lectures_2-3 - 1/7/2010 Thermodynamics We will review: Some...

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Unformatted text preview: 1/7/2010 Thermodynamics We will review: Some Thermodynamic Terms Internal Energy: a state function heat and work The First Law of Thermodynamics Absolute Internal Energy of an ideal gas: E=3/2 R.T Isothermic and adiabatic expansion/compression of an ideal gas Enthalpy: a state function Internal Energy and Enthalpy Changes Standard States and Standard Enthalpy Changes Heat capacities and heat flow Definitions: 2 1) System - The part of the universe a System scientist is interested in. 2) Surroundings - Everything else in the Surroundings universe that is outside of the system. 3) Boundary - A real or imaginary barrier between Boundary real or imaginary barrier between the system and its surroundings through which THERMAL ENERGY may flow, work may be done by the system or on the system, and matter may or may not be exchanged. 4) Universe - The system plus the surroundings is called the universe 1 1/7/2010 Closed System - A system that does not exchange matter with its surroundings but can exchange both heat and work Open System - A system that may exchange THERMAL ENERGY and MATTER with its surroundings. Isolated System - A system that does not exchange MATTER or THERMAL ENERGY with its surroundings. Energy - The ability to do work. Potential Energy – Stored energy that is related to an object relative position. examples: gravitational, electrical, chemical Kinetic Energy The energy of Kinetic Energy - The energy of motion. K.E. = ½ mv2 Units of energy (MKS system) K.E. = ½(2 kg)(1 m/s)2 = 1kg1kgm2/s2 = 1 Joule Example gravitational potential energy: A body of mass m at a height h over the earth’s over surface is attracted towards de center of the earth with a gravitational force equal to its weight W : Weight Weight = m. g m where g is the gravity constant g= 9.8 m/s2 and has a Potential Energy P.E.: P.E. P.E. = Weight x h = m .g. h W h Check that the Unit of P.E. (MKS) is the Joule (MKS) is P.E. = m .g. h = (1 / 9.8) Kg. (9.8) m/s2.. 1m m/s = 1 Joule = 1 Kg . m2 / s2 2 1/7/2010 A Process in which potential energy is transferred. A and B are identical balls A was released and allowed to strike B. Why did B end up with a lower potential energy than that lost by A? What happened in this process step by step? Combustion of methane releases chemical potential energy as heat Energy diagram for the reaction of nitrogen and oxygen 3 1/7/2010 Types of energy a) Thermal b) Electrical c) Radiant Radiant d) Chemical e) Mechanical f) Nuclear Temperature, Thermal Energy, Heat Temperature - An indirect measure of the average kinetic energy of the molecules, atoms, or ions in the material; A measure of the “hotness” or “coldness” of a material; an INTENSIVE property of matter. Thermal Energy Thermal Energy - The energy that is transferred energy that is transferred from hotter objects to colder objects due to the kinetic energy of the molecules, atoms, or ions; an EXTENSIVE property of matter. Heat - The transfer of thermal energy that results from a difference in temperature; the process of transferring thermal energy from hotter objects to colder objects. Physical Definition of Work Work = Force x distance m.g h work = w = F.d work = F.d = m g h = P.E When the object falls the force of gravity performs work on it equal to its initial gravitational potential energy A force performs work only if the body on which it acts moves moves!! 4 1/7/2010 The First Law of Thermodynamics First Energy is neither created nor destroyed. The energy of the universe is constant Δ E = Efinal - Einitial ΔEuniverse = ΔEsystem + ΔEsurroundings Other Units of Energy Joule (J) 1 J = 1 kg*m2/s2 Calorie (cal) 1 cal = 4.18J British Thermal Unit 1 Btu = 1055 J The First Law of Thermodynamics First ΔE = q + w Energy can be transferred as heat or as work Example of a system transferring energy as heat only. A piston moving a distance performs work (PV work) 5 1/7/2010 Pressure-volume work. F/A W = - Pext ΔV Is this expression equal to w = F . d ? Pressure = Force/ Area = F/A Δ h Volume = Area x h ΔV= Area x Δh = A . Δh Pext . ΔV W = - (F/A) . A. Δh Area cancels W = - F . Δh W= - Pext ΔV If Thermal Energy is given up by the system, it must be absorbed by the surroundings. This type of change is EXOTHERMIC type of change is EXOTHERMIC. If Thermal Energy is absorbed by the system must come from the surroundings. This type of the change is ENDOTHERMIC. Two different paths for the energy change of a system. The internal energy is a state function (independent of the path) Neither heat nor work are state functions, however the change in internal energy which is the sum of work and heat exchanged is a state function! 6 1/7/2010 A system losing energy as work only. Sign Conventions* for q, w and ΔE q + w = ΔE + + - + + - + depends on sizes of q and w depends on sizes of q and w - * For q: + means system gains heat; - means system loses heat. * For w: + means word done on system; - means work done by system. Internal Energy, E , The total energy of a system. Components of Internal Energy Contributions to the kinetic energy: •The molecule moving through space, Ek(translation) • The molecule rotating, Ek(rotation) • The bound atoms vibrating, Ek(vibration) • The electrons moving within each atom, Ek(electron) Contributions to the potential energy: • Forces between the bound atoms vibrating, Ep(vibration) • Forces between nucleus and electrons and between electrons in each atom, Ep(atom) • Forces between the protons and neutrons in each nucleus, Ep(nuclei) • Forces between nuclei and shared electron pair in each bond, Ep(bond) 7 1/7/2010 Components of internal energy (E) The total internal energy of one mol of ideal gas is: E = 3/2 R.T For more than one mol: E = 3/2 n. R. T Note that If the temperature does not change (isothermic process) the total internal energy does not change. For one molecule: E= 3/2 R/Na . T Na = Avogadro’s # = 6.02 x 1023 The gas constant for one molecule = R/Na = k (the Boltzman constant) R/N therefore the internal energy of one molecule of ideal gas: E= 3/2 . k . T Ideal gas expansion at constant temperature ΔE=q+w If Δ T = 0 ΔE=0 w = - Pext . ΔV q This process can be performed at constant temperature (isothermic) if heat is provided so slowly that the system is allowed to loose as much internal energy, as work on the surroundings, as internal energy is gained as heat from the surroundings. 8 1/7/2010 The isothermal expansion of an ideal gas The temperature does not change, therefore the internal energy must remain constant and Δ E = 0 ΔE=q+w=0 q=-w Exactly as much work is done on the surroundings as heat is provided by the surroundings. This is a great machine : it converts heat into work with 100% efficiency!!!!! ……. Well… it is only an ideal machine Adiabatic Condition (either thermal insulation or a change so fast that heat exchange has no time to occur): Adiabatic means no thermal exchange between system and surroundings , therefore q=0 By definition an isolated system is adiabatic. In an adiabatic expansion or compression: ΔE = w In an adiabatic expansion, work is performed on the surroundings. Therefore the internal energy of the system must decrease. If the system is an ideal gas a decrease in E must be accompanied by a decrease in temperature. This is the process used in refrigeration where a compressed gas (freon in the past) is forced to expand rapidly against atmospheric pressure. The opposite occurs in an adiabatic compression! Examples Examples of work and heat exchanges in ideal gases Already covered: •Isothermal expansion/compression of an ideal gas. ΔE = 0 q = -w ΔE = w •Adiabatic expansion/compression of an ideal gas q=0 Two more cases: •Heating ideal gases at constant volume The Molar heat capacity at constant volume : CV= 3/2 R •Heating ideal gases at constant pressure The molar heat capacity at constant Pressure: CP= 5/2 R 9 1/7/2010 Heat Capacity: the amount of thermal energy exchange (heat), q, necessary to change the temperature of a given mass, m, by one degree Kelvin or Celsius. Specific heat capacity, c : is the heat necessary to change the temperaure of 1 g of substance by 1 degree C (or K)* c (Joule/g.K) = q (Joule) / Mass (g) . ΔT (K) Therefore the amount of heat can be calculated: q (Joule) = c (Joule/g.K) . Mass (g) . ΔT (K) Molar heat capacity,C : is the heat necessary to change the temperature of one mole of substance( by 1 degree C (or K) Using Molar heat capacities the heat exchanged is: q (Joule) = C (Joule/mol.K) . n (mol) . ΔT (K) (ΔT in 0C is the same as for K). Heating (cooling) an ideal gas at constant volume: An ideal gas not allowed to expand (or contract) as it is heated (cooled) it is said that is undergoing a heat exchange at constant volume. This way ALL the heat is used to increase (decrease) its internal energy (no work on the surroundings is allowed). ΔE = q + w = q + (-Pext . ΔV) At constant volume ΔV=0 and q = qV qV = Δ E The heat exchanged at constant volume equals the change in internal energy of the ideal gas If the initial internal energy of one mol of ideal gas is: Ei = 3/2 R.Ti And its temperature changes to Tf due to a heat exchange with the surroundings, its internal energy will change to its final value: Ef = 3/2 R. Tf The change in internal energy: Ef – Ei = 3/2 R (Tf – Ti) In simpler notation: simpler notation: Δ E = 3/2 R . ΔT and Δ E/ ΔT = 3/2 R 3/2 and 3/2 but qV = Δ E at constant volume The Molar heat capacity at constant volume: CV (How much heat needs to be exchanged at constant volume to change the temperature of one Mole of ideal gas by 1 K) CV = qV / ΔT = Δ E / ΔT = 3/2 R CV = 3/2 R 10 1/7/2010 Enthalpy definition: H = E + P.V Because Because E, P, and V are state functions, H must be a state function The enthalpy change: ΔH = Hfinal - Hinitial In general ΔH is: Δ H = Δ E + P. Δ V + V . Δ P If neither P or V are constant 0 At constant P : Δ H = Δ E + P. Δ V + V . Δ P Δ H = Δ E + P. Δ V The Meaning of Enthalpy Typically we carry out processes at constant pressure Therefore the work is w = - PΔV Cases when The enthalpy change is: ΔH is nearly equal to ΔE: ΔH = ΔE + PΔV ΔH = qp + w + PΔV ΔH = qp + (- PΔV) + PΔV qp = Δ H 1. Reactions that do not involve gases. 2. Reactions in which the number of moles of gas does not change moles of gas does not change. 3. Reactions in which the number of moles of gas does change but q is >>> PΔV. The enthalpy change is the thermal energy gained or lost by a system when the system undergoes a change under constant pressure. pressure. Heating (cooling) an ideal gas at constant pressure: •(allowed to expand (or contract) and therefore to exchange both heat and work with the surroundings) We have just shown that the heat exchanged at constant pressure is: q p = CP Δ T = Δ H Therefore: and ΔH = ΔE + P ΔV So, P ΔV = R ΔT qP = CP ΔT = ΔE + P ΔV for one mole of ideal gas P V = RT q P = CP ΔT = ΔE + R ΔT divide by ΔT CP = qP/ ΔT = (ΔE + R ΔT ) /ΔT = ΔE / ΔT + R ΔT/ ΔT CP = ΔE / ΔT + R = CV + R = 3/2 R + R = 5/2 R The ideal gas molar heat capacities CV = 3/2 R CP = 5/2 R CP – CV = R 11 ...
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This note was uploaded on 03/10/2010 for the course CHEM 7650 taught by Professor Bogo during the Spring '10 term at UCSC.

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