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Our Chemistry Road trip so far … Moles, molecules and reactions Molecules and Moles acting alone.

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Presentation on theme: "Our Chemistry Road trip so far … Moles, molecules and reactions Molecules and Moles acting alone."— Presentation transcript:

1 Our Chemistry Road trip so far … Moles, molecules and reactions Molecules and Moles acting alone

2 Chapter 5:Gases (pp. 189-230) an exploration of `aggregate’ behavior Next up: the Chemistry of the Crowd

3 Pressure (P) concepts & measures in “Physics-speak “(see p 192 ) P = Force Area Physics speak Physics units P = newtons m 2 = kg*m/s 2 m*m = kg m s 2

4 Pressure (P) concepts & measures in “Physics-speak “(see p 191 ) 1 kg = 1 Pascal (Pa) m s 2 

5 Some lowbrow chem insights into Pressure Getting hammered Bed of nails

6 Pressure expresses how much a (constant) impressed force is spread out… MORE AREA, LOWER PRESSURE LESS AREA, HIGHER PRESSURE What the bed of nails demonstration illustrates….

7 outer space Earth ~ 20 mile tall column of air in Earth’s atmosphere creates pressure we all feel Pressure =1 atm at sea level Pressure in `lowbrow’ Chem speak: barometers Edge of Earth’s atmosphere

8 Pressure in `lowbrow’ Chem speak (continued): barometers ( see fig. 5.3, 5.4 of text) Evacuated space P in mm Hg=760 External atmospheric pressure, P Hg A Torricelli barometer

9 Atmosphere demonstrated the Italian Way ( BIG time Torricelli Barometers) Height of Hg columns is independent of cross- sectional areas of tubes Lesson from 17 th century picture Vat of mercury (Hg)

10 1 atmosphere =760 mm Hg 1 atmosphere=10336 mm H 2 O =33.9 ft H 2 O = 20 miles of air = 15 pounds/in 2 (psi) = 760 torr Pressure units in CHEMISTRY LAND =29.92 inches Hg Other common measures of an atmosphere

11 1 atmosphere =760 mm Hg = 760 torr =29.92 inches Hg = 15 pounds/in 2 (psi) 28 inches Hg= atm = torr = psi 0.936 711 14.0 P conversions

12 Physics speak 1 atm = 101.3 kPa Confusing Compromise 1 bar =100 kPa= 0.986 atm Pressure units in PHYSICS COUNTRY WE ARE GOING TO STICK WITH ATMOSPHERES (atm) IN THIS COURSE

13 Empiric gas law derivations: the common sense way T down  ?? P, T, V, n n P V down V = bT T up  ?? V up CHARLES’ LAW (p. 195) V 1 /T 1 = V 2 /T 2 V 1 = T 1 V 2 T 2 => T, V vary V T What variables do we follow when describing a gas ???

14 Empiric gas law derivations: the common sense way (cont.) n T V down ??  P up  P down V up ?? BOYLE’S LAW () P = k/V P 1 V 1 =P 2 V 2 P 1 = 1/V 1 P 2 1/V 2 => P, V vary V P

15 Empiric gas law derivations: the common sense way (cont.) n V T up  ?? P up T down  ?? P down P = aT GAY-LUSSAC’S LAW (not in text) P 1 /T 1 = P 2 /T 2 T 1 = P 1 T 2 P 2 => T, P vary P T T in K

16 How Boyle, Gay-Lussac and Charles Laws are reflected in the Combined Gas Law (when n is constant) P 1 V 1 = P 2 V 2 T 1 T 2 constant n,P P 1 V 1 = P 2 V 2 T 1 T 2 Charles’ Law (P 1 =P 2 ) constant n, T Boyle’s Law (T 1 =T 2 ) constant n, V Gay-Lussac’s Law (V 1 =V 2 ) Combined Gas Law constant n P 1 V 1 = P 2 V 2 T 1 T 2 P 1 V 1 = P 2 V 2 T 1 T 2 ConditionsName of Gas Law Gas Law Equation

17 An ideal gas at constant V and P=2 atm is heated from 300 to 600 K. What is the final P ? A.1 atm B.2 atm C.3 atm D.4 atm

18 1.A sample of oxygen gas is expanded from 20 to 50 liters at constant temperature. The final pressure is 4 atm. What was the initial pressure ? P 1 =10 atm COMBINED GAS LAW PROBLEMS…BOARD WORK

19 A sample of ideal gas arrives at 300K when expanded from 3 to 9 L at constant P. What was the original temperature ? A.900 K B.100 K C.600 K D.150 K

20 2. A child’s balloon originally occupies 5 liters at sea level (P=1 atm) and room temperature (300 K). It is released and is allowed to rise to an altitude where the pressure is 0.25 atm and the temperature is 150 K. What is the balloon’s new volume ? 10 L COMBINED GAS LAW PROBLEMS…BOARD WORK (CONT.)

21 An ideal gas in a fixed volume and an initial pressure of 10 atm and initial T of 177 C has a final pressure of 3.33 atm. What is the final T(K) (K=C+273) A.150 K B.59 K C.450 K D.531 K

22 3.The volume of a piston at fixed pressure changes as it is cooled from 500 o C to 250 o C. If the final volume is 6.76 L, what is the initial volume ? V 1 =10 L COMBINED GAS LAW PROBLEMS…BOARD WORK (CONT.)

23 4. Autoclaves are essentially pressure cookers. At 1 atm, steam has a temperature of 100 o C. Would you expect the pressure to double if the autoclave to attains a steam temperature of 200 o C ? a) NO…must convert C  K…ratio is not 200/100 What pressure do you actually expect to reach at 200 o C? 473.15 =1.73 atm 373.15 COMBINED GAS LAW PROBLEMS…BOARD WORK (CONT.)

24 When n varies…. Blowing up a balloonWhat varies ? n, V (1st time n changes) => P and T are constant n up ? => V up n down ? => V down n V Avogadro’s Law (pp. 289-90) V=an

25 Trickier problem 2.00 L H 2 at 0.625 atm 1.00 L N 2 at 0.200 atm Stopcock closed Stopcock open Final P = ??? 0.483 atm

26 Combined Gas Law Reviewed As long as n=moles is held constant: P 1 V 1 = P 2 V 2 T 1 T 2

27 What happens if we let n vary too ??

28 P (piston head) n V (varies T Heating/cooling coils piston walls Hypothetical Gas Property testing apparatus Ideal Gas Law: letting T,P,V, n and gas ID all vary (pp. 198-203) GASGAS GAS VALVE insulatio n

29 0 o 1 1 22.414 Vary gas and fix three out of four gas variables… Gas varied H 2 (2) He(4) N 2 (28) CO 2 (44) SF 6 (146) T( O C) P(atm) n(moles) V(obs) Variables fixed at constant values (`STP’) STP =Standard Temperature & Pressure ( and n=1)..see what happens Gas ID (and size) not important Ideal Gas derived: Why gas identity not important 0 o 1 1 22.414

30 What happens if we hold different sets of three variables constant and watch the fourth for a given gas ? P(atm) V(L) T(K)n(moles) 24001 1 300 1 5 2 300 5 5 3 16.43 24.65 0.406 101 PV/nT Anything constant ?? 0.08206 Ideal Gas Law derived (cont.): origin of R

31 PV = 0.08205746 nT =R (atm L) (mol K ) PV =nRT Or…

32 The ideal gas law lead to : molecular masses verification of stoichiometries

33 1. An 11 gram sample of a gas occupies 2.0 liters at 2.0538 atm and 200 K. What is the molecular mass of the gas ? (R=0.08206) 44 g/mol 2.What volume is occupied by 0.09 g of gas phase H 2 O (MW=18 g/mol) at 300 K and 0.123 atm ? 1.00 L

34 3. A 1.5 liter can of gas reaches a pressure of 45.45 atm at 500 K. How many molecules of gas are in the can ? (R=0.08206 atm L/K mol) 1.0*10 24 1 mol count =6.02*10 23


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