Chapter 5 Physical States of Matter 9th Chemistry Notes

Chapter 5 Physical States of Matter 9th Chemistry Notes Notes. Easy notes that contain questions, exercises,s and key points of the chapter.

Short Questions Chapter 5 Physical States of Matter 9th Chemistry Notes

Can you give a reason why it takes a longer time to cook at high altitudes?

Boiling points depend upon the external pressure over the surface of the liquid. At sea level external pressure is 1 atm, the boiling point of water is 100oC. As we go at higher altitudes, the external pressure decreases, and the water boils at lower temperatures. Because the water is boiling at lower temperatures and less heat is being transferred, more time is required to cook the same amount of food.

Q.2) Glass softens over a wide range of temperatures. Ice melts at a specific temperature. Explain the reason for this difference.

Answer:
Glass is an amorphous solid in which the particles are not regularly arranged. Because of this, the intermolecular forces among its particles vary in strength within a sample. Melting thus occurs at different temperatures for different portions of the same sample as the intermolecular forces are overcome. Therefore, glass does not exhibit sharp melting point but softens over a wide range of temperature.
Ice contains regularly repeating arrangements of particles. The intermolecular forces among particles are same everywhere within a sample. Therefore, it melts at a specific temperature.

Q.3)   Explain why it happens that on a hot summer day when there is sweat on the body of a person, one feels cool under fast moving fan?

Answer:
For evaporation to occur, the molecules need energy which they get from the body by taking away heat and body gets cooler by losing heat. If there is no wind, the evaporating molecules keep hovering near the surface of the body and impede the evaporation of more molecules and even push some back to the body. A fast moving fan blows air around, which speeds the evaporation by taking the vapour away. Since evaporation is a cooling process, more evaporation causes more cooling due to fan air.

Q.4) Why are the densities of gases lower than that of liquids?

Answer:
Gases are light in mass and their molecules occupy more volume than liquids. Moreover, molecules of gases are farther apart. As the liquid molecules have strong intermolecular forces, they are closely packed and the spaces between them are negligible. Therefore, densities of gases are lower than that of liquids.

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Read more: Structure of Atoms Cha 2 Chemistry 9 class notes

Q.5) What is the relationship between the atmospheric pressure and boiling point of a liquid?

Answer:
Boiling point of a liquid depens on the atmospheric pressure. With the increase in atmospheric pressure, the boiling point also increases. Similarly, the decrease in atmospheric pressure causes decrease in boiling point.
For example, at sea level atmoshperic pressure is 1 atm, the boiling point of water is 100oC. As we go at higher altitude, the atmospheric pressure decreases, and the water boils at lower temperatures.

Q.6) Why gas is compressible but a solid is not compressible? Give reason(s).

Answer:
Gas can be easily compressed because the molecules are far away from each other and very weak intermolecular forces are present between them.
Solid is incompressible because the particles are closely packed in a fixed pattern and there are strong forces of attraction between the molecules.

Long Questions Chemistry 9 class notes

Q.1) Define Boyle’s law and verify it experimentally.

Answer:
Boyle’s Law
Boyle’s law states that the pressure and volume of a gas have an inverse relationship, when temperature is held constant.
It means as pressure increases, the volume of the gas decreases in proportion. Similarly, as pressure decreases, the volume of the gas increases.
Mathematically it can be written as:
P ∝ 1 / V    or    PV = k
where ‘P’ is the pressure of the gas, ‘V’ is the volume of the gas, and ‘k’ is a constant.
The equation above states that product of pressure and volume is constant for a given mass of confined gas at constant temperature.
If P1V1 = k                    then P2V2 =k
Where P1 = initial pressure                  V1 = initial volume
            P2 = final pressure                   V2 = final volume
As both equations have same constant, their variable have same value.
Thus P1V1 = P2V2
Experimental verification of Boyle’s law
Suppose there’s a gas confined in a cylinder with a piston at the top. The initial state of the gas has a volume equal to 8.0 dm3 and the pressure is 1.0 atm. Temperature and number of moles is constant. Weights are slowly added to the top of the piston to increase the pressure. When the pressure is 2 atm the volume decreases to 4.0 dm3.  The product of pressure and volume remains a constant i.e 8.
P1V1 = P2V2
1 x 8 = 2 x 4

Experimental Verification of Boyles Law
Chapter 5 Physical States of Matter 9th Chemistry Notes 7

Q.2) Differentiate between:

a. Evaporation and Boiling Point
b. Effusion and Diffusion of gases
c. Condensation and Evaporation

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Answer:
a) Difference between evaporation and boiling

Read more: Class 9 Chemistry Cha 2 Periodic Table and Periodicity of Properties

EvaporationBoiling
Evaporation is a process in which a substance changes its state from liquid state to gaseous state without boiling.Boiling is a process in which a substance changes its state from liquid state to gaseous state with boiling.
It is a slow process.It is a rapid process.
No bubbles form in evaporation because vapour pressure is less than atmospheric pressure.Bubbles form in boiling because vapour pressure is equal to the atmospheric pressure.
It takes place at all temperatures.It occurs at a particular temperature.
Evaporation takes place from the exposed surface.Boiling occurs throughout the liquid.
Some particles move fast and some move slowly in evaporation.All the particles move very rapidly in the process of boiling.
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b) Difference between diffusion and effusion of gases

Read more: Chapter 4 Structure of Molecules Chemistry Class 9 Notes

DiffusionEffusion
This process of mixing of gases by random motion of molecules is called diffusion.The escape of molecules in the gaseous state one by one without collision through a hole of molecular dimension is called effusion.
It is the ability of a gas to travel through a small opening.It is the ability of gases to mix with each other usually in the absence of a barrier.
Diffusion occurs due to difference in concentrations.Effusion is facilitated by a difference of pressures.
The rate at which diffusion occurs is limited by the size and kinetic energy of the other particles.Effusion typically transports particles more quickly.


c) Difference between condensation and evaporation

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CondensationEvaporation
Condensation is the change from a vapor to a condensed state (solid or liquid).Evaporation is the change of a liquid to a gas.
Condensation occurs at a constant temperature.Evaporation does not occur at a constant temperature.
Condensation is a phase change regardless of the temperature.Evaporation occurs before a liquid reaches its boiling point.
Condensation happens mainly at higher altitudes.Evaporation usually takes place in low altitudes.
In the process of condensation, energy is released.In the process of evaporation, energy is consumed.

Q.3) Define the term allotropy with examples. Explain the three allotropic forms of carbon in detail.

Answer:
Allotropy
Allotropy is the property of elements to exist in two or more different crystalline forms, in the same physical state. They are known as the allotropes of these elements.
Examples:
Allotropes of oxygen are oxygen (O2) and ozone (O3).
Allotropes of sulphur are orthorhombic and monoclinic.
Allotropes of tin are white tin and grey tin.
Allotropes of Carbon
Carbon exists in crystalline allotropic form as well as non-crystalline or amorphous allotropic form.
Carbon exists in three crystalline allotropic form namely diamond, graphite and bucky balls.
Carbon also exists in non- crystalline or amorphous allotropic form such as coal,coke, charcoal, lamp black etc.
Crystalline Allotropic Form of Carbon
i) Diamond
In diamond each carbon atom is covalently bonded to four others, creating a rigid compact array. This makes diamond the hardest known substance. Diamonds are used for cutting and polishing hard surface. It is bad conductor of electricity because the valence electrons are tightly held by covalent bonds.

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Chapter 5 Physical States of Matter 9th Chemistry Notes 8

ii) Graphite
In diamond each carbon atom is covalently bonded to three other carbon atoms rather than to four atoms asin diamond.
In graphite carbon atoms are arranged in layers of hexagonal arrays. Weak bonds exist between the layers that allow them to slide over one another. This makes graphite soft. Graphite is used as electrode, lubricant in machines, and black pigment.

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Chapter 5 Physical States of Matter 9th Chemistry Notes 9

iii) Bucky Ball
In bucky ball, 40 to 100 carbon atoms are arranged in a hollow cage like structure. Carbon atoms are arranged in pentagons (five member ring) and hexagons (six member ring) just like a soccer ball. In bucky ball, the carbon atoms joined together making pentagonal, hexagonal structures.
Bucky balls are used as semiconductors, superconductors and lubricants.

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Chapter 5 Physical States of Matter 9th Chemistry Notes 10

Q.4) What are solids? Differentiate between amorphous and crystalline solids.

Answer:
In solid state of matter the particles are closely packed in a fixed pattern. In solids there occurs a strong force of attraction between the solid particles, which hold them firmly together, so that they cannot leave their position. Solid particles possess only vibrational motion. Hence, solid cannot diffuse like gases and liquids.
Difference between amorphous and crystalline solids

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Crystalline solidsAmorphous solids
A crystalline solid is a solid that is composed of orderly, repeating three-dimensional arrangement of particles.Amorphous solid does not have a well-defined arrangement of its particles.
Crystalline solids have sharp melting points.Amorphous solids do not melt at a definite temperature but gradually soften when heated.
Crystalline solids give clean cleavage.Amorphous solids give irregular cut.
Crystalline solids are anisotropic i.e. the properties like electrical conductance, refractive index, thermal expansion, etc., have different values in different directions.Amorphous solids are isotropic i.e. the properties like electrical conductance, refractive index, thermal expansion, etc., have the same values in all directions.
Examples:
Diamond, sodium chloride etc
Examples:
Glass, plastic, rubber etc

Q.5) Define Charle’s Law and verify it graphically and diagrammatically.

Answer:
Charles’s Law
Charles’s law states that , the volume is directly proportional to the absolute  temperature at constant pressure. As the temperature of the gas increases, the volume increases.
Mathematically it can be written as:
V ∝ T 
V = kcT
or    V / T = k
where ‘V’ is the volume of the gas, ‘T’ is the temperature of the gas, and ‘kc’ is a constant.
The equation above states that ratio of volume to temperature of a given mass of a gas is constant at constant pressure.
If V1 / T1 = k        then V2 / T2 = k
Where V1 = initial volume                  T1 = initial temperature
            V2 = final volume                    T2 = final temperature
As both equations have same constant, their variable have same value.
Thus, V1 / T1 = V2 / T2
Graphic representation
If the values of volume ‘V’ is plotted against temperature ‘T’, a straight line is obtained, which shows that the volume is directly proportional to the absolute temperature.

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Chapter 5 Physical States of Matter 9th Chemistry Notes 11

Diagrammatic representation
The Charles’s law can be diagrammatically represented as,

Diagrammatic representation of Charles law
Chapter 5 Physical States of Matter 9th Chemistry Notes 12

Numericals Chemistry 9 class chapter 5

Q.1) Calculate the final pressure of a sample of a gas that is changed at constant temperature to 14.3 dmfrom 7.55 dm3 at 828 torr.

Answer:
Given that
P1 = 828 torr                V1 = 7.55 dm3
P2 =?                             V2 = 14.3 dm3
Formula applied
P1V1 = P1V2
828 x 7.55 = P2 x 14.3
P2 = 6251.4 / 14.3
     = 437.160 torr

Q.2) Calculate the final volume at 302 K of a 5.41 dmsample of a gas originally at 353 K if the pressure does not change.

Answer:
Given that
V1 = 5.41 dm3                T1 = 353 K
V2 =?                              T2 = 302 K
Formula applied
V1 / T1 = V2 / T2
5.41 / 353 = V2 / 302
V2 = 0.0153 x 302
     = 4.628 dm3

Q.3) Calculate the initial volume at 0oC of a sample of gas that is changed to 731 cm3 by cooling to -14oC at constant pressure.

Answer:
Given that
V1 =?                    T1 = 00C = 0 + 273 = 273 K
V2 = 731 cm3        T2 = -14oC = -14 + 273 = 259 K
Formula applied
V1 / T1 = V2 / T2
V1 / 273 = 731 / 259
V1 = 2.822 x 273
    = 770.406 cm3

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Q.4) A sample of a gas at room temperature occupies 0.80 dm3 at 1.5 atm. What will be its volume when the pressure of the gas is raised to 2.1 atm?

Answer:
Given that
P1 = 1.5 atm                                                                V1 = 0.80 dm3
P2 = 2.1 atm                                                                V2 =?
Formula applied
P1V1 = P2V2
1.5 x 0.80 = 2.1 x V2
V2 = 1.2 / 2.1
     = 0.571 dm3

Q.5) Calculate the final volume of 319oC of a sample of gas original 5.13 dm3 at 171oC, if the pressure does not change.

Answer:
Given that
V1 = 5.13 dm3                    T1 = 319oC = 319 + 273 = 592 K
V2 =?                                 T2 = 1710C = 171 + 273 = 444 K
Formula applied
V1 / T1 = V2 / T2
5.13 / 592 = V2 / 444
V2 = 0.0086 x 444
    = 3.847 dm3  ≈  4 dm3

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