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notes06-07-page296 - ¢ H μ(the Joule-Thomson coe ffi...

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The work done on the left is w L = P 1 4 V = P 1 (0 V 1 ) = P 1 V 1 . (42.46) The work done on the right is w R = P 2 4 V = P 2 ( V 2 0) = P 2 V 2 . (42.47) Now, 4 U = U 2 U 1 = w L + w R = P 1 V 1 P 2 V 2 (42.48) Thus U 2 + P 2 V 2 = U 1 + P 1 V 1 H 2 = H 1 (42.49) For Joule-Thomson expansion the enthalpy is constant. We want to fi nd ¡ ∂T ∂V
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Unformatted text preview: ¢ H ≡ μ. (the Joule-Thomson coe ffi cient). Identity: μ ∂T ∂P ¶ H = − μ ∂T ∂H ¶ P | {z } 1 /C P μ ∂H ∂P ¶ T = 1 C P μ ∂H ∂P ¶ T = μ (42.50) 284...
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