Basic Chemistry-I State of Matter PDF

ITM Vocational University, Vadodara BASIC CHEMISTRY-I Sub code: Faculty’s Name – Batch Name – B.Sc._Batch 2023 Unit-2 Atomic structure 1 9/26/2023 ...

ITM Vocational University, Vadodara BASIC CHEMISTRY-I Sub code: Faculty’s Name – Batch Name – B.Sc._Batch 2023 Unit-2 Atomic structure 1 9/26/2023  ITM Vocational University, Vadodara Learning Outcomes To learn about various states of matter To gain insight about physical and chemical properties of matter To understand difference between elements, mixtures and compounds Atomic structure 2 9/26/2023  ITM Vocational University, Vadodara Unit – II State of Matter: Introduction and Classification of states of matter: (a) gas; (b) solid; (c) liquid; (d) plasma, physical and chemical properties of matter, substance and mixture, elements and compound, Gay Lussac’s law of gaseous volume. Atomic structure 3 9/26/2023  ITM Vocational University, Vadodara Matter Matter is anything that takes up space Atomic structure 4 9/26/2023  ITM Vocational University, Vadodara Three states of matter are: Solids Liquids Gases Atomic structure 5 9/26/2023  ITM Vocational University, Vadodara Plasma as the 4th state of Matter Plasma is formed by providing heat to the gas It is ionized gas consisting of positive ions and free electrons typically at low pressures or at very high temperatures Example: Lightning, Solar wind, Aurora, Fluroscent light, Nuclear fireball Atomic structure 6 9/26/2023  ITM Vocational University, Vadodara Atomic structure 7 9/26/2023  ITM Vocational University, Vadodara Atomic structure 8 9/26/2023  ITM Vocational University, Vadodara Atomic structure 9 9/26/2023  ITM Vocational University, Vadodara Plasma  A plasma is a gaslike mixture of + and – charged particles  A plasma is a very good conductor of electricity  ex. Fluorescent lights, stars  Plasma, like gases have an indefinite shape and an indefinite volume.  Most common state of matter in the universe. Atomic structure 10 9/26/2023  ITM Vocational University, Vadodara Plasma Particles The negatively charged electrons (yellow) are freely streaming through the positively charged ions (blue). Atomic structure 11 9/26/2023  ITM Vocational University, Vadodara Physical and Chemical properties of matter  All properties of matter are either extensive or intensive and either physical or chemical. Extensive properties, such as mass and volume, depend on the amount of matter that is being measured.  Intensive properties, such as density and color, do not depend on the amount of matter.  Both extensive and intensive properties are physical properties, which means they can be measured without changing the substance’s chemical identity.  For example, the freezing point of a substance is a physical property: when water freezes, it’s still water (H2O)—it’s just in a different physical state. Atomic structure 12 9/26/2023  ITM Vocational University, Vadodara Physical Properties Physical properties are properties that can be measured or observed without changing the chemical nature of the substance. Some examples of physical properties are:  color (intensive)  density (intensive)  volume (extensive)  mass (extensive)  boiling point (intensive): the temperature at which a substance boils  melting point (intensive): the temperature at which a substance melts Atomic structure 13 9/26/2023  ITM Vocational University, Vadodara Chemical Properties  Remember, the definition of a chemical property is that measuring that property must lead to a change in the substance’s chemical structure. Here are several examples of chemical properties:  Heat of combustion is the energy released when a compound undergoes complete combustion (burning) with oxygen. The symbol for the heat of combustion is ΔHc.  Chemical stability refers to whether a compound will react with water or air (chemically stable substances will not react). Hydrolysis and oxidation are two such reactions and are both chemical changes.  Flammability refers to whether a compound will burn when exposed to flame. Again, burning is a chemical reaction—commonly a high- temperature reaction in the presence of oxygen.  The preferred oxidation state is the lowest-energy oxidation state that a metal will undergo reactions in order to achieve (if another element is present to accept or donate electrons). Atomic structure 14 9/26/2023  ITM Vocational University, Vadodara Atomic structure 15 9/26/2023  ITM Vocational University, Vadodara 1 – Pure substance The substances that contain only one type of particle and they are free from any mixture, are known as pure substances. Gold, silver, iron, and aluminium are pure substances, to name a few. By their chemical composition, pure substances get divided into two types – elements and compounds.  Elements – An element is a pure substance which contains a single kind of atom. Such a substance cannot get broken into two or a simpler substance by physical or chemical means. For instance, when you can break down gold, what you get is gold again. Elements can be metals, non-metals, and metalloids.  Compounds – Compound is a pure substance; it contains two or more elements which get chemically combined in a predetermined proportion. For example, water is a compound as it has two elements hydrogen and oxygen, which get combined in a fixed proportion. Atomic structure 16 9/26/2023  ITM Vocational University, Vadodara 2 – Mixture  Mixtures refer to substances that contain two or more forms of matter. A solution of salt and water, sugar and water, air, different gases, etc. is examples of mixtures. Based on their compositions, mixtures can get divided into two types – homogeneous and heterogeneous mixtures.  Homogeneous Mixtures – Those mixtures having a uniform composition throughout their bodies are homogeneous mixtures. Soda water, lemonade, a mixture of salt in water are a few examples of homogeneous mixtures.  Heterogeneous Mixtures – Those mixtures which do not have a uniform composition entirely are heterogeneous mixtures. A mixture of oil and water, soil and sand are heterogeneous mixtures as they don’t have a uniform composition. Atomic structure 17 9/26/2023  ITM Vocational University, Vadodara Atomic structure 18 9/26/2023  ITM Vocational University, Vadodara Elements and Atoms Atomic structure 19 9/26/2023  ITM Vocational University, Vadodara Compounds Atomic structure 20 9/26/2023  ITM Vocational University, Vadodara Atomic structure 21 9/26/2023  ITM Vocational University, Vadodara Molecules Atomic structure 22 9/26/2023  ITM Vocational University, Vadodara Chemical Formulas Atomic structure 23 9/26/2023  ITM Vocational University, Vadodara Gay Lussac’s law of gaseous volume Gay-Lussac’s law is a gas law which states that the pressure exerted by a gas (of a given mass and kept at a constant volume) varies directly with the absolute temperature of the gas. In other words, the pressure exerted by a gas is proportional to the temperature of the gas when the mass is fixed and the volume is constant. This law was formulated by the French chemist Joseph Gay-Lussac in the year 1808. The mathematical expression of Gay-Lussac’s law can be written as follows: P ∝ T ; P/T = k Where:  P is the pressure exerted by the gas  T is the absolute temperature of the gas  k is a constant. Atomic structure 24 9/26/2023  ITM Vocational University, Vadodara The relationship between the pressure and absolute temperature of a given mass of gas (at constant volume) can be illustrated graphically as follows. Atomic structure 25 9/26/2023  ITM Vocational University, Vadodara From the graph, it can be understood that the pressure of a gas (kept at constant volume) reduces constantly as it is cooled until the gas eventually undergoes condensation and becomes a liquid. Atomic structure 26 9/26/2023  ITM Vocational University, Vadodara Gay-Lussac’s law implies that the ratio of the initial pressure and temperature is equal to the ratio of the final pressure and temperature for a gas of a fixed mass kept at a constant volume. This formula can be expressed as follows: (P1/T1) = (P2/T2) Where:  P1 is the initial pressure  T1 is the initial temperature  P2 is the final pressure  T2 is the final temperature Atomic structure 27 9/26/2023  ITM Vocational University, Vadodara This expression can be derived from the pressure-temperature proportionality for gas. Since P ∝ T for gases of fixed mass kept at constant volume: P1/T1 = k (initial pressure/ initial temperature = constant) P2/T2 = k (final pressure/ final temperature = constant) Therefore, P1/T1 = P2/T2 = k Or, P1T2 = P2T1 Atomic structure 28 9/26/2023  ITM Vocational University, Vadodara Examples of Gay-Lussac’s Law When a pressurized aerosol can (such as a deodorant can or a spray-paint can) is heated, the resulting increase in the pressure exerted by the gases on the container (owing to Gay-Lussac’s law) can result in an explosion. This is the reason why many pressurized containers have warning labels stating that the container must be kept away from fire and stored in a cool environment. Atomic structure 29 9/26/2023  ITM Vocational University, Vadodara Atomic structure 30 9/26/2023  ITM Vocational University, Vadodara An illustration describing the increase in pressure which accompanies an increase in the absolute temperature of a gas kept at a constant volume is provided above. Another example of Gay-Lussac’s law can be observed in pressure cookers. When the cooker is heated, the pressure exerted by the steam inside the container increases. The high temperature and pressure inside the container cause the food to cook faster. Examples of Gay-Lussac’s Law Atomic structure 31 9/26/2023  ITM Vocational University, Vadodara Exercise 1 The pressure of a gas in a cylinder when it is heated to a temperature of 250K is 1.5 atm. What was the initial temperature of the gas if its initial pressure was 1 atm.  Given, Initial pressure, P1 = 1 atm Final pressure, P2 = 1.5 atm Final temperature, T2 = 250 K As per Gay-Lussac’s Law, P1T2 = P2T1 Therefore, T1 = (P1T2)/P2 = (1*250)/(1.5) = 166.66 Kelvin. Atomic structure 32 9/26/2023  ITM Vocational University, Vadodara Question: At a temperature of 300 K, the pressure of the gas in a deodorant can is 3 atm. Calculate the pressure of the gas when it is heated to 900 K.  Initial pressure, P1 = 3 atm Initial temperature, T1 = 300K Final temperature, T2 = 900 K Therefore, final pressure (P2) = (P1T2)/T1 = (3 atm*900K)/300K = 9 atm. Atomic structure 33 9/26/2023  ITM Vocational University, Vadodara Atomic structure 34 9/26/2023