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Pressure in `lowbrow’ Chem speak (continued) : barometers ( see p 284)

Pressure in `lowbrow’ Chem speak (continued) : barometers ( see p 284). Evacuated space. External atmospheric pressure, P. P in mm Hg=760. A Torricelli barometer . Hg. http://catalogue.museogalileo.it/multimedia/TorricellisBarometricExperiment.html.

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Pressure in `lowbrow’ Chem speak (continued) : barometers ( see p 284)

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  1. Pressure in `lowbrow’ Chem speak (continued): barometers ( see p 284) Evacuated space External atmospheric pressure, P P in mm Hg=760 A Torricelli barometer Hg http://catalogue.museogalileo.it/multimedia/TorricellisBarometricExperiment.html

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

  3. in CHEMISTRY LAND Pressure units 1 atmosphere =760 mm Hg = 760 torr =29.92 inches Hg Other common measures of an atmosphere 1 atmosphere=10336 mm H2O =33.9 ft H2O = 20 miles of air = 15 pounds/in2 (psi)

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

  5. Other `unit’ issues: temperature scales OnlyT(K) varies directly with V, P* T(K) =T(oC)+ 273.15 *The Bill Gates money in his pocket analogy….

  6. Empiricgas law derivations:the common sense way What variables do we follow when describing a gas ??? P, T, V, n n P Tup  ?? Vup => T, V vary Tdown ?? Vdown CHARLES’ LAW (p. 287-8) V V = bT V1 = T1 V2T2 T V1/T1= V2/T2

  7. Empiricgas law derivations:the common sense way (cont.) nT V down ?? Pup => P, V vary Vup ??  Pdown P BOYLE’S LAW (p. 285-6) P = k/V P1 = 1/V1 P21/V2 V P1V1=P2V2

  8. Empiricgas law derivations:the common sense way (cont.) n V Tup ?? Pup => T, P vary Tdown ?? Pdown P GAY-LUSSAC’S LAW (not in text) P = aT T in K T1= P1 T2P2 T P1/T1= P2/T2

  9. How Boyle, Gay-Lussac and Charles Laws are reflected in the Combined Gas Law (when n is constant) Conditions Gas Law Equation Name of Gas Law P1V1 = P2 V2 T1 T2 constant n Combined Gas Law constant n,P P1V1 = P2 V2 T1 T2 Charles’ Law (P1=P2) P1V1 = P2 V2 T1 T2 constant n, T Boyle’s Law (T1=T2) P1V1 = P2 V2 T1 T2 Gay-Lussac’s Law (V1=V2) constant n, V

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

  11. 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 ? P1=10 atm • A sample of gas at fixed volume is heated from 300 K • to 600 K. If the initial pressure is 5 atm, what is the • final pressure ? P2= 10 atm • The volume of a piston at fixed pressure changes as it is cooled • from 500 oC to 250oC. If the final volume is 6.76 liters, • what is the initial volume ? V1=10 L

  12. 4. 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 5. Autoclaves are essentially pressure cookers ( fixed volume and water vapor) used to sterilize medical and biological tools like scalpels and sample bottles. At 1 atm, steam has a temperature of 100o C.Would you expect the pressure to double if the autoclave to attains steam temperature of 200oC ? a) NO…must convert C K…ratio is not 200/100 What pressure do you actually expect to reach at 200 oC? 473.15=1.27 atm 373.15

  13. Trickier Text problem 59 page 322 Stopcock closed 2.00 L H2 at 0.625 atm 1.00 L N2 at 0.200 atm Final P = ??? Stopcock open 0.483 atm

  14. Ideal Gas Law: letting T,P,V, n and gas ID all vary (pp. 290-299) P (piston head) Hypothetical Gas Property testing apparatus piston walls insulation T n V (varies GAS GAS VALVE Heating/coolingcoils

  15. Ideal Gas derived: Why gas identity not important Vary gas and fix three out of four gas variables… Variables fixed at constant values (`STP’) ..see what happens Gas varied T(OC) P(atm) n(moles) V(obs) H2(2) He(4) N2(28) CO2(44) SF6 (146) 0o 1 1 22.414 0o 1 1 22.414 0o 1 1 22.414 0o 1 1 22.414 0o 1 1 22.414 STP =Standard Temperature & Pressure ( and n=1) Gas ID (and size) not important

  16. Ideal Gas Law derived (cont.): origin of R What happens if we holddifferent sets of three variables constantand watch the fourth for a given gas ? Anything constant ?? PV/nT P(atm) V(L) T(K) n(moles) 16.43 2 400 1 0.08206 1 0.08206 24.65 300 1 0.08206 5 2 300 0.406 0.08206 3 5 5 101 Or… PV = 0.08205746=R (a constant) nT (atm L/K mol) PV =nRT

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