1. Gas Laws Day 3

2. Gas Law Foldable
Fold the left and right to the middle.

3. Dalton’s Law of Partial Pressure
The pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases.
PTotal = P1 + P2 + P3….

4. A balloon is filled with air (O2, CO2, & N2) at a pressure of 1.3 atm.
Example
A balloon is filled with air (O2, CO2, & N2) at a pressure of 1.3 atm.
If PO2 = 0.4 atm and PCO2 = 0.3 atm, what is the partial pressure of the nitrogen gas?

5. PTotal = P1 + P2 + P3….
Ptotal = PO2 + PCO2 + PN2
1.3 atm = 0.4 atm atm + PN2
PN2 = 0.6 atm

6. Boyle’s Law: Pressure vs. Volume
At a constant temperature, the volume of a fixed mass of gas varies inversely with the pressure
P1V1 = P2V2
*use if temperature is constant

7. Inverse
Indirect

8. Pressure and Volume with constant Temperature
Changing the volume of the container changes the amount of space between the particles. The less space, the more the particles collide with each other and the walls.

9. Uses of Boyle’s Law
Testing materials stability and ability to maintain their shape under force.
Compressing gases for use in cooking cylinders, SCUBA tanks, and shaving cream.
Used to describe density relationships between gases.

10. Example
If a gas expands from a volume of 5 L to 25 L at an initial pressure of 3.5 atm, what will be its new pressure?
P1 = 3.5 atm
V1= 5 L
P2 = ?
V2 = 25 L

11. Charles’ Law: Volume vs Temperature
At constant pressure, the volume of a fixed mass of gas varies directly with the Kelvin temperature
V1 = V2
T1 T2
*use if pressure is constant

12. Direct

13. Charles’ Law Graph Temp decreases Vol decreases. Temperature (K)
Volume (mL)
546
1092
373
746
283
566
274
548
273
272
544
200
400 50
100
Temp decreases
Vol decreases.

14. Temperature and Volume with constant Pressure
Changing the temperature but requiring the pressure to stay the same causes the volume to increase.

15. Uses of Charles’ Law
Describing the properties of gases, liquids, and solids at extremely low temperatures.
Hot air ballooning
Used to describe density relationships of gases.

16. Example
If 22.4 L of oxygen is heated from 23˚C to 50 ˚C, what is its new volume?
V1 = 22.4 L
T1 = = 296 K
V2 = ?
T2 = = 323 K

17. Absolute Zero Temperature at which all molecular motion stops.
It is defined by 0 K or -273C.
Scientists used Charles Law to extrapolate the temperature of absolute zero.

18. Avogadro’s Law: Volume vs. Moles
At a constant temperature & pressure, the volume of a gas varies directly with the moles
V1 = V2
n1 n2
*use if temperature and pressure
are constant

19. Avogadro’s Law
As the number of moles increases, the volume expands to make room for the additional gas

20. Example
If 4.65 L of CO2 increases from 0.8 moles to 3.75 moles, what is the new volume of the gas?
V1 = 4.65 L
n1 = 0.8 moles
V2 = ? L
n2 = 3.75 moles
V1 = V2
n1 n2

21. V1 = V2 n1 n2 V2 = V1n2 n1 V2 = (4.65 L)(3.75 mol) = 21.79 L

22. The Combined Gas Law
P1 V1 = P2 V2
n1T1 n2T2

23. The Combined Gas Law
Expresses a relationship between pressure, volume, and temperature (and moles) of a fixed amount of gas.
It takes all three gas laws: (Boyle’s, Charles’s, & Avogadro’s) and combines them form one usable equation.

24. Example 1
A gas is cooled from 45˚C to 20˚C. The pressure changes from 103 kPa to kPa as the volumes settles to 16.0 L. What was the initial volume?
P1 = 103 kPa
V1 = ?
T1 = 45oC = 318 K
P2 = kPa
V2 = 16.0 L
T2 = 20oC = 293 K

25. P1 V1 = P2 V2 n1T1 n2T2 (103 kPa)(V1) = (101.3 kPa)(16.0 L)
(n1) (318 K) (n2) (293 K)
V1 = 17.1 L

26. Example 2 (on back of foldable)
A 1.5 mole sample of methane was originally L at 25˚C at 1.1 atm. If we decreased the volume of the container to 0.25 L, increased the pressure to 2.0 atm and added 2.5 moles, what would be the new temperature in ˚C?
P1 = 1.1 atm
V1 = 0.5 L
T1 = = 298 K
n1 = 1.5 moles
P2 = 2.0 atm
V2 = 0.25 L
T2 = ?
n2 = = 4 moles

27. P1 V1 = P2 V2 n1T1 n2T2 P1 V1n2T2 = P2 V2n1T1 P1 V1n2 P1 V1n2
*** easier if you solve for T2 first THEN plug in the #’s ***
P1 V1n2T2 = P2 V2n1T1
P1 V1n P1 V1n2
= (2.0 atm)(0.25 L)(1.5 mol)(298K)
(1.1 atm)(0.5 L)(4.0 mol)
T2 = K – 273
= ˚C