Airbags fill with N 2 gas in an accident. Gas is generated by the decomposition of. 2 NaN 3 ---> > 2 Na + 3 N 2
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2 2 Airbags fill with N 2 gas in an accident. Gas is generated by the decomposition of sodium azide,, NaN 3. 2 NaN 3 ---> > 2 Na + 3 N 2
3 3
4 4 There is a lot of free space in a gas. Gases can be expanded infinitely. Gases occupy containers uniformly and completely. Gases diffuse and mix rapidly.
5 5 Gas properties can be modeled using math. Model depends on V = volume of the gas (L) T = temperature (K) n = amount (moles) P = pressure (atmospheres)
6 6! Pressure of air is measured with a BAROMETER (developed by Torricelli in 1643)
7 ! 7 Hg rises in tube until force of Hg (down) balances the force of atmosphere (pushing up). P of Hg pushing down related to Hg density column height
8 8! Column height measures P of atmosphere 1 standard atm = 760 mm Hg = 29.9 inches = about 34 feet of water SI unit is PASCAL, Pa, where 1 atm = kpa
9 "# #$ 9 % Brings together gas properties. Can be derived from experiment and theory.
10 10 &' #( If n and T are constant, then PV = (nrt( nrt) ) = k This means, for example, that P goes up as V goes down. Robert Boyle ( ). 1691). Son of Early of Cork, Ireland.
11 &' #( 11 A bicycle pump is a good example of Boyle s law. As the volume of the air trapped in the pump is reduced, its pressure goes up, and air is forced into the tire.
12 12 ' #( If n and P are constant, then V = (nr( nr/p)t = kt V and T are directly related. Jacques Charles ( ). Isolated boron and studied gases. Balloonist.
13 Charles s original balloon 13 Modern long-distance balloon
14 ' #( 14
15 15 )*+' & Equal volumes of gases at the same T and P have the same number of molecules. V = n (RT/P) = kn V and n are directly related. twice as many molecules
16 16 )*+' & The gases in this experiment are all measured at the same T and P.
17 , * % 17 How much N 2 is req d to fill a small room with a volume of 960 cubic feet (27,000 L) to P = 745 mm Hg at 25 o C? R = L atm/k mol Solution 1. Get all data into proper units V = 27,000 L T = 25 o C = 298 K P = 745 mm Hg (1 atm/760 mm Hg) = 0.98 atm
18 , * % 18 How much N 2 is req d to fill a small room with a volume of 960 cubic feet (27,000 L) to P = 745 mm Hg at 25 o C? R = L atm/k mol Solution 2. Now calc. n = PV / RT n = (0.98 atm)(2.7 x 10 4 L) ( L atm/k mol)(298 K) n = 1.1 x 10 3 mol (or about 30 kg of gas)
19 19 + & 2 H 2 O 2 (liq) ---> > 2 H 2 O(g) + O 2 (g) Decompose 1.1 g of H 2 O 2 in a flask with a volume of 2.50 L. What is the pressure of O 2 at 25 o C?? Of H 2 O? Bombardier beetle uses decomposition of hydrogen peroxide to defend itself.
20 20 + & 2 H 2 O 2 (liq) ---> > 2 H 2 O(g) + O 2 (g) Decompose 1.1 g of H 2 O 2 in a flask with a volume of 2.50 L. What is the pressure of O 2 at 25 o C?? Of H 2 O? Solution Strategy: Calculate moles of H 2 O 2 and then moles of O 2 and H 2 O. Finally, calc.. P from n, R, T, and V.
21 21 + & 2 H 2 O 2 (liq) ---> > 2 H 2 O(g) + O 2 (g) Decompose 1.1 g of H 2 O 2 in a flask with a volume of 2.50 L. What is the pressure of O 2 at 25 o C?? Of H 2 O? Solution 1.1 g H 2 O 2 1 mol 34.0 g = mol mol H 2 O 2 1 mol O 2 2 mol H 2 O 2 = mol O 2
22 + & 22 2 H 2 O 2 (liq) ---> > 2 H 2 O(g) + O 2 (g) Decompose 1.1 g of H 2 O 2 in a flask with a volume of 2.50 L. What is the pressure of O 2 at 25 o C?? Of H 2 O? Solution P of O 2 = nrt/v = (0.016 mol)( L atm/k mol)(298 K) 2.50 L P of O 2 = 0.16 atm
23 + & 23 What is P of H 2 O? Could calculate as above. But recall Avogadro s hypothesis. V P 2 H 2 O 2 (liq) ---> > 2 H 2 O(g) + O 2 (g) n at same T and P n at same T and V There are 2 times as many moles of H 2 O as moles of O 2. P is proportional to n. Therefore, P of H 2 O is twice that of O 2. P of H 2 O = 0.32 atm
24 24 "' #(! 2 H 2 O 2 (liq) ---> > 2 H 2 O(g) + O 2 (g) 0.32 atm 0.16 atm What is the total pressure in the flask? P total total in gas mixture = P A + P B +... Therefore, P total = P(H 2 O) + P(O 2 ) = 0.48 atm Dalton s Law: total P is sum of PARTIAL pressures.
25 25 "' #( John Dalton
26 "-. /0 Low density 26 High density
27 "-. /0 27 PV = nrt n V = P RT m M V = P RT where M = molar mass d = m V = PM RT d and M proportional
28 28,- "-. The density of air at 15 o C and 1.00 atm is 1.23 g/l. What is the molar mass of air? 1. Calc. moles of air. V = 1.00 L P = 1.00 atm T = 288 K n = PV/RT = mol 2. Calc. molar mass mass/mol = 1.23 g/ mol = 29.1 g/mol
29 1- #,# Theory used to explain gas laws. KMT assumptions are Gases consist of molecules in constant, random motion. P arises from collisions with container walls. No attractive or repulsive forces between molecules. Collisions elastic. Volume of molecules is negligible. 29
30 1! & 30 Because we assume molecules are in motion, they have a kinetic energy. KE = (1/2)(mass)(speed) 2 4 * ) )* 1/ As T goes up, KE also increases and so does speed.
31 1! & 31 At the same T, all gases have the same average KE. As T goes up, KE also increases and so does speed.
32 1! & 32 Maxwell s equation u2 3RT M root mean square speed where u is the speed and M is the molar mass. speed INCREASES with T speed DECREASES with M
33 33 " 5!! +
34 &! range of Molecules of a given gas have a range speeds. 34
35 &! Average velocity decreases with increasing mass. 35
36 36 ",- - ",- diffusion is the gradual mixing of molecules of different gases. effusion is the movement of molecules through a small hole into an empty container.
37 ",- - ",- Molecules effuse thru holes in a rubber balloon, for example, at a rate (= moles/time) that is proportional to T inversely proportional to M. Therefore, He effuses more rapidly than O 2 at same T. He 37
38 ",- - ",- 38 Graham s law governs effusion and diffusion of gas molecules. Rate for A Rate for B M of B M of A Rate of effusion is inversely proportional to its molar mass. Thomas Graham, Professor in Glasgow and London.
39 "! +! HCl and NH 3 diffuse from opposite ends of tube. Gases meet to form NH 4 Cl 39 HCl heavier than NH 3 Therefore, NH 4 Cl forms closer to HCl end of tube.
40 40, * 1,++ #( Recall that KMT assumptions are Gases consist of molecules in constant, random motion. P arises from collisions with container walls. No attractive or repulsive forces between molecules. Collisions elastic. Volume of molecules is negligible.
41 )*+' & + 1! & 41 P proportional to n
42 42!4!4 + 1! & P proportional to T
43 43 &' #( + 1! & P proportional to 1/V
44 ") + #( 44 Real molecules have volume. There are intermolecular forces. Otherwise a gas could not become a liquid. Fig
45 ") + #( 45 Account for volume of molecules and intermolecular forces with intermolecular forces with VAN DER WAAL S EQUATION. Measured P Measured V = V(ideal) n 2 a ) (P V 2 V - nb nrt vol. correction intermol. forces J. van der Waals, , 1923, Professor of Physics, Amsterdam. Nobel Prize 1910.
46 ") + #( Measured P Measured V = V(ideal) 46 n 2 a P V 2 intermol. forces V - nb nrt vol. correction Cl 2 gas has a = 6.49, b = For 8.0 mol Cl 2 in a 4.0 L tank at 27 o C. P (ideal) = nrt/v = 49.3 atm P (van der Waals) ) = 29.5 atm
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