sandler_sm_06_15

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

This preview shows pages 1–2. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 10/22/2011 for the course ENG 101 taught by Professor Yukov during the Spring '11 term at UCLA.

### Page1 / 2

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online