1. A Reminder… assume ideal
We that we live in an world where:
Gas particles have no mass
Gas particles have no volume
Gas particles have elastic collisions
These assumptions are used when trying to calculate the AMOUNT of a gas we have!

2. Why are these assumptions important?
PV = nRT
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3. PV = nRT P V n R T The Ideal Gas Law RESSURE OLUME MOLES OF GAS
GAS CONSTANT
EMPERATURE
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4. The MysteRious R 62.4 mmHg · L mol · K 8.31 kPa · L mol · K
R is a constant (doesn’t change).
Number value of R depends on other units.
Units of R are a combination of many units.
62.4 mmHg · L
mol · K
8.31 kPa · L
mol · K
atm · L
mol · K
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5. Ummm… What? PV = nRT P V R = n T (kPa) (mm Hg) (L) (atm) R = (mol) (K)
Solve for R:
P V
R =
n T
Plug in units:
(kPa)
(mm Hg) (L)
(atm)
R =
(mol) (K)

6. Gas Laws, Gas Laws Everywhere!T1 = V2 T2
Charles' Law
Boyle's Law
P1 x V1 = P2 x V2
P1 V1
P2 V2 = T1 T2
Combined Gas Law
Ideal Gas Law
Used with CHANGING CONDITIONS
P V = n R T
Used with only ONE SET OF CONDITIONS

7. When to Use PV = nRT Calculating amount of gas in moles
Calculating P, V, or T if moles of gas are known.
IMPORTANT! We must have 3 out of 4 pieces of information: P V n T

8. Practice with the Ideal Gas Law
A gas sample occupies 2.62 L at 285ºC and 3.42 atm. How many moles are present in this sample?

9. But Let’s Be Practical…
We don’t usually measure in moles!
We usually measure quantities in GRAMS!
PV = nRT
PVM = gRT

10. PVM = gRT P V M g R T RESSURE OLUME OLAR MASS OF GAS (g/mol)
RAMS OF GAS
GAS CONSTANT
EMPERATURE
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11. Practice with the Ideal Gas Law
A balloon is filled with g of helium to a pressure of 1.26 atm. If the desired volume of the balloon is L, what must the temperature be in ºC?

12. PV=nRT vs. PVM=gRT Use PV=nRT when: Use PVM=gRT when:
You are given moles in the problem.
You are searching for moles as an answer.
Use PVM=gRT when:
You are given grams in the problem.
You are searching for grams as an answer.

13. What Else Happens Under Unchanging Conditions?
At constant V and T, pressure is easy to calculate!
“The sum of the individual pressures is equal to the total pressure.”
Total Pressure = Pressure of gas 1 + Pressure of gas Pressure of gas 3 + Pressure of gas 4 …
Ptotal = P1 + P2 + P3 + …
Dalton's Law of Partial Pressures

14. Partial Pressures Practice
A sample of hydrogen gas is collected over water at 25ºC. The vapor pressure of water at 25ºC is mmHg. If the total pressure is mmHg, what is the partial pressure of the hydrogen?

15. Diffusion and Effusion
Diffusion: movement of gas particles from an area of high concentration to an area of low concentration
-occurs in a space where other gas particles are present
Effusion: movement of gas particles from an area of high concentration to a vacuum
-moves to a space where NO other gas particles are present

16. Calculations with Diffusion/Effusion
Graham’s Law of Effusion: The rate of effusion of a gas is inversely proportional to the square root of its molecular weight.
Rate1 =
MM2
Rate2
MM1
“Rate” refers to molecules that effuse per second or liters of gas that effuse per second (it is not time!)
This graph shows the time required for 25ml of each substance to effuse. What conclusion can you make about the rate of effusion compared to the molecular mass of a substance?

17. Effusion Practice
Calculate the ratio of effusion rates for hydrogen and nitrogen.

18. Effusion Practice
2. What is the molar mass of butane? Butane’s rate of effusion is 3.8 times slower than that of Helium.

19. What do Changing Conditions Affect?
We have learned that we can change 3 variables: Temperature, Volume, and Pressure.
If MASS remains constant…
…But VOLUME changes…
Then DENSITY CHANGES! D = M V

20. Two Types of Density Problems:
At STP:
Not at STP:
molar volume of any gas at STP =
D = PM RT
22.4 Liters
Density at STP =
molar mass
molar volume
22.4 Liters

21. Practice with Density Problems:
Determine the density of ethane (C2H6) at STP:
Determine the density of C2H6 at 3.0 atm and 41ºC.