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Unformatted text preview: Lecture 11 1 As we have already noted for any ideal gas U=U(T) depends on temperature only. So, for an isothermal process: ΔU=0 for an ideal gas. This implies q+w=0, or q=w, or  q= w δ δ rev Furthermore, if the process is reversible:  q ln rev w PdV nRTd V δ δ = =  =  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 δq=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 (19.419.9) Lecture 11 2 Fig 19.5 Isobaric: ΔP=0, path D Isothermal: ΔT=0, path A Adiabatic: δq=0, path B Note for adiabatic path δq=0 which means dU= δw. Adiabatic: dU= δw V ext nRT C dT P dV PdV dV V =  =  =  Ideal gas reversible Ideal gas 2 2 1 1 ln ln V T V V T V dT dV C R T V C d T d V R =  =  ∫ ∫ Lecture 11 3 Remember for anything other than a monatomic ideal gas, the heat capacity has a temperature dependence. ( 29 2 1 2 1 1 2 ln ln ln T V T C T V V d T R V V =  = ∫ V 3 for a monatomic gas, C 2 R = 3/2 2 1 2 1 1 2 1 2 3 So: ln ln or 2 T V T V T V T V = = For an adiabatic expansion V 2 >V 1 so T 2 <T 1 ; the gas cools! 3/2 5/2 2 1 1 2 2 1 2 2 1 1 Since , T T P P T nRT V P T T P P T = = → = Question: If argon expands from a pressure of 10...
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This note was uploaded on 04/08/2010 for the course CHEM 444 taught by Professor Jameslis during the Spring '08 term at University of Illinois at Urbana–Champaign.
 Spring '08
 JamesLis
 Physical chemistry, pH

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