CHAPTER 5 - STATES OF MATTER

Example 5:

Calculate the mass of KClO3 is required to produce 2.40 L
oxygen gas that measured at a pressure of 1 atm and a
temperature of 26oC.

2KClO3(s) → 2KCl(s) + 3O2(g)

(Ans : 7.99 g)

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ANSWER
PV  nRT

n  PV
RT

 1atm x 2.04 L
0.08206 Latmmol 1K1 (299.15K )
 nO2

0.0978 mol

3 mol O2  2 mol KClO3

0.0978 mol O2  0.0978 x 2
3

 0.0625 mol

mass KClO3  0.0625 mol x122.5 gmol 1
 7.987 g

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e) Dalton’s Law of Partial Pressure

Partial pressure is the pressure of an individual
gas component in a mixture.

Dalton’s law of partial pressure states that the
total pressure of a mixture of non reacting gases
is equal to the sum of the pressures that each gas
would exert if it were present alone.

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For a mixture of 3 gases, A,B and C :

PT = PA + PB + PC

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According to ideal gas equation:

In the mixture of gas A and gas B:

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Mole fraction and pressures

Dalton’s theory allows us to form a relationship between
mole fraction, partial pressure and a total pressure.

XA = mol fraction of gas A 57

Example 1
A gaseous mixture of 7.00 g N2 and 3.21 g CH4 is placed in
a 12.0 L cylinder at 25oC.
a) What is the partial pressure of each gas?
b) What is the total pressure in the cylinder?

Ans: a) 0.51 atm , 0.41 atm
b) 0.92atm

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ANSWER

7g n CH 4  3.21g 1 0.200 mol
gmol 16 gmol
a) nN2  1 0.25 mol
28

n RT
PCH4  CH4
 n N2 RT 
PN 2 V
V
 0.2006 x 0.08206 x 298.15
 0.25 x 0.08206 x 298.15 12.0 L
12.0 L
0.409 atm

 0.51atm

b) PT  PN2  PCH4
0.51  0.409

0.92 atm

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Example 2
A mixture of gases contains 4.53 moles of neon, 0.82 moles
of argon and 2.25 moles of xenon. Calculate the partial
pressure of the gases if the total pressure is 2.15 atm at a
certain temperature.

(Ans : P Ne = 1.28 atm,, P Ar = 0.232 atm, P Xe = 0.63 atm)

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ANSWER

nT  n Ne  nAr  nXe
 4.53  0.82  2.25

7.6 mol

PNe  n Ne PT  PXe  nXe PT 
nT nT

 4.53 2.15  2.25 2.15
7.6
7.6
1.28 atm
0.64 atm

PAr  n Ar PT 
nT

 0.82 2.15
7.6

0.23 atm

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Example 3
A sample of gas at 5.88 atm contains 1.2 g CH4, 0.4 g H2 and
0.1 g He. Calculate :
(a) The partial pressure of CH4, H2 and He in the mixture.
(b) What is the partial pressure of CH4 and H2 if He is

removed?

Ans : a) P CH4= 1.47 atm,P H2 = 3.92 atm, P He= 0.49 atm
b) P CH4= 1.47 atm,P H2 = 3.92 atm

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ANSWER

nH2  0.4 0.2 mol n CH 4  1.2 0.075 mol
2 1.6

n He  0.1  0.025 mol nT  nCH4  nH2  nHe
4 0.075 mol  0.2 mol  0.025 mol

0.3 mol

 nCH4  nHe PT   nH2 PT 
nT nT nT
 PCH4 PT PHe PH 2

 0.075 5.88  0.025 5.88  0.2 5.88
0.3 0.3
0.3

1.47 atm 0.49 atm  3.92 atm

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ANSWER

b) If He isremoved, nT  0.3  0.025
 0.275 mol

PT  5.88  0.49
 5.39 mol

 nH2 PT   nCH4
nT nT
PH 2  PCH4 PT

 0.2 5.39  0.075 5.39
0.275 0.275

3.92 atm 1.47 atm

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Example 4
4.00 dm3 of oxygen at a pressure of 2 atm and 1.00 dm3
nitrogen at a pressure of 1 atm are introduced into a 2.00 dm3
vessel.

a) Calculate the total pressure in the vessel.
b) What is the mole fraction of oxygen in the vessel?

Ans : (a) 4.5 atm (b) 0.89

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ANSWER

a) P1V1  P2V2 b) PO2  XO2 (PT )
 P2
(O2 )  1atm 4 dm3 XO2  4 atm
2 dm3 4.5 atm

 4 atm  0.89

P1V1  P2V2
 P2
(N2 )  1atm 1dm3
2 dm3

 0.5 atm

PT  PN2  PO2
 4.5 atm

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 One of the applications of Dalton’s Law is to calculate the pressure
of a gas collected over water ( for gases that neither react nor
soluble in water).

gas gas + water vapour

 The gas collected is actually a mixture of the gas and water vapour.

PT  Pgas  PH2O

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Example 1
Consider the reaction below :

2KClO3 2KCl + 3O2

A sample of 5.45 L of oxygen is collected over water at a
total pressure of 735.5 torr at 25 °C. How many grams of
oxygen have been collected? (at 25°C, Pwater = 23.8 torr)

(Ans : 6.68 g)

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ANSWER

a) PT  PO2  PH2O
735.5  PO2  23.8
PO2 711.7 torr
 0.936 atm

T  25  273.15
 298.15 K

PO2 V  nO2 RT

nO2  PO2 V
RT
0.936 atm5.45 K
 0.08206 Latmmol 1K1298.15 K

0.2086 atm

mass of O2  0.2086 x32
 6.68g

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Example 2
Excess amount of hydrochloric acid is added to 2.5 g of
pure zinc. The gas produced is collected over water in a gas
cylinder at 25 oC and 100.0 kNm-2. Calculate :

a) the number of mole of gas produced in the reaction.
b) the volume of gas collected in the cylinder.

Ans :
(a) 0.038 mol (b) 0.93 L

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ANSWER

a) 2HCl  Zn  ZnCl2  H2

n Zn  2.5
65.5

 0.0385 mol

1 mol Zn  1 mol H2
moles of H2 produce 0.0382 mol

b) PO2 V  nO2 RT
V  nO2 RT
PH 2

 0.0382 mol x 0.08206 Latmmol 1K1301.15 K

1.0132 atm
0.93 L

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Ideal and non-ideal behaviors of gases

Ideal gas - any gas that obeys the ideal gas equation and
has the properties as outlined by the Kinetic
Molecular Theory of gases

Real gas (non-ideal gas) - gases which do not obey
ideal gas properties

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Deviation from Ideal Behavior

Real gases do not show ideal behavior and said as that
they deviate from the ideal behaviour
The deviation is due to gas molecules do have:
i. its own volume and they occupy some space.
ii. intermolecular forces acting between them .

73

Real gases do not behave ideally because:

Real gases do not behave ideally because:

i. Gas molecules have volumes

The free volume where the molecules move about is
smaller than the volume of the container.

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ii. There are attractive forces between molecules

• Collision between molecules and the wall of container are
less frequent

• The actual pressure exerted by the gas is lower than
calculated by ideal gas equation

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The deviation is more significant at high pressure and
low temperature.
At high pressure:
1. The volume of container decreases, thus the molecules of
gas are closer to each other and begin to occupy a sizeable
portion of the container.
 The volume of gas molecules is significant.

2. As the volume of container decreases at high pressure, the
molecules are so close to each other
 The intermolecular forces between molecules become
important.

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At low temperature
• The kinetic energies of the molecules decreases
• The molecules move at low speed

The intermolecular forces between molecules become
significant.

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However, real gases behave almost ideally at a low pressure
and high temperature.

At low pressure and high temperature, the deviations are small
and ideal gas equation can be used without generating serious
error because:

(a) at low pressure:

• To achieve a low pressure, the volume of a container is
increased.

• When the volume of a container increases the molecules will
be far apart from one another, hence the intermolecular forces
can be neglected.

• At a low pressure the volume of the container is extremely

large compared to the size of the molecules, thus the volume

of molecules can be neglected. 78

(b) at high temperature:
• The gas molecules have high kinetic energies and move

at high speed
• The molecules are able to free themselves from the

intermolecular forces that act between them.
• The intermolecular forces can be neglected, thus they

behave almost ideally.

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Variation of PV vs P of 1 mole of N2 gas at different T
RT

PV 203 K 293 K
RT

673 K

1.0 1000 K
Ideal gas
The value approaches 1.0 at a
very low pressure P/atm

• the line converge to 1.0 mol when p≈ 0 • The lines approaching the ideal line

• gas behave ideally at a very low when T increases
pressure i.e high volume • gas behave ideally at a high

temperature

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The van der Waals Equation

• Since real gases deviate from ideal behavior especially at
high pressure and low temperature

• the ideal gas equation (PV=nRT) cannot be used to
accurately describe the gas variables of real gases.

• Ideal gas equation needs to be adjusted
• Adjusting the equation, two parameters need to

be reconsidered :
• attractive forces between the gas molecules
• volume of the gas molecules

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a) Attractive Forces Between Molecules

Attractive forces which act between the gas molecules will :
● make the molecules move slower
● give less impact to the wall
● pressure exerted by the real gas is less compared to the ideal
gas

● therefore,Ptnhr2eeaalt<erPmideparlessure need to be corrected by adding
coefficient V2

P ideal = P real + n2a n = nn=unmubmebreorfomf moloelses
V2
a=correcting factor for

a = aprpeossuitrieve constant which depend on the
sVtr=evnogltuhmeofotfhceoantttariancetrive forces acting
between the molecules

V = volume of the container 82

b) Volume of Gas Molecules

Since the gas molecules occupy a sizeable portion of a container,
the space in which the molecules are able to move are less than
the volume of the container.

Vcontainer * Vreal = the space in which the molecules
able to move

nb

Vreal < Vcontainer
The correction factor to the volume is Vreal = Vcontainer – nb

where ;

b = a constant representing the volume
occupied by the gas molecules

n = number of moles 83

 Thus, the pressure of gas equation becomes,

P= nRT - n2a
V-nb V2
real gas Correction
pressure Correction for for molecular
volume of attraction
molecules

* a and b = van der Waals constant

84

Combining both factors into the ideal gas equation :
PV = nRT

 n2a   V- nb = nRT van der Waals equation
P + 
 
V2

where P = pressure of real gas
V = volume of container
n = mole of gas

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High value of a and b show great deviation from ideal gas.
(The gas tend to show the real gas character)

Value of a
• Molecules with strong attractive forces has high value of a

Value of b
• Large molecular size have high value of b due to large

volume of the gas molecules.

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5.2 LIQUID

92

LEARNING OUTCOMES

At the end of this lesson, students should be able to :

 Relate the properties of liquids to intermolecular

forces, molecular arrangement and molecular

motion in explaining shape, volume, surface

tension, viscosity, compressibility and diffusion.

 Explain : Based on Kinetic Molecular

- vaporization process Theory & Intermolecular
- condensation process Forces

 Define vapour pressure and boiling point

illustrate :

- intermolecular forces and vapour pressure

- vapour pressure and boiling point 93

5.2 Liquid

The properties of liquids

1. Volume and Shape
Liquid has a definite volume but not a definite shape.
• the particles are arranged closely but not rigidly
• held together by a strong intermolecular forces but not

strong enough to hold the particles firmly in place
• particles are able to move freely
• thus, a liquid flows to fit the shape of its container and is

confined to a certain volume

94

2. Surface tension

•molecules within a liquid are pulled in all direction by
intermolecular forces (net attraction = 0)

•molecules at the surface are only pulled downward and
sideways (net attraction = into the liquid)

a liquid surface tends to have the smallest
possible area
 liquid surface tighten like an elastic film

95

•surface tension is the energy required to increase the
surface area of a liquid by a unit amount
e.g: surface tension of water = 7.29102 J m2
•The stronger the forces between particles in a liquid, the
greater the surface tension

96

Fig. 11.8

molecule at
the surface

molecule within
the liquid

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3. Viscosity
• Viscosity is the resistance of a liquid to flow.
• The higher the viscosity, the more slowly the liquid flows.

The factors affecting the viscosity :
a) The strength of intermolecular forces

Liquids that have strong intermolecular forces have
higher viscosity
b) The size of the molecules
Liquids with larger size (higher molar mass) is
more viscous because it has stronger intermolecular forces

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c) The temperature of the liquid
Increase in temperature, decrease the viscosity of a liquid
because :

• At higher temperature, molecules have higher kinetic
energy,

• the molecules move faster
• molecules can overcome the intermolecular attractive

forces more easily
• resistance to flow decrease
• Viscosity decrease

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4. Compressibility
• in liquid, the particles are packed closely together
• thus, there is very little empty space between the

molecules
• liquids are much more difficult to compress than gas

5. Diffusion
• Diffusion is the movement of liquid from one fluid through

another.
• Diffusion can occur in a liquid because molecules are not

tightly packed and can move randomly around one another.

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