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.