Lect+15+08 - Lecture 15 Chapter 12 First LAW Chapter 13...

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Lecture 15 Chapter 12 - First LAW Chapter 13 - Second LAW Adiabatic reversible processes in ideal gases An example of the adiabatic ideal gas process Second Law - The Entropy of the Universe (system+surroundings) Increases during all (spontaneous) processes Equilibrium is achieved when the entropy of the Universe is maximized Study Chapter 13 before next lecture, entropy is often a confusing topic the first time through
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Adiabatic Ideal Gas Processes Adiabatic - no heat transfer in or out of system. No heat transfer is accomplished by using an insulated container with vacuum being the best insulation (Dewar flasks). Remember that with q = 0 then E = w So q = o (adiabatic process, ideal gas)
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Isothermal and adiabatic reversible processes
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Adiabatic processes in ideal gases No heat transfer; q = 0 so E = w When volume changes dV then T changes by dT and since E depends only on T (ideal gas) dE = nC v dT and since this is reversible P ext = P; and w = -P ext dV; So nC v dT = -PdV = -nRTdV/V for changes so If C v does not depend on T over T 1 to T 2 then C v ln T 2 /T 1 = -R ln (V 2 /V 1 ) (T 2 /T 1 ) Cv = (V 1 /V 2 ) R = (V 1 /V 2 ) Cp-Cv C v dT/T = T1 T2 - R dV/V V1 V2
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Adiabatic process in ideal gases (T 2 /T 1 ) Cv = (V 1 /V 2 ) R = (V 1 /V 2 ) Cp-Cv Define a new term: γ = Cp /Cv (T 2 /T 1 ) = (V 1 /V 2 ) (Cp/Cv)-1 = (V 1 /V 2 ) γ -1 T 1 V 1 γ -1 = T 2 V 2 γ -1 If you know or can estimate Cv and Cp If you know T 1 ,V 1 , and T 2 can calculate V 2 If you know T 1 ,V 1 , and V 2 can calculate T 2 Similarly if you know P 1 ,V 1 and P 2 or V 2 can calculate the unknown using the equation P 1 V 1 γ = P 2 V 2 γ
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