TMK Chapter 5 States of Matter

SESSION 2020/2021 TOPIC 5: STATES OF MATTER
CHEMISTRY SK015

TUTORIAL 5

5.1: Gas

1. a) Basic assumptions of the Kinetic Molecular Theory of Gases:
An ideal gas obeys all gas laws and fits the assumptions of the Kinetic Molecular
Theory:
i. The size of gas particles is extremely small compared with the volume of the
container. Therefore, the volume of gas particles is negligible.
ii. Gas particles are in constant and random motion. They frequently collide with
one another and also with the wall of the container.
iii. Collisions between gas particles are elastic: No loss of kinetic energy.
iv. Intermolecular forces (attractive and repulsive forces) between gas particles are
negligible.
v. The average kinetic energy of the gas particles is proportional to the absolute
temperature.

Boyle’s Law
Boyle’s Law states that at constant temperature, the volume occupied by a fixed
amount of gas is inversely proportional to the applied (external) pressure.

Thus, qualitatively, as stated by Boyle’s Law, whenever the volume occupied by gas
is decreasing, it will lead to the rapid collisions between gas particles and the wall of
the container. This will eventually cause the pressure exerted by the gas particles to
be increased.

b) Given: P2 = 0.541 atm
P1 = 0.970 atm V2 = ?
V1 = 725 mL

By applying Boyle’s Law:
P1V1 = P2V2
0.970 x 725 = 0.541x V2

V2 = 1300 mL

c) Given: P2 = ?
P1 = 1.2 atm V2 = 0.075 L
V1 = 3.8 L

By applying Boyle’s Law:
P1V1 = P2V2
1.2 x 3.8 = P2 x 0.075

P2 = 60.8 atm

2. a) Definition of Charles’ Law
Charles’ Law states that at a constant pressure, the volume occupied by a fixed
amount of gas is directly proportional to the absolute temperature of the gas (in
Kelvin).

b) How temperature affects the volume occupied by gas particles:
• as the temperature increases, the kinetic energy of the gas particles increases
• the frequency of collisions between gas particles increases
• thus, pressure increases

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• in order to maintain the pressure, volume occupied by the gas particles must be
increased

c) Given: T2 = 22.0 oC = 295.15 K
T1 = 145.0 oC = 418.15 K V2 = ?
V1 = 93.0 L

By applying Charles’ Law:

V1 = V2
T1 T2

( (93.0 L) ) = ( V2 K )
295.25
418.15 K

V2 = 65.6 L

d) Given: T2 = ?
T1 = 88 oC = 362.15 K V2 = 3.4 L
V1 = 9.6 L

By applying Charles’ Law:

V1 = V2
T1 T2

(9.6 L) ) = (3.4 L )
(362.15 K
T2

T2 = 127.9 K

3. a) Given: T2 = ?
T1 = 24 oC = 297.15 K V2 = 15.00 L
V1 = 8.70 L P2 = 876 torr
P1 = 895 torr

By using combined gas law:

P1V1 = P2V2
T1 T2

(895 torr ) (8.70 L) = (876 torr ) (15.00 L)
( 297.15 K )
T2

T2 = 501.5 K

b) Given:

T1 = 25 oC = 298.15 K T2 = 45.0 oC = 318.15 K

V1 = 1 L V2 = 2 L

P1 = 2.50 atm P2 = ?

By using combined gas law:

P1V1 = P2V2
T1 T2

( 2.50 atm ) (1L ) = P2 (2L)
( 298.15 K ) ( 318.15 K
)

P2 = 1.33 atm

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4. a) Given:
T = 27 oC = 300.15 K
V = 5.0 L
P = 228 torr = 0.30 atm

By using ideal gas equation, PV=nRT

PV = nRT

n = PV
RT

(0.30 atm)(5.0 L)

( )= 0.08206 atm L mol−1K−1 (300.15 K )

= 0.061 mol

b) Given:
T = 45 oC = 319.15 K
V = 207 mL = 0.207 L
P =699 mmHg = 0.919 atm

By using ideal gas equation, PV=nRT

PV = mRT
Mr

m = PVMr
RT

( )(0.919 atm)(0.207 L) 92.5 g mol−1
( )= 0.08206 atm L mol−1K−1 (319.15 K )

= 0.67 g

c) Given:
T = 123.0 oC = 396.15 K
m = 245 g
P = 5.33 atm

By using ideal gas equation, PV=nRT

PV = mRT
Mr

V = mRT
PMr

( )(0.919 atm)(0.207 L) 92.5 g mol−1
( )= 0.08206 atm L mol−1K−1 (319.15 K )

= 83.0 L

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5. a) Derivation of an equation that relates density of gas to its pressure:

PV = nRT

PV =  m RT P = pressure of gas (atm)
 M  V = volume of gas (L)
m = mass of gas (g)
P = mRT M = molar mass of gas (gmol -1 )
MV R = gas constant = 0.08206L.atm.mol −1.K −1
n = no. of mole of gas (mol)
P =  m  RT  = density of gas (gL-1 )
 V  M

P = RT
M

  = PM
RT

b) Given:
T = 25 oC = 298.15 K
= 2.60 g L−1
P = 101 kPa = 0.997 atm

From ideal gas equation :

PV = nRT

P = RT
M

M = RT
P

( )( )2.60 g L−1 0.08206 atm L mol−1K−1 (298.15 K)

= (0.997 atm)

= 64 g mol-1

c) Given:
T = 446 oC = 719.15 K, = 4.33 g L−1, P = 100.8 kPa = 0.9948 atm

From ideal gas equation :

PV = nRT

P = RT
M

M = RT
P

( )( )4.33 g L−1 0.08206 atm L mol−1K−1 (719.15 K)

= (0.9948 atm)

= 257 g mol-1

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6. a) Gas D

b) Gas A

c) By using Dalton’s Law of Partial Pressure:

Partial pressure of gas D, PD = XDPT

= 5 (0.75 atm)
16

= 0.23 atm

7. Given:

PT = 6.0 atm
Mass of CH4 = 6.20 g
Mass of H2 = 1.60 g
Mass of N2 = 8.20 g

a) Calculations for determining the partial pressure of each gas:

n CH4 = 0.3875 mol
n H2 = 0.8 mol
n N2 = 0.293 mol

By applying Dalton’s Law of Partial Pressures:

Pg = XgPT

PCH4 = XCH4 PT

= (0.3875) ( 6.0 atm )
(1.4805)

= 1.6 atm

PH2 = XH2 PT

= (0.8) ( 6.0 atm )
(1.4805)

= 3.2 atm

PN2 = XN2 PT

= (0.293) (6.0 atm)
(1.4805)

= 1.2 atm

b) Given:
T = 25oC = 298.15 K
PT = 6.0 atm
nT = 1.4805 mol

Using ideal gas equation,

PTV = nTRT

Volume of container = Volume occupied by gas, V= n T RT
PT

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( )(1.4805 mol) 0.08206 atm L mol-1 K-1 (298.15 K)
V = (6.0 atm) = 6.0 L

8. Given: P2 = ?
For O2 gas V2 = 8.0 L
P1 = 24 atm
V1 = 5.0 L

For N2 gas P2 = ?
P1 = 32 atm V2 = 8.0 L
V1 = 3.0 L

a) partial pressure of each gas.
O2
P1V1 = P2V2

(24 atm)(5.0 L) = P2(8.0 L)

P2 = 15 atm

N2
P1V1 = P2V2
(32 atm)(3.0 L) = P2(8.0 L)
P2 =12 atm

b) the total pressure.
PT = 15 atm + 12 atm = 27 atm

c) the mole fraction of each gas
Pg = XgPT O2

XO2 = (15 atm)
( 27 atm)

= 0.56

XN2 = (12 atm)
( 27 atm)

= 0.44

9. Given:
T = 24oC = 297.15 K
PT = 762.00 mmHg, Pwater = 22.38 mmHg
V = 128 mL

According to Dalton’s law of partial pressure:

PT = PO2 + PH2O

 PO2 = PT − PH2O

= (762.00 − 22.38)mmHg = 739.62mmHg = 0.973atm

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From ideal gas equation:

Mass of oxygen gas collected, m= PVMr
RT

( )(0.973 atm)(0.128 L) 32 g mol−1
( )= 0.08206 atm L mol−1K−1 (297.15 K )

= 0.163 g

10. Given:
T = 26oC = 299.15 K
P = 758 torr = 0.997 atm
V = 28.50 mL

Equation of the reaction:
Mg(s) + 2HCl(aq) → MgCl2 (aq) + H2 (g)

Using ideal gas equation:
Mole of hydrogen gas produced,

n = PV
RT

(0.997 atm)(0.0285 L)

( )= 0.08206 atm L mol−1K−1 (299.15 K )

=1.157 × 10−3

From the balanced equation:
1 mol of H2 gas produced by 1 mol of Mg
Therefore,
1.157 × 10−3 of H2 gas produced by 1.157 × 10−3 of Mg

( )( ) mass of Mg = 1.15710−3 mol 24.3gmol −1

= (0.0281g ) 1mg g 
110−3

= 28.1 mg

11. a) Since real gas does not exhibit ideal gas behavior at high pressure and low
temperature, the ideal gas equation need to be adjusted by correcting two
parameters:
i. Corrective term for volume

• In actual situation, the real gas particles do occupy sizeable portion of the
container

• Space occupied by the gas particles have to be taken into consideration
• The space restricted the movement of the gas particles
• The volume occupied by the gas particles must be subtracted from the total

volume
• The volume is decreased by the factor of nb, which accounts for the finite

volume occupied by the gas particles

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V = Vcontainer − Vgas
= Vcontainer − nb

Where: n = moles of gas particles,
b = constant representing the volume occupied by
the gas particles
(larger molecular size, higher value of b)

ii. Corrective term for pressure

• The intermolecular forces do exist between real gas particles and

should not be neglected

• These attraction forces have an effect on the speed of the moving

gas particles

• Gas particles that experience these forces will move slowly

• This will lead to fewer collisions of gas particles with the wall of the

container and lower pressure exerted by the collisions as compared

to the ideal gases

 Preal  Pideal

• The exerted pressure is decreased by the factor of (n2a/V2) which

accounts for the intermolecular forces between the gas particles

• The term pressure is corrected by adding the (n2a/V2) coefficient

 P = Preal + n2a
V2

Where: n = moles of gas particles

a = constant relates to intermolecular forces between

gas particles

(stronger intermolecular forces, higher value of a)

b) H2S has larger a value than H2 gas.

REASON: H2S is polar molecule; hence it has stronger intermolecular forces.

5.2: Liquid

1. a) Ethanol has lower surface tension compare to water.
• Both can form hydrogen bond.
• However, water molecules can form stronger hydrogen bond compare to ethanol
because water molecule is more polar than ethanol.
• Therefore ethanol has lower surface tension.

b) Oil flows faster through a narrow tube
• Increase in temperature decreases the viscosity of oil.
• As temperature increases kinetic energy of oil particles increases and moves
faster.
• This weakens the intermolecular forces.
• Thus decreases the viscosity and oil flows faster.

2. a) ice to water (melting)
• In ice, particles are arranged closely packed. The particles vibrate and rotate at
their positions.
• When temperature increases, the particles vibrates and rotates faster hence
overcome the interparticle bonding between them.

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• As they melt, the particles in liquid are free to move and arranged in cluster
arrangement.

b) Wet clothes dry (vapourisation)
• In liquid phase, particles are arranged in cluster.
• As temperature increases, they gain energy and moves faster.
• They are free to move randomly as they overcome the interparticle bonding.
• Particles far apart; water evaporates from the wet clothes.

c) Droplets appear on glass of cold water
• Because of the low temperature of the glass, it’s surrounding becomes cooler
than it was before.
• Due to which, the moisture in air around the outer surface of the glass
condensates, settles down and hence turns into liquid form outside the glass.

3. a) False
• CBr4 bigger size than CCl4
• Van der Waals forces in CBr4 stronger than CCl4
• Thus CBr4 should be less volatile

b) True
• CBr4 bigger size than CCl4
• Van der Waals forces in CBr4 stronger than CCl4
• Thus boiling point CBr4 is higher

c) False
• CBr4 bigger size than CCl4
• Van der Waals forces in CBr4 stronger than CCl4

d) False
• Van der Waals forces in CBr4 stronger than CCl4
• Thus CBr4 less volatile than CCl4

4. Vapour pressure of CH3CH2OH < CH3COCH3
• Hydrogen bond in CH3CH2OH is stronger than Van der Waals forces in CH3COCH3
• CH3COCH3 molecules can easily escape to form vapour as compared to CH3CH2OH
• CH3CH2OH less volatile
• CH3CH2OH has lower vapour pressure
• CH3CH2OH has higher boiling point

5.3: Solid

1. Describe phase changes that occur in process A

• Process A is sublimation in which solid changes directly to gas without becoming a
liquid.

• Molecules in solid phase are arranged closely together in fix position. The
movements of the molecules are restricted to vibration and rotation.

• When a solid absorbs heat, the molecules vibrate faster. The kinetic energy of some
molecules is strong enough to overcome the attractive forces and leave the solid
phase as vapour.

• In the gaseous phase, the molecules move at very high speed and are very far away
from one another.

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2. a) Differentiate between amorphous and crystalline solids.

Types of solid Crystalline Solid Amorphous Solid

Arrangement of • Has a well-defined shape • Poorly defined shape
• Lack in orderly arrangement of their
particles • Particles (atoms, molecules or particles
ions) are in orderly
arrangement

Formation Formed when liquid is cooled Formed when liquid is cooled rapidly
slowly

Physical Sharp and well-defined melting No definite melting point
properties point

Example Metal, salt , diamond Charcoal, rubber, glass

b) State the interparticle forces involved in each of the following crystalline solids:
i. metallic : metallic bond
ii. ionic: ionic bond
iii. molecular covalent : Van der Waals forces
iv. giant covalent : covalent bond

5.4: Phase Diagram

1. a) Indicate the phases presents in the region labeled A, B and C.
A: Solid
B: Liquid
C: Vapour

b) A sample of solid phosphorus cannot be melted in an open container at atmospheric
pressure.
▪ Melting involves changes from solid to liquid
▪ As the phase diagram shows, the lowest pressure at which liquid exists is at the
triple point pressure is 43 atm.
▪ However, 1.00 atm is far below 43 atm.
▪ Thus, liquid cannot exist at this pressure
▪ Therefore, solid phosphorus sublimes to a gas instead

2. a) Name point A, and B.
A: Critical point
B: Triple point

b) Physical properties of water at point A and B.
A: liquid and vapour indistinguishable (super critical fluid).
B: solid, liquid and vapour in equilibrium.

c) Curve represents the equilibrium between ice and water vapour
BC

d) Phase changes when a sample at point E is heated at constant pressure until point F.
solid → solid and liquid in equilibrium → liquid → liquid and vapour in
equilibrium → vapour

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e) BC line intersects 1 atm line.
Normal melting point.

f) The BC line has a negative slope.
This is because ice is less dense than water, while most solids denser than
their liquids.

3. a) Define critical point and triple point.
• Critical point is the temperature and pressure where liquid and gas are at
equilibrium and above this point gaseous phase and liquid phase cannot be
distinguish.
• Triple point is the temperature and pressure where the three phase’s solid,
liquid, and gas coexist at equilibrium.

b) Sketch the phase diagram of X.

P/atm

normal melting point

solid liquid
1.0 normal boiling point

0.5

triple point gas

18 20 300 T/oC

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MEKA 5

1. (a) Boyle’s law state that the volume of a fixed amount of gas is inversely proportional to
the gas pressure at a constant temperature.

(b) 748 torr = 0.984 atm
P1V1 = P2V2
(0.984 atm)(10.3 L) = (1.88 atm)V2
V2 = 5.39 L

2. (a) PV = nRT

PV = ; where

P=

P = ; where ρ =

(b) ρ = 1.50 gL-1 Mr C6H6= 78 gmol-1
P = 800 torr
= 1.0536 atm

P= ; 1.0536 =

T = 667.65 K
= 394.5 °C

3. (a) At STP; T = 0 °C or 273.15 K

P = 1 atm

(b) V1 = 255 mL P1= 0.85 atm T1 = 25 °C
= 298.15 K
At STP; P2 = 1.00 atm
V2= ? T2 = 273.15 K

V2 = 198.58 mL

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4. (a) The total pressure of a mixture of non-reacting gases in a container is the sum of the
partial pressure exerted by each individual gas in the mixture.

(b) i. Nitrogen;
ii.
iii. P1V1 = P2V2

P2 = P1V1 = ( 4 ) ( 400 ) = 800 kNm-2
V2 (2)

Argon;

P1V1 = P2V2

P2 = P1V1 = (1) ( 200 ) = 100 kNm-2
V2 (2)

PT = + PAr

PT = 800 + 100 = 900 kNm-2

5. (a) n Ag2O =

= 7.12 x 10-3 mol
(b) From the balanced equation:

2 mol of Ag2O produce 1 mol of O2
Therefore,
7.12 x 10-3 mol Ag2O produce 3.56 x 10-3 mol O2

(c) PO2= 750.00 -23.76 = 726.24 mmHg = 0.956 atm

(d) T = 298.15 K

PV = nRT

( )nRT 3.5610−3 (0.08206)(298.15)
V= P = =0.0911 L
(0.956)

6. (a) Molecules in solid phase are arranged closely together in fix position. The movements
of the molecules are restricted to vibration and rotation. When a solid absorbs heat, the
molecules vibrate faster. At a certain temperature – the melting point, the kinetic energy
of the molecules is strong enough to overcome the attractive forces. The molecules
start to move freely and are no longer confined to fixed position, thus solid turns to
liquid. However, kinetic energy of the molecules in this phase is not strong enough for
them to move far away from one another. They are still in contact and just slide past
one another.

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(b) Molecules in the gaseous phase have high kinetic energy, move at very high speeds
and are very far away from one another. When some of these molecules loose energy,
they are not able to overcome the intermolecular forces and start to move closer to one
another to form liquid. Also due to the random motion, some of vapour molecules may
strike the surface of the liquid and be captured there. In this phase, they are in contact
and just slide past one another.

(c) Molecules in solid phase are arranged closely together in fix position. The movements
of the molecules are restricted to vibration and rotation. When a solid absorbs heat, the
molecules vibrate faster. The kinetic energy of some molecules is strong enough to
overcome the attractive forces and leave the solid phase as vapour.
In the gaseous phase the molecules, move at very high speeds and are very far away
from one another.

7. (a) No effect.
(b) -Weak intermolecular forces.
-Easy to escape as vapour.
-Vapour pressure increase.
(c) -Temperature increase.
-Kinetic energy increase.
-More molecule liquid can escape as vapour.
--Vapour pressure increase.

8. Interparticles forces: covalent bond
Diamond is extremely hard because each carbon atom in diamond is tetrahedrally bonded
to four other carbon atoms infinitely. High energy required to break the infinite amount of
covalent bond in diamond.

9. (a) Vapour → vapour and solid in equilibrium → solid
(b) Vapour → vapour and liquid in equilibrium → liquid → liquid and solid in equilibrium →
solid

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KUMBE 5

1. Mass of gas Y = 2.54 g

V vessel = 933.16 mL = 0.9332 L

P = 0.9800 atm

T = 304.15 K

Molar mass of Y = mRT = ( 2.54) ( 0.08206) ( 304.15) = 69.3 gmol-1
PV ( 0.98) ( 0.9332 )

1 mol of Y molecules weighs 69.3 g

1 molecule of Y weighs 69.3 a.m.u

Therefore,

Relative molecular mass of Y = mass of one Y molecule
112 mass of one 12C atom

= 69.3 a.m.u
11212 a.m.u

= 69.3

2. (a) 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l)

(b) i. GIVEN:
Mole of available NH3 = 2.0 mol, Mole of available O2 = 6.0 mol

From the balanced equation:
4 mol of NH3 react with 5 mol of O2
Therefore,
2 mol of NH3 should react with 2.5 mol of O2
Mole of O2 needed is less than mole of available O2
O2 is the excess reactant
∴ NH3 is limiting reactant

Since NH3 is the limiting reactant, it limits the amount of both products, NO and
H2O, formed
From the balanced equation:
4 mol of NH3 produce 4 mol of NO and 6 mol of H2O
Therefore,
2 mol of NH3 produce 2 mol of NO and 3 mol of H2O

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iii. T = 373.15 K

PT = nRT = (8.5) (0.08206) ( 373.15) = 52 atm
V (5)

P NO = (82.5)(52) =12 atm

P H2O = ( 3 )(52) = 18 atm
8.5

P O2 = (83..55)(52) = 21 atm

3. (a) a: intermolecular forces
b: volume of gas molecule

(b) Value of a would be higher since propane is a large molecule with stronger van der
Waals forces.
Value of b also would be higher due to the large size of propane molecule.

(c)

Ideal Gas Real Gas

The volume of gas molecules are negligible Gas molecules have a specific volume

@ The gas molecules have no volume Therefore Vreal = Vcontainer - nb

Therefore Vreal = Vcontainer

The attractive forces between molecules are There are weak attractive forces between

negligible @ There are no attractive forces molecules

between molecules. Therefore Pideal = Preal +

Therefore the attractive forces do not affect

the gas pressure

Pideal = Preal

(d) Very 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.

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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.

4. (a) Three properties of a liquid:
i. It conforms to the shape and volume of the container, volume is limited by
surface
ii. Moderate viscosity

iii. Very low compressibility

(b) Boiling point of ethanal < methanol < ethanol.
• Ethanol and methanol has hydrogen bond act between their molecules but
ethanal only has weak Van der Waals forces.
• Ethanol has bigger molecular size than methanol; hence it has stronger Van
der Waals forces than methanol.
• Strength of intermolecular forces of ethanal < methanol < ethanol.
• Vapour pressure of ethanal > methanol > ethanol.
• Boiling point of ethanal < methanol < ethanol.

5. (a)

Molecules/ atom Types of crystalline solid Interparticles forces
NaCl Ionic crystal Ionic bond
Mg
SiO2 Metallic crystal Metallic bond
S8 Gigantic covalent crystal Covalent bond
Molecular covalent crystal Van der Waals forces

(b) i.

P/atm

73.0 C
liquid
solid

5.2
T gas

-57 31 T/oC

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ii. CHEMISTRY SK015

Triple point of CO2 is 5.2 atm and -57oC that is above 1 atm and -78oC.
Liquid phase of CO2 does not exist below triple point. Only solid and vapour
phases exist at ordinary temperature and pressure (below triple point).
Therefore, solid CO2 does not melt at ordinary pressure and temperature but it
sublimes.

iii. When a sample of carbon dioxide in a closed vessel at -78oC and 1 atm was
pressurized isothermally to 10 atm, the phase changed from vapour and solid
in equilibrium to solid phase completely.

At 10 atm, when the sample went through isobaric cooling to 20oC, the solid
phase of carbon dioxide change to solid and liquid phases in equilibrium, and
then changed completely to liquid phase. As the cooling process continues,
the sample completely changed to a phase where the liquid and vapour phases
exist in equilibrium.

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