2. No, it’s not related to R2D2
Combined Gas Law
The good news is that you don’t have to remember all three gas laws! Since they are all related to each other, we can combine them into a single equation. BE SURE YOU KNOW THIS EQUATION!
P1 V P2 V2 =
T T2
No, it’s not related to R2D2

3. Combined Gas Law
If you should only need one of the other gas laws, you can cover up the item that is constant and you will get that gas law! = P1 V1 P2
Boyle’s Law
Charles’ Law
Gay-Lussac’s Law V2 T1 T2

4. Combined Gas Law Problem
A sample of helium gas has a volume of L, a pressure of atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm?
Set up Data Table
P1 = atm V1 = 180 mL T1 = 302 K
P2 = atm V2= 90 mL T2 = ??
*Make volume units the same!

5. Calculation
P1 = atm V1 = 180 mL T1 = 302 K
P2 = atm V2= 90 mL T2 = ??
P1 V P2 V2 =
T T2
T2 = P2 V2 T1
P1 V1
T2 = 3.20 atm x mL x 302 K atm x mL
T2 = 604 K = °C
= K

6. Learning Check
A gas has a volume of 675 mL at 35°C and atm pressure. What is the temperature in °C when the gas has a volume of L and a pressure of 802 mm Hg?
Note: Volumes must be the same unit and pressures must be the same unit!

7. 1. List variables:
V1 = 675 mL = L
T1 = 35°C = 308 K
P1 = atm
V2 = .315 L
P2 = 802 mmHg = 1.06 atm
Decide on the appropriate gas law:
Everything’s changing, so Combined!
3. Rearrange to solve for unknown:
T2 =
(P2) (V2) (T1)
= K =
(178.4K)
(P1) (V1)

8. Ideal gas law
IF WE COMBINE ALL OF THE LAWS we’ve looked at TOGETHER - INCLUDING AVOGADRO’S LAW - WE GET: PV
=nRT

9. Ideal gas constant(R) R IS A CONSTANT THAT CONNECTS THE 4 VARIABLES
R IS DEPENDENT ON THE UNITS OF THE VARIABLE FOR PRESSURE
TEMP IS ALWAYS IN KELVIN
VOLUME IS ALWAYS IN LITERS
PRESSURE IS IN EITHER atm OR mmHg OR kPa

10. L•atm R=.0821 mol•K L•mmHg R=62.4 mol•K L•kPa R=8.314 mol•K
Because of the different pressure units we use there are 3 different values for “R””
R=.0821
L•atm
mol•K
IF PRESSURE IS GIVEN IN atm
R=62.4
L•mmHg
mol•K
IF PRESSURE IS GIVEN IN mmHg
R=8.314
L•kPa
mol•K
IF PRESSURE IS GIVEN IN kPa

11. Learning Check
Dinitrogen monoxide (N2O), laughing gas, is used by dentists as an anesthetic. If 2.86 mol of gas occupies a 20.0 L tank at 23°C, what is the pressure (in atm) in the tank in the dentist office?

12. T  R  P  V  Using Ideal gas law ? n = 2.86 moles 0.0821 L•atm
1. List variables:
n = 2.86 moles
0.0821
L•atm
mol•K
R 
P  ?
T 
V 
20.0 L
296K
2. Rearrange to solve for unknown:
PV = nRT
P = nRT V
(2.86 mol)(.0821)(296K)
20.0 L
3, Plug & Chug:
=3.48 atm

13. GAS DIFFUSION AND EFFUSION
Effusion is the movement of molecules through a small hole into an empty container.
Diffusion is the movement of molecules to fill a container

14. Graham’s Law M = molar mass & Gas B is the heavier gas!
Rates of effusion of gases are inversely proportional to the square root of their molar masses, at constant temp. & pressure.
Thomas Graham, Professor in Glasgow and London.
M = molar mass & Gas B is the heavier gas!

15. He Graham’s Law Lighter gases effuse faster than heavier ones
Molecules effuse thru holes in a rubber balloon – that’s the main reason they get ‘whimpy’ after awhile! They do this at a rate that is inversely proportional to molar mass. Therefore, He (4 g/mol) effuses more rapidly than O2 (32 g/mol) at same T. (It’s lighter!) He
Lighter gases effuse faster than heavier ones

16. Graham’s Law
We can use the entire equation to calculate the actual speed of gas particles, however…
We will just use the square root side to COMPARE rates of effusion (speeds)
Ex. Compare the rates of effusion of oxygen gas & hydrogen gas.
1st – find their molar masses!
O2 = 32.0 g/mol H2 = 2.0 g/mol
2nd -Put the heavier gas (Gas B) in the numerator! 4
32 g
You’re not done yet!
2.0 g

17. Graham’s Law The number 4 is not much of a “comparison”!
You must put your answer in sentence form!
Try this:
“_______ gas travels (or effuses) at a rate
___ times faster than _________ gas.”
So, the answer is…
“Hydrogen gas travels (or effuses) at a rate
4 times faster than oxygen gas.”
(Lighter gas)
(#)
(heavier)

18. Graham’s Law
You try it!
Compare the rates of effusion of Ar and nitrogen gas (N2)
39.9 g/mol
28.0 g/mol
39.9
28.0
1.19
“Nitrogen gas travels (or effuses) at a rate 1.19 times faster than argon gas.”

19. All of these gas laws work just ducky
assuming the gases are ‘ideal”
(Points with “no volume” & “no mutual
attraction”
Most of the time, gases conform to
ideal conditions.
So, when are gases not “ideal”?
Under conditions of low temperature
& high pressure
(force molecules close enough to
affect each other!)

20. Deviations from Ideal Gas Law
Real Molecules have volumes and attractive forces between them.
Gases are not “Ideal” under conditions of high pressure & low temperature which bring particles close enough together to affect each other!

21. STOP HERE