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Homework 10 posted and due Friday 13 April

Homework 10 posted and due Friday 13 April. When n is constant…use the Combined Gas Law:. P 1 V 1 = P 2 V 2 T 1 T 2. T must be in Kelvin (K). T(K)= [T( o C)+273]. COMBINED GAS LAW PROBLEMS…BOARD WORK (CONT.). The volume of a piston at fixed pressure changes as it is

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Homework 10 posted and due Friday 13 April

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  1. Homework 10 posted and due Friday 13 April

  2. When n is constant…use the Combined Gas Law: P1V1 =P2V2 T1 T2 T must be in Kelvin (K) T(K)= [T(oC)+273]

  3. COMBINED GAS LAW PROBLEMS…BOARD WORK (CONT.) • 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 L, what is the initial volume ? T(K)= [T(oC)+273] V1=10 L

  4. 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)? T(K)= [T(o C)+273] • 150 K • 59 K • 450 K • 531 K

  5. COMBINED GAS LAW PROBLEMS…BOARD WORK (CONT.) 4. Autoclaves are essentially pressure cookers. At 1 atm, steam has a temperature of 100o C. Would you expect the pressure to double if the autoclave to attains a 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? Recall: [T(K) =273 +T(oC)] 473.15 =1.73 atm 373.15

  6. Gas at constant volume and moles is initially at 27 oC and a pressure of 1 atm. What temperature (in C) must we reach to attain a pressure of 2 atm ? [T(K) =273 +T(oC)] • 54 oC • 127 oC • 300 oC • 427 oC

  7. Not all combined gas law problems are easy…

  8. Trickier problem 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

  9. Need more practice with Combined Gas Law ????

  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. What happens if we let n vary too ??

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

  13. Ideal Gas derived: Why gas identity not important Vary gas and fix three out of four gas variables… ..see what happens Variables fixed at constant values (`STP’) 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 (!!)

  14. Ideal Gas Law derived (cont.): origin of R What happens if we hold different sets of three variables constant and 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 24.65 0.08206 300 1 0.08206 0.406 5 2 300 0.08206 101 3 5 5

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

  16. The ideal gas law leads to : • molecular masses • verification of stoichiometries

  17. 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

  18. 0.9 grams of an unknown material occupies 2 L at 1.23 atm at a temperature of 600 K. What is the molecular mass of the unknown? (R=0.082) • 18 g/mol • 12.3 g/mol • 2.46 g/mol • 36 g/mol

  19. 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 mol count =6.02*1023 1.0*1024

  20. A 4 liter cylinder at 300 K reaches a pressure of 3.075 atm. How many gas molecules are in the cylinder ? (R=0.082 atm L/K mol. 1 mol=6*1023 molecules) • 0.042*1023 • 3*1023 • 3*1022 • 12*1023

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