Thermodynamics pt. 1.1 PDF

CHEM 217 INTRODUCTION TO THERMODYNAMICS 1  SYSTEM, BOUNDARY, SURROUNDINGS A system is that part of the universe which is under thermodynamic study and the rest of the universe is surroundings. The real or imaginary surface separating the system from the surroun...

CHEM 217 INTRODUCTION TO THERMODYNAMICS 1  SYSTEM, BOUNDARY, SURROUNDINGS A system is that part of the universe which is under thermodynamic study and the rest of the universe is surroundings. The real or imaginary surface separating the system from the surroundings is called the boundary 2  TYPES OF THERMODYNAMIC SYSTEMS ✓ An isolated system is one that cannot transfer neither matter nor energy to and from its surroundings ✓ A closed system is one which cannot transfer matter but can transfer energy in the form of heat, work and radiation to and from its surroundings. ✓ An open system is one which can transfer both energy and matter to and from its surroundings 3  Homogeneous and Heterogeneous Systems ✓Homogeneous System. When a system is uniform throughout Examples are : a pure single solid, liquid or gas, -made of one phase only. ✓Heterogeneous system is one which consists of two or more phases. It is not uniform throughout. Example: ice in contact with water. ✓A phase is defined as a homogeneous, physically distinct and mechanically separable portion of a system 4  Intensive and Extensive Properties ❖Intensive Properties A property which does not depend on the quantity of matter present in the system Examples: pressure, temperature, density, and concentration ❖Extensive Properties A property that does depend on the quantity of matter present in the system Examples: volume, number of moles, enthalpy, entropy, and 5 Gibbs’ free energy  State of a System A thermodynamic system is said to be in a certain state when all its properties are fixed. The fundamental macroscopic properties which determine the state of a system are pressure (P), temperature (T), volume (V), mass and composition. 6  Thermodynamics The study of the flow of heat or any other form of energy into or out of a system as it undergoes a physical or chemical transformation First law of thermodynamics: Energy of the universe is constant – Known as the law of conservation of energy – Energy cannot be created or destroyed; it can only be converted from on form to another. Thermochemistry is the study of heat changes associated with chemical reactions. 7  Work and Energy Energy is the capacity of a system to do work (and ultimately, to lift a weight). If a system can do a lot of work, we say that it possesses a lot of energy the total store of energy in a system is called its internal energy, U or E. We cannot measure the absolute value of the internal energy of a system, because that value includes the energies of all the atoms, their electrons, and the components of their nuclei. Work is the transfer of energy to a system by a process that is equivalent to raising or lowering a weight. For work done on a system, w is positive, for work done by a system w is negative. The internal energy of a system maybe changed by doing work: When energy is transferred to a system by doing work and in the absence of other changes, ∆U = w. 8  The First Law of Thermodynamics The total internal energy of an isolated system is constant. if an isolated system has a certain internal energy at one instant and we inspect it again later, then we shall find that it still has exactly the same internal energy, no matter how much time has passed. In practice, is not to truly isolate a chemical reaction from its surroundings. Thus is important to measure the energy that leaves or enters a system. Therefore we measure change in the internal energy(E or U) of system. ΔE = E final - E Initial or ΔE = EProducts - EReactants ΔESystem + ΔESurrounding = 0 ΔESystem = - ΔESurrounding 9  The First Law of Thermodynamics heat transfer in heat transfer out (endothermic), +q (exothermic), -q SYSTEM ∆E = q + w w transfer in w transfer out (+w) (-w) 10  Work and Energy The work required to move an object a certain distance against an opposing force is calculated by multiplying the opposing force by the distance moved against it Work = opposing force X distance moved Unit is the joule, J, with 1 J = 1 kgm2s-2 11  EXPANSION WORK Types of work associated with a chemical process – Work done by a gas through expansion – Work done to a gas through compression Example - Motion of a car – In an automobile engine, heat from the combustion of gasoline expands the gases in the cylinder to push back the piston, and this results in the motion of the car 12  Deriving the Equation for Work Consider a gas confined to a cylindrical container with a movable piston – F is the force acting on the piston of area A – Pressure is defined as force per unit area F P= A 13  Deriving the Equation for Work – Consider that the piston moves a distance of Δh Work = force  distance = F  h Work = F  h = P  A  h – Volume of the cylinder equals the area of the piston times the height of the cylinder Change in volume ΔV resulting from the piston moving a distance Δh is: V = final volume − initial volume = A  h 14  Deriving the Equation for Work (Continued 2) – Substitute the expression derived for ΔV into the expression for work Work = P  A  h = PV Since the system is doing work on the surroundings, the sign of work should be negative – ΔV is a positive quantity since volume is increasing – Therefore, w = − PV 15  Deriving the Equation for Work For a gas expanding against an external pressure P, w is a negative quantity as required – Work flows out of the system When a gas is compressed, ΔV is a negative quantity (the volume decreases) – This makes w a positive quantity (work flows into the system) 16  Calculating the work of reversible isothermal expansion of gas dw = - PΔV At each stage of the expansion pressure is related to volume by the ideal gas law. PV = nRT = P = nRT/V 𝑛𝑅𝑇 dw = - ⅆ𝑣 𝑣 w = -nRT ln (Vfinal / Vinitial ) 17  Examples Calculate the work associated with the expansion of a gas from 46 L to 64 L at a constant external pressure of 15 atm Ans: w = − PV ΔV = 64 L – 46 L = 18 L w = -15 atm x 18 L = -270 L.atm 1L.atm = 101.325J -270 L.atm = 270 x 101.325 = -27,357.75J = -27.4kJ 18 Calculating the work done when a gas expands Suppose a gas expands by 500. mL (0.500 L) against a pressure of 1.20 atm and no heat is exchanged with the surroundings during the expansion. (a) How much work is done in the expansion? (b) What is the change in internal energy of the system? Ans: w = -PΔV = - 1.20 atm x 0.500 L = -0.600 L.atm = 1 L.atm = 101.325J 0.600 =.600 x 101.325J = -60.8J (B) No heat ⸫ w = ΔU 19 ΔU = -60.8J  Calculating the work of isothermal expansion of gas A piston confines 0.100 mol Ar(g) in a volume of 1.00 L at 25°C. Two experiments are performed. (a) The gas is allowed to expand through an additional 1.00 L against a constant pressure of 1.00 atm. (b) The gas is allowed to expand reversibly and isothermally to the same final volume. Which process does more work? Ans: (a) Irreversible path w = - PΔV = - 1.00 atm x 1.00 L = 1.00 L.atm = -101.3 J (b) Reversible w = -nRT ln (Vfinal - Vinitial ) -1 -1 w = - 0.100 mol x 8.3145J.K mol x 298K x ln (2/1) = -172J Thus, gas does more work in the reversible 20