sandler_sm_06_15

sandler_sm_06_15 - 6 af 6.15(a From table TB ~ 320 K i.e B...

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6.15 (a) From table: , i.e., T B ~320K BT B B af = ( ) = 320 K 0 The inversion temperature is the temperature at which T PC VT V T T PT RT P B R P dB dT V T RT P B RT P T dB dT dB dT HP PP P F H G I K J == F H G I K J L N M O Q P F H G I K J =+ L N M O Q P F H G I K J −− = 0 1 P Thus, T inv is the temperature at which dB dt −= 0 . Plot up B vs. T , obtain dB dT either graphically, or numerically from the tabular data. I find T inv ~ 600 K . Also, dB dT decreases with increasing temperature (i.e., dB dT ~. 456 cm mol K 3 at 87.5 K and 0.027 cm mol K 3 at 650 K. Presumably it is negative at even higher temperature!) (b) Generally µ = F H G I K J =− F H G I K J R S T U V W R S T U V W T V TC dB dT 11 Using the data in the table it is easy to show that for T T < inv , dB dT −< > 0 0 , while for T T > inv , dB dT −> < 00 . (c) Since Fig. 2.4-3 for nitrogen is an H - P plot is easiest to proceed as follows ∂∂ T PH T P H HT T T P F H G I K J = = 1 Since C P a f = P is > 0 and less than [Except at a phase transition—see Chap. 6 and Problem 6.1—however, has no meaning in the two-phase region], if dT dP H is to be zero, then dH dP T a f must equal zero. That is, an inversion point occurs when isotherms are parallel
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This note was uploaded on 10/22/2011 for the course ENG 101 taught by Professor Yukov during the Spring '11 term at UCLA.

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sandler_sm_06_15 - 6 af 6.15(a From table TB ~ 320 K i.e B...

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