula

2.00 x103 g Also recall that the unit

number of moles n = = 124.61 mol
16.05 g/mol for mole is mol

Recall that the linking conversion factor is 22.4 L/mol at STP and using
our usual relationship using arrows we have
1 mol 22.4 L

124.61 mol xL

Cross multiply and make x the subject of the formula

22.4 L ×124.61 mol
𝑥= =2791.264 L = 2.8 ×10 3 L
1 mol
16
 Gay-Lussac’s Law
Gay-Lussac’s law states that the pressure of a fixed amount of gas
varies directly with the Kelvin temperature when the volume remains
constant. It can be expressed mathematically as follows.

PµT or P= dT where P= pressure, T=Kelvin P1 P2

temperature and d = proportionality constant =
T1 T2

17
Example
The pressure of the oxygen gas inside a canister is 5.00 atm at 25.0°C.
The canister is located at a camp high on Mount Everest. If the
temperature there falls to -10.0°C, what is the new pressure inside the
canister?
Solution
What do we know ? P1= 5 atm, T1= 298 K, T2= 263 K. What we are
looking for is P2.
P1 P2 5 atm P2 Make P2 the subject of the
= ⟹ =
T1 T2 298 K 263 K formula and solve

5 atm × 263 K
P2 = = 4.41 atm
298 K

18
 Combined law and Ideal Gas Law
In a number of applications involving gases, such as the weather
balloons, pressure, temperature, and volume might all change. Boyle’s,
Charles’s, and Gay-Lussac’s laws can be combined into a single law.
This combined gas law states the relationship among pressure,
temperature, and volume of a fixed amount of gas. All three variables
have the same relationship to each other as they have in the other gas
laws: pressure is inversely proportional to volume and directly
proportional to temperature, and volume is directly proportional to
temperature. The combined gas law can be expressed mathematically
as follows.
P1 V1 P2 V2
=
T1 T2

Example
A gas at 110 kPa and 30.0°C fills a flexible container with an initial volume of 2.00 L. If the
temperature is raised to 80.0°C and the pressure increases to 440 kPa, what is the new
volume?
19
Solution
As in previous examples, we need to have a quick look at what
information we have and what is being asked for. Remember this
approach to solve many questions
So we have P1=110 kPa , T1= 30.0°C , V1=2.00 L and T2= 80.0°C,
P2= 440 kPa. We need V2.

Remember temperature is
always to be in KELVINS

K = OC +273

P1 V1 P2 V2 110 kPa × 2.0 L 440 kPa × V2

= ⟹ =
T1 T2 303 K 353 K

So work out the rest and make sure your solution is 0.58 L

20
Recap on the three laws (ABC)
Avogadro’s Law V= an
k
Boyle’s Law PV=k or V=
P
Charles’s Law V=bT

So combining all these laws expressed as volume we have

k
V= =bT =an
P
(k b a)n T nT
V= =R or PV = n RT
P P
Where R = k b a = combined proportionality constant called Universal
gas constant. R value depends on pressure units used so be conversant
with various values given for this constant.

PV = n RT is called the IDEAL GAS LAW

21
22
23
It is important to recognize that the ideal gas law is an empirical
equation—it is based on experimental measurements of the properties
of gases. A gas that obeys this equation is said to behave ideally.

Example
Calculate the number of moles of ammonia gas (NH3) contained in a
3.0-L vessel at 3.00 × 102 K with a pressure of 1.50 atm.

Solution

PV = n RT Make n the subject of the formula and solve for it

PV 1.50 atm ×3.0 L

n= ⟹
L.atm
=0.18 mol
RT 0.0821 ×3.00×10 2 K
mol.K
Make sure you understand the dimension analysis of units and cancel out
appropriately.
24
 Kinetic molecular theory of gases

Postulates of kinetic theory:

(i) Gases consist of small particles called molecules. These moles are
constant motion.
(ii) Moles collide with each other and the walls of the container.
(iii) These collisions are elastic collisions ( No energy is lost)
(iv) Gas exert pressure due to the collision with the walls of the
container.
(v) Particles are assumed to exert no forces of attraction on each other.
(vi) The average kinetic energy is assumed to be directly proportional to
the Kelvin temperature of the gas.

What does the word Postulate mean?

25
 Gas Stoichiometry
Recall that Molar volume of a gas at STP = 22.4 L (1 atm, 0 °C)

nRT 1 mol ×0.08206 atm L mol−1K−1x 273.15 K

Molar volume of gas = =
P 1.00 atm

>
= 22.414689 L = 22.4 L Remember
22.4 L/mol

Remember that this was stated at high school as 1 mol of any gas
occupies 22.4 L at STP OR 24.0 L at RTP
Example
If a sample of nitrogen occupies 1.75 L at STP. How many moles of
nitrogen are present?

Solution
PV 1.00 atm×1.75 L
Number of moles in 1.75 L = =
RT 0.08206 atm L mol−1K−1x 273.15 K
= 0.07807 = 7.81 x 10-2 mol
29
Example
What mass of sodium azide is required to inflate an airbag to 70.0 L at
STP?
Balanced reaction is: 2 NaN3(s)®2 Na(s) + 3 N2(g)

Solution
PV 1.00 atm×70 L
nN = = = 3.1229 mol
2 RT −1 −1
0.08206 atm L mol K x 273.15 K

3 mol N2 2 mol NaN3 Cross multiply, make

x the subject and
3.1229 mol N2 x mol NaN3 solve

2 mol NaN3 ×3.1229 mol N2

𝑥 mol NaN3 = =2.08 mol
3 mol N2

m NaN3 = nNaN ×M =2.082 mol×65.02g/mol = 135 g

3
30
Example
Consider the reaction between 50.0 mL of liquid Methanol, CH3OH,
( density 0.850 g/ mL) and 22.8 L of oxygen at a pressure of 2.00
atm and 27OC. The products of the reaction are CO2 and H2O.
Calculate the number of moles of water formed if the reaction goes
to completion.
Solution
First write a balanced chemical equation and find the number of
moles for each reactant and ultimately the number of moles of
water. REMEMBER OTHER CONCEPTS LIKE LIMITING
REAGENT HERE AND ELSEWHERE.
2CH3OH(l) + 3O2 (g) ®2CO2(g) + 4H2O(g)
mass 50.0mL ×0.850 g/mL 42.5 g
nCH3OH = = = =1.326 mol
M 12.01+ 4×1.01 +16.00 32.05 g/mol

PV 2.00 atm × 22.8 L

nO = = = 1.851mol
2 RT atm L
0.08206 273.15 + 27 K
mol K 31
 Let us learn another way of determining the limiting reactant by using
mole ratios. From the balanced equation, 3 mol O2 reacts with 2 mol
CH3OH expressed as
nO n
2 = CH3OH which translates to 1.851 = 1.326
3 2 3 2
0.613 O2 : 0.663 CH3OH

Therefore the Limiting reactant is oxygen since 0.613 is less than 0.663.

Remember, the limiting reagent determines how much product forms so

all of the products will depend on the moles of oxygen.
nO nH O n
2= 2 which translates to 1.851 = H2O
3 4 3 2

Therefore moles of water = nH O =2.47 mol

2

See how we apply other concepts we learnt earlier to different problems.

So learnt to link all these other concepts 32
 Molar mass of gases
Ideal gas law can be used to determine the molar mass of a gas or a volatile
liquid. The volatile liquid sample is placed in the previously weighted flask. This
flask is then heated in a water bath. The liquid evaporates and the vapour
completely fills the flask at 100 °C. The flask is cooled and weighed again. The
mass of the vapour is given by the difference in mass. The volume of the gas is
the volume of the flask and the pressure is atmospheric pressure. Molecular
weight can be calculated from ideal gas equation as follows:
mass Don’t forget this
Number of moles, n =

And from Ideal gas law n =

molar mass
PV
=
mass
RT molar mass
> formula. Learn it by
heart

mass×RT mass
Molar mass = and = density
PV V
density×RT
Molar mass =
P
33
Example
A chemist has synthesized a greenish-yellow gaseous compound of chlorine and
oxygen and finds that its density is 7.71 g/L at 36 OC and 2.88 atm. Calculate the
molar mass of the compound.

Solution
We can calculate the molar mass of a gas if we know its density,
temperature, and pressure. The molecular formula of the compound
must be consistent with its molar mass. What temperature unit should we
use if you can remember?
Develop the habit of
Molar mass =
density×RT
P > stating the formula
you are going to use.
This costs marks!

7.71 g/L × 0.08206 L.atm K−1.mol−1 ×T309.15 K

Molar mass = = 67.9 g/mol
2.88 atm

Try using the ideal gas equation to solve this problem given the
information above. You should arrive at the same solution.
34
 Dalton’s Law of Partial Pressure
States that for a mixture of gases in a container, the total pressure
exerted is the sum of the pressures that each gas would exert if it
were alone.

P total = P1 + P2 +P3 +P4 +…Px (where Px = partial pressure of gas x)

P1 = n1 RT/ V, P2 = n2 RT/ V, P3 = n3 RT/ V, P4 = n4 RT/ V, Px = nx RT/ V

Ptotal = n1RT + n2 RT + n3 RT + n4 RT + …nx RT

V V V V V

Ptotal = (n1+ n2 + n3 + n4 + ..nx )( RT )

V

Note that RT/V is constant for all gases as they occupy the same
container, experience the same temperature and have the same R

Partial Pressure: Partial pressure refers to the pressure exerted by an

individual gas in a mixture of gasses. OR The partial pressure of a gas
is the pressure that gas would exert if it occupied the container by itself.
35
 67*89: 2; *239< 2; ( =(<
Mole fraction =
)2)(3 67*89: 2; *239< 2; =(<9< >6 )?9 *>@)7:9

67*89: 2; *239< 2; ( =(<

Partial pressure = ×𝑇𝑜𝑡𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒
)2)(3 67*89: 2; *239< 2; =(<9< >6 )?9 *>@)7:9

Example:
Mixture of Helium and oxygen can be used in scuba diving tanks to help prevent “the
bends”. For a particular dive, 46 L of He at 25 °C and 1.0 atm and 12 L of oxygen at
25 °C and 1.0 atm were pumped into the tank with a volume of 5.0 L. Calculate the
partial pressure of each gas and the total pressure in the tank at 25 °C.
Solution:
!" %.' ()* ×,-.' .
Moles of helium = = = 1.9 𝑚𝑜𝑙
#$ '.'/0'-..()* 1 !"*23 !"×04/.%5 1
!" %.' ()* ×%0.' .
Moles of oxygen = #$ = '.'/0'-..()* 1!"*23!"×04/.%5 1 = 0.49 𝑚𝑜𝑙

6#$ %.4×'.'/0'-..()* 1 !"*23 !"×04/.%5 1

Pressure exerted by He = "
= 5.' .
= 9.29 atm

6#$ '.,4×'.'/0'-..()* 1 !"*23 !"×04/.%5 1

Pressure exerted by O2 = = = 2.398 atm
" 5.' .

Total pressure = 9.3 + 2.4 = 11.7 atm

36
 Lecture 4- Collecting a gas over water
Dalton’s law of partial pressures is most commonly encountered when a gas
is collected by displacement of water. Because the gas has been bubbled
through water, it contains some water molecules and is said to be “wet.” The
total pressure of this wet gas is the sum of the partial pressure of the gas
itself and the partial pressure of the water vapor it contains. The partial
pressure of water vapors is called the vapor pressure of water. It depends
only on the temperature of the experiment.
Example
A sample of solid potassium chlorate was heated in a test tube and
decomposed as follows:
2 KClO3 ® 2 KCl + 3O2
The oxygen was collected by displacement of water at 22°C and a total
pressure of 754 torr. The volume of the gas collected was 0.650 L , and
vapour pressure of water at 22 °C is 21 torr. Calculate the partial pressure of
oxygen collected and the mass of potassium chlorate in the sample that was
decomposed.
37
 Total pressure = partial pressure of oxygen + vapour pressure of water
𝑃"#$%& = 𝑃'! + 𝑃(!'
partial pressure of oxygen = 754 – 21 =733 torr
PO ×V
2 733 torr×0.650 L
nO = =
2 RT 62.364 L.torr K−1mol−1×295.15

=0.02588 mol = 2.59 x 10-2 mol

nO nKClO 0.0259 nKClO

2= 3
3 2 which translates to = 3
3 2
nKClO =0.0173 mol
3

Mass = 0.0173 mol x 122.6 g mol-1= 2.12 g of potassium chlorate

Other important application
Gas exchange between living organisms and the environment depends on the
properties of gases, in particular, partial pressure and solubility. Gaseous exchange
during RESPIRATION in lungs also depends on Partial pressure.
38
 Diffusion and effusion
Diffusion: Inter mixing of two or more gases to form a homogeneous
mixture without any chemical change is called "DIFFUSION OF GASES".
Diffusion is purely a physical phenomenon. Gases diffuse very quickly
due to large empty spaces among molecules. Different gases diffuse with
different rates (depending on velocities). It takes relatively long time to
complete.
Effusion: is the term used to describe the passage of a gas through a tiny
orifice into an evacuated chamber, as shown in the figure below.

The rate of effusion measures the speed at which the gas travels
through the tiny hole into a vacuum.
39
 Root Mean Square Velocity
Root Mean Square (rms) Velocity is defined as the square root of the
average of the squares 𝑢) of velocities of particles. See that the u
has a bar on top to signify average velocity

urms = u' 2
We can get expression of urms from the KMT equation by making u' 2
the subject
1 3
(KE)avg=NA mu' 2 = R T
2 2

3 2 3RT
u' 2 = R T× = Then take square root on both sides
2 m NA m NA
Where m = mass in kilograms of a single
3RT 3RT gas particle and M=NA x m to give the
u' 2 = = mass of a mole of gas particles in
m NA M
kilograms. 40
Example
Calculate the root-mean-square speeds of helium atoms and nitrogen
molecules in m/s at 25°C.

Solution
To calculate urms, the units of R should be 8.314 JK-1 mol-1 and,
because 1 J =1 kg m2 /s2 , the molar mass must be in kg/mol. The
molar mass of He is 4.003 g/mol, or 4.003 x10-3 kg/mol.

𝑚0 A%
3RT 3×8.314 𝑘𝑔 0 . 𝐾 . 𝑚𝑜𝑙A% ×298 𝐾
𝑢:*< = = 𝑠 = 1362 𝑚/𝑠
M 4.003×10AB 𝑘𝑔/𝑚𝑜𝑙

Do the same for N2 and prove that the Urms= 515 m/s

41
 Graham’s Law: Rate of effusion is inversely proportional to the square
root of the mass of the particles (density of the gas) under the same
conditions.
1 𝑅𝑎𝑡𝑒 𝑜𝑓 𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑜𝑓 𝑔𝑎𝑠 1 √𝑀)
𝑟 Or for two gases =
√𝑑 𝑅𝑎𝑡𝑒 𝑜𝑓 𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑜𝑓 𝑔𝑎𝑠 2 √𝑀 *

Derivation from Kinetic Molecular Theory:

Rate of effusion of gas depends on the velocity of the gas molecules at any
temperature. Gases at the same temperature have same average kinetic energy.

1 1 >!! ?"
KE1 = KE2 or 𝑚* 𝑣*) = 𝑚) 𝑣)) which simplifies to =
2 2 >"! ?!
3𝑅𝑇
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑜𝑓 𝑔𝑎𝑠 1 𝑢+,- 𝑜𝑓 𝑔𝑎𝑠 1 √ 𝑀)
𝑀*
= = =
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑜𝑓 𝑔𝑎𝑠 2 𝑢+,- 𝑜𝑓 𝑔𝑎𝑠 2 √3𝑅𝑇 𝑀*
𝑀)

This is same as Graham’s Law.

Graham's law provides a basis for separating isotopes by diffusion.
42
Example
A given volume of N2 requires 68.3 seconds to effuse from a hole into a
vacuum chamber. Under the same conditions another unknown gas
required 85.6 seconds for the same volume to effuse. What is the molar
mass of the unknown gas and what is the gas?

This is another form of

Graham’s law expressed

>
!"#$ %& $&&'()%* %& '*+*%,* -"( #DC √0DC
= = in time units. See that the
!"#$ %& $&&'()%* %& .C #E √0E ratios are making sense
i.e rate is inversely
$
)1.) %&' proportional to the time
$"! √/"! 01.3 -
= ⟹ = required for diffusion
$# √/# 14.0 -
Mx

Taking squares on either side and

making Mx the subject, we have
𝑔 𝑔
28.2 28.2 The gas is carbon
0.79789 0 = 𝑚𝑜𝑙 ⟹ 𝑀 = 𝑚𝑜𝑙 = 44.02 𝑔/𝑚𝑜𝑙
@
Mx 0.63663 dioxide
43
 Lecture 5-Real Gases
Ideal gas is a hypothetical concept in which gas molecules do not interact
with each other, and do not occupy any space. These assumptions are
not always valid for real gases. Real gases follow these ideal gas laws at
low pressure and high temperature.
Experimentally observed graphs are shown below
203 K

CH4
N2
2.0 8.0 293 K
H2
PV CO2 PV 1.4
Ideal 673 K
nRT 1.0 gas nRT
1.0 Ideal
gas

0 0
0 200 400 600 800 1000 0 200 400 600 800
P (atm) P (atm)
Adopted from Zumdhal and Zumdhal 8th ed
44
Real gases consist of atoms or molecules which occupy space and
attract each other. van der Waals made the following correction in
pressure and the volume.
Effect of pressure:
In real gases at low temperatures molecules attract each other which
causes a delay in reaching the walls of the container. This also reduces
the number of collisions in a given time. The impact is further reduced
due to the attraction by other molecules.
n 2
Pobs = Pideal − a
V
Effect of Volume:
According to KMT gas molecules are so small that the volume of gas
is volume of the container. However at high pressure the volume of the
molecules is significant fraction of the volume. Therefore, the free
volume for the molecules is less than the volume of the container.

45
 Memorize
V container = Videal + nb this
formula
Videal = V container - nb
If we substitute in ideal gas equation

H J
[𝑃EFG + 𝑎 ]× 𝑉()*+,-*./ − 𝑛𝑏 = 𝑛𝑅𝑇
I

where a and b are constants for a gas. This equation is called van
der Waal’s equation. The values of a and b are determined
experimentally.
Example
Calculate the pressure exerted by 10.000 mol of helium gas at 25OC in
a volume of 1.00 L using van der Waals equation. Also calculate the
pressure deal gas would exert under these condition.

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 Non ideal behavior.
$ &
[𝑃!"# + 𝑎 ]× 𝑉 − 𝑛𝑏 = 𝑛𝑅𝑇
%

a =0.034 L2 atm/ mol2 and b = 0.0237 L/ mol

'(.((( *!+ &

Pobs + 0.034 L2 atm/ 2
mol [ ](1.00 – 10.00x 0.0237 L/ mol) = nRT
'.(( ,

(Pobs + 3.4) x 0.768 = 10.0 x 0.08206 x 298.15/1.00

Pobs = 316 atm

Ideal pressure P = nRT/V

P = 10.0 mol x 0.08206 L.atm.K-1.mol-1 x
298.15 K/1.00 L
Pressure = 245 atm

PRACTICE WITH MORE EXAMPLES IN ZUMDHAL AND ZUMDHAL.

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