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P 1 V 1 = P 2 V 2 y = A / x Pressure = A Volume PV = constant Boyle’s Law Inverse relationship ↑P ↓V ↓P ↑V

P 1 V 1 = P 2 V 2

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Boyle’s Law. y = A / x. Pressure = A Volume. PV = constant. P 1 V 1 = P 2 V 2. ↑ P ↓ V. Inverse relationship. ↓P ↑V. Practice Problem A gas occupies a volume of 3.86 L at 0.750 atm . At what pressure will the volume be 4.86 L ?. V 1. P 1. ? P 2. V 2. - PowerPoint PPT Presentation

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Page 1: P 1 V 1  = P 2 V 2

P1V1 = P2V2

y = A / x

Pressure = A Volume

PV = constant

Boyle’s Law

Inverse relationship ↑P ↓V

↓P ↑V

Page 2: P 1 V 1  = P 2 V 2

Practice Problem

A gas occupies a volume of 3.86 L at 0.750 atm.At what pressure will the volume be 4.86 L?

V1 P1

? P2 V2

P1V1 = P2V2

(0.750 atm)(3.86 L) = P2(4.86 L)

(0.750 atm)(3.86 L) = P2

(4.86 L) = 0.596 atm

Page 3: P 1 V 1  = P 2 V 2

Charles’Law

V1 = V2

T1 T2 K = 0C + 273

V = constant

T

y = mx + b

Direct relationship

↑T ↑V

-273

Page 4: P 1 V 1  = P 2 V 2

Practice Problem

A 4.50 L container of nitrogen gas at 28.0 0C is heated to 56.0 0C. Assuming the volume of the container can vary, what is the new volume of the gas?

V1

? V2

T1

T2

V1 = V2

T1 T2 K = 0C + 273

4.50 L = V2

301 K 329 K V2 = 4.92 L

Page 5: P 1 V 1  = P 2 V 2

-273 0C

Gay-Lussac’s Lawy = mx + b

P1 = P2

T1 T2

P = constant

T

Direct relationship

K = 0C + 273

↑T ↑P

Page 6: P 1 V 1  = P 2 V 2

Practice Problem

A gas cylinder contains 40.0 L of gas at 45.0 0C and has a pressure of 650. torr. What will the pressure be if the temperature is changed to 100. 0C?

T1

T2P1

? P2

P1 = P2

T1 T2 K = 0C + 273

650. torr = P2 318 K 373 K

P2 = 762 torr

Page 7: P 1 V 1  = P 2 V 2

P, V PV = constantP, T P = constant

TV, T V = constant

T

Combined Gas LawPV = constant T

P1V1 = P2V2

T1 T2

K = 0C + 273

STP = Standard Temperature & Pressure1 atm00C, 273 K

Page 8: P 1 V 1  = P 2 V 2

Practice Problem

15.00 L of gas at 45.0 0C and 800. torr is heatedto 400.0C, and the pressure changed to 300. torr.What is the new volume?

V1

? V2

T1

T2

P2

P1

P1V1 = P2V2 T1 T2

K = 0C + 273

(800. torr)(15.00 L) = (300. torr) V2 318 K 673 K

V2 = 84.7 L

Page 9: P 1 V 1  = P 2 V 2

Practice Problem

To what temperature must 5.00 L of oxygen at 50. 0C and 600. torr be heated in order to have a volume of 10.0 L and a pressure of 800. torr?

T1

? T2 V1

P1 P2V2

P1V1 = P2V2 T1 T2

K = 0C + 273

(600. torr)(5.00 L) = (800. torr)(10.0 L) 323 K T2

T2 = 861 K = 588 0C

Page 10: P 1 V 1  = P 2 V 2

Do problems: 1 & 2 pg 4434-6 pg 4468 & 9 pg 44811& 12 pg 450