1. Boyle’s Law
y = A / x
Pressure = A
Volume
PV = constant
P1V1 = P2V2
↑P ↓V
Inverse relationship
↓P ↑V

2. Practice Problem
A gas occupies a volume of 3.86 L at 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)
= atm

3. Charles’Law
V = constant T
y = mx + b
V1 = V2
T1 T2
K = 0C + 273
Direct relationship
↑T ↑V
-273

4. Practice Problem
A 4.50 L container of nitrogen gas at C is
heated to C. Assuming the volume of the
container can vary, what is the new volume of
the gas? T1 V1 T2
? V2
V1 = V2
T1 T2
4.50 L = V2
301 K K
K = 0C + 273
V2 = L

5. Gay-Lussac’s Law
y = mx + b
P = constant T
P1 = P2
T1 T2
K = 0C + 273
Direct relationship
-273 0C
↑T ↑P

6. A gas cylinder contains 40.0 L of gas at 45.0 0C and T1
Practice Problem
A gas cylinder contains 40.0 L of gas at C and
has a pressure of 650. torr. What will the pressure be
if the temperature is changed to C? T1
? P2 P1 T2
P1 = P2
T1 T2
650. torr = P2
318 K K
K = 0C + 273
P2 = 762 torr

7. P, V PV = constant
Combined Gas Law
P, T P = constant T
PV = constant T
V, T V = constant T
P1V1 = P2V2
T T2
K = 0C + 273
STP =
Standard
Temperature
& Pressure
00C, 273 K
1 atm

8. Practice Problem
15.00 L of gas at C and 800. torr is heated
to 400.0C, and the pressure changed to 300. torr.
What is the new volume? V1 T1 P1 P2 T2
? V2
P1V1 = P2V2
T T2
K = 0C + 273
(800. torr)(15.00 L) = (300. torr) V2
318 K K
V2 = 84.7 L

9. 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?
? T2 V1 T1 P1 V2 P2
P1V1 = P2V2
T T2
K = 0C + 273
(600. torr)(5.00 L) = (800. torr)(10.0 L)
323 K T2
T2 = 861 K = 588 0C

10. Do problems: 1 & 2 pg 443
4-6 pg 446
8 & 9 pg 448
11& 12 pg 450