9.24b

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

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

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