The Combined Gas Law by Abhishek Jaguessar

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.