

11.4 (Avogadro's and Ideal)
Presentation
•
Science
•
9th - 12th Grade
•
Medium
+5
Standards-aligned
Stacy King
Used 4+ times
FREE Resource
18 Slides • 10 Questions
1
Gas Laws
Class Announcements:
5 days to makeup or retake exams!
School Announcements:
Check the Finals Schedule
No Tutoring 22nd
Boyle's Law, Charles’s Law
Combined Law
Avogadro and Ideal
Dalton's Law Partial Pressures
2
Avogadro's Principle
22.4 Liters in one mole
Avogadro's Law: volumes of gases at the same temperature and pressure contain the same number of moles
There are 22.4 L : 1 mol
3
Avogadro's Law Calculation
What Volume does 0.0685 mol of gas occupy at STP?
Step one: convert moles of gas to liters
0.685 mol
1 mol
22.4 L
4
Fill in the Blank
5
Multiple Choice
How many liters will 3.59 moles of N2 contain?
80.42 L N2
67.2 L N2
79 L N2
22.4 L N2
6
Avogadro's Law Calculation
What Quantity of gas, in moles, is contained in 2.21 L at STP?
Step one: convert liters of gas to moles
2.21 L
22.4 L
1 mol
7
Fill in the Blank
Round to three Decimal places
8
Mass to Mole Review
Calculating Molar Mass
Separate each element of the Chemical Formula.
Write out the subscripts and find the molar mass for each element
C6H12O6
C 6 x 12.01 = 72.06
H 12 x 1.01 = 12.12
O 6 x 16.00 = 96.00
C6H12O6 molar mass = 180.18 g/mol
Add them together to get the molecular molar mass
9
Multiple Choice
What is the correct molar mass of KCl?
74.55 g/mol
119.00 g/mol
39.10 g/mol
79.90 g/mol
10
Review Stoichiometry
How many moles are in 12.5 grams of C4H10?
C: 4 x 12.01 = 48.04
H: 10 x 1.01 = 10.1
58.14 g/ mol = 58.14 grams : 1 mol
12.5 g C4H10
1 mol
58.14 g
11
Multiple Choice
How many moles are in 12.5 grams of C4H10?
0.21 g
0.069 g
726.75 g
12
Avogadro and Stoichiometry
How many Liters are in 12.5 grams of C4H10?
58.14 g/ mol = 58.14 grams : 1 mol
12.5 g C4H10
1 mol C4H10
58.14 g C4H10
1 mol C4H10
22.4 L C4H10
13
Avogadro and Stoichiometry
How many grams are in 35 L of C4H10?
58.14 g/ mol = 58.14 grams : 1 mol
35 L C4H10
1 mol C4H10
58.14 g C4H10
1 mol C4H10
22.4 L C4H10
14
Ideal Gas Law
An ideal gas is defined as one in which all collisions between atoms or molecules are perfectly elastic and in which there are no intermolecular attractive forces.
The ideal gas law can be viewed as arising from the kinetic (movement) and pressure of gas molecules colliding with the walls of a container.
15
A Picky Law
The Ideal Gas Law is considered picky because there are specific units depending on pressure unit.
If a variable is not in the correct unit, you must convert it!
16
Pressure unit | R Constant |
|---|---|
Pa | 8.314462 |
atm | 0.0821 |
Variable | Symbol | unit |
|---|---|---|
Pressure | P | Pa or atm |
Volume | V | L |
moles | n | moles |
Gas constant | R | L (pressure) |
Temperature | T | K |
Equation PV=nRT
17
Ideal Gas Law Units
Pressure (P) measured in Pa (pascal)
Volume (V) is L (Liters)
moles (n) mols
Temperature (K) Kelvin
Universal gas constant (R) is 8.314462 (constants never change)
Equation: PV = nRT
18
Ideal Gas Law Units
Pressure (P) measured in atm
Volume (V) is L (Liters)
moles (n) mols
Temperature (K) Kelvin
Universal gas constant (R) is 0.0821 (constants never change)
Equation: PV = nRT
19
Match
Match the following
P
T
V
R
n
Pa or atm
K
L
8.314462 or 0.0821
mol
Pa or atm
K
L
8.314462 or 0.0821
mol
20
Ideal Gas Law
PV = nRT
where...
21
Dropdown
22
Ideal Gas Law Example
If 0.867 moles of nitrogen trioxide is collected at 305.2 K and 0.949 atm, what volume of nitrogen trioxide is present?
23
Ideal Gas Law Example
P = 0.949 atm
V = ?
n = 0.867 moles
R = 0.0821 L*atm/mol*K
T = 305.2 K
24
Labelling
Place the following values in the correct spot
0.949 atm
0.867 mole
305.2 K
25
Ideal Gas Law Example
Let's try another one!
A sample of sulfur trioxide is collected when the atmospheric pressure was 1.21 atm and the room temperature was 22.7 oC. It occupies 0.2598L. Calculate the amount of grams of sulfur trioxide present.
26
Multiple Choice
What is our unknown value?
P
V
n
R
T
27
Ideal Gas Law Example #2
After converting from Celsius to Kelvin,
P = 1.21 atm
V = 0.2598 L
n = ?
R = 0.0821 L*atm/mol*K
T = 295.85 K
28
Labelling
Place the following values in the correct spot
1.21 atm
0.2598 L
295.86 K
Gas Laws
Class Announcements:
5 days to makeup or retake exams!
School Announcements:
Check the Finals Schedule
No Tutoring 22nd
Boyle's Law, Charles’s Law
Combined Law
Avogadro and Ideal
Dalton's Law Partial Pressures
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