9.24b - Recall: Ideal Gas Law PV = n RT S h o u ld a p p ly...

Info iconThis preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: Recall: Ideal Gas Law PV = n RT S h o u ld a p p ly : • W h e n n /V i s s m a ll • W h e n K E > > P E ( b ig T ) • L o w p re s s u re I d e a l g a s e s b e h a v e i d e n t i c a l l y - - t h e y “ d o n o t i n t e r a c t ”, b y d e fin itio n , a n d th e r e fo r e d o s o i d e n tic a lly . K in e tic e n e r g y o n ly m o le c u la r d e ta il, a n d d e p e n d s o n T b u t n o t M o r o th e r fe a tu re . R e c a ll: V p r o p o r tio n a l to n --> A v o g a d r o’ s in s ig h t! How to Measure Pressure? equal How to Measure Pressure? F ( r ig h t) = P R • A F(left) = PL•A + m•g •A m = V•d ensity = h•A•density F ( r ig h t) = F ( le ft) h•density = PR - PL 1 How to Measure Pressure? h•density = PR - PL h d e p e n d s o n flu id ! e .g ., “ m m H g ” ( t o r r) 1 a tm = 7 6 0 t o rr Observations M a n y o b s e r v a tio n s a r e c o n s is te n t w ith th e id e a l g a s la w . In fa c t, m a n y o b s e r v a tio n s p r e d a te th e id e a l g a s la w : B o y le : a d d e d m e r c u r y to c o m p r e s s gas. P re s s u re o f g a s = 7 6 0 t o rr + h . Boyle’s Data V o lu m e ~ 1 /P P V = n R T ( h e re , c o n s ta n t) 2 Boyle vs. Other (PV = k?) Boyle vs. Other (PV = k?) C o m p a re d a ta ra n g e to B o y le ’ s d a ta . T h is d a ta a t c o n s ta n t n, R , and T. Not “ i deal ” . Lyles: Effect of T (const. V) • E g g a c c e le r a te d - - m u s t h a v e b e e n a fo r c e • Vacuums don’t pull, it’s just that they push less • F o r c e fr o m lo ts o f little a ir m o le c u le s b e a tin g th e h e c k o u t o f th e e g g a n d c a n • A t c o n s ta n t v o lu m e , P ~ T ( in fa c t, is lin e a r ) P V = n R T ( V , n , R c o n s ta n t) • C o u ld B o y le h a v e d o n e th is ? 3 Charles: Effect of T (const. P) A t c o n s ta n t p re s s u re ! N o te th e in te r c e p t a t -2 7 3 .2 ° C W h y s lo p e s d iffe r e n t? Lyles: Effect of T (const. P) • • • • • • • • • • B a llo o n fille d w ith n itr o g e n ~16 cm diameter. V = 4/3 π (8 cm)3 = ~2.1 L R o o m te m p e ra tu re 2 9 8 K L iq u id n itr o g e n 7 7 K V(77K) = 4/3 π (5 cm)3 = ~0.5 L J u s t fo r c o m p le te n e s s : B a llo o n fille d w ith a ir ~16 cm diameter. V = 4/3 π (8 cm)3 = ~2.1 L R o o m te m p e ra tu re 2 9 8 K L iq u id n itr o g e n 7 7 K V(77K) = 4/3 π (0 cm)3 = ???? Important Points • • • • B e c a r e fu l w h e n y o u u s e liq u id n itr o g e n ! A s g a s c o o ls , m o le c u le s b e g in to in te r a c t K E v s . P E : K E ~ T , P E is ( r e la tiv e ly ) in d e p e n d e n t o f T In fa c t, g a s “ n o t id e a l ” lo n g b e fo r e b e c o m in g liq u id 0 °C 1 a tm 4 “Real” Gases “Real” Gases In te r a c tio n s b e tw e e n m o le c u le s re d u c e s th e p r e s s u r e o n th e w a ll. Pobs = P - a(n/V)2 P o w e r o f 2 r e fle c ts p a ir w is e i n te r a c tio n s . “Real” Gases a a n d b r e f le c t p r o p e r t ie s o f m o le c u le s ! H o w b ig t h e y a r e . How (stongly) they interact with each other. N o te , h o w e v e r , th a t th e v a n d e r W a a ls e q u a tio n is n o t “ t h e T r u th ” - - ju s t a n im p r o v e d a p p r o x im a tio n ( m o d e l) o f r e a lity 5 Mixtures of Gases • D a lto n : w h e n y o u m ix g a s e s , th e ir p r e s s u r e s a r e a d d itiv e • N o te th a t th is m u s t b e th e c a s e fo r ideal gases (it’s in the assumptions) Ptot = P1 + P2 + … = n1RT/V + n2 R T/V + … = (n1 + n2 + … ) RT/V = n totRT/V Gas Stoichiometry • H o w to g e t a g o o d b u ta n e fla m e ? In a te n n is b a ll c o n ta in e r ? H o w m u c h b u ta n e d o y o u w a n t to s e e ? • Butane: C4H 10 • C o m b u s tio n : 2 C4H 10(g) + 13 O2(g) 8 CO2(g) +10 H2O (g) • A ir is 2 1 % o x y g e n • T e n n is b a ll c a n is 8 5 0 m L Solution: • • • • • • • • V butane + V air = 850 m L V air = 850 m L - Vbutane V oxygen = 0.21 V air = 0.21 (850 m L - V butane) = 178.5 mL – 0.21Vb utane 1 3 v o lu m e s o f o x y g e n a r e r e q u ir e d fo r e v e r y 2 v o lu m e s o f b u ta n e o r 6 .5 to 1 . V oxygen = 6.5Vbutane 6 .5Vbutane = 178.5 m L – 0.21Vbutane 6 .71Vb utane = 178.5 mL V butane = 26.6 mL 6 Why did butane displace air? • C le a r ly , g a s e s c o llid e & c o llid e o fte n • H o w o fte n ? Intermolecular Collisions • V o lu m e “ s w e p t” b y g a s in o n e s e c o n d V = Area•distance = Area*rate*time = (π d 2)(u avg)(1 s) ( N o te : s a m e a v e r a g e v o lu m e p e r s e c o n d , e v e n if m o le c u le c o llid e s in m id d le ) # o f c o llis io n s p e r s e c o n d = # o f p a r tic le s in th a t v o lu m e A c tu a lly , w a n t th e r e la tiv e v e lo c ity o f tw o p a r tic le s , which is 21/2 uavg. Collisions rate = 4 (N/V) d 2 ( π R T/M)1/2 O xygen at 300 K, 1.0 a tm - -> 4•109 s-1 Im p o r ta n t in th e r a te s o f g a s p h a s e r e a c tio n s Mean Free Path H o w fa r b e tw e e n c o llis io n s ? Distance = rate • time = u avg • (1/collision rate) ______ 1 _ ___ 21/2(N/V)(π d 2) m o r e p a r tic le s , l e s s d is ta n c e b e tw e e n c o llis io n s . b ig g e r p a r tic le s , le s s d is ta n c e b e tw e e n c o llis io n s . A ls o ig n o r e s in te r a c tio n s /a ttr a c tio n s b e tw e e n m o le c u le s ! 7 ...
View Full Document

This note was uploaded on 05/31/2009 for the course CHEM 21L taught by Professor Roy during the Fall '08 term at Duke.

Ask a homework question - tutors are online