The Gaseous State PDF

The Gaseous State by dr Manal ALSoub  Gas State Owing to the vigorous and rapid motion of the gas molecules, they travel in random paths and collide with each other and with the wall of the container, therefore they exert pressure a force per unit a...

The Gaseous State by dr Manal ALSoub  Gas State Owing to the vigorous and rapid motion of the gas molecules, they travel in random paths and collide with each other and with the wall of the container, therefore they exert pressure a force per unit area. So it is measured in dyne/cm2, atm, or mm (cm) of mercury (mm Hg). (1 atm ≡ 760 mmHg ≡ 1.0133*106 dynes/cm2).  Gas State = 760 mmHg Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 5–3  Gas state A gas uniformly fills any container and assumes its shape (volume). Its volume (V) is usually expressed in liters or mL=cm3).  Gas is the only state that is compressible.  A gas mixes completely with any other gas.  The Empirical Gas Laws Avogadro’s Law: Equal volumes of any two gases at the same temperature and 22.4 L/mol pressure contain the same number of molecules. The volume of one mole of gas is called the molar gas volume, Vm. Volumes of gases are often compared at standard temperature and pressure (STP), chosen to be 0 oC and 1 atm pressure. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 5–5  The Empirical Gas Laws – Avogados Law – At STP, the molar volume, Vm, that is, the volume occupied by one mole of any gas, is 22.4 L/mol Vn – So, the volume of a sample of gas is directly proportional to the number of moles of gas, nGas molecules exert relatively small forces on each other (mo). Number of gas molecules by moles (n).. lecules try to act independently of one another Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 5–6  Gas state The temperature (T) is another important factor that affects the gas behavior, and is given in absolute or kelvin degrees in gas equations. K= ºC+273.15 Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 5–7  The Empirical Gas Laws Boyle’s Law: The volume of a sample of gas at a given temperature varies inversely with the applied pressure Pf Vf Pi Vi V  1/P (constant moles and T) Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 5–8  The Empirical Gas Laws Charles’s Law: The volume occupied by any sample of gas at constant pressure is directly proportional to its absolute temperature. V  Tabs (constant moles and P) or Vf Vi Tf  Ti Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 5–9  The Empirical Gas Laws Gay-Lussac’s Law: The pressure exerted by a gas at constant volume is directly proportional to its absolute temperature. P  Tabs (constant moles and V) or Pf Pi Tf Ti Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 5–10  The Empirical Gas Laws Combined Gas Law: In the event that all three parameters, P, V, and T, are changing, their combined relationship is defined as follows: Pi Vi Pf Vf Ti  Tf Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 5–11  The Ideal Gas Law From the empirical gas laws, we See that volume varies in proportion to pressure, absolute temperature, and moles. V  1/P Boyle' s Law V  Tabs Charles' Law Vn Avogadro' s Law The Ideal Gas Law PVα nT  The Ideal Gas Law This implies that there must exist a proportionality constant governing these relationships. – Combining the three proportionalities, we can obtain the following relationship. V " R" ( nTabs P ) where “R” is the proportionality constant referred to as the ideal gas constant. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 5–13  The Ideal Gas Law The numerical value of R can be derived using Avogadro’s law, which states that one mole of any gas at STP (i.e. at 0ºC (273.15 K) and 760 mmHg (1 atm) found by experiment to be 22.4 liters. R  nTVP (22.4 L)(1.00 atm) R  (1.00 mol)(273 K) 0.0821 molK Latm Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 5–14  The Ideal Gas Law Thus, the ideal gas equation, is usually expressed in the following form: PV nRT P is pressure (in atm) V is volume (in liters) n is number of atoms (in moles) T is temperature (in Kelvin) R is universal gas constant 0.0821 L.atm/K.mol Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 5–15  A Problem to Consider An experiment calls for 3.50 moles of chlorine, Cl2. What volume would this be if the gas volume is measured at 34°C and 2.45 atm? Gas behave idealy since V nRT P (3.50 mol)(0.0821 mol K )(307 K) then V Latm  2.45 atm then V 36.0 L Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 5–16  Molecular Weight Determination This Eq. showed the relationship between moles and mass. moles  mass molecular mass or n m M m Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 5–17  Molecular Weight Determination If we substitute this in the ideal gas equation, we obtain PV ( M m )RT m If we solve this equation for the molecular mass, we obtain mRT Mm  PV Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 5–18  A Problem to Consider A 15.5 gram sample of an unknown gas occupied a volume of 5.75 L at 25°C and a pressure of 1.08 atm. Calculate its molecular mass. mRT Since Mm  PV (15.5 g)(0.0821 mol K )(298 K) Latm then Mm  (1.08 atm)(5.75 L) M m 61.1 g/mol Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 5–19  Density Determination If we look again at our derivation of the molecular mass equation, PV ( M m )RT m we can solve for m/V, which represents density. m PM m D  V RT Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 5–20  A Problem to Consider Calculate the density of ozone, O3 (Mm = 48.0g/mol), at 50°C and 1.75 atm of pressure. PM m Since D  RT (1.75 atm)(48.0 g/mol) then D  (0.0821 mol K )(323 K) Latm D 3.17 g/L Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 5–21  Kinetic-Molecular Theory A simple model based on the actions of individual atoms or molecules Volume of particles is negligible Particles are in constant motion No inherent attractive or repulsive forces Collisions are elastic, and thus no energy is exchanged upon collisions The average kinetic energy of a collection of particles is proportional to the temperature (K) Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 5–22  Kinetic-Molecular Theory A simple model based on the actions of individual atoms or molecules The kinetics molecular theory was developed to explain the behavior of gases and to lend additional support to the validity of the ideal gas law. Some of the important statements include: 1- Gases are composed of particles called atoms or molecules, the total volume is so small as to be negligible in relation to the volume of the space in which the molecules are confined. This condition is approximated in actual gases only at low pressures and high temperatures, in which case the molecules of the gas are far apart. 2- The particles of the gas do not attract one another but rather quickly move with complete independence; again, this statement only applies at low pressures.  Kinetic-Molecular Theory A simple model based on the actions of individual atoms or molecules 3- The particles exhibit continuous random motion owing to their kinetic energy (the average kinetic energy is directly proportional to the absolute temperature of the gas). 4- The molecules exhibit perfect elasticity, that is, there is no net loss of speed or transfer of energy after they collide with one another and with the molecules in the walls of the confining vessel, which latter effect accounts for the gas pressure. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 5–24  Real gases. Real gases are not composed of infinitely small and perfectly elastic nonattracting spheres. Instead, they are composed of molecules of finite volume that tend to attract one another. These factors affect the volume and pressure terms in the ideal equation. Van der Waals made some changes in the ideal gas law to account for the behaviour of the real gas. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 5–25  Real Gases Real gases do not follow PV = nRT perfectly. The van der Waals equation corrects for the nonideal nature of real gases. (P  2 )( V n 2a V - nb) nRT a corrects for interaction between atoms. b corrects for volume occupied by atoms. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 5–26  Real Gases In the van der Waals equation, V becomes ( V - nb) where “nb” represents the volume occupied by “n” moles of molecules. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 5–27  Real Gases Also, in the van der Waals equation, P becomes ( P  2 ) n 2a V where “n2a/V2” represents the effect on pressure to intermolecular attractions or repulsions. Tables gives values of van der Waals constants for various gases. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 5–28  A Problem to Consider If sulfur dioxide were an “ideal” gas, the pressure at 0°C exerted by 1.000 mol occupying 22.41 L would be 1.000 atm. Use the van der Waals equation to estimate the “real” pressure. Table lists the following values for SO2 a = 6.865 L2.atm/mol2 b = 0.05679 L/mol Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 5–29  A Problem to Consider First, let’s rearrange the van der Waals equation to solve for pressure. 2 nRT n a P - 2 V - nb V R= 0.0821 L. atm/mol. K T = 273.2 K V = 22.41 L a = 6.865 L2.atm/mol2 b = 0.05679 L/mol Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 5–30  A Problem to Consider 2 nRT n a P - 2 V - nb V 2 (1.000 mol)(0.082 06 mol )( 273. 2 K ) (1.000 mol) (6.865 L atm ) 2 Latm P K - mol 2 22.41 L - (1.000 mol)(0.056 79 L/mol) ( 22.41 L) 2 P 0.989 atm The “real” pressure exerted by 1.00 mol of SO2 at STP is slightly less than the “ideal” pressure. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 5–31  Real gases It is observed from the equation that a/V2 & b become less significant at large volumes (V) at low pressures or high temperatures and hence the gas behaviour comes closer to the ideal one. Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 5–32 Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 5–33 Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 5–34