11.4 (Avogadro's and Ideal)

​

​

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

​

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

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

Fill in the Blanks

What Volume does 0.0685 mol of gas occupy at STP? put L

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

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

Fill in the Blanks

What Quantity of gas, in moles, is contained in 2.21 L at STP?

Round to three Decimal places

Mass to Mole Review

Calculating Molar Mass

1. Separate each element of the Chemical Formula.

2. Write out the subscripts and find the molar mass for each element

   1. C₆H₁₂O₆

      1. C 6 x 12.01 = 72.06

      1. H 12 x 1.01 = 12.12

      2. O 6 x 16.00 = 96.00

​C₆H₁₂O₆ molar mass = 180.18 g/mol

Add them together to get the molecular molar mass

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

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

Multiple Choice

How many moles are in 12.5 grams of C₄H₁₀?

• 0.21 g

• 0.069 g

• 726.75 g

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

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

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.

​

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!

​

​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

​

​

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

​

​

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

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

Ideal Gas Law

PV = nRT

where...

Dropdown

In the ideal gas law, P = ​

Select answer

, V = ​

Select answer

, n = ​

Select answer

, R = ​

Select answer

, and T = ​

Select answer

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?

Ideal Gas Law Example

P = 0.949 atm

V = ?

n = 0.867 moles

R = 0.0821 L*atm/mol*K 

T = 305.2 K

Labeling

Place the following values in the correct spot

Drag labels to their correct position on the image

0.949 atm

0.867 mole

305.2 K

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.

Multiple Choice

What is our unknown value?

• P

• V

• n

• R

• T

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

Labeling

Place the following values in the correct spot

Drag labels to their correct position on the image

1.21 atm

295.86 K

0.2598 L