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

1 / 15

# T h e C o m b i n e d G a s L a w - PowerPoint PPT Presentation

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

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'T h e C o m b i n e d G a s L a w' - Patman

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.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### 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)

(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

(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)

P1T2V1

P1T2V1

P2T1V2

P1T2V1

P2T1V2

P2T1V2

P2

T1

T2

V1

P1

V2

=

=

=

=

=

=

P2V2

T1V2

P1V1

P2T1

T2P1

T2V1

Combined Gas Law Equations

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)

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

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

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

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.

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

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