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The Combined Gas Law

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Manipulating Variables in equations Often in an equation we want to isolate some variable, usually the unknown From math: what ever you do to one side of an equation you have to do to the other side Doing this keeps both sides the same E.g. x + 5 = 7, what does x equal? We subtract 5 from both sides … x + 5 – 5 = 7 – 5, thus x = 2 Alternatively, we can represent this as 5 moving to the other side of the equals sign … x + 5 = 7 becomes x = 7 – 5 or x = 2 Thus, for addition or subtraction, when you change sides you change signs

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Multiplication and division We can do a similar operation with multiplication and division E.g. 5x = 7, what does x equal? We divide each side by 5 (to isolate x) … 5x/5 = 7/5 … x = 7/5 … x = 1.4 Alternatively, we can represent this as 5 moving to the other side of the equals sign … 5x = 7 becomes x = 7/5 Thus, for multiplication and division, when you change sides you change position (top to bottom, bottom to top)

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Multiplication and division Lets look at a more complicated example: (x) (y) 5 = 7a7a b Isolate a in the equation: Move b to the other side (from bottom to top) 5b (x) (y) = 7a7a (x)(y)(b) 5 = 7a7a (5)(7) =a or Move 7 to the other side (from top to bottom) (x)(y)(b) (35) =a

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Multiplication and division This time, isolate b in the equation: (x) (y) 5 = 7a b Move b to the other side (it must be on top) … (x) (y) 5 = 7a b Move everything to the other side of b 35a xy =b (b)(x)(y) 5 = 7a Q - Rearrange the following equation to isolate each variable (you should have 6 equations) P 1 V 1 P 2 V 2 T 1 T 2 =

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Combined Gas Law Equations P1P1 = P2T1V2P2T1V2 T2V1T2V1 V1V1 = P2T1V2P2T1V2 T2P1T2P1 T2T2 = P2T1V2P2T1V2 P1V1P1V1 T1T1 = P1T2V1P1T2V1 P2V2P2V2 P2P2 = P1T2V1P1T2V1 T1V2T1V2 V2V2 = P1T2V1P1T2V1 P2T1P2T1

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Combining the gas laws So far we have seen two gas laws: Jacques Charles Robert Boyle P1V1P1V1 =P2V2P2V2 V1V1 T1T1 = V2V2 T2T2 These are all subsets of a more encompassing law: the combined gas law P1P1 T1T1 = P2P2 T2T2 Read pages 437, 438. Do Q 26 – 33 (skip 31) P 1 V 1 P 2 V 2 T 1 T 2 = Joseph Louis Gay-Lussac

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Q 26 V 1 = 50.0 ml, P 1 = 101 kPa V 2 = 12.5 mL, P 2 = ? T 1 = T 2 P1V1P1V1 T1T1 = P2V2P2V2 T2T2 (101 kPa)(50.0 mL) (T 1 ) = (P 2 )(12.5 mL) (T 2 ) (101 kPa)(50.0 mL)(T 2 ) (T 1 )(12.5 mL) =(P 2 ) =404 kPa Notice that T cancels out if T 1 = T 2

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Q 27 V 1 = 0.10 L, T 1 = 298 K V 2 = ?, T 2 = 463 P 1 = P 2 P1V1P1V1 T1T1 = P2V2P2V2 T2T2 (P 1 )(0.10 L) (298 K) = (P 2 )(V 2 ) (463) (P 1 )(0.10 L)(463 K) (P 2 )(298 K) =(V 2 ) =0.16 L Notice that P cancels out if P 1 = P 2

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Q 28 P 1 = 150 kPa, T 1 = 308 K P 2 = 250 kPa, T 2 = ? V 1 = V 2 P1V1P1V1 T1T1 = P2V2P2V2 T2T2 (150 kPa)(V 1 ) (308 K) = (250 kPa)(V 2 ) (T 2 ) (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

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Q 29 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) (293 K) = (90 kPa)(V 2 ) (308 K) (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.

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Q 30 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) (303 K) = (100 kPa)(V 2 ) (298 K) (800 kPa)(1.0 L)(298 K) (100 kPa)(303 K) =(V 2 ) =7.9 L

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Q 32 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) (283 K) = (0.95 atm)(V 2 ) (297 K) (6.5 atm)(2.0 mL)(297 K) (0.95 atm)(283 K) =(V 2 ) =14 mL 33. The amount of gas (i.e. number of moles of gas) does not change. For more lessons, visit

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