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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 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 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 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 = − = −
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