CHAPTER 5 - STATES OF MATTER

CHAPTER 5
STATES OF MATTER

CHEMISTRY UNIT
KOLEJ MATRIKULASI MELAKA

SHARED BY MISS DALINA BINTI DAUD

CHAPTER 5.0

STATES OF MATTER

5.1 Gas
5.2 Liquid
5.3 Solid
5.4 Phase Diagram

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3

5.1 GAS

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LEARNING OUTCOMES

At the end of the lesson, student should be able to :

a) Explain qualitatively the basic assumptions of the
kinetic molecular theory of gases for an ideal gas.

b) Define gas laws
(i) Boyle’s Law
(ii) Charles’s Law
(iii) Avogadro’s Law

c) Sketch and interpret the graphs of Boyle’s and
Charles’s laws

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d) Perform calculations involving gas laws and ideal
gas equation

e) Define and perform calculation using Dalton’s Law

f) Explain the ideal and non-ideal behaviours of
gases in terms of intermolecular forces and
molecular volume.

g) explain the conditions at which real gases
approach the ideal behaviour.

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General properties of gases:
1. Occupy more space than their solids or liquids.
2. Have much lower densities then solids or liquids.
3. Expand to fill their containers –assume the volume

and shape of their containers.
4. Easily compressible compared to solids or liquids

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General properties of gases:
5. Flows readily
6. Mix completely and evenly when put together in the

same container
7. 1 mole of gas occupies a volume of 24 L at room

conditions and 22.4 L at STP

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Kinetic Molecular Theory of Gases

• Describes the behavior of an ideal gas.
• Ideal gas : gases which obey the ideal gas equation

(PV = nRT)

The Kinetic Molecular Theory of Gases
begins with five postulates that describe
the behavior of molecules in a gas.

a gas that obeys the ideal gas law

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Kinetic Molecular Theory of Gases

Theory that explains the behavior of gases.
The theory is based on the following assumptions:

1. A gas consists of a very large number of extremely
small particles(molecules or atoms) in constant,
random, straight-line motion.

2. Collisions among molecules are perfectly elastic.
(energy can be transferred from one molecule to
another during collision but the total energy of all
molecules in the system remain the same)

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3. The volume of gas molecules are negligible
compared to the volume of its container.

4. Attractive and repulsive forces between molecules
are negligible.

5. The average kinetic energy of molecules is
proportional to the absolute temperature.

A gas corresponding to these assumptions is called
an ideal gas.

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The Gas Laws

a) Boyle’s Law :

Boyle’s law states that at constant temperature, the
volume of a fixed amount of gas is inversely
proportional to the gas pressure.

V 1 (at constant T & n)

P

where V = volume
P = pressure
T = temperature
n = number of moles

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A molecular description of Boyle’s Law

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V = k1 Therefore, PV = k1
P1V1 = P2V2

V  k 1  1  OR P  k 1  1 
 P   V 

Therefore, PV  k
1
P1V1 P2V2

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Graphs based on Boyle’s Law

Graph of P versus V Graph of P versus 1

P V
P

V 1
V
pressure is inversely
proportional to volume pprreospsourtrieonisadl tiorectvloyl1ume

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Graph of PV versus P

PV

PV = constant

P
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Example 1

A sample of chlorine gas occupies a volume of 2 L at a pressure
of 1 atm. Calculate the pressure of the gas if the volume is
increased to 5 L at constant temperature.

(ans : 0.4 atm)

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ANSWER

P1V1  P2V2

1atm2L  P25L

P2  1atm 2L 

5L

 0.4 atm

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Example 2

The pressure of a sample of hydrogen gas in a 50.0 mL
container is 765 mmHg. The sample is then transferred into
another container and the measured pressure is 825 mmHg.
What is the volume of the second container?

(Ans :46.36 mL)

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ANSWER

P1V1  P2V2

V2  765 mmHg 50 mL

825 mmHg

 46.36 mL

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b) Charles’s Law
Charles observed that at constant pressure, the
volume of a gas expands when heated and
contracts when cooled.

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Plot of Volume versus Temperature (C)

V

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Plot of Volume versus Temperature (K)

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Charles’s law states that at constant pressure, the
volume of a fixed amount of gas is directly
proportional to the absolute temperature of the gas.

V  T (at constant P & n)

where V = volume
T = absolute temperature (K)
P = pressure
n = number of moles

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V k T
2

Therefore, V  k
T2

V1  V2
T1 T2

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Example 1

A sample of carbon monoxide gas occupies 3.2 L at 125 °C.
The sample is then cooled at constant pressure until it
contracts to 1.54 L. Calculate the final temperature in degree
celsius.

(Ans : -81.54 °C)

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ANSWER

V1  V2
T1 T2

3.2 L K  1.54 L
T2
125 273.15

T2 191.61K

  81.54 C

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Example 2

A sample of gas trapped in a capillary tube by a
plug of mercury at 22 oC has a volume of 4.5 mL.
Calculate the volume of the gas when the capillary
tube is heated to 60 oC.
(Ans : 5.08 mL)

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ANSWER

V1  V2
T1 T2

4.5 mL K  V2 K

22  273.15 60  273.15

V2 5.08 mL

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The Combination of Boyle’s and Charles’s Law

Boyle’s law: V 1
Charles’ law: P

V T

VT
P

PV  k
T3

P1V1 = P2V2
T1 T2

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Example 1

A sample of methane gas occupies 25.5 L at 298.15 K and
153.3 kPa. Find its volume at STP.

(Ans : 35.35 L)

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ANSWER

P1V1  P2V2
T1 T2

1.513 atm25.5 L  1x V2
 298.15 K 273.15
 K

V2 35.35 mL

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Example 2

A gas–filled weather balloon with a volume of 55.0 L is
released at sea-level conditions of 759 mmHg and 27 oC.
Calculate the volume of the balloon when it rises to an altitude
at which the temperature is -5 oC and the pressure is 0.066
atm.

(Ans : 743.52 L)

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ANSWER

P1V1  P2V2
T1 T2

0.9987 atm55.0 L   0.066 atm x V2
 27  273.15 K
 5  273.15 K

V2 743.52 L

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c) Avogadro’s Law

Avogadro’s law states that at constant temperature and
pressure, the volume of a gas is directly proportional to
the number of moles of the gas.

V  n (at constant T & P)

V k n
4

V k V1 = V2
n4 n1 n2

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A molecular description of Avogadro’s Law

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Example 1

2 moles of chlorine gas kept in a cylinder with piston
occupies a volume of 49 L. When another 3 moles of
chlorine gas is pumped into the cylinder at constant
temperature and pressure the piston moves upwards to
accommodate the gas. Calculate the final volume of the
gas.

(Ans :122.5L)

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ANSWER

V1  V2
n1 n2
49 L  V2
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V2 122.5 L

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(d) The Ideal Gas Equation

Combination of Boyle's law, Charles's law and Avogadro's law :

Boyle's Law: V 1
P

Charles' Law: VT

Avogadro's Law: V  n

V  nT Where :
P R is the ideal gas constant

V = R nT
P

PV = nRT

Ideal gas equation

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The value and unit of R depend on the unit of pressure
and volume used in the equation.

unit of unit of value of R unit of R
pressure volume 0.08206
L atmmol1 K1
atm L
N m mol1 K1
N m2 or Pa m3 mol1 K1
or
Pa dm3 J mol1 K1
mmHg L mol1 K1
mmHg m3 8.314
or torr L mol1 K1
torr L
or 62.36
dm3

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Further Application oMf Tr he Ideal Gas Law
 Density and molar Pma=ssmoRf aTgas can be calculated

by rearranging the IdeaVl GMasr Equation:
dRT

PV  nRT P = Mr

d = PMr
RT

Mr = molar mass of a gas
n = mol of gas
m = mass of the gas

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PV  nRT

PV=  m  RT
 Mr 
 

P = mRT (d = density of a gas)
VMr
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P = dRT
Mr

d = PMr
RT

Example 1:

A steel gas tank has a volume of 275 L and is filled with
0.485 kg of O2. Calculate the pressure of O2 if the
temperature is 29 oC.

(Ans : 1.37 atm)

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ANSWER

n CO2  0.485 x1000g
32 gmol 1

15.156 mol

PV  nRT

 PO2
15.156 mol 0.08206 Latmmol 1K1 (309.15K )
2.88 atm

1.37 atm

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Example 2:

A sample of chlorine gas is kept in a 5.0 L container at 228 torr
and 27oC. How many moles of gas are present in the sample?

(Ans : 0.06 mol)

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ANSWER

T  27  273.15

300.15 K

PV  nRT

n  PV
RT

0.3atm x 5 L
0.08206 Latmmol 1K1
 nCl  (300.15K )

0.061 mol

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Example 3:
A chemist has synthesized a greenish-yellow compound of
chlorine and oxygen and finds that its density is 7.71 g L-1
at 36°C and 2.88 atm. Calculate the molar mass of the
compound.

(Ans : 67.9 gmol-1)

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ANSWER

T 36  273.15
309.15 K

PV  nRT

  PMr
RT

M r  RT
P

 Mr
 7.71gL1 x 0.08206 Latmmol 1K1 309.15 K
2.88 atm

Mr  67.9 gmol 1

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Example 4:

Calculate the density (in gL-1) of CO gas at:

a) room conditions (Ans : 1.14 gL-1)
b) STP (Ans : 1.25 gL-1)

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ANSWER

a)   PMr
RT

1atm x 28 gmol 1
0.08206 Latmmol 1K1 298.15 K
  CO 

1.14 gL1

b)   PMr
RT

1atm x 28 gmol 1
0.08206 Latmmol 1K1 273.15 K
  CO 

1.25 gL1

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