v1 = p v combined gas lawcombined gas law boyle, charles, and lussac •boyle’s law: p1v1 = p2v2...
TRANSCRIPT
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)