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T h e C o m b i n e d “ G a s L a w ”. 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 : - PowerPoint PPT Presentation
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The Combined“Gas Law”
Various Gas Laws• Boyles Law:
– initial pressure equals final pressure times final volume P1V1 P2V2
• Charles Law:
– the ratio of volume to temperature of a given gas at fixed pressure is constant V1/T1 = V2/T2
• Gay-Lussac’s Law:
– the ratio of pressure to temperature of a given gas at fixed volume is constant P1/T1 = P2/ T2
• Avogadro's Law:
– at fixed pressure and temperature, the ratio of volume to moles (n) of a gas is constant V1/n1 V2/n2
• 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 Ptotal= P1+P2 etc.
• Combined Gas Law: – the combined law incorporates all of the laws:
Boyles, Charles, Avogadro's, and Lussac’s P1V1/n1T1 = P2V2/n2T2 n= moles
Combining the gas laws
Jacques CharlesRobert Boyle
P1V1 = P2V2V1
T1
=V2
T2These are all subsets of a more encompassing law:
the combined gas law
P1
T1
=P2
T2
P1V1 P2V2
T1 T2
=
Joseph Louis Gay-Lussac
•P1V1 P2V2
• T1 T2
P1V1 T2 = P2V2T1
P1V1 T2 = P2V2T1
V1 T2 V1T2
P1 = P2V2T1
V1T2
Manipulating Variables in Equations
Combined Gas Law Equations
P1 =P2V2T1
V1T2
V1 =P2V2T1
P1T2
T2 =P2V2T1
P1V1
T1 =P1V1T2
P2V2
P2 =P1V1T2
V2T1
V2 =P1V1T2
P2T1
P1V1 T2 = P2V2T1
P1V1 P2V2
T1 T2
=
• What is the initial pressure of a system if the final volume is 250 ml.
• Initial pressure (P1)= 101 Kpa
• Initial volume was 500 ml
• All temperatures are at STP= 273 OK
• P1= ?
• V1= 250 mls
• P2= 101kPa
• V2= 500 ml
*T1 and T2 are similar
P1 =P2V2T1
V1T2
P1V1 P2V2
T1 T2P1V1 T2 = P2V2T1
P1 =P2V2
V1
P1 =(101kPa)(250mls)
500 mls= 50.5 kPa
P1 = 150 kPa, T1 = 308 K
P2 = 250 kPa, T2 = ? V1 = V2
P1V1
T1
=P2V2
T2
(250 kPa)(V2)(308 K)(150 kPa)(V1)
=(T2) = 513 K= 240 °C
Notice that V cancels out if V1 = V2
T2 =P2V2T1
P1V1
P1V1 T2 = P2V2T1
K at STP= 273, therefore, 513 K- 273= 240OC
P1 = 100 kPa, V1 = 5.00 L, T1 = 293 K
P2 = 90 kPa, V2 = ?, T2 = 308 K
P1V1
T1
=P2V2
T2
(100 kPa)(5.00 L)(308 K)(90 kPa)(293 K)
=(V2) = 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.
P1V1 T2 = P2V2T1
V2 =P1V1T2
P2T1
P1 = 800 kPa, V1 = 1.0 L, T1 = 303 K
P2 = 100 kPa, V2 = ?, T2 = 298 K
P1V1
T1
=P2V2
T2
(800 kPa)(1.0 L)(298 K)(100 kPa)(303 K)
=(V2) = 7.9 L
V2 =P1V1T2
P2T1
P1 = 6.5 atm, V1 = 2.0 mL, T1 = 283 K
P2 = 0.95 atm, V2 = ?, T2 = 297 K
P1V1
T1
=P2V2
T2
(6.5 atm)(2.0 mL)(297 K)(0.95 atm)(283 K)
=(V2) = 14 mL
The amount of gas (i.e. number of moles of gas) does not change.
For more lessons, visit www.chalkbored.com
V2 =P1V1T2
P2T1
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