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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**

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

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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) P1V1 P2V2 T T2 =

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**Combined Gas Law Equations**

P1 = P2T1V2 T2V1 P2 = P1T2V1 T1V2 T1 = P1T2V1 P2V2 T2 = P2T1V2 P1V1 V1 = P2T1V2 T2P1 V2 = P1T2V1 P2T1

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**These are all subsets of a more encompassing law: the combined gas law**

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

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**Q 26 V1 = 50.0 ml, P1 = 101 kPa V2 = 12.5 mL, P2 = ? T1 = T2 P1V1 T1 =**

(101 kPa)(50.0 mL) (T1) = (P2)(12.5 mL) (T2) (101 kPa)(50.0 mL)(T2) (T1)(12.5 mL) = 404 kPa = (P2) Notice that T cancels out if T1 = T2

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**Q 27 V1 = 0.10 L, T1 = 298 K V2 = ?, T2 = 463 P1 = P2 P1V1 T1 = P2V2**

(P1)(0.10 L) (298 K) = (P2)(V2) (463) (P1)(0.10 L)(463 K) (P2)(298 K) = 0.16 L = (V2) Notice that P cancels out if P1 = P2

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**Q 28 P1 = 150 kPa, T1 = 308 K P2 = 250 kPa, T2 = ? V1 = V2 P1V1 T1 =**

(150 kPa)(V1) (308 K) = (250 kPa)(V2) (T2) (250 kPa)(V2)(308 K) (150 kPa)(V1) = 513 K = 240 °C = (T2) Notice that V cancels out if V1 = V2

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Q 29 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) (293 K) = (90 kPa)(V2) (308 K) (100 kPa)(5.00 L)(308 K) (90 kPa)(293 K) = 5.84 L = (V2) 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 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) (303 K) = (100 kPa)(V2) (298 K) (800 kPa)(1.0 L)(298 K) (100 kPa)(303 K) = 7.9 L = (V2)

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**For more lessons, visit www.chalkbored.com**

Q 32 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) (283 K) = (0.95 atm)(V2) (297 K) (6.5 atm)(2.0 mL)(297 K) (0.95 atm)(283 K) = 14 mL = (V2) 33. The amount of gas (i.e. number of moles of gas) does not change. For more lessons, visit

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The Gas Laws Chapter 14 Chemistry.

The Gas Laws Chapter 14 Chemistry.

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