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L20ThermoI145s10 - Lecture 20 First Law of Thermodynamics...

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Lecture 20, spring 2010 ENGR 145, Chemistry of Materials Case Western Reserve University Reading assignment : OGC §12.2-12.3; C&R §17.2 Learning objectives: Understand how a material can exchange energy with its surroundings: as heat as work Understand the first law of thermodynamics Understand how changes in enthalpy are related to: changes in temperature changes in state Understand how heat capacity is related to the bonding, state, and type of a material Know the role that phonons and the Debye temperature play w.r.t. heat capacity Lecture 20: First Law of Thermodynamics; Calorimetry 1
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Lecture 20, spring 2010 ENGR 145, Chemistry of Materials Case Western Reserve University add thermal energy U T separation r oscillates about eq ʼ m spacing r 0 U T k B T C&R Figs. 2.8b & 17.3 Adding Heat to a Material 2
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Lecture 20, spring 2010 ENGR 145, Chemistry of Materials Case Western Reserve University Doing Work on a Material In compressing a material (constant pressure from all directions), … work is done on it by the surroundings … its energy is raised C&R Figs. 2.8b compression Compressing a material raises its energy 3
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Lecture 20, spring 2010 ENGR 145, Chemistry of Materials Case Western Reserve University Heat and work are forms in which energy is transferred between a system and its surroundings The first law of thermodynamics describes the net change in internal energy U of a system q : heat into system w : work done on system One type of work: change in volume V surr (= – V sys ) at constant pressure P : Net change in energy of system at constant pressure: Δ U sys = q + w First Law of Thermodynamics [OGC §12.2] w = P V surr = P V sys Δ U sys = q P + P Δ V surr = q P P Δ V sys OGC eq.12.3 4 subscript P denotes “at constant pressure”
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Lecture 20, spring 2010 ENGR 145, Chemistry of Materials Case Western Reserve University Definition of enthalpy H : Change in enthalpy H of a system is At constant pressure, P sys = 0 From previous slide, at constant pressure: Rearranging: Δ U sys = q p P Δ V sys q p H U + PV Δ H sys = Δ U sys + Δ PV ( ) sys Enthalpy [OGC §12.3] Δ H sys = Δ U sys + P sys Δ V sys = Δ U sys + P Δ V sys = Δ U + PV ( ) P = H P OGC eq.12.7a OGC eq.12.7b = Δ U sys + P sys Δ V sys + V sys Δ P sys =0 5 subscript P denotes “at constant pressure”
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