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.
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
Q.2) Differentiate between:
a. Evaporation and Boiling Point b. Effusion and Diffusion of gases c. Condensation and Evaporation
Answer: a) Difference between evaporation and boiling
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
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.
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.
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.
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
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.
Diagrammatic representation The Charles’s law can be diagrammatically represented as,
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 dm3 from 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 dm3 sample 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
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