Homework 10 posted and due Friday 13 April

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

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

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

4. Autoclaves are essentially pressure cookers. At 1 atm, 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

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

Not all combined gas law problems are easy…

Final P = ??? 0.483 atm Trickier problem 2.00 L H2 at 0.625 atm 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

Need more practice with Combined Gas Law ????

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

What happens if we let n vary too ??

Heating/cooling coils 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/cooling coils

Gas ID (and size) not important (!!) 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 (!!)

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 0.406 0.08206 5 2 300 101 0.08206 5 5 3

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

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

1. An 11 gram sample of a gas occupies 2. 0 liters at 2 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

9 grams of an unknown material occupies 2 L at 1 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

  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

A 4 liter cylinder at 300 K reaches a pressure of 3. 075 atm 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