1. Gases and Gas Laws Chemistry– Unit 11: Chapter 14

2. Kinetic Molecular Theory (THIS IS IMPORTANT!!)
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 avg. KE directly related to Kelvin temperature.

3. Real Gases Particles in a REAL gas… Gas behavior is most ideal…
have their own volume
attract each other
Gas behavior is most ideal…
at low pressures
at high temperatures
in nonpolar atoms/molecules

4. Characteristics of Gases
Gases expand to fill any container.
random motion, no attraction
Gases are fluids (like liquids).
no attraction
Gases have very low densities.
no volume = lots of empty space

5. Characteristics of Gases
Gases can be compressed.
no volume = lots of empty space
Gases undergo diffusion & effusion.
random motion

6. The Meaning of Temperature
(KE) 3 2
avg = RT
Kelvin temperature is an index of the random motions of gas particles (higherT means greater motion.)

7. Kinetic Energy of Gas Particles
At the same conditions of temperature, all gases have the same average kinetic energy. 2 1 mv KE =
m = mass
v = velocity

8. Temperature K = ºC + 273 ºF -459 32 212 ºC -273 100 K 273 373
Always use absolute temperature (Kelvin) when working with gases. ºF
-459 32
212 ºC
-273
100 K
273
373
K = ºC + 273

9. ºF =
ºC = 32 C 5 9 o +
32) - F ( 9 5 o
Absolute Zero: _____________________________________
Absolute Zero = ____K = _____ºC
Temperature at which motion stops.
- 273

10. 32) - (355 9 5 = Example Conversions: (a) Convert 355 ºF to ºC and K
(b) Convert -40 ºC to K and ºF
32) -
(355 9 5 = oC
= 179oC
K = 179 oC + 273
= 452 K 9 oF o
= -40 oF = - 40 C + 32 5
K = - 40 oC + 273
= 233 K

11. Pressure
Which shoes create the most pressure?

12. Measuring Pressure
The first device for measuring atmospheric pressure was developed by Evangelista Torricelli during the 17th century.
The device was called a “barometer”.
Baro = weight
Meter = measure

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

14. Pressure 101.3 kPa (kilopascal) 1 atm 760 mm Hg 760 torr 14.7 psi
KEY UNITS AT SEA LEVEL
101.3 kPa (kilopascal)
1 atm
760 mm Hg
760 torr
14.7 psi
** All of these amounts are equal to each other (conversion factors), just in different units!

15. Pressure Conversions 475 mm Hg x = 0.625 atm 29.4 psi x
A. What is 475 mm Hg expressed in atm?
1 atm
760 mm Hg
B. The pressure of a tire is measured as 29.4 psi.What is this pressure in mm Hg?
14.7 psi
475 mm Hg x
= atm
29.4 psi x
= 1.52 x 103 mm Hg

16. Pressure Conversions A. What is 2 atm expressed in torr?
B. The pressure of a tire is measured as 32.0 psi.What is this pressure in kPa?

17. IQ 1 List the properties of an ideal gas. What is this theory called?
How is an ideal gas different from a real gas?

18. IQ #3
What are 2 ways to increase the pressure of a gas (Think about the formula for pressure)?
What effect does temperature have on kinetic energy?
What do you expect would happen to the volume of a balloon if the gas inside it made more collisions with it?

19. Boyle’s Law
Robert Boyle ( ). Son of Earl of Cork, Ireland. P V
PV = k

20. Boyle’s Law
The pressure and volume of a gas are inversely related
at constant mass & temp P V
PV = k

21. Boyle’s Law P1V1 = P2 V2 For example, P goes up as V goes down.
This means Pressure and Volume are INVERSELY PROPORTIONAL if moles and temperature are constant (do not change).
For example, P goes up as V goes down.
P1V1 = P2 V2
As the volume of the air trapped in the bicycle pump is reduced, its pressure goes up, and air is forced into the tire.

22. Gas Law Problems BOYLE’S LAW P V
A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa.
BOYLE’S LAW
GIVEN:
V1 = 100. mL
P1 = 150. kPa
V2 = ?
P2 = 200. kPa P V
WORK:
P1V1= P2V2 V2= P1V1 P2
(150.kPa)(100.mL)=(200.kPa)V2
V2 = 75.0 mL

23. Charles’ Law V T

24. Charles’ Law
The volume and absolute temperature (K) of a gas are directly related
at constant mass & pressure V T

25. Charles’s Law V and T are directly proportional. V1 V2 = T1 T2
If one temperature goes up, the volume goes up!
Jacques Charles ( ). Isolated boron and studied gases. Balloonist.

26. Modern long-distance balloon
Charles’s original balloon
Modern long-distance balloon

27. Gas Law Problems CHARLES’ LAW T V
A gas occupies 473 cm3 at 36°C. Find its volume at 94°C.
CHARLES’ LAW
GIVEN:
V1 = 473 cm3
T1 = 36°C = 309K
V2 = ?
T2 = 94°C = 367K T V
WORK:
V1 = V2
T1 T2
V2 = V1T2 T1
(473 cm3 x 367 K)/309 K =V2
V2 = 562 cm3

28. Gay-Lussac’s Law P T

29. Gay-Lussac’s Law
The pressure and absolute temperature (K) of a gas are directly related
at constant mass & volume P T

30. Gay-Lussac’s Law P and T are directly proportional. P1 P2 = T1 T2
If one temperature goes up, the pressure goes up!
Joseph Louis Gay-Lussac ( )

31. Gas Law Problems GAY-LUSSAC’S LAW P T
A gas’ pressure is 765 torr at 23°C. At what temperature will the pressure be torr?
GAY-LUSSAC’S LAW
GIVEN:
P1 = 765 torr
T1 = 23°C = 296K
P2 = torr
T2 = ? P T
WORK:
P1 = P2
T1 = T2
T2 = T1P2 P1
T2 = (560.0 torr)(296K)/765 torr
T2 = 217 K = -56°C

32. Combined Gas Law = T1 T2 P1 V1 P2 V2
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

33. 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! =
Boyle’s Law
Charles’ Law
Gay-Lussac’s Law P1 V1 T1 P2 V2 T2

34. Gas Law Problems COMBINED GAS LAW P T V V1 = 7.84 cm3 P1 = 71.8 kPa
A gas occupies 7.84 cm3 at 71.8 kPa & 25°C. Find its volume at STP.
COMBINED GAS LAW
GIVEN:
V1 = 7.84 cm3
P1 = 71.8 kPa
T1 = 25°C = 298 K
V2 = ?
P2 = kPa
T2 = 273 K
P T V
WORK:
P1V1T2 = P2V2T1
(71.8 kPa)(7.84 cm3)(273 K)
=( kPa) V2 (298 K)
V2 = 5.09 cm3

35. Combined Gas Law Problem #2
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 = ??

36. Calculation T2 = 604 K - 273 = 331 °C = 604 K T2 = P2 V2 T1 P1 V1
P1 = atm V1 = 180 mL T1 = 302 K
P2 = atm V2= 90 mL T2 = ??
P1 V P2 V2
= P1 V1 T2 = P2 V2 T1
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

37. Learning Check (Group)
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?

38. IQ 2 1. Covert -56 °C to Kelvin. 2. How many torr in 37.8 psi?
3. Covert 0 °C into °F.

39. One More Practice Problem (Group)
A balloon has a volume of 785 mL on a fall day when the temperature is 21°C. In the winter, the gas cools to 0°C. What is the new volume of the balloon?

40. Try This One on Your Own!
A sample of neon gas used in a neon sign has a volume of 15 L at STP. What is the volume (L) of the neon gas at 2.0 atm and –25°C?

41. Standard Temperature & Pressure
STP
STP allows us to compare amounts of gases between different pressures and temperatures
Standard Temperature & Pressure
0°C K
1 atm kPa
-OR-

42. Avogadro’s Hypothesis
Equal volumes of gases at the same T and P have the same number of molecules.
V = n (RT/P) = kn
V and n are directly related.
twice as many molecules

43. Avagadro’s Hypothesis and the Kinetic Molecular Theory
The gases in this experiment are all measured at the same T and V.
P proportional to n

44. IDEAL GAS LAW P V = n R T Brings together gas properties.
Can be derived from experiment and theory.
BE SURE YOU KNOW THIS EQUATION!

45. Using PV = nRT P = Pressure V = Volume T = Temperature
n = number of moles
R is a constant, called the Ideal Gas Constant
R =
L • atm
Mol • K

46. Sample Problem #1
How much N2 is required to fill a small room with a volume of 960 cubic feet (27,000 L) to 745 mm Hg at 25 oC?
Solution
1. Get all data into proper units
V = 27,000 L
T = 25 oC = 298 K
P = 745 mm Hg (1 atm/760 mm Hg) = 0.98 atm
And we always know R, L atm / mol K

47. Using PV = nRT PV = nRT RT RT
How much N2 is req’d to fill a small room with a volume of 960 cubic feet (27,000 L) to P = 745 mm Hg at 25 oC?
Solution
2. Now plug in those values and solve for the unknown.
PV = nRT
RT RT
n = 1.1 x 103 mol (or about 30 kg of gas)

48. 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 (mm Hg) in the tank in the dentist office?

49. Learning Check #2
A 5.0 L cylinder contains oxygen gas at 20.0°C and 735 mm Hg. How many grams of oxygen are in the cylinder?

50. IQ #3 What are “STP” conditions? What is the volume of an ideal gas?
A 5.0 L cylinder contains oxygen gas at 20.0°C and 735 mm Hg. How many grams of oxygen are in the cylinder?
273 K/ 1 atm
An ideal gas has no volume!!
6.4 g

51. Deviations from Ideal Gas Law
Real molecules have volume.
The ideal gas consumes the entire amount of available volume. It does not account for the volume of the molecules themselves.
There are intermolecular forces.
An ideal gas assumes there are no attractions between molecules. Attractions slow down the molecules and reduce the amount of collisions.
Otherwise a gas could not condense to become a liquid.

52. Gas Stoichiometry Moles  Liters of a Gas: STP - use 22.4 L/mol
Non-STP - use ideal gas law
Non-STP
Given liters of gas?
start with ideal gas law
Looking for liters of gas?
start with stoichiometry conv.

53. Gas Stoichiometry Problem #1
What volume of CO2 forms from 5.25 g of CaCO3 at 1.02 atm & 25ºC?
CaCO3  CaO CO2
5.25 g
? L non-STP
Looking for liters: Start with stoich and calculate moles of CO2.
5.25 g
CaCO3
1 mol
CaCO3
100.09g
1 mol
CO2
CaCO3
= mol CO2
Plug this into the Ideal Gas Law to find liters.

54. Gas Stoichiometry Problem #1
What volume of CO2 forms from 5.25 g of CaCO3 at 1.02 atm & 25ºC?
GIVEN:
P = 1.02 atm
V = ?
n = mol
T = 25°C = 298 K
R = Latm/molK
WORK:
PV = nRT
(1.02 atm)V =( mol)( Latm/molK) (298K)
V = 1.26 L

55. Gas Stoichiometry Problem #2
How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C?
4 Al O2  2 Al2O3
15.0 L
non-STP
? g
GIVEN:
P = 97.3 kPa= .961atm
V = 15.0 L
n = ?
T = 21°C = 294 K
R = Latm/molK
WORK:
PV = nRT
(.961 atm) (15.0 L) = n ( Latm/molK) (294K)
n = mol O2
Given liters: Start with Ideal Gas Law and calculate moles of O2.
NEXT 

56. Gas Stoichiometry Problem #2
How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C?
4 Al O2  2 Al2O3
15.0L
non-STP
? g
Use stoich to convert moles of O2 to grams Al2O3.
2 mol Al2O3
3 mol O2
g Al2O3
1 mol
Al2O3
0.597 mol O2
= 40.6 g Al2O3

57. Gas Stoichiometry Problem #3
2 H2O2 (l) ---> 2 H2O (g) + O2 (g)
Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. What is the volume of O2 at STP?
Bombardier beetle uses decomposition of hydrogen peroxide to defend itself.

58. Gas Stoichiometry Problem #3
2 H2O2 (l) ---> 2 H2O (g) + O2 (g)
Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. What is the volume of O2 at STP?
Solution
1.1 g H2O mol H2O mol O L O g H2O mol H2O2 1 mol O2
= 0.36 L O2 at STP!!

59. Gas Stoichiometry: Practice!
A. What is the volume at STP of 4.00g of CH4?
How many grams of He are present in 8.0 L of gas at STP?

60. Ptotal = P1 + P2 + ... Dalton’s Law
The total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases.
Ptotal = P1 + P
When a H2 gas is collected by water displacement, the gas in the collection bottle is actually a mixture of H2 and water vapor.

61. Dalton’s Law GIVEN: PH2 = ? Ptotal = 94.4 kPa PH2O = 2.72 kPa WORK:
Hydrogen gas is collected over water at 22.5°C. Find the pressure of the dry gas if the atmospheric pressure is 94.4 kPa.
The total pressure in the collection bottle is equal to atmospheric pressure and is a mixture of H2 and water vapor.
GIVEN:
PH2 = ?
Ptotal = 94.4 kPa
PH2O = 2.72 kPa
WORK:
Ptotal = PH2 + PH2O
94.4 kPa = PH kPa
PH2 = 91.7 kPa
Look up water-vapor pressure for 22.5°C.
Sig Figs: Round to least number of decimal places.

62. Dalton’s Law GIVEN: Pgas = ? Ptotal = 742.0 torr PH2O = 42.2 torr
A gas is collected over water at a temp of 35.0°C when the barometric pressure is torr. What is the partial pressure of the dry gas?
The total pressure in the collection bottle is equal to barometric pressure and is a mixture of the “gas” and water vapor.
GIVEN:
Pgas = ?
Ptotal = torr
PH2O = 42.2 torr
WORK:
Ptotal = Pgas + PH2O
742.0 torr = PH torr
Pgas = torr
Look up water-vapor pressure for 35.0°C.
Sig Figs: Round to least number of decimal places.

63. Graham’s Law Diffusion Effusion
Spreading of gas molecules throughout a container until evenly distributed.
Effusion
Passing of gas molecules through a tiny opening in a container

64. KE = ½mv2 Graham’s Law Speed of diffusion/effusion
Kinetic energy is determined by the temperature of the gas.
At the same temp & KE, heavier molecules move more slowly.
Larger m  smaller v
KE = ½mv2

65. Graham’s Law Graham’s Law
Rate of diffusion of a gas is inversely related to the square root of its molar mass.
The equation shows the ratio of Gas A’s speed to Gas B’s speed.

66. Graham’s Law
Determine the relative rate of diffusion for krypton and bromine.
The first gas is “Gas A” and the second gas is “Gas B”.
Relative rate mean find the ratio “vA/vB”.
Kr diffuses times faster than Br2.

67. Put the gas with the unknown speed as “Gas A”.
Graham’s Law
A molecule of oxygen gas has an average speed of m/s at a given temp and pressure. What is the average speed of hydrogen molecules at the same conditions?
Put the gas with the unknown speed as “Gas A”.

68. Graham’s Law
An unknown gas diffuses 4.0 times faster than O2. Find its molar mass.
The first gas is “Gas A” and the second gas is “Gas B”.
The ratio “vA/vB” is 4.0.
Square both sides to get rid of the square root sign.