the combined gas law by abhishek jaguessar
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THE COMBINED GAS LAWBY ABHISHEK JAGUESSAR
The CombinedGas
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)5
=7ab
• Isolate a in the equation:• Move b to the other side (from bottom to top)
5 b(x) (y)
=7a
(x)(y)(b)5
=7a
(x)(y)(b)(5)(7)
= a or
• Move 7 to the other side (from top to bottom)
(x)(y)(b)(35)
=a
Multiplication and division• This time, isolate b in the equation:
(x) (y)5
=7ab
• Move b to the other side (it must be on top) …(x) (y)
5=
7ab
• Move everything to the other side of b35axy
=b(b)(x)(y)5
=7a
Q - Rearrange the following equation to isolate each variable (you should have 6 equations)
P1V1 P2V2
T1 T2
=
Combined Gas Law Equations
P1 =P2T1V2
T2V1
V1 =P2T1V2
T2P1
T2 =P2T1V2
P1V1
T1 =P1T2V1
P2V2
P2 =P1T2V1
T1V2
V2 =P1T2V1
P2T1
Combining the gas laws• So far we have seen two gas laws:
Jacques Charles
Robert Boyle
P1V1 = P2V2V1
T1
=V2
T2These are all subsets of a more encompassing law:
the combined gas law
P1
T1
=P2
T2
Read pages 437, 438. Do Q 26 – 33 (skip 31)
P1V1 P2V2
T1 T2
=
Joseph Louis Gay-Lussac
Q 26V1 = 50.0 ml, P1 = 101 kPa
V2 = 12.5 mL, P2 = ? T1 = T2
P1V1
T1
=P2V2
T2
(101 kPa)(50.0 mL)(T1)
=(P2)(12.5 mL)
(T2)
(101 kPa)(50.0 mL)(T2)
(T1)(12.5 mL)=(P2) =404 kPa
Notice that T cancels out if T1 = T2
Q 27V1 = 0.10 L, T1 = 298 K
V2 = ?, T2 = 463 P1 = P2
P1V1
T1
=P2V2
T2
(P1)(0.10 L)(298 K)
=(P2)(V2)
(463)
(P1)(0.10 L)(463 K)
(P2)(298 K)=(V2) = 0.16 L
Notice that P cancels out if P1 = P2
Q 28P1 = 150 kPa, T1 = 308 K
P2 = 250 kPa, T2 = ? V1 = V2
P1V1
T1
=P2V2
T2
(150 kPa)(V1)(308 K)
=(250 kPa)(V2)
(T2)
(250 kPa)(V2)(308 K)
(150 kPa)(V1)=(T2) = 513 K
= 240 °C
Notice that V cancels out if V1 = V2
Q 29 P1 = 100 kPa, V1 = 5.00 L, T1 = 293 K
P2 = 90 kPa, V2 = ?, T2 = 308 KP1V1
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)
=(V2) = 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.
Q 30P1 = 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)
=(V2) = 7.9 L
Q 32P1 = 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)
=(V2) = 14 mL
33. The amount of gas (i.e. number of moles of gas) does not change.
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