1. The Combined Gas 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. 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

5. 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 =

6. Combined Gas Law EquationsP1 =
P2T1V2
T2V1 P2 =
P1T2V1
T1V2 T1 =
P1T2V1
P2V2 T2 =
P2T1V2
P1V1 V1 =
P2T1V2
T2P1 V2 =
P1T2V1
P2T1

7. 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)

8. 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

11. 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

12. 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

13. 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.

14. 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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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.
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