Tiêu đề 1A.GASEOUS STATE FINAL DVK SR A ( 01 - 31 ).pdf ✅

NISHITH Multimedia India (Pvt.) Ltd., 1 JEE MAINS - CW - VOL - I JEE ADVANCED - VOL - I GASEOUS STATE NISHITH Multimedia India (Pvt.) Ltd., SYNOPSIS * THERMAL ENERGY Thermal energy is the energy of a body arising from motion of its atoms or molecules. It is directly proportional to the temperature of the substance. It is the measure of average kinetic energy of the particles of the matter and is thus responsible for movement of particles. This movement of particles is called thermal motion. * Intermolecular forces vs thermal energy Three states of matter are the result of balance between intermolecular forces and the thermal energy of the molecules. When molecular interactions are very weak, molecules do not cling together to make liquid or solid unless thermal energy is reduced by lowering the temperature. Predominance of thermal energy and the molecular interaction energy of a substance in three states is depicted as follows : Gas Liquid Solid (Predominance of intermolecular interactions) Gas Liquid Solid (Predominance of thermal energy) * THE GASEOUS STATE In the case of the gaseous state there are negligible attraction forces - between the molecules allowing them to move freely hence there is maximum random motion in the gaseous state as compared to other states therefore they have neither a fixed volume nor a fixed geometry. The different measurable quantities of any gas are Pressure, Volume, Temperature, and amount of the gas. The gases are characterised by: (i) Volume : The volume of a gas is the space that it occupies. They do not have fixed volume but occupy the volume of container in which they kept. It is expressed in mL (cm3 ), L(dm3 ) or m3 (1 m3 =1000L). (ii) Pressure : Gases exert pressure on the walls of the container in all directions by collision of molecules. It is defined as the force per unit area and is uniform in all the directions. It is measured in N/m2 or mm of Hg or atm or torr or bar. The device used to measure the pressure of the atmosphere is called barometer. 1 Pascal = 1 kg/m. s2 1 atm = 1.01325 x 105 Pa  100 K Pa, 1 bar = 1,00,000 Pa = 0.987 atm * Measurement of gas pressure. Pressures other than atmospheric pressure are measured b y device called manometer. There are two types of manometer: open end and closed end manometers * There are three possibilities described in pen end manometer 1) If the level of the Hg in the two limbs is same, gas pressure (Pgas ) = atmospheric pressure (Patm) P P gas atm  2) If the level of the Hg in the longer limb is higher, gas pressure (Pgas ) = (Patm) + (differenece between two two levels) P P P gas atm h   3) If the level of the Hg in the shorter limb is higher, gas pressure (Pgas ) = (Patm) - (differenece between two two levels) P P P gas atm h   * Closed end manometer generally used to measure low gas pressures. P P gas atm  * Barometic distribution: In the case of ordinary gases, pressure in the container is unaffected by the gravitational field. But in high molecular mass polymeric gases, pressure varies with height Let P0 be the pressure at ground level and P be the pressure at height ‘h’; then 10 0 2.303log P Mgh P R               GASEOUS STATE
NISHITH Multimedia India (Pvt.) Ltd., 2 JEE MAINS - CW - VOL - I JEE ADVANCED - VOL - I GASEOUS STATE NISHITH Multimedia India (Pvt.) Ltd., Similarly for density and number of moles the equation may be given as: 10 0 2.303log d Mgh d R               10 0 2.303log n Mgh n R               These relations are valid under isothermal conditions for density, pressure and number of moles. If temperature is not constant then the relations of barometric distribution may be given as 10 0 2.303log P Mgh P RT               10 0 2.303log n Mgh n RT               (iii) Temperature : It is measured in terms of hotness or coldness. The measurement is based on the expansion of certain material (mostly it is mercury) with increasing temperature. It is measured in celsius scale or absolute scale (Kelvin scale). For gases-, STP (Standard Temperature and Pressure) conditions are 273 K (0°C) and 1 atm pressure. A gas at this temperature & pressure is said to be at S.T.P. Note: that the standard temperature for gas measurements (0°C or 273 K) is not the same as that usually assumed for thermodynamic measurements (25°C or 298 K is said to be SATP condition or Standard Ambient Temperature and Pressure). K = 0C + 273.150C = 5 32 9   F      * Kelvin or absolute scale: A new temperature scale was introduced known as Kelvin scale or absolute scale named after the British physicist and mathematician Lord Kelvin. The lower limit of the scale is called absolute zero which corresponds to –273°C. At absolute zero or – 273°C, all molecular motions would stop and the volume of the gas would become zero. The gas would become a liquid or solid. Thus, absolute zero is that temperature at which no substance exists in the gaseous state. The temperature in absolute scale is always obtained by adding 273 to the temperature expressed in °C. K = (t°C + 273) The concept of absolute zero temperature can also be derived from the laws of thermodynamics. According to the first law of thermodynamics, the absolute zero of temperature is the lowest possible temperature on the thermodynamic scale and according to the second law of thermodynamics, this temperature can’t be attained. (iv) Expansibility: Gases have limitless expansibility. They expand to fill the entire vessel they are placed in and take the shape and volume of the vessel. (v) Compressibility: Gases are easily compressed by application of pressure through a movable piston fitted to the container since the molecules are very far from each other. (vi) Diffusibility: Non-reacting gases can diffuse rapidly through one another to form a homogeneous mixture. * GAS LAWS * Boyle’s Law: It relates the volume and the pressure of a given mass of a gas at constant temperature. The relationship between the volume and the pressure of a gas was studied by Robert Boyle in 1662. He found that increasing the pressure at constant temperature on a sample of a gas causes the volume of the gas to decrease proportionately, i.e., if the pressure is doubled, the volume becomes half and so on. Boyle’s law states that, “At constant temperature, the volume of a sample of a gas varies inversely with the pressure”.  1 P V  (when temperature and number of moles are kept constant) The proportionality can be changed into an equality by introducing a constant k, i.e., k P or PV k V   Alternatively, Boyle’s law can also be stated as follows:
NISHITH Multimedia India (Pvt.) Ltd., 3 JEE MAINS - CW - VOL - I JEE ADVANCED - VOL - I GASEOUS STATE NISHITH Multimedia India (Pvt.) Ltd., “Temperature remaining constant, the product of pressure and volume of a given mass of a gas is constant”. The value of the constant depends upon the amount of a gas and the temperature. Mathematically, it can be written as, P V P V P V ... 1 1 2 2 3 3    PnVn Boyle’s law can be verified by any one of the following three ways graphically. The first curve shows the variation of volume of a given mass of gas with pressure at constant temperature. The shape of the curve is rectangular hyperbola. This curve is also called isotherm. The second curve showing the relationship between volume and reciprocal of pressure is a straight line. It confirms the statement that at constant temperature, volume of a given mass of gas is inversely proportional to the pressure. The third curve shows a straight line parallel to pressure-axis. This confirms that the product of pressure and volume of a given mass of a gas at constant temperature is constant. Location of straight line and curve changes with temperature in the isotherm shown in the following figures. According to Boyle’s law, PV = Constant at constant temperature  log P + log V = constant log P = –log V + constant Illustration 1: A balloon is filled with hydrogen at room temperature. It will burst if pressure exceeds 0.2 bar. If at 1 bar pressure the gas occupies 2.27 L volume, upto what volume can the balloon be expanded. (A) 12.3L (B) 11.35 L (C) 20.5 L (D) 6.0 L Solution: (B) According to Boyle’s Law p1V1 = p2V2 If p1 is 1 bar; V1 will be 2.27 L If p2 = 0.2 bar; then 1 1 2 2 p V V p  2 1 bar 2.27L V 11.35L 0.2 bar     Since balloon bursts at 0.2 bar pressure, the volume of balloon should be less than 11.35L.
NISHITH Multimedia India (Pvt.) Ltd., 4 JEE MAINS - CW - VOL - I JEE ADVANCED - VOL - I GASEOUS STATE NISHITH Multimedia India (Pvt.) Ltd., * Charle’s Law: It relates the volume and temperature of a given mass of a gas at constant pressure. Experiments have shown that when 273 mL sample of a gas at 0°C is heated to 1°C, its volume increases by 1 mL, i.e., it becomes 274 mL. At 5°C, the volume increases to 278 mL if the pressure remains constant in both cases. Similarly, when 273 mL sample of gas at 0°C is cooled to –1°C, its volume decreases to 272 mL while at –5°C, the volume decreases to 268 mL if the pressure remains constant. Thus, all gases expand or contract by the same fraction of their volume at 0°C per degree change of temperature, i.e., for each degree change of temperature, the volume of a sample of a gas changes by the fraction 1 273 of its volume at 0°C. Let the volume of a given amount of gas be V0 at 0°C. The temperature is increased by t°C and the new volume becomes Vt Thus, 0 t 0 0 V t V V t V 1 273 273            or t 0 273 t V V 273         ... (i) Now kelvin scale may be used for deducing Charles’ law. By substituting T for (273 + t) and T0 for 273 in Eq. (i). 0 t 0 V T V T   or t 0 0 V V T T  or V T  constant, if pressure is kept constant For two or more conditions of temperature of a fixed amount of gas at constant pressure Charle’s law can be realated as: 1 2 n 1 2 n V V V = =....... =constant T T T Volume (litre) 1.0 2.0 0 -200 73 0 273 200 473 100 373 300 573 °C K Temperature -100 173 Now the Charles’ law can be stated as follows: “The volume of a given amount of a gas at constant pressure varies directly as its absolute temperature”. V T  (if pressure is kept constant) Volume Temp. ( C) 0 0 P1 P2 P3 P3>P2> P1 Charles’ law can be verified experimentally by plotting the values of volumes of a given amount of a gas under respective absolute temperature at constant pressure. The straight line confirms the above statement. Illustration 2 : On a ship sailing in pacific ocean where temperature is 23.4°C, a balloon is filled with 2 L air. What will be the volume of the balloon when the ship reaches Indian ocean, where temperature is 26.1°C ? (A) 2.018 L (B) 4.036 L (C) 8.72 L (D) 9.5 L Solution: (A) V1 = 2L T1 = (23.4 + 273 ) K = 296.4 K T2 = 26.1 + 273 = 299.1 K From Charles law 1 2 1 2 V V T T   1 2 2 1 V T V T   2 2L 299.1K V 2.018L 296.4K    * Gay lussac’s law (Pressure-Temperature Relationship): t relates the pressure and absolute temperature of a given mass of a gas at constant volume. “Volume remaining constant, the pressure of given mass of a gas increases or decreases by 1 273 of its pressure at 0°C per degree change of temperature”. 0 t 0 P t P P 273    or t 0 t P P 1 273        