1. GAS LAWS

2. Kinetic Molecular Theory
Particles in an ideal gas…
have no volume.
have elastic collisions.
are in constant, random, straight-line motion.
don’t attract or repel each other.
have an average KE directly related to Kelvin temperature.

3. Real Gases Particles in a REAL gas… Gas behavior is most ideal…
have their own volume
attract and repel each other
Gas behavior is most ideal…
at low pressures
at high temperatures
***Most real gases act like ideal gases except under high pressure and low temperature.

4. Characteristics of Gases
Gases expand to fill any container.
Take the shape and volume of their container.
Gases are fluids (like liquids).
Little to no attraction between the particles
Gases have very low densities.
= lots of empty space between the particles

5. Characteristics of Gases
Gases can be compressed.
lots of empty space between the particles
Indefinite density
Gases undergo diffusion.
random motion
scatter in all directions

6. Pressure
Which shoes create the most pressure?

7. Pressure- how much a gas is pushing on a container.
Atmospheric pressure- atmospheric gases push on everything on Earth
UNITS AT SEA LEVEL
1 atm =101.3 kPa (kilopascal)= 760 mmHg =760 torr

8. Pressure Barometer measures atmospheric pressure Aneroid Barometer
Mercury Barometer
Aneroid Barometer

9. Pressure Manometer measures contained gas pressure U-tube Manometer
Bourdon-tube gauge
C. Johannesson

10. Temperature= how fast the molecules are moving
Always use absolute temperature (Kelvin) when working with gases. ºF ºC K
-459 32
212
-273
100
273
373
K = ºC + 273
C. Johannesson

11. Standard Temperature & Pressure
STP
Standard Temperature & Pressure
0°C K
1 atm kPa
760 mm Hg
-OR-
-OR-

12. Volume = how much space a gas occupies
Units
L, mL, cm3
1000 mL = 1 L
1 mL = 1 cm3

13. BASIC GAS LAWS P V T

14. Charles’ Law V1 = V2 T1 T2 T is always in K V T
T  V (temperature is directly proportional to volume)
T ↑ V↑ & T↓ V↓
V1 = V2
T T2 T is always in K
K = °C + 273
P and n = constant V T

15. Charles’ Law V1 V2 = T1 T2 (Pressure is held constant)
The Kelvin temperature of a gas is directly related to the volume of the gas when there is no change in pressure or amount.
T T2
V V2 =
(Pressure is held constant)
Timberlake, Chemistry 7th Edition, page 259

16. MECHANICS OF BREATHING
Charles’ Law
MECHANICS OF BREATHING
Gas travels from high pressure to low pressure. This is also responsible for all weather patterns.
Timberlake, Chemistry 7th Edition, page 254

17. Charles’ Law The egg out of the bottle
Hot air rises and gases expand when heated.
Charles carried out experiments to quantify the relationship between the temperature and volume of a gas and showed that a plot of the volume of a given sample of gas versus temperature (in ºC) at constant pressure is a straight line.
Gay-Lussac showed that a plot of V versus T was a straight line that could be extrapolated to –273.15ºC at zero volume, a theoretical state.
The slope of the plot of V versus T varies for the same gas at different pressures, but the intercept remains constant at –273.15ºC.
Plots of V versus T for different amounts of varied gases are straight lines with different slopes but the same intercept on the T axis.
Significance of the invariant T intercept in plots of V versus T was recognized by Thomson (Lord Kelvin), who postulated that –273.15ºC was the lowest possible temperature that could theoretically be achieved, and he called it absolute zero (0 K).
Charles’s and Gay-Lussac’s findings can be stated as: At constant pressure, the volume of a fixed amount of a gas is directly proportional to its absolute temperature
(in K).
This relationship is referred to as Charles’s law and is stated mathematically as
V = (constant) [T (in K)] or V  T (in K, at constant P).
Courtesy Christy Johannesson

18. Charles’ Law Problem Substitute and Solve T1 = 27.0oC +273= 300 K
Mrs. Rodriguez inflates a balloon for a party. She is in an air-conditioned room at 27.0oC, and the balloon has a volume of 4.0 L. Because she is a curious and intrepid chemistry teacher, she heats the balloon to a temperature of 57.0oC. What is the new volume of the balloon if the pressure remains constant?
Given Unkown Equation
Substitute and Solve
T1 = 27.0oC +273= 300 K
V1 = 4.0 L
T2 = 57.0oC +273= 330 K
P1V1 = P2V2
T V1T2
V2 = ? L
4.0 L = V2 =
300 K K
4.4 L

19. Charles’ Law Learning Check
A 25 L balloon is released into the air on a warm afternoon (42º C). The next morning the balloon is recovered on the ground. It is a very cold morning and the balloon has shrunk to 22 L. What is the temperature in º C?
Given Unkown Equation
Substitute and Solve
V1 = 25 L
T1 = 42 oC +273= 315 K
V2 = 22 L
P1V1 = P2V2
T V1T2
T2 = ? ºC
25 L = 22 L =
315 K T2
277.2 K = 4.2 ºC

20. Boyle’s Law
P↓ V ↑ & P↑ V ↓
P  1/V (pressure is inversely proportional to volume)
P1V1 = P2V2
T and n = constant P V

21. Boyle’s Law P1V1 = P2V2 (Temperature is held constant)
When the volume of a gas decreases, its pressure increases as long as there is no change in the temperature or the amount of the gas.
Timberlake, Chemistry 7th Edition, page 253

22. MECHANICS OF BREATHING
Boyle’s Law
Marshmallows in a vacuum
MECHANICS OF BREATHING
Gas travels from high pressure to low pressure. This is also responsible for all weather patterns.
Timberlake, Chemistry 7th Edition, page 254

23. Mechanics of Breathing
Boyle’s Law
Mechanics of Breathing
MECHANICS OF BREATHING
Gas travels from high pressure to low pressure. This is also responsible for all weather patterns.
Timberlake, Chemistry 7th Edition, page 254

24. Boyle’s Law Problem Given Unkown Equation Substitute and Solve
A balloon is filled with 30.L of helium gas at 1.00 atm. What is the volume when the balloon rises to an altitude where the pressure is only 0.25 atm?
Given Unkown Equation
Substitute and Solve
P1V1 = P2V2
T T2
V1 = 30 L
P1 = 1 atm
P2 = .25atm
V2 = ? L
V atm = 30 L x 1.0 atm = 120 L

25. Boyle’s Law Learning Check
A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa.
Given Unkown Equation
Substitute and Solve
V1 = 100. mL
= L
P1 = 150. kPa
P2 = kPa
P1V1 = P2V2
T T2
V2 = ? L
V2 x 200. kPa = L x 150. kPa= 75.0 mL L

26. AVOGADRO’S LAW Vn Vn V n (direct) V1 = V2 n1 n2 T & P Constant VV n

27. Avogadro’s Law Problem
A 3.0 liter sample of gas contains 7.0 moles. How much gas will there be, in order for the sample to be 2.3 liters? P & T do not change
Given Unkown Equation
Substitute and Solve
P1V1 = P2V2
n1T n2T2
V1 = 3.0 L
n1 = 7.0 mol
V2 = 2.3 L
n2 = ? mol
3.0 L = 2.3 L =
7.0 mol n2 mol
5.4 mol

28. Gay-Lussac’s Law P1 = P2 T1 T2 Direct relationship PT PT  P
V & n constant
Direct relationship
PT PT 

29. Gay-Lussac Law
Collapsing Barrel

30. Gay-Lussac Law
Tank car implosion

31. COMBINED IDEAL GAS LAW P1V1 = P2V2 n1T1 n2T2
If P, V, n, or T are constant then they cancel out of the equation.
n usually constant (unless you add or remove gas), so
T T2

32. Combined Gas Law Problem
Ms. Evans travels to work in a hot air balloon from the Rocky Mountains. At her launch site, the temperature is 5.00 °C, the atmospheric pressure is atm, and the volume of the air in the balloon is L. When she lands in Plano, the temperature is 28.0 °C and the atmospheric pressure is kPa. What is the new volume of the air in the balloon?
Given Unkown Equation
Substitute and Solve
T1 = 5.0oC +273= 278 K
P1 = atm
V1 = L
T2 = 28.0oC +273= 301 K
P2 = kPa = 1 atm
V2 = ? L
V1 x P1 = V 2 x P2
T T 2
V2 x 1 atm = L x atm = 104 L
301K K

33. Combined Gas Law Learning Check
Nitrogen gas is in a 7.51 L container at 5.C and 0.58 atm. What is the new volume of the gas at STP?
Given Unkown Equation
Substitute and Solve
T1 = 5.0oC +273= 278 K
P1 = atm
V1 = 7.51 L
T2 = 273 K
P2 = 1 atm
V2 = ? L
V1 x P1 = V 2 x P2
T T 2
V2 x 1.0 atm = 7.51L x 0.58 atm = 4.3 L
273 K K

34. Ideal Gas Law (“Pivnert”)
PV=nRT
R = The Ideal Gas Constant
R = (L*atm) R = 62.4 (L*mm Hg)
(mol*K) (mol*K)
R = (L*kPa)
(mol*K)
V has to be in Liters, n in Moles, T in Kelvin,
P can be in atm, kPa or mmHg
* Choose which R to used based on the units of your pressure.
P V = n R T
(atm) (L) = (moles) (L*atm/mol*K) (K)
(kPa) (L) = (moles) (L*kPa/mol*K) (K)
mm Hg (L) = (moles) (L*mmHg/mol*K) (K)

35. Ideal Gas Law Problem Substitute and Solve
A rigid steel cylinder with a volume of 20.0 L is filled with nitrogen gas to a final pressure of atm at 27.0 oC. How many moles of gas does the cylinder hold?
Given Unkown Equation
Substitute and Solve
V = 20.0 L
P = atm
T =27.0oC +273= 300 K
moles of nitrogen?
PV=nRT
R= atm L/K Mole
n 0821 atm L/K Mole x 300 K = atm x 20.0L= 162 moles

36. Ideal Gas Law Learning Check
A balloon contains 2.00 mol of nitrogen at a pressure of atm and a temperature of 37C. What is the volume of the balloon?
Given Unkown Equation
Substitute and Solve
n = 2.00 mol
P = atm
T =37.0oC +273= 310 K
V in L?
PV=nRT
R= atm L/K Mole
0.980 atm x V= 2.00 mol x atm L/K Mole x 310 K = 51.9 L

37. Dalton’s Law of Partial Pressure
The total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases.
Ptotal = Pgas 1 + Pgas 2 + P­gas 3 + …
A metal container holds a mixture of 2.00 atm of nitrogen, 1.50 atm of oxygen and 3.00 atm of helium. What is the total pressure in the canister?
6.5 atm

38. Welcome to Mole Island
1 mol =
6.02 x 1023 particles

39. Welcome to Mole Island
1 mol = molar mass

40. Welcome to Mole Island
1 mole = 22.4 L
@ STP

41. Gas Stoichiometry Moles  Liters of a Gas:
2C4H10 (g) + 13O2(g) ͢ 8CO2(g) H2O(g)
2 mol mol ͢ mol mol
2 L L ͢ L L
Recall: The coefficients in a chemical reaction represent molar amounts of substances taking part in the reaction.
Avogadro’s principle states that one mole of any gas occupies 22.4 L
at STP.
Thus when gases are involved, the coefficients in a balanced chemical equation represent not only molar amounts but also relatives volumes
Courtesy Christy Johannesson

42. Gas Stoichiometry Problem
In the following combustion reaction, what volume of methane (CH4) is needed to produce 26 L of water vapor?
CH4 (g) + 2O2(g) ͢ CO2(g) + 2H2O(g)
x L ͢ L
1 mol ͢ mol
1 L ͢ L
x L = 26 L
1L L
x= 13 L
Courtesy Christy Johannesson

43. Gas Stoichiometry use ideal gas law
PV=nRT
Looking for grams or moles of gas?
Step 1: start with ideal gas law to find moles of gas
Step 2: 1change to grams of gas
Grams/mol? 1) Use Ideal Gas Law
2) Do stoichiometry calculations
Courtesy Christy Johannesson

44. Example 1 4 Al(s) + 3 O2(g)  2 Al2O3(s)
How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C? PV=nRT
4 Al(s) O2(g)  2 Al2O3(s)
Given Unkown
V O2 = 15.0 L O grams of Al2O3? R= atm L/K Mole
P O2 = 97.3 kPa= atm
T O2 =21oC +273= 294 K
Step 1: Calculate moles of O2
n = PV = atm x 15.0 L = mol O2
RT atm L/K Mole 294 K
Given liters: Start with Ideal Gas Law and calculate moles of O2.
Use stoich to convert moles of O2 to grams Al2O3.
Step 2: Calculate mass of Al2O3
mol O2 = X mol Al2O3 = mol Al2O3
3moleO mole Al2O3
mol Al2O3 x g Al2O3=
1 mol Al2O3
41 g Al2O3
Courtesy Christy Johannesson

45. Gas Stoichiometry use ideal gas law
PV=nRT
Looking for volume of gas?
Step 1: start with stoichiometry conversion to find moles of gas
Step 2: use ideal gas law to find the volume
Liters ? ) Do stoichiometry calculations ) Use Ideal Gas Law
Courtesy Christy Johannesson

46. Example 2 CaCO3  CaO + CO2 Step 2: Calculate volume of CO2
What volume of CO2 forms from 5.25 g of CaCO3 at kPa & 25ºC?
CaCO3  CaO CO2
Given Unkown PV=nRT
m = 5.25 g CaCO volume of CO2? R= atm L/K Mole
P = kPa = 1 atm T =25.0oC +273= 298 K
Step 1: Calculate moles of CO2
5.25 g CaCO3 x 1 mole CaCO3 = mol CaCO3
100 g CaCO3
1 mole CO2 = 1mole CaCO3 ; mol CO2
Looking for liters: Start with stoich and calculate moles of CO2.
Plug this into the Ideal Gas Law to find volume.
Step 2: Calculate volume of CO2
V = nRT = mol CO2 x atm L/K Mole x 298 K = 1.28 L
P 1 atm