Combined Gas LawBoyle, Charles, and Lussac

• Boyle’s Law: p1v1 = p2v2

• Charles’ Law: v1/T1 = v2/T2

• Gay-Lussac’s Law: p1/T1 = p2/T2

The Gas Laws

Let’s Combine Them

P1V1 P2V2= T1 T2

Sample ProblemA balloon containing hydrogen gas at 200 C and a pressure of 100 kPa has a volume of 7.5 L. Calculate the volume of the balloon after it

rises 10 km into the upper atmosphere, where the temperature is -360 C and the outside air

pressure is 28 kPa.

Sample SolutionUnknown:v2: ?

Known:T1: 200 C p1: 100.0 kPav1: 7.50 L

T2: -360 Cp2: 28 kPa

+ 273 = 293 K

+ 273 = 237 K

Sample SolutionP1V1 P2V2= T1 T2

P1V1T2 V2 = T1P2

Sample Solutionv2 = 100.0 kPa x 7.50 L x 237 K

28 kPa x 293 K

v2 = 22 L

Practice/Homework

Page 438: #s 26 - 33

Unknown(what you’re solving for): P2?Known: V1 = 50.0 ml, P1 = 101 kPa, V2

P1V1T1

= P2V2T2

(101 kPa)(50.0 mL)(T2)(T1)(12.5 mL)=(P2) = 404

kPa

Question 26 (p. 438)

=(P2) (P1) (V1) (T2)(T1) (V2)

Unknown: V2?Known: V1 = 0.10 ml, T1 = 298 K, T2 = 463 K

P1V1T1

= P2V2T2

(P1) (0.10 L) (463 K)(298 K) (P2)

=(V2) = 0.16 L

Homework Questions

=(V2) (P1) (V1) (T2)(T1) (P2)

#27

Unknown: T2?Known: P1 = 150 kPa, T1 = 308 K, P2 = 250 kPa

P1V1T1

= P2V2T2

(250 kPa) (V2) (308 K)(150 kPa) (V1)

=(T2) = 513 K

Homework Questions

=(T2) (P2) (V2) (T1) (P1) (V1)

#28

=2400 C

Unknown: V2?Known: P1 = 100 kPa, V1 = 5.00 L, T1 = 293 K, P2 = 90 kPa kPa, T2 = 308 K

P1V1T1

= P2V2T2

(100 kPa) (5.00 L) (308 K)(293 K) (90 kPa)=(V2) =5.84 L

Homework Questions

=(V2)

#29

(P1) (V1) (T2)(T1) (P2)

Unknown: V2?Known: P1 = 800 kPa, V1 = 1.0 L, T1 = 303 K, P2 = 100 kPa kPa, T2 = 298 K

(800 kPa) (1.0 L) (298 K)(303 K) (100 kPa)=(V2) = 7.9 L

Homework Questions

=(V2)

#30

(P1) (V1) (T2)(T1) (P2)

Unknown: V2?Known: P1 = 6.5 atm, V1 = 2.0 mL, T1 = 283 K, P2 = 0.95 atm, T2 = 297 K

(6.5 atm) (2.0 mL) (297 K)(283 K) (0.95 atm)=(V2) = 14 mL

Homework Questions

=(V2)

#32

(P1) (V1) (T2)(T1) (P2)