1. Some Review and Reminders

2. Kinetic Molecular Theory
The average kinetic energy of the particles is directly related to the temperature of the system.
Fluids- Liquids and gases (because they flow)
Diffusion- process by which the particles of a substance become evenly distributed as a result of their random movements

3. Let’s talk some more about gases
Physical properties are all similar.
Composed mainly of nonmetallic elements with simple formulas and low molar masses.
Unlike liquids and solids, gases
expand to fill their containers.
are highly compressible.
have extremely low densities.
Two or more gases form a homogeneous mixture.

4. IB – Volume Measurements

5. Molar Volume
According to Avogadro, the volume occupied by one mole of any gas will be the same under the same temperature an pressure conditions
At standard temperature and pressure (STP)
1 mol gas = 22.7 dm3 mol-1

6. Avogadro’s Law
The volume of a gas at constant temperature and pressure is directly proportional to the number of moles of the gas.
Also, at STP, one mole of gas occupies 22.7 dm3.
Mathematically: V = constant  n

8. Properties Which Define the State of a Gas Sample
Temperature
Pressure
Volume
Amount of gas, usually expressed as number of moles

9. 1) Temperature Kelvin v. Celsius
Temperature (K) = temperature (°C)
Standard Temperature = 0°C or K

10. 2) Pressure
Pressure is the amount of force applied to an area:
P = F A
Atmospheric pressure is the weight of air per unit of area.

11. Units of Pressure Pascals: 1 Pa = 1 N/m2 (SI unit of pressure)
Bar: 1 bar = 105 Pa = 100 kPa
mm Hg or torr: These units are literally the difference in the heights measured in mm of two connected columns of mercury, as in the barometer in the figure.
Atmosphere:
1.00 atm = 760 torr = 760 mm Hg = 100 kPa

12. Standard Pressure
Normal atmospheric pressure at sea level is referred to as standard atmospheric pressure.
It is equal to
1.00 atm.
760 torr (760 mmHg).
kPa.

15. Charles’s Law -As temperature increases, volume increases
Volume and Temperature
Remember degrees Kelvin = degrees Celsius
V/T = constant relationship
V1/T1 = V2/T2
-As temperature increases, volume increases
-As temperature decreases, volume decreases
-Direct relationship

16. Charles’s Law Example
A sample of Neon gas occupies a volume of 752mL at 25°C (298K). What will the volume of the gas be at 50°C (323K)?
V1/T1 = V2/T2
V2 = V1T2/T1 = (752 mL)*(323K)/(298K)
V2 = 815 mL Ne

17. Boyle’s Law Pressure and Volume PV = k K = constant relationship
P1V1= P2V2
As pressure increases, volume decreases
As volume increases, pressure decreases
Inverse relationship

18. Boyle’s Law Example
A balloon is filled with 150 mL of Helium gas with initial pressure of atm. What will the volume be if the pressure is increased to atm?
V1P1 = V2P2
Solving for V2
V2 = (V1P1)/P2 = (150mL*0.947 atm)/(0.987 atm)
V2 = 144 mL He
Check it- does this make sense?

19. Gay-Lussac’s Law Pressure and Temperature P = kT or P/T = k
P1/T1 = P2/T2
As pressure increases, temperature increases
As pressure decreases, temperature decreases
Direct relationship

20. Gay-Lussac’s Law Example
A sample of gas under 3.00 atm of pressure is heated from 25°C (298K) to 52°C (325K). What is the resulting pressure?
P1/T1 = P2/T2
P2 = P1*T2/T1 = (3.00 atm)*(325K)/(298K)
P2 = 3.27 atm

21. Combined Gas Law

22. Example
A 40.0 L balloon is filled with air at sea level (1.00 atm, 25°C). It’s tied to a rock and thrown into a cold body of water, and it sinks to the point where the temperature is 4.0°C and the pressure is atm. What will it’s new volume be?

23. Example
A 40.0 L balloon is filled with air at sea level (1.00 atm, 25°C). It’s tied to a rock and thrown into a cold body of water, and it sinks to the point where the temperature is 4.0°C and the pressure is atm. What will it’s new volume be?
3.38 L

24. What is an “ideal gas”?
Described as consisting of largely empty space containing free moving particles of negligible volume having no inter-particle forces.
No gas fits this model exactly, but it is useful for predicting and interpreting how gases behave under different conditions

25. Real Gases
In the real world, the behavior of gases only conforms to the ideal-gas equation at relatively high temperature and low pressure.
Even the same gas will show wildly different behavior under high pressure at different temperatures.

26. Deviations from Ideal Behavior
The assumptions made in the kinetic-molecular model (negligible volume of gas molecules themselves, no attractive forces between gas molecules, etc.) break down at high pressure and/or low temperature.

27. Corrections for Nonideal Behavior
The ideal-gas equation can be adjusted to take these deviations from ideal behavior into account.
The corrected ideal-gas equation is known as the van der Waals equation.
The pressure adjustment is due to the fact that molecules attract and repel each other.
The volume adjustment is due to the fact that molecules occupy some space on their own.

28. Ideal-Gas Equation nT V  P So far we’ve seen that
V  1/P (Boyle’s law).
V  T (Charles’s law).
V  n (Avogadro’s law).
Combining these, we get
V  nT P
Finally, to make it an equality, we use a constant of proportionality (R) and reorganize; this gives the Ideal-Gas Equation: PV = nRT.

29. The Ideal Gas Law
Combines Boyle’s, Charles’s, Gay-Lussac’s and Avogadro’s Laws
PV = nRT
R = gas constant
n = number of moles

30. The Gas Constant
Value depends on the units of temperature and pressure being used

31. Density of Gases n/V = P/RT.
If we divide both sides of the ideal-gas equation by V and by RT, we get
n/V = P/RT.
Also: moles  molecular mass = mass
n  M = m.
If we multiply both sides by M, we get
m/V = MP/RT
and m/V is density, d; the result is:
d = MP/RT.

32. Density & Molar Mass of a Gas
To recap:
One needs to know only the molecular mass, the pressure, and the temperature to calculate the density of a gas.
d = MP/RT
Also, if we know the mass, volume, and temperature of a gas, we can find its molar mass.
M = mRT/PV

33. PV = nRT Example P(10.0) = (0.500)(0.0821)(298) P= (12.2264)/(10.0)
What is the pressure in atm exerted by a mol sample of nitrogen gas in a 10.0L container at 298K?
PV = nRT
P(10.0) = (0.500)(0.0821)(298)
P= ( )/(10.0)

34. Example
What is the pressure in atm exerted by a mol sample of nitrogen gas in a 10.0L container at 298K?
1.22 atm

35. PV = nRT Example (90.0) V = (0.100)(8.314)(300.15) V= (2.7727)/(90.0)
To find the volume of a flask, it was first evacuated so that it contained no gas at all. When 4.40g of carbon dioxide was introduced, it exerted a pressure of 90.0 kPa at 27.0°C. Determine the volume of the flask.
n=4.40g/44g/mol=0.1mol
T=27.0g = K
PV = nRT
(90.0) V = (0.100)(8.314)(300.15)
V= (2.7727)/(90.0)

36. Example
To find the volume of a flask, it was first evacuated so that it contained no gas at all. When 4.40g of carbon dioxide was introduced, it exerted a pressure of 90.0 kPa at 27.0°C. Determine the volume of the flask.
2.77 L = 2.77 dm3