1. Chapter 10
Gases

2. Gases
What gases are important for each of the following: O2, CO2 and/or He?
A B C D.

3. Gases
What gases are important for each of the following: O2, CO2 and/or He?
A. CO B. O2/CO C. O2 D. He

4. Pressure
SI units = Newton/meter2 = 1 Pascal (Pa)
1 standard atmosphere = 101,325 Pa
1 standard atmosphere = 1 atm = 760 mm Hg = 760 torr
1 atm = 760 torr
1 atm = 101, 325 Pa

7. Pressure Conversions: An Example
The pressure of a gas is measured as 2.5 atm. Represent this pressure in both torr and pascals.
How do we get there?

8. Exercise
The vapor pressure over a beaker of hot water is measured as 656 torr. What is this pressure in atmospheres?
a) atm
b) atm
c) atm
d) atm
The correct answer is b.
656 torr × (1 atm/760 torr) = atm

9. Four Physical Quantities for Gases
Phys. Qty.
Symbol
SI unit
Other common units
pressure P
Pascal (Pa)
atm, mm Hg, torr, psi
volume V m3
dm3, L, mL, cm3
temp. T K
°C, °F
moles n
mol

10. Boyle’s Law (1st )
In constant temperature (T), and number of moles of gas (n).
PV = k (k is a constant for a given sample of air at a specific temperature)

12. P x V = constant P1 x V1 = P2 x V2 P1 = 726 mmHg P2 = ? V1 = 946 mL
A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL?
P x V = constant
P1 x V1 = P2 x V2
P1 = 726 mmHg
P2 = ?
V1 = 946 mL
V2 = 154 mL
P1 x V1 V2
726 mmHg x 946 mL
154 mL =
P2 =
= 4460 mmHg

13. Exercise
A sample of helium gas occupies 12.4 L at 23°C and atm. What volume will it occupy at 1.20 atm assuming that the temperature stays constant?
9.88 L
P1V1 = P2V2
(0.956 atm) (12.4 L) = (1.20 atm) (V2)
V2 = 9.88 L
(0.956 atm)(12.4 L) = (1.20 atm)(V2)
The new volume is 9.88 L.

14. (constant Pressure and number of moles).
Charles’s Law (2nd)
(constant Pressure and number of moles).
V=bT (b is a proportionality constant)
K = °C + 273
0 K is called absolute zero.

15. Variation in Gas Volume with Temperature at Constant Pressure
As T increases
V increases

16. A sample of carbon monoxide gas occupies 3. 20 L at 125 0C
A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant?
V1 /T1 = V2 /T2
V1 = 3.20 L
V2 = 1.54 L
T1 = K
T2 = ?
T1 = 125 (0C) (K) = K
V2 x T1 V1
1.54 L x K
3.20 L =
T2 =
= 192 K

17. Exercise
Suppose a balloon containing 1.30 L of air at 24.7°C is placed into a beaker containing liquid nitrogen at –78.5°C. What will the volume of the sample of air become (at constant pressure)?
0.849 L
The new volume will become L. Remember to convert the temperatures to Kelvin.
(1.30 L) / ( ) = V2 / ( )

18. V = an (a is a proportionality constant)
Avogadro’s Law (3rd)
(constant T and P).
V = an (a is a proportionality constant)

19. The Relationship Between Volume and Moles

20. Exercise
If 2.45 mol of argon gas occupies a volume of 89.0 L, what volume will 2.10 mol of argon occupy under the same conditions of temperature and pressure?
76.3 L
The new volume is 76.3 L.
(2.45 mol) / (89.0 L) = (2.10 mol) / (V2)

21. Joseph Louis Gay-Lussac (1778-1850)
Gay-Lussac’s Law
If n and V are constant, then P α T
P and T are directly proportional.
If one temperature goes up, the pressure goes up!
Joseph Louis Gay-Lussac ( )

22. Gay Lussac’s Law Summary
The pressure and temperature of a gas are directly related, provided that the volume remains constant.

23. Example P1 P2 T1 T2
3.00 atm V2
298 K
325 K
3.27 L

24. Gay-Lussac’s Law Practice #1
1.8 atm
1.9 atm
273 K T2
288 K or
15.2 oC

25. Gay-Lussac’s Law: Pressure and Temperature
We can simplify this relationship by the formula: Where, P1, P2 = pressure in any unit (atm, kPa, or mmHg), BUT they must match! T1, T2 = temperature is always in Kelvin! (Recall, just add °C)
P1 P2
T1 T2 =

26. Gay-Lussac’s Law: Example
A gas has a pressure of 103kPa at 25°C. What will the pressure be when the temperature reaches 928°C? P1= T1= P2= T2=
103kPa
P1 P2
T1 T2 =
25°C +273= 298K ?
928°C+273= 1201K
(103kPa)
(P2) =
(1201K)
(298K) P2 =
415kPa

27. We can bring all of these laws together into one comprehensive law:
V = bT (constant P and n)
V = an (constant T and P)
V = (constant T and n)
PV = nRT
(where R = L·atm/mol·K, the universal gas constant)

28. The Combined Gas Law P1 V1 P2 V2 = T1 T2 P2 467mmHg = P1= V1= T1= P2=
A gas occupies 3.78L at 529mmHg and 17.2°C. At what pressure would the volume of the gas be 4.54L if the temperature is increased to 34.8°C?
P1=
V1=
T1=
P2=
V2=
T2=
P1 V1 P2 V2
T T2 =
529mmHg
3.78L
17.2°C + 273= 290.2K
(529mmHg)
(3.78L)
(P2)
(4.54L) ? =
4.54L
(290.2K)
(307.8K)
467mmHg
34.8°C + 273= 307.8K P2 =

29. Using PV = nRT Solution 2. Now calc. n = PV / 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?
R = L•atm/K•mol
Solution
2. Now calc. n = PV / RT
n = 1.1 x 103 mol (or about 30 kg of gas)

30. Gases and Stoichiometry
2 H2O2(liq) ---> 2 H2O(g) + O2(g)
Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. What is the pressure of O2 at 25 oC? Of H2O?
Solution
Strategy:
Calculate moles of H2O2 and then moles of O2 and H2O.
Finally, calc. P from n, R, T, and V.

31. Gases and Stoichiometry
2 H2O2(liq) ---> 2 H2O(g) + O2(g)
Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. What is the pressure of O2 at 25 oC? Of H2O?
Solution

32. Gases and Stoichiometry
2 H2O2(liq) ---> 2 H2O(g) + O2(g)
Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. What is the pressure of O2 at 25 oC? Of H2O?
Solution
P of O2 = 0.16 atm

33. PV = nRT nRT V = P 1.37 mol x 0.0821 x 273.15 K V = 1 atm V = 30.7 L
What is the volume (in liters) occupied by 49.8 g of HCl at STP?
T = 0 0C = K
P = 1 atm
PV = nRT
n = 49.8 g x
1 mol HCl
36.45 g HCl
= 1.37 mol
V =
nRT P
V =
1 atm
1.37 mol x x K
L•atm
mol•K
V = 30.7 L

34. PV = nRT n, V and R are constant nR V = P T = constant P1 T1 P2 T2 =
Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)?
PV = nRT
n, V and R are constant nR V = P T
= constant
P1 = 1.20 atm
T1 = 291 K
P2 = ?
T2 = 358 K P1 T1 P2 T2 =
P2 = P1 x T2 T1
= 1.20 atm x
358 K
291 K
= 1.48 atm

35. (of Partial Pressures)
Dalton’s Law
(of Partial Pressures)

36. on the left exerts a pressure
Partial Pressures:
The gas in each tank
on the left exerts a pressure

37. Ptotal = Pgas1 + Pgas2 + Pgas3 …
Dalton’s Law states: The total pressure of a mixture of gases is equal to the sum of the partial pressures of the gases in the mixture.

38. Example Problem 1
An automobile tire contains a mixture of nitrogen, oxygen, and carbon dioxide with partial pressures of 25 psi, 7 psi, and 3 psi respectively. What is the total pressure inside the tire?

39. Example Problem 1
An automobile tire contains a mixture of nitrogen, oxygen, and carbon dioxide with partial pressures of 25 psi, 7 psi, and 3 psi respectively. What is the total pressure inside the tire? Ptotal = 25 psi + 7 psi + 3 psi = 35 psi

40. Example Problem 2
A football has a mixture of nitrogen and oxygen gases. The pressure inside the football is 760. mmHg. The partial pressure of nitrogen (PN2) is 600. mmHg. What is the partial pressure of oxygen (PO2)?

41. Example Problem 2
A football has a mixture of nitrogen and oxygen gases. The pressure inside the football is 760. mmHg. The partial pressure of nitrogen (PN2) is 600. mmHg. What is the partial pressure of oxygen (PO2)? Ptot = PN2 + PO2 760mmHg - 600mmHg = PO2 160mmHg = PO2

42. Example 3
A gas mixture containing oxygen, nitrogen, and carbon dioxide has PO2 = 20.1 kPa, PN2 = 18.3 kPa, and PCO2 = 34.4 kPa. What is Ptotal?
Ptotal = P1 + P2 + P3
Ptotal = PO2 + PN2 + PCO2
Ptotal = 20.1 kPa kPa kPa
Ptotal = 72.8 kPa

43. Example 4
A gas mixture containing oxygen, nitrogen, and argon has a total pressure of 50.2 kPa. If PO2 = 20.1 kPa and PN2 = 18.3 kPa what is PAr?
Ptotal = P1 + P2 + P3
Ptotal = PO2 + PN2 + PAr
PAr = Ptotal - PO2 - PN2
PAr = 50.2 kPa kPa kPa
Ptotal = 11.8 kPa

44. Mole fraction = Pressure fraction

45. E. g. A 1. 00L sample of dry air at 25°C and 786mmHg contains 0
E.g. A 1.00L sample of dry air at 25°C and 786mmHg contains 0.925g of N2, plus other gases. (a) What is the partial pressure of N2 in the air sample? (b) What is the mole fraction and mole percent of N2 in the air?
0.925gN2 x 1mol N2 = molN2
28.0g N2
PN2 = nN2RT/V
= 0.330molx Latm/(K mol) x 298K
1.00L
= atm (=613mmHg)
(b) Mole fraction of N2 = PN2 = 613mmHg = 0.780
P mmHg
Air contains 78.0 mole percent of N2

46. Pi = Xi PT PT = 1.37 atm 0.116 8.24 + 0.421 + 0.116 Xpropane =
A sample of natural gas contains 8.24 moles of CH4, moles of C2H6, and moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)?
Pi = Xi PT
PT = 1.37 atm
0.116
Xpropane = =
Ppropane = x 1.37 atm
= atm

47. GAS DENSITY Screen 12.5
PV = nRT
d and M proportional

48. dRT P M = d = m V = = 2.21 1 atm x 0.0821 x 300.15 K M = M =
A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and C. What is the molar mass of the gas?
dRT P
M =
d = m V
4.65 g
2.10 L =
= 2.21 g L
2.21 g L
1 atm
x x K
L•atm
mol•K
M =
M =
54.5 g/mol

49. Kinetic Energy of Gas Particles
At the same conditions of temperature, all gases have the same average kinetic energy.
m = mass
v = velocity
 At the same temperature, small molecules move FASTER than large molecules

50. The Meaning of Temperature
Kelvin temperature is an index of the random motions of gas particles (higher T means greater motion.)

51. Kinetic Molecular Theory
Maxwell’s equation
where u is the speed and M is the molar mass.
speed INCREASES with T
speed DECREASES with M

52. Diffusion
Diffusion describes the mixing of gases. The rate of diffusion is the rate of gas mixing.
Diffusion is the result of random movement of gas molecules
The rate of diffusion increases with temperature
Small molecules diffuse faster than large molecules

53. Effusion
Effusion: describes the passage of gas into an evacuated chamber.

54. GAS DIFFUSION AND EFFUSION
Diffusion - The rate at which two gases mix. 
Effusion - The rate at which a gas escapes through a pinhole into a vacuum. 

55. GAS DIFFUSION AND EFFUSION
Graham’s law governs effusion and diffusion of gas molecules.
Rate of effusion is inversely proportional to its sq. root molar mass.
Thomas Graham, Professor in Glasgow and London.

56. Graham’s Law of Effusion
M1 = Molar Mass of gas 1
M2 = Molar Mass of gas 2

57. Graham’s Law of Effusion
The rate of effusion of a gas is inversely proportional to the square root of its molar mass.”
Rate = Rate of effusion
M=Molar Mass of gas
Example: What is the relative rate of effusion of H2 vs. O2?

58. Ideal Gas laws – exercise
At what temperature does 16.3 g of nitrogen gas have a pressure of 1.25atm in a 25.0 L tank??

59. Ideal Gas laws – exercise
Calculate the molecular weight of a gas if g of the gas stored in a 7.50 L tank that exerts a pressure of atm at a constant temperature of 35.5°C.

60. Ideal Gas laws – exercise
An unknown gas has a density of g/L at STP. What is the identity of the gas? (Ar, O2, Cl2, HF, H2O)?

61. Problem example: O2 generated in the decomposition of KClO3 is collected over water. The volume of the gas collected at 24 oC and at an atmospheric pressure of 762 torr is 128 ml. Calculate the number of moles of O2 obtained. The vapor pressure of H2O at 24 oC is 22.4 torr.

62. First step: Calculate the partial pressure of O2.
(Dalton’s law)

63. = 762 torr – 22.4 torr
= torr (extra sig. fig)
= atm

64. From the ideal gas equation PV = nRT,
(0.973 atm) (0.128 l) =
( l atm mol-1K-1)(297 K)
= mols