1. The Ideal Gas Law & Co.

2. 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!

3. Why are these assumptions important?
PV = nRT
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4. PV = nRT P V n R T The Ideal Gas Law RESSURE OLUME MOLES OF GAS
GAS CONSTANT
EMPERATURE
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5. 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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6. 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)

7. 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

8. 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

9. 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?
PV = nRT
P V n =
R T
(3.42 atm)
(2.62 L) n = =
0.196 mol
L · atm
mol · K
(558 K)

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

11. 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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12. 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?
P V M
PVM = gRT T =
g R
4.00 g
mol
(1.26 atm)
(1.250 L) T =
308 K =
- 273
L · atm
mol · K
( g)
35 ºC

13. 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.

14. 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

15. 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?
Ptotal = PH PH2O =
523.8 mm Hg
PH2 +
23.8 mm Hg =
PH2
500.0 mm Hg
Source: 2003 EOC Chemistry Exam

16. 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

17. Two Types of Density Problems:
At STP:
Not at STP:
molar volume of any gas at STP =
Determine new volume (V2 ) using Combined Gas Law
22.4 Liters
P1 V1
P2 V2 = T1 T2
STP Values
non-STP
Density at STP =
Density at non-STP =
molar mass
molar mass
molar volume
22.4 Liters V2

18. Practice with Density Problems:
Determine the density of ethane (C2H6) at STP:
Determine the density of C2H6 at 3.0 atm and 41ºC.
P1 V1
P2 V2 = T1 T2
V2 = L
molar mass
D (at STP) =
molar volume
D = molar mass V2
V2 = P1 V1 T2
molar mass = g
T1 P2
P1 = 1.0 atm
V1 = 22.4 L
T1 = 273 K
P2 = 3.0 atm
V2 = ?
T2 = 314 K
molar volume = L
V2 = (1.0 atm) (22.4 L) (314 K)
D = g
8.6 L =
3.5 g/L
(273 K) (3.0 atm)
30.08 g
D = =
1.34 g/L
22.4 L
V2 = 8.6 L

19. What do Changing Conditions Affect?
Density
Stoichiometry problems
So Far:
Now…
STP
Non-STP

20. Mass-Volume at Non-STP
Two parts to solving these problems:
Use Stoichiometry
Use Gas Law