Lecture 11 - More Thermodynamics (19.4-19.9) As we have...

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Lecture 11 1 As we have already noted for any ideal gas U=U(T) Å depends on temperature only. So, for an isothermal process: Δ U=0 for an ideal gas. This implies q+w=0, or -q=w, or - q= w δ rev Furthermore, if the process is reversible: - q ln rev w PdV nRTd V δδ == = 2 rev 1 and: q ln V nRT V = w rev <0 i.e. work done on surroundings For V 2 >V 1 q rev >0 heat entered system to keep gas at constant temperature. Now, consider an adiabatic process to the same final volume. We define adiabatic to be δ q=0. If there is no heat flow and we go to the same final volume: T 2 <T 1 (gas must cool!). But by how much? More Thermodynamics (19.4-19.9)
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Lecture 11 2 Fig 19.5 Isobaric: Δ P=0, path D Isothermal: Δ T=0, path A Adiabatic: δ q=0, path B Note for adiabatic path δ q=0 which means dU= δ w. Adiabatic: dU= δ w V ext nRT CdT PdV dV V =− Ideal gas reversible Ideal gas 22 11 ln V TV V dT dV CR C dT R ∫∫
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Lecture 11 3 Remember for anything other than a monatomic ideal gas, the heat capacity has a temperature dependence. () 2 1 21 12 ln T V T CT VV dT R =− = V 3 for a monatomic gas, C 2 R = 3/2 2 1 1 2 3 So: ln ln or 2 TV T V T V ⎛⎞ == ⎜⎟ ⎝⎠ For an adiabatic expansion V 2 >V 1 so T 2 <T 1 ; the gas cools! 5/2 2 2 1 1 Since , TT P P T nRT V PT T P P T = Question: If argon expands from a pressure of 10 5 Pa (1 bar) to 50 mbar and had an initial temperature of 298K, what is the final temperature?
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This note was uploaded on 04/01/2012 for the course CHEM 444 taught by Professor Gruebele,m during the Spring '08 term at University of Illinois, Urbana Champaign.

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Lecture 11 - More Thermodynamics (19.4-19.9) As we have...

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