1 / 15

T h e C o m b i n e d G a s L a w

T h e C o m b i n e d G a s L a w. 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

Patman
Download Presentation

T h e C o m b i n e d G a s L a w

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. The CombinedGas Law

  2. 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

  3. 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)

  4. (x) (y) 7a = 5 b (x) (y) 7a = 5 b (x)(y)(b) 7a = 5 (x)(y)(b) (x)(y)(b) = a a = (35) (5)(7) Multiplication and division • Let’s look at a more complicated example: • Isolate a in the equation: • Move b to the other side (from bottom to top) • Move 7 to the other side (from top to bottom) or

  5. (x) (y) 7a = 5 b (x) (y) 7a = 5 b 35a (b)(x)(y) 7a b = = xy 5 P1V1 P2V2 T1 T2 = Multiplication and division • This time, isolate b in the equation: • Move b to the other side (it must be on top) … • Move everything to the other side of b Q - Rearrange the following equation to isolate each variable (you should have 6 equations)

  6. P1T2V1 P1T2V1 P2T1V2 P1T2V1 P2T1V2 P2T1V2 P2 T1 T2 V1 P1 V2 = = = = = = P2V2 T1V2 P1V1 P2T1 T2P1 T2V1 Combined Gas Law Equations

  7. V1 P1 V2 P2 P1V1 = P2V2 = = T1 T1 T2 T2 P1V1 P2V2 T1 T2 = Combining the gas laws • So far we have seen two gas laws: Robert Boyle Jacques Charles Joseph Louis Gay-Lussac These are all subsets of a more encompassing law: the combined gas law Read pages 437, 438. Do Q 26 – 33 (skip 31)

  8. P1V1 P2V2 = T1 T2 (101 kPa)(50.0 mL) (P2)(12.5 mL) = (T1) (T2) (101 kPa)(50.0 mL)(T2) = 404 kPa (P2) = (T1)(12.5 mL) Q 26 V1 = 50.0 ml, P1 = 101 kPa V2 = 12.5 mL, P2 = ? T1 = T2 Notice that T cancels out if T1 = T2

  9. P1V1 P2V2 = T1 T2 (P1)(0.10 L) (P2)(V2) = (298 K) (463) (P1)(0.10 L)(463 K) = 0.16 L (V2) = (P2)(298 K) Q 27 V1 = 0.10 L, T1 = 298 K V2 = ?, T2 = 463 P1 = P2 Notice that P cancels out if P1 = P2

  10. P1V1 P2V2 = T1 T2 (150 kPa)(V1) (250 kPa)(V2) = (308 K) (T2) (250 kPa)(V2)(308 K) = 513 K = 240 °C (T2) = (150 kPa)(V1) Q 28 P1 = 150 kPa, T1 = 308 K P2 = 250 kPa, T2 = ? V1 = V2 Notice that V cancels out if V1 = V2

  11. P1V1 P2V2 = T1 T2 (100 kPa)(5.00 L) (90 kPa)(V2) = (293 K) (308 K) (100 kPa)(5.00 L)(308 K) = 5.84 L (V2) = (90 kPa)(293 K) Q 29 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.

  12. P1V1 P2V2 = T1 T2 (800 kPa)(1.0 L) (100 kPa)(V2) = (303 K) (298 K) (800 kPa)(1.0 L)(298 K) = 7.9 L (V2) = (100 kPa)(303 K) Q 30 P1 = 800 kPa, V1 = 1.0 L, T1 = 303 K P2 = 100 kPa, V2 = ?, T2 = 298 K

  13. P1V1 P2V2 = T1 T2 (6.5 atm)(2.0 mL) (0.95 atm)(V2) = (283 K) (297 K) (6.5 atm)(2.0 mL)(297 K) = 14 mL (V2) = (0.95 atm)(283 K) Q 32 P1 = 6.5 atm, V1 = 2.0 mL, T1 = 283 K P2 = 0.95 atm, V2 = ?, T2 = 297 K 33. The amount of gas (i.e. number of moles of gas) does not change. For more lessons, visit www.chalkbored.com

More Related