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Fundamentals of Physical Chemistry Dr. Jaroslava Miksovska CP 328 [email protected] Textbook Lecture 1 Introduction Aim of the lecture: 1) Physical state of matter 2) Properties of matter 3) Properties of gases (ideal gas) • Physical chemistry A) physical properties of matter (composition, structure, energy) B) basis for modern methods of analysis Energy capture and storage Ligand binding to proteins Structure of proteins • States of matter: A) gas B) liquid Macroscopic Random Fill container movement No strong intermolecul ar forces Well defined surface C) solid Microscopic Particles are in contact, move in Fixed volume restricted manner Well Particles defined trapped in a shape single Fixed position and volume oscillate Physical state and its characterization - Properties of matter in terms of state (gas, liquid solid ), volume, temperature, pressure and amount Mass – measure of quantity ( 1 kg = 103 g) Volume – amount of space the sample occupies (liters (L), cm3) Temperature – property that determine the direction of flow of energy (Kelvin, oC), T/K = θ/oC + 273.15 Objects are in thermal equilibrium if there is no flow of energy between them ZERO LAW OF THERMODYNAMICS if object A is in thermal equilibrium with object B, and object B is in thermal equilibrium with object C, then object C is also in thermal equilibrium with object A. • Chemical amount ( # of moles) – amount of substance (mol) n = m/M (m- mass, M – molar mass) n = N/NA ( N – number of particles, NA – Avogadro constant; NA = 6.022x1023 mol-1 for H2O (M= 18 g mol-1) the mass of one mole is 18 g and the mass of one molecule is (18 g)/(6.02 × 1023) ≈ 3 × 10−23 g. The Avogadro constant NA is related to the Faraday constant. The Faraday constant F is the absolute value of the charge on one mole of electrons. Therefore F = NAe, where e is the absolute value of the charge on one electron. Each year in America, from 6:02 AM to 6:02 PM on 10/23, the mole is celebrated by chemists, mathematicians and physicists. • Concentration Molar concentration [J] = cJ = nJ/VJ (mol/L, M) (VJ – volume of solution = Vsolute + Vsolvent) Molality bJ = nJ/msolvent (mol/kg) Molar fraction: fJ= nJ/n (unitless) Force (F) Mercury barometer F = ma ( m= mass; a = acceleration); vector quantity Units: 1 N = 1 kg ms-2 Pressure (p) p = force per area; p = F/A (A – area) Units: kg m-1 s-2 1 Pa = 1 Nm-2 = 1 kg m-1 s-2 1 bar = 105 Pa 1 atm = 101.325 kPa = 1.013 25 bar 760 Torr = 1 atm 1 Torr ~ Hg Mercury barometer p = gh (-density; h – height, g – acceleration of free fall – 9.81 ms-2) • Energy – capacity to do work 1 J = kg m2 s-2 a) kinetic energy – energy of a body due to the motion Ek = ½ mv2 (m – mass, v – speed) b) potential energy – energy of a body due to its position; depend on the forces acting on a body in the gravitation field: Ep = mgh for change particles Ep = Q1Q2/(4or) Q – charge; o – vacuum permitivity (8.854 x 10-12 J-1C2m-1) r- distance between particles radiation energy E = Nh (h – Planc constant = 6.6252x10- 34Js; - frequency) E = Ek + Ep (energy can be converted between different form but total energy remind constant) Energy conversion and energy exchange Solar energy is: absorbed by desert; water and converted to heat absorbed by solar panels and converted to electric energy (use to do work) absorbed by plants and converted to chemical energy (chemical bond) Thermodynamics deal with interchanges between different form of energy OPEN SYSTEM Open systems can exchange both matter and energy with an outside system. CLOSE SYSTEM Closed systems exchange energy but not matter with an outside system. ISOLATED SYSTEM Isolated systems can exchange neither energy nor matter with an outside system. Equation of state State of matter is characterized by: Volume, Temperature, Pressure and amount of substance (n) These parameters are NOT independent p = f(V, n, p, T) Equation of state - equation that describes the state of matter Perfect gas law: pV = nRT R – gas constant R = 0.08206 atm L K-1 mol-1 R = 8.314 J K-1 mol-1 Boyles law V 1/P (at constant T) Charles and Gay-Lussac law V 1/T (at constant p) Avogadro’s law V n (at constant P and T) p1V1/T1 = p2V2/T2 Combined gas equation Perfect gas: consisting of identical particles Particles: occupy negligible volume undergo perfect elastic collision with each other no intermolecular forces no intramolecular storage of energy energy consist of kinetic energy t = -273.15 oC = 0 K Charles law V p1 Absolute zero The lowest p3 attainable temperature p2 SATP – standard ambient p & T p = 1 barr and T = 298 K STP - standard p & T p = 1 atm T = 273 K Extensive properties: properties whose values depend on amount of substance Intensive properties: values does not depend on the amount of substance Molar volume (Vm) Vm = V/n = RT/p t θ (oC) • Dalton law of partial pressure Total pressure of a system composed of two or more gases the total pressure correspond to the sum of individual pressures each gas would exert if it occupies the same volume pT = p1+p2+p3+... = pi Partial pressure of i-th component can be expressed as: p i = x i pT See the derivation • Real gases Perfect gas law – holds for real gases at low p (p<10 atm) and high T At high pressure – deviation from ideal behavior For a perfect gas: pVm = RT and thus pVm/RT = 1 Real gases: Compressibility factor Z Z 1 Z = pVm/RT = Vm/Vm with Vom = molar volume of perfect gas Z > 1 Vm of gas is larger than Vm of perfect gas due to the repulsion between molecules Z < 1 attractive forces are dominant • Equation of state for real gas Z = 1 + B/Vm + C/Vm2 + …. B – second virial coefficient C- third virial coefficient (Virial: comes from that Latin word vis, viris, meaning force) Determined experimentally from the plot of p V T plots for real gases pVm/RT = 1 + B/Vm + C/Vm2 +… p= RT/Vm (1 + B/Vm + C/Vm2 +…) p = nRT/V (1 + B/Vm + C/Vm2 +…) B/Vm >> C/V2m Virial equation of state • Van der Waal equation of state p= nRT/(V-nb) – a(n/V)2 Gas a (Pa m3) b(m3/mol) nb term - is the very small approximate volume occupied by the molecules themselves Helium 3.46 x 10-3 23.71 x 10-6 Neon 2.12 x 10-2 17.10 x 10-6 a(n/V)2 term - a is a positive proportionality constant, taking into account that the pressure is reduced due to the attractive forces Hydrogen 2.45 x 10-2 26.61 x 10-6 (p + a(n/V)2)(V-nb) ) = nRT Van der Waals equation is valid over the wider range of volume and pressure Provides molecular interpretation of equation of state Carbon dioxide Water vapor 3.96 x 10-1 42.69 x 10-6 5.47 x 10-1 30.52 x 10-6 Value of a parameter correlates with boiling point • Homework 1. At STP 0.28 L of a gas weights 0.400 g. Calculate the molar mass of the gas. 2. Dissolving 3.00 g of an impure sample of CaCO3 in an excess of HCl produces 0.656 L of CO2 (measured at 20 oC and 792 mm Hg). Calculate the percent by mass of CaCO3 in the sample. 3. Nitrogen forms several gaseous oxides. One of them has a density of 1.27 g L-1 measured at 764 mm Hg and 150 oC. Write the formula of the compound. 4. Nitrogen dioxide can not be obtained in a pure form in the gas phase because it exists as a mixture of NO2 and N2O4. At 25 oC and 0.98 atm, the density of this gas mixture is 2.7 gL-1. What is the partial pressure of each gas. 5. Two bulbs of volumes Va and Vb are connected by a stopcock. The number of mole of gases are na and nb, and initially the gases are at the same pressure, P, and temperature, T. Show that the final pressure of the system, after the stopcock has been open is equal P. Assume idea-gas behavior. 6. A flask contains a mixture of two ideal gases, A and B. Show graphically how the total pressure of the system depends on the amount of A present; that is plot the total pressure versus the mole fraction of A. Do the same for B on the same graph. The total number of moles of A and B is constant. 7. Problems from the textbook: 1.3 (not included); 1.6 (1.10); 1.8 (1.12); 1.30; (1.35) 1.31 (1.36);.32 (1.37), 1.33 (1.39)