The Combined
Gas
Law
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
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)
Multiplication and division
• Let’s look at a more complicated example:
(x) (y)
7a
=
5
b
• Isolate a in the equation:
• Move b to the other side (from bottom to top)
(x) (y)
7a
=
5
b
• Move 7 to the other side (from top to bottom)
(x)(y)(b)
7a
=
5
(x)(y)(b)
(x)(y)(b)
= a or a =
(35)
(5)(7)
Multiplication and division
• This time, isolate b in the equation:
(x) (y)
7a
=
5
b
• Move b to the other side (it must be on top) …
(x) (y)
7a
=
5
b
• Move everything to the other side of b
35a
(b)(x)(y)
7a
b =
=
xy
5
Q - Rearrange the following
P1V1 P2V2
equation to isolate each variable
=
T
T
1
2
(you should have 6 equations)
Combined Gas Law Equations
P2T1V2
P1 =
T2V1
T1
P1T2V1
=
P2V2
P2T1V2
V1 =
T2P1
P1T2V1
P2 =
T1V2
T2
P2T1V2
=
P1V1
P1T2V1
V2 =
P2T1
Combining the gas laws
• So far we have seen two gas laws:
Robert Boyle
Jacques
Charles
V
Joseph Louis Gay-Lussac
V2
P1
P2
=
=
T1
T2
T1
T2
These are all subsets of a P V
P2V2
1 1
more encompassing law:
=
T1
T2
the combined gas law
Read pages 437, 438. Do Q 26 – 33 (skip 31)
P1V1 = P2V2
1
Q 26
V1 = 50.0 ml, P1 = 101 kPa
V2 = 12.5 mL, P2 = ?
P1V1
=
T1
(101 kPa)(50.0 mL)
=
(T1)
(P2) =
T1 = T2
P2V2
T2
(P2)(12.5 mL)
(T2)
(101 kPa)(50.0 mL)(T2)
(T1)(12.5 mL)
Notice that T cancels out if T1 = T2
= 404 kPa
Q 27
V1 = 0.10 L, T1 = 298 K
V2 = ?, T2 = 463
P1V1
T1
(P1)(0.10 L)
(298 K)
(V2) =
=
P1 = P2
P2V2
T2
=
(P2)(V2)
(463)
(P1)(0.10 L)(463 K)
(P2)(298 K)
Notice that P cancels out if P1 = P2
= 0.16 L
Q 28
P1 = 150 kPa, T1 = 308 K
P2 = 250 kPa, T2 = ?
P1V1
T1
(150 kPa)(V1)
(308 K)
(T2) =
=
=
V1 = V2
P2V2
T2
(250 kPa)(V2)
(T2)
(250 kPa)(V2)(308 K)
(150 kPa)(V1)
Notice that V cancels out if V1 = V2
= 513 K
= 240 °C
P1 = 100 kPa, V1 = 5.00 L, T1 = 293 K
P2 = 90 kPa, V2 = ?, T2 = 308 K
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
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.
Q 30
P1 = 800 kPa, V1 = 1.0 L, T1 = 303 K
P2 = 100 kPa, V2 = ?, T2 = 298 K
P1V1
=
T1
(800 kPa)(1.0 L)
=
(303 K)
(V2) =
P2V2
T2
(100 kPa)(V2)
(298 K)
(800 kPa)(1.0 L)(298 K)
(100 kPa)(303 K)
= 7.9 L
Q 32
P1 = 6.5 atm, V1 = 2.0 mL, T1 = 283 K
P2 = 0.95 atm, V2 = ?, T2 = 297 K
P1V1
=
T1
(6.5 atm)(2.0 mL)
=
(283 K)
P2V2
T2
(0.95 atm)(V2)
(297 K)
(6.5 atm)(2.0 mL)(297 K) = 14 mL
(V2) =
(0.95 atm)(283 K)
33. The amount of gas (i.e. number of moles of
gas) does not change.
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