1. Homework # 11
Dalton’s Law WS

2. Aim: How can we calculate the total pressure exerted by several confined gases? (1st day)

3. Why did you select homogeneous mixture?
Air is a ____________
Homogeneous mixture
Why did you select homogeneous mixture?

4. What gases are present in the air?
78.08% O2
20.95% Ar
0.94%
CO2
0.03%

5. What is the atmospheric pressure in mmHg?

6. Which gas exert more pressure? Why?N2
78.08% O2
20.95% Ar
0.94%
CO2
0.03%

7. The % of Gases in Air : Partial Pressure (STP)
78.08%
593.4 mmHg O2
20.95%
159.2 mmHg Ar
0.94%
7.1 mmHg
CO2
0.03%
0.2 mmHg
PAIR = PN2 + PO2 + PAr + PCO2 = 760 mmHg

8. A. Dalton’s Law of Partial Pressures
Dalton discovered that when gases are mixed in the same container and they have the same temperature, the total pressure exerted by the gases is equal to the sum of the pressures exerted by the individual gases:
Ptotal = Pgas 1 + Pgas Pgas 3………

9. Dalton’s Law of Partial Pressures
V and T are constant P1 P2
Ptotal = P1 + P2

10. 1. Consider a container that has a mixture of N2 and O2 gas
1. Consider a container that has a mixture of N2 and O2 gas. If the pressure of the oxygen gas is 400 torr and the pressure of the nitrogen gas is 360 torr, what is the total pressure?
Ptotal = Pgas 1 + Pgas Pgas 3………
Ptotal = PN2 + PO2
Ptotal = 360 torr torr
Ptotal = 760 torr

11. 2. Refer to the example given before
2. Refer to the example given before. Write the equation for the partial pressure of N2 and O2. If the total pressure in the container is 900 torr and the partial pressure of the oxygen gas is 560, what is the partial pressure of the nitrogen?
Ptotal = PO PN2
900 torr = 560 torr + PN2
900 torr – 560 torr = PN2
PN2 = 340 torr

12. PN2= 0.65 x 760 mm Hg = 494 mm Hg PO2 = 0.15 x 760 mm Hg = 114 mm Hg
3. A mixture of gases at 760 mm Hg pressure contains 65 %nitrogen,15 % oxygen, and 20 % carbon dioxide by volume. What is the partial pressure of each gas?
Ptotal = PO PN2 + P CO2
PN2= x 760 mm Hg = 494 mm Hg
PO2 = x 760 mm Hg = 114 mm Hg
PCO2 = x 760 mm Hg = 152 mm Hg

13. 4. An equilibrium mixture contains H2 at 560 torr, N2 at 180 torr and O2 at 250 torr pressure. What is the total pressure of the gases, in mm Hg and atm, in the system?

14. Ptotal = PH2 + PN2 + PO2
Ptotal = 560 mm Hg mm Hg mm Hg
Ptotal = 990 mm Hg

15. Ptotal = 990 mm Hg X
1 atm
760 mm Hg
Ptotal = 1.3 atm

16. Mole fraction Partial pressure
AND
Partial pressure

17. B. Mole Fractions of Gases from Partial Pressures
The mole fraction of a gas A in a mixture of gases is defines as:
XA = number of moles A_______
total number of moles of gases
XA= mole fraction

19. PA = XAPT The partial pressure of a gas is related to
its mole fraction and can be determined using the following equation:
PA = XAPT
XA = mole fraction
PA = partial pressure
PT = total pressure

20. First: calculate the mole fraction:
5. A vessel contains 0.75 mol of nitrogen, 0.20 mol of hydrogen, and mol of fluorine at a total pressure of 2.5 atm. What is the partial pressure of each gas?
First: calculate the mole fraction:
XN2 = mole
1.00 mole
= 0.75
XH2 = 0.20 mole 1.00 mole
= 0.20
XF2 = mole
1.00 mole
= 0.050

21. Second: Calculate the partial pressure: PA= XAPT
PN2= (0.75)(2.5 atm)
= atm
PH2= (0.20)(2.5 atm)
= 0.50 atm
PF2= (0.050)(2.5 atm)
= 0.13 atm

22. Calculate the mole fraction of each gas in the mixture
6. A mixture of a gas consist of 3.00 moles of helium , 4.00 moles of argon, and 1.00 mole of neon. The total pressure of the mixture is torr.
Calculate the mole fraction of each gas in the mixture
Calculate the partial pressure of each gas in the mixture.

23. XHe = 3.00 mole 8.00 mole = 0.375 = 0.500 XAr = 4.00 mole 8.00 mole
A mixture of a gas consist of 3.00 moles of helium , 4.00 moles of argon, and 1.00 mole of neon. The total pressure of the mixture is torr. a. Calculate the mole fraction of each gas in the mixture
XHe = mole
8.00 mole
= 0.375
XAr = 4.00 mole 8.00 mole
= 0.500
XNe = mole
8.00 mole
= 0.125

24. PA= XAPT PHe= (0.375)(1200. torr) = 450. torr PAr= (0.500)(1200. torr)
A mixture of a gas consist of 3.00 moles of helium , 4.00 moles of argon, and 1.00 mole of neon. The total pressure of the mixture is torr. b. Calculate the partial pressure of each gas in the mixture.
PA= XAPT
PHe= (0.375)( torr)
= 450. torr
PAr= (0.500)(1200. torr)
= 600. torr
PNe= (0.125)(1200) torr
= 150. torr

25. Second Day:

26. C. Collecting a gas over water : https://www. youtube. com/watch

27. How can we find the pressure of the gas collected alone?

28. MnO2
2KClO3 (s)  2KCl (s) + 3O2 (g)
PT = PO2 + PH2O

29. Total pressure is the pressure of the gas + the vapor pressure of the water.
P total = P gas + P water
P gas = P total - P water

30. Table O ºC
Torr
(mmHg)
4.6 26
25.2 5
6.5 27
26.7 10
9.2 28
28.3 15
12.8 29
30.0 16
13.6 30
31.8 17
14.5 40
55.3 18
15.5 50
92.5 19
16.5 60
149.4 20
17.5 70
233.7 21
18.7 80
355.1 22
19.8 90
525.8 23
21.1
100
760.0 24
22.4
105
906.1 25
23.8
110
1074.6

31. = 764 torr – 25.2 torr (see table O)
7. CH4 gas is collected in a eudiometer at 26.0 ºC. When the water level is adjusted, the volume of gas inside the eudiometer is 78.0 mL. If the atmospheric pressure is 764 torr, how many grams of CH4 were collected?
P total = P gas + P water
PCH4
= Patm – P water
PCH4
= 764 torr – 25.2 torr (see table O)
PCH4
= torr

32. PV = nRT
n = m M
PV = mRT M
Solve m:
m = PVM RT

33. m = PVM RT
P = torr = 0.97 atm V = 78.0 mL = L MCH4= 16 g/mole T= 26 ºC = 299 K
0.97 atm x L x 16 _g__
mole
L • atm x 299 K
mole•K
m =

34. m = PVM RT 0.97 atm x 0.078 L x 16 _g__ mole 0.0821 L • atm x 299 K
mole•K
m =
m = g CH4

35. 1. Calculate the partial pressure:
8. When oxygen gas is collected over water at 30 ºC and the total pressure is 645 mm Hg: a) What is the partial pressure of O2? b) What are the mole fraction of oxygen and water vapor?
1. Calculate the partial pressure:
Pt = PO2 + P H2O
PO2 = Pt - P H2O
PO2 = 645 mm Hg – 31.8 mm Hg (see table O)
PO2 = 613 mm Hg

36. b. What are the mole fraction of oxygen and water vapor?
PA= XAPT
The partial pressure of O2 and H2O are related to their mole fractions.
PO2= XO2PT
PH2O = XH2OPT
Mole fraction
Mole fraction

37. PO2= XO2PT PH2O = XH2OPT PO2 = XO2 PT PH2O = XH2O PT 613 = XO2 645
The partial pressure of O2 and H2O are related to their mole fractions.
PO2= XO2PT
PH2O = XH2OPT
PO = XO PT
PH2O = XH2O PT
613 = XO
31.8 = XH2O
645
XO2 = 0.950
XH2O =

38. X O2 + XH2O = = 0.999

39. End of the show!!!