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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 :
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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
V1 P1 V2 P2 P1V1 = P2V2 = = T1 T1 T2 T2 P1V1 P2V2 T1 T2 = Combining the gas laws Robert Boyle Jacques Charles Joseph Louis Gay-Lussac These are all subsets of a more encompassing law: the combined gas law
Manipulating Variables in Equations • P1V1 P2V2 • T1T2 P1V1T2 = P2V2T1 P1V1 T2 = P2V2T1 V1 T2 V1T2 P1 = P2V2T1 V1T2
P1V1T2 P1V1T2 P2V2T1 P1V1T2 P2V2T1 P2V2T1 T1 P2 T2 V1 P1 V2 = = = = = = V2T1 P1V1 P1T2 V1T2 P2V2 P2T1 P1V1 P2V2 T1 T2 = Combined Gas Law Equations P1V1 T2 = P2V2T1
(101kPa)(250mls) P2V2T1 P2V2 P1 P1 P1 = = = V1T2 V1 500 mls • 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 P1V1 P2V2 T1 T2 P1V1 T2 = P2V2T1 = 50.5 kPa
P2V2T1 T2 = P1V1 P1V1 P2V2 = T1 T2 (250 kPa)(V2)(308 K) = 513 K = 240 °C (T2) = (150 kPa)(V1) P1 = 150 kPa, T1 = 308 K P2 = 250 kPa, T2 = ? V1 = V2 P1V1 T2 = P2V2T1 K at STP= 273, therefore, 513 K- 273= 240OC Notice that V cancels out if V1 = V2
P1V1T2 V2 = P2T1 P1V1 P2V2 = T1 T2 (100 kPa)(5.00 L)(308 K) = 5.84 L (V2) = (90 kPa)(293 K) P1V1 T2 = P2V2T1 P1 = 100 kPa, V1 = 5.00 L, T1 = 293 K P2 = 90 kPa, V2 = ?, T2 = 308 K 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.
P1V1T2 V2 = P2T1 P1V1 P2V2 = T1 T2 (800 kPa)(1.0 L)(298 K) (V2) = (100 kPa)(303 K) 7.9 L = P1 = 800 kPa, V1 = 1.0 L, T1 = 303 K P2 = 100 kPa, V2 = ?, T2 = 298 K
P1V1T2 V2 = P2T1 P1V1 P2V2 = T1 T2 (6.5 atm)(2.0 mL)(297 K) = 14 mL (V2) = (0.95 atm)(283 K) P1 = 6.5 atm, V1 = 2.0 mL, T1 = 283 K P2 = 0.95 atm, V2 = ?, T2 = 297 K The amount of gas (i.e. number of moles of gas) does not change. For more lessons, visit www.chalkbored.com