1. Gas Laws and Behavior of Gases
Mrs. Paparella
February

2. Boyle’s Law
At constant temperature the product of Pressure and Volume is a constant PV=k
P1V1=P2V2
Indirect relationship between variables
As pressure on the gas increases , the volume decreases due to gas molecules being forced into smaller space.
Pressure units (atm, kPa, torr, mm Hg) must be the same on both sides of the equation
Volume units ( cm3, ml, L) must also be the same on both sides.

3. Charles’ Law V1 = V2 T1 T2 Temperature MUST be in Kelvin
At constant pressure, Volume/Kelvin Temp = k
V1 = V2
T1 T2
Temperature MUST be in Kelvin
Direct relationship between T and V
As Kelvin Temp increases so does the volume
Increased Temp increases the Kinetic Energy and gas molecules move faster thus increasing the volume of the gas.

4. Gay-Lussac’s Law T1 T2 This law relates Pressure and Temperature
P = P2
T T2
Again you need to make sure to use Kelvin for the Temperature.
When pressure increases , the temperature would also increase.

5. Combined Gas Law P1V1 = P2V2 T1 T2
All 3 variables are in use; must use Kelvin Temp
STP = Standard Temperature and Pressure
Standard Temp = 273 K
Standard Pressure = 1 atm or Kpa or 760 torr or 760 mm Hg

6. Dalton’s Law of Partial Pressures
In a mixture of gases , the total pressure is equal to the sum of the individual partial pressures of the gases.
Ptotal = P1 + P2 + P3 + etc.

7. Dalton’s Law Continued
If you have more moles of one gas than another, that gas will have a greater partial pressure than the others.
Example: A mixture of 3 moles of neon, 2 moles of argon and 1 mole of helium has a total pressure of torr. What is the partial pressure of each gas?
1200 torr = 3x x x
1200 torr = 6x
200 = x
Therefore Ne= 600 torr; Ar = 400 torr; He= 200 torr

8. Dalton’s Law Practice
Solve the following problem: A mixture of gases contains 0.55 mol N2, 0.2 mol O2 and 0.1 mol CO2. If the total pressure of the mixture is 1.32 atm, what is the partial pressure of each component?
Ptotal = P1 + P2 +P3
1.32 atm = 0.55 x x x
1.32 atm = 0.85 x
1.55 atm = x
1.32 atm = 0.55(1.55 atm) (1.55 atm) (1.55 atm)
1.32 atm = atm atm atm

9. Kinetic Molecular Theory of Gases
Gas Molecules move in constant, random straight-line motion.
When gas molecules collide, there is an elastic collision; therefore no energy is lost. Energy can be transferred, but not lost.
The actual volume of the gases themselves is so small in comparison to the volume of the container, that the gases volume is said to be negligible. (ignored) There is a great distance separating the molecules from each other.
Gas molecules do not attract each other.

10. Ideal Gas Behavior Gases like to spread out far and move fast.
What conditions favor this ability?
high temperature– move fast
low pressure- spread out
The examples of ideal gases that you need to remember are Hydrogen and Helium.
They are light weight and easy to move fast and spread out.

11. Kinetic Energy of Gas Molecules
KE = ½ mv2
m = mass of the molecule or atom in Kg
v = velocity of the molecule or atom in m/s
KE = kg m2/s2 = Nm = Joule ( you will use this in Physics many times)
Notice that kinetic energy has the unit of Joules (J), just like Heat energy.
The faster the molecule moves, the greater the KE it possesses.
Remember that Temperature is a measure of the average KE of the molecules.
The higher the temp, the faster they move, the greater the KE.

12. Ideal Gas Law PV = nRT (must memorize) n= number of moles of gas
R = Ideal gas constant = L-atm/mol-K
T= Kelvin Temp
P= Pressure in atm
V= volume in L
You can find any of the variables given the others

13. Dalton’s Law applied with Ideal Gas Law
Rearrange Ideal gas Law for P
P total = (nRT/V) (nRT/V) (nRT/V)3
Solve the following problem: 0.5 mol O2 and 0.3 mol of Ne at -20OC fills a volume of 25.0 L at what pressure?
First convert -20OC to Kelvin ( = 253K)
Ptotal = (0.5mol x L-atm/mol-K x 253K /25 L) +
(0.3 mol x L-atm/mol-K x 253K/25L)

14. Solution: P O2 = 0.415 atm P Ne =0.25 atm P total = 0.665 atm
Try this one: 0.2 mol of Ne and 0.4 mol Kr occupy a volume of 15 Liters at -8 OC . What is the partial pressure of each gas and the total pressure?

15. PV = nRt ; P = nRT/V P Ne = 0.29 atm
P Ne = (0.2 mol x L-atm/mol-K x 265K /15L)
P Ne = 0.29 atm
P Kr = (0.4 mol x L-atm/mol-K x 265K /15L)
P Kr = 0.58 atm
P total = 0.29 atm atm = .87 atm

16. Graham’s Law of Diffusion or Effusion
Effusion-escape through a tiny hole, like the pores in a balloon
The speed at which a gas moves is related to its molecular mass.
Faster gases will have lower masses and diffuse (spread out) faster than heavy gases.
The rate of diffusion or effusion of two gases is proportional to the inverted square root of their molecular masses
Example: Rate of H2 to He = √4/ √ /1.414 = 1.414
The H2 moves times faster than the He

17. Solve these:
Place the following gases in order of increasing average molecular speed at 300 K: CO2, N2O, HF, F2, H2
If you had 4 separate balloons filled with equal volumes of Neon, Helium, Oxygen and Krypton, which balloon would deflate over time the fastest ? What order would the others deflate in?
Compare the rate of diffusion of Neon and Argon.
.6 mole of He and .4 mole of Ar are at Standard Temperature in a container with a volume of 20 L. What is the partial pressure of each gas and the total pressure on the container?
What volume does 2.5 L of gas at STP occupy when the pressure is changed to atm and 80 OC?

18. Solutions to Solve These:
1. CO2, N2O, HF, F2, H2
CO2 = = 44
N2O = =44
HF = 1+19 =20
F2 = 19+19=38
H2 =1+1=2
So the order would be CO2=N2O, F2,HF,H2
2. Helium is the lightest gas so it would deflate 1st, followed by neon, oxygen and krypton.
3. Ne/Ar =√Ar/ √Ne
= √40/ √20
=6.32/4.47
=1.41
Neon moves times faster than Argon

19. P = nRT/V 4. P Ne = (0.6 mol x .0821 L-atm/mol-K x 273K /20L)
P Ar = (0. 4 mol x L-atm/mol-K x 273K /20L)
=0.45 atm
P total = 0.67 atm atm = 1.12 atm

20. P1V1 = P2V2 T1 T2 5. 1.0 atm x 2.5 L = 2.0 atm x V2 273 K 353 K
What volume does 2.5 L of gas at STP occupy when the pressure is changed to atm and 80 OC?
atm x 2.5 L = 2.0 atm x V2
273 K K
V2 =1.6 L

21. Further Application of the Ideal-Gas Equation
See section for details of the equation derivation (Brown , p )
We can use the Ideal gas equation to also relate density according to the molecular mass of the gas.
D= PM/RT ( must memorize)
where M = molecular mass; D= density in g/L ; R= ideal gas constant ; T = Kelvin Temp; P = pressure in atm
Try this: Find the molecular weight of 0.5 g of a gas that occupies a volume of 200 mL at a pressure of 2.0 atm and a temperature of 20°C.
 1st find the density in g/L: 0.5 g/0.200L =2.5 g/L
2nd convert 20°C to Kelvin; 20°C =293K
3rd solve for M

22. Solution to Try This:
Find the molecular weight of 0.5 g of a gas that occupies a volume of 200 mL at a pressure of 2.0 atm and a temperature of 20°C.
 1st find the density in g/L: 0.5 g/0.200L =2.5 g/L
2nd convert 20°C to Kelvin; 20°C =293K
3rd solve for M
D = PM/RT; DRT = PM ; DRT/P = M
2.5 g/L x L-atm/mol –K x 293 K / 2.0 atm = M
30.07 g/mole =Molecular Mass

23. One More for the Road……
Find the molecular mass of a substance that weighs 2.0 g and occupies a volume of 500 ml at Standard Temp and 0.8 atm.

24. Find the molecular mass of a substance that weighs 2
Find the molecular mass of a substance that weighs 2.0 g and occupies a volume of 500 ml at Standard Temp and 0.8 atm.
D= 2.0 g/0.500 L = 4.0 g/L
D =PM/RT
M= DRT/P
M= 4.0 g/L x L-atm/mol-K x 273 K/ 0.8 atm
M =112.1 g/mole