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The Combined “ Gas Law ”. Various Gas Laws Boyles Law: –initial pressure equals final pressure times final volume  P 1 V 1  P 2 V 2 Charles Law: –the.

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Presentation on theme: "The Combined “ Gas Law ”. Various Gas Laws Boyles Law: –initial pressure equals final pressure times final volume  P 1 V 1  P 2 V 2 Charles Law: –the."— Presentation transcript:

1 The Combined “ Gas Law ”

2 Various Gas Laws Boyles Law: –initial pressure equals final pressure times final volume  P 1 V 1  P 2 V 2 Charles Law: –the ratio of volume to temperature of a given gas at fixed pressure is constant  V 1 /T 1 = V 2 /T 2 Gay-Lussac’s Law: –the ratio of pressure to temperature of a given gas at fixed volume is constant  P 1 /T 1 = P 2 / T 2 Avogadro's Law: –at fixed pressure and temperature, the ratio of volume to moles (n) of a gas is constant  V 1 /n 1  V 2 /n 2

3 The Ideal Gas Law: –relates the amount of gas produced in a reaction  PV=nRT where n= moles R= 8.31 L X kPa/mol x K – This is a constant number Dalton’s Law of Partial Pressures: –the total pressure of a mixture of gases equals the sum of the pressures  P total = P 1 +P 2 etc. Combined Gas Law: –the combined law incorporates all of the laws: Boyles, Charles, Avogadro's, and Lussac’s  P 1 V 1 /n 1 T 1 = P 2 V 2 /n 2 T 2 n= moles

4 Combining the gas laws Jacques CharlesRobert Boyle P1V1P1V1 =P2V2P2V2 V1V1 T1T1 = V2V2 T2T2 These are all subsets of a more encompassing law: the combined gas law P1P1 T1T1 = P2P2 T2T2 P 1 V 1 P 2 V 2 T 1 T 2 = Joseph Louis Gay-Lussac

5 P 1 V 1 P 2 V 2 T 1 T 2 P 1 V 1 T 2 = P 2 V 2 T 1 V 1 T 2 V 1 T 2 P 1 = P 2 V 2 T 1 V 1 T 2 Manipulating Variables in Equations

6 Combined Gas Law Equations P1P1 = P2V2T1P2V2T1 V1T2V1T2 V1V1 = P2V2T1P2V2T1 P1T2P1T2 T2T2 = P2V2T1P2V2T1 P1V1P1V1 T1T1 = P1V1T2P1V1T2 P2V2P2V2 P2P2 = P1V1T2P1V1T2 V2T1V2T1 V2V2 = P1V1T2P1V1T2 P2T1P2T1 P 1 V 1 T 2 = P 2 V 2 T 1 P 1 V 1 P 2 V 2 T 1 T 2 =

7 What is the initial pressure of a system if the final volume is 250 ml. Initial pressure (P 1 )= 101 Kpa Initial volume was 500 ml All temperatures are at STP= 273 O K P 1 = ? V 1 = 250 mls P 2 = 101kPa V 2 = 500 ml *T 1 and T 2 are similar P1P1 = P2V2T1P2V2T1 V1T2V1T2 P 1 V 1 P 2 V 2 T 1 T 2 P 1 V 1 T 2 = P 2 V 2 T 1 P1P1 = P2V2P2V2 V1V1 P1P1 = (101kPa)(250mls) 500 mls = 50.5 kPa

8 P 1 = 150 kPa, T 1 = 308 K P 2 = 250 kPa, T 2 = ? V 1 = V 2 P1V1P1V1 T1T1 = P2V2P2V2 T2T2 (250 kPa)(V 2 )(308 K) (150 kPa)(V 1 ) =(T 2 ) =513 K = 240 °C Notice that V cancels out if V 1 = V 2 T2T2 = P2V2T1P2V2T1 P1V1P1V1 P 1 V 1 T 2 = P 2 V 2 T 1 K at STP= 273, therefore, 513 K- 273= 240 O C

9 P 1 = 100 kPa, V 1 = 5.00 L, T 1 = 293 K P 2 = 90 kPa, V 2 = ?, T 2 = 308 K P1V1P1V1 T1T1 = P2V2P2V2 T2T2 (100 kPa)(5.00 L)(308 K) (90 kPa)(293 K) =(V 2 ) =5.84 L Note: although kPa is used here, any unit for pressure will work, provided the same units are used throughout. The only unit that MUST be used is K for temperature. P 1 V 1 T 2 = P 2 V 2 T 1 V2V2 = P1V1T2P1V1T2 P2T1P2T1

10 P 1 = 800 kPa, V 1 = 1.0 L, T 1 = 303 K P 2 = 100 kPa, V 2 = ?, T 2 = 298 K P1V1P1V1 T1T1 = P2V2P2V2 T2T2 (800 kPa)(1.0 L)(298 K) (100 kPa)(303 K) =(V 2 ) = 7.9 L V2V2 = P1V1T2P1V1T2 P2T1P2T1

11 P 1 = 6.5 atm, V 1 = 2.0 mL, T 1 = 283 K P 2 = 0.95 atm, V 2 = ?, T 2 = 297 K P1V1P1V1 T1T1 = P2V2P2V2 T2T2 (6.5 atm)(2.0 mL)(297 K) (0.95 atm)(283 K) =(V 2 ) =14 mL The amount of gas (i.e. number of moles of gas) does not change. For more lessons, visit www.chalkbored.com www.chalkbored.com V2V2 = P1V1T2P1V1T2 P2T1P2T1


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