1. Gases Chapter 5 E-mail: benzene4president@gmail.com
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2. Gases – Ch. 5
1. Determine if the following are directly or inversely proportional – assume all other variables are constant
a. Pressure and volume
b. Pressure and temperature

3. Gases – Ch. 5
2. The valve between two tanks is opened. See below. Calculate the ratio of partial pressures (O2:Ne) in the container after the valve is opened. a b c d e
3.25-L
8.64 atm O2
2.48-L
5.40 atm Ne

4. Gases – Ch. 5 Combined Gas Law
If one variable (P, V, n or T) of a gas changes
at least one other variable must change
P1V1 n1T1 = P2V2 n2T2

5. Gases – Ch. 5
3. In an experiment 300 m3 of methane is collected over water at 785 torr and 65 °C. What is the volume of the dry gas (in m3) at STP? The vapor pressure of water at 65 °C is 188 torr.

6. Gases – Ch. 5
4. A gaseous mixture containing 1.5 mol Ar, 6 mol He and 3.5 mol Ne has a total pressure of 7.0 atm. What is the partial pressure of Ne?
a atm
b atm
c atm
d atm
e atm

7. Gases – Ch. 5
5. A mixture of oxygen and helium is 92.3% by mass oxygen. What is the partial pressure of oxygen if the total pressure is 745 Torr?

8. Gases – Ch. 5
6. A sample of oxygen gas has a volume of 4.50 L at 27°C and torr. How many oxygen molecules are in the sample?
a × 1023
b × 1022
c × 1024
d × 1022
e. none of these

9. Gases – Ch. 5 Ideal gas law Considering one set of variables
for a gas under “ideal” conditions
PV = nRT
R ⇒ Universal gas constant
R = atmL/molK
R = torrL/molK
R = KPaL/molK or J/molK

10. Gases – Ch. 5 7. Consider the combustion of liquid hexane:
2 C6H14 (l) + 19 O2 (g) → 12 CO2 (g) + 14 H2O (l)
1.52-g of hexane is combined with 2.95 L of oxygen at 312K and 890 torr. The carbon dioxide gas is collected and isolated at 297 K and atm. What volume of carbon dioxide gas will be collected, assuming 100% yield?
a L
b L
c L
d L
e L

11. Gases – Ch. 5
8. A 3.54-g sample of lead(II) nitrate (molar mass = 331 g/mol) is heated in an evacuated cylinder with a volume of 1.60 L. The salt decomposes when heated, according to the following equation:
2 Pb(NO3)2 (s)  2 PbO (s) + 4 NO2 (g) + O2 (g)
Assuming complete decomposition, what is the pressure (in atm) in the cylinder after decomposition and cooling to a temperature of 300. K? Assume the PbO(s) takes up negligible volume.

12. Gases – Ch. 5
mol of O2 gas and 3.0 mol of solid carbon, C (s) are put into a 3.50-liter container at 23°C. If the carbon and oxygen react completely to form CO (g), what will be the final pressure (in atm) in the container at 23°C?

13. Gases – Ch. 5
10. The density of an unknown gas at STP is g/L. Identify the gas.
a. NO
b. Ne
c. CH4
d. O2

14. Gases – Ch. 5 Molar mass (M) can be used to
identify an unknown substance
M = DRT P or
M = mRT PV

15. Gases – Ch. 5
11. Air is 79% N2 and 21% O2 by volume. Calculate the density of air at 1.0 atm, 25°C.
a g/L
b g/L
c g/L
d g/L
e. none of these

16. Gases – Ch. 5
12. These plots represent the speed distribution for 1.0 L of oxygen at 300 K and 1000 K. Identify which temperature goes with each plot.

17. Gases – Ch. 5
Average Speed
𝑈𝑎𝑣𝑒= 8𝑅𝑇 𝜋𝑀

18. Gases – Ch. 5
13. These plots represent the speed distribution for 1.0 L of He at 300 K and 1.0 L of Ar at 300 K. Identify which gas goes with each plot.

19. Gases – Ch. 5
14. Calculate the temperature at which the average velocity of Ar (g) equals the average velocity of Ne (g) at 25°C.
a. 317°C
b. 151°C
c °C
d. 25°C
e. none of these

20. Gases – Ch. 5
15. It takes 12 seconds for 8 mL of hydrogen gas to effuse through a porous barrier at STP. How long will it take for the same volume of carbon dioxide to effuse at STP?

21. effusing under the same conditions
Gases – Ch. 5
Graham’s Law
If two or more gases are
effusing under the same conditions
𝑡𝑖𝑚𝑒1 𝑡𝑖𝑚𝑒2 = 𝑀1 𝑀 𝑜𝑟 𝑟𝑎𝑡𝑒1 𝑟𝑎𝑡𝑒 𝑀2 𝑀1

22. Gases – Ch. 5
16. The diffusion rate of H2 gas is 6.45 times faster than that of a certain noble gas (both gases are at the same temperature). What is the noble gas?
a. Ne
b. He
c. Ar
d. Kr
e. Xe

23. Gases – Ch. 5
17. Consider two 5 L flasks filled with different gases. Flask A has carbon monoxide at 250 torr and 0 °C while flask B has nitrogen at 500 torr and 0 °C.
a. Which flask has the molecules with the greatest average kinetic energy?
b. Which flask has the greatest collisions per second?

24. Gases – Ch. 5 Useful equations KEavg = 3/2RT KE = 1/2mu2
𝐶𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛 𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦⇒ 𝑍= 4𝑁 𝑑 2 𝑉 𝜋𝑅𝑇 𝑀

25. Gases – Ch. 5
18. Under what conditions will a gas behave the most like an ideal gas?

26. Gases – Ch. 5

27. Gases – Ch. 5
19. Which of the following gases will have the lowest molar volume at STP?
a. He
b. CH2Cl2
c. CO2

28. Gases – Ch. 5
The molar volume 𝑉 𝑛 can be derived from the ideal gas law:
𝑉 𝑛 = 𝑅𝑇 𝑃
At STP the molar volume of an ideal gas is L/mol
As a gas deviates from ideal behavior the molar volume decreases

29. Gases – Ch. 5 [ 𝑃 𝑜𝑏𝑠 + a 𝑛 𝑉 2](V – nb) = nRT 𝑣𝑎𝑛 𝑑𝑒𝑟 𝑊𝑎𝑎𝑙𝑠 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛
a ⇒ compensates for the attractive forces between gas particles
b ⇒ compensates for the volume of the gas particles

30. Gases – Ch. 5 Gas a (atmL2/mol2) b (L/mol) He 0.034 0.0237 Ne 0.211
0.0171 Ar
1.35
0.0322 Kr
2.32
0.0398 Xe
4.19
0.0511 N2
1.39
0.0391
CO2
3.59
0.0427
CH4
2.25
0.0428
NH3
4.17
0.0371
H2O
5.46
0.0305

31. Gases – Ch. 5
You have completed ch. 5

32. Answer key – Ch. 5
1. Determine if the following are directly or inversely proportional – assume all other variables are constant
a. Pressure and volume inversely
b. Pressure and temperature directly
PV = nRT
If you multiply variables ⇒ inversely proportional
If you divide variables ⇒ directly proportional

33. Answer key – Ch. 5
2. The valve between two tanks is opened. See below. Calculate the ratio of partial pressures (O2:Ne) in the container after the valve is opened. a b c d e
Each gas is affected by the valve opening
P1V1 n1T1 = P2V2 n2T where n and T are constant
solving for P2 ⇒ P2 = P1V1 V2
for O2 ⇒ P2 = (8.64 atm)(3.25 L) (3.25L+2.48L)
P2 = 4.9 atm
for Ne ⇒ P2 = (5.4 atm)(2.48 L) (3.25L+2.48L) = 2.3 atm
The ratio of partial pressures ⇒ 𝑃 𝑂2 𝑃 𝑁𝑒 = atm 2.3 atm = 2.1

34. Answer key – Ch. 5
3. In an experiment 300 m3 of methane is collected over water at 785 torr and 65 °C. What is the volume of the dry gas (in m3) at STP? The vapor pressure of water at 65 °C is 188 torr.
When a gas is collected over water there will be water vapor in the collection chamber. A dry gas implies that the water vapor has been removed.
The PCH4 = Ptotal – PH2O = 785 torr – 188 torr = 597 torr
At STP the temperature and pressure are 273K and 760 torr respectively.
Using the combined gas law where n is constant
P1V1 n1T1 = P2V2 n2T2
solving for V2 ⇒ V2 = P1V1T2 P2T1 ⇒ V2 = (597 torr)(300 m3)(273 K) (760 torr)( K)
V2 = 190 m3

35. Answer key – Ch. 5
4. A gaseous mixture containing 1.5 mol Ar, 6 mol He and 3.5 mol Ne has a total pressure of 7.0 atm. What is the partial pressure of Ne?
a atm
b atm
c atm
d atm
e atm
PNe = XNePtotal
XNe = nNe ntotal
XNe = 3.5mol 11mol = 0.318
PNe = (0.318)(7.0 atm) = 2.2 atm

36. Answer key – Ch. 5
5. A mixture of oxygen and helium is 92.3% by mass oxygen. What is the partial pressure of oxygen if atmospheric pressure is 745 Torr? If you had a 100 g sample ⇒ 92.3 g of O2 and 7.7 g of He 92.3 g O g/mol = 2.88 mol O2 7.7 g of He g/mol = 1.92 mol He PO2 = XO2Ptotal = 2.88 mol O total mol 745 torr = 447 torr
4.80 total moles

37. Answer key – Ch. 5
6. A sample of oxygen gas has a volume of 4.50 L at 27°C and torr. How many oxygen molecules does it contain?
a × 1023
b × 1022
c × 1024
d × 1022
e. none of these
# of molecules can be derived from moles
PV = nRT
n = 𝑃𝑉 𝑅𝑇
n = (800torr)(4.5L) (62.37torrL/molK)(300K)
n = mol
0.192 mol x 6.022x1023 molecules/mol =
1.16x1023 molecules

38. Answer key – Ch. 5 7. Consider the combustion of liquid hexane:
2 C6H14 (l) + 19 O2 (g) → 12 CO2 (g) + 14 H2O (l)
1.52-g of hexane is combined with 2.95 L of oxygen at 312K and 890 torr. The carbon dioxide gas is collected and isolated at 297 K and atm. What volume of carbon dioxide gas will be collected, assuming 100% yield?
a L
b L
c L
d L
e L
Need to determine the limiting reagent
nC6H14 = g g/𝑚𝑜𝑙 = mol vs.
nO2 = PV RT (890 torr/760 torr/atm)(2.95L) ( atmL/molK)(312K) = mol
0.135/19 < /2 so O2 is the LR
0.135 mol O2 (12 mol CO2) (19 molO2) = mol CO2
V = nRT P = ( mol)( atmL/molK)(297K) (0.93atm) = 2.23 L

39. Answer key – Ch. 5
8. A 3.54-g sample of lead(II) nitrate (molar mass = 331 g/mol) is heated in an evacuated cylinder with a volume of 1.60 L. The salt decomposes when heated, according to the following equation:
2 Pb(NO3)2 (s)  2 PbO (s) + 4 NO2 (g) + O2 (g)
Assuming complete decomposition, what is the pressure (in atm) in the cylinder after decomposition and cooling to a temperature of 300. K? Assume the PbO(s) takes up negligible volume.
The reaction produces 2 gases so the pressure in the container
is the total pressure ⇒ Ptotal = ntotalRT V
3.54 g Pb(NO3)2 (1mol Pb(NO3)2 ) (331 g Pb(NO3)2) 𝑥 (4 mol NO2+1mol O2) 2 mol Pb(NO3)2 = mol gas
Ptotal = ( mol)( atmL/molK)(300K) (1.6L) = 0.41 atm

40. Determine limiting reagent
Answer key – Ch. 5
mol of O2 gas and 3.0 mol of solid carbon, C (s) are put into a 3.50-liter container at 23°C. If the carbon and oxygen react completely to form CO (g), what will be the final pressure (in atm) in the container at 23°C?
2 C (s) + O2 (g)  2 CO (g)
Determine limiting reagent
C ⇒ 3 mol C/2 = 1.5
O2 ⇒ 2.5 mol/1 = 2.5 ⇒ C is the LR
Since C is the LR ⇒ in addition to the CO formed there will be excess O2 in the container so the pressure will be the total pressure ⇒ Ptotal = ntotalRT/V
…continue to next slide

41. Answer key – Ch. 5 9. …continued 2 C (s) O2 (g) → 2 CO (g) 3 mole-3
-3( )
+3( )
1 mole
Since there is 4 mol of gas in the container
Ptotal = (4 𝑚𝑜𝑙)( 𝑎𝑡𝑚𝐿 𝑚𝑜𝑙𝐾 )(23+273𝐾) 3.5 𝐿
Ptotal = 27.8 atm

42. Answer key – Ch. 5
10. The density of an unknown gas at STP is g/L. Identify the gas.
a. NO
b. Ne
c. CH4
d. O2
Molar mass can be useful to identify a substance
M = DRT P
M = (0.715 g/L)( atmL/molK)(273K) (1atm)
M = 16 g/mol
Unknown gas is CH4

43. Answer key – Ch. 5
11. Air is 79% N2 and 21% O2 by volume. Calculate the density of air at 1.0 atm, 25°C.
a g/L
b g/L
c g/L
d g/L
e. none of these
Density is in the equation ⇒
M = DRT P
D = MP RT
Since we have 2 gases ⇒
D = (MN2 PN2 )+(MO2 PO2) RT
D= (28g/mol)(0.79atm)+(32g/mol)(0.21 atm) ( atmL/molK)(298K)
D = 1.18 g/L

44. Answer key – Ch. 5
12. These plots represent the speed distribution for 1.0 L of oxygen at 300 K and 1000 K. Identify which temperature goes with each plot.
According to the average speed
equation ⇒ uavg = (8RT/π M)1/2
we can see the relationship between
average speed and temperature
as T ↑ uavg ↑
since uavg B > uavg A ⇒ TB>TA
Plot A ⇒ 300K
Plot B ⇒ 1000K
Average
Speed of A
Average
Speed of B

45. Answer key – Ch. 5
13. These plots represent the speed distribution for 1.0 L of He at 300 K and 1.0 L of Ar at 300 K. Identify which gas goes with each plot.
According to the average speed
equation ⇒ 𝑈𝑎𝑣𝑒= 8𝑅𝑇 𝜋𝑀
we can see the relationship between
average speed and molar mass
as molar mass ↑ uavg ↓
since uavg B > uavg A ⇒ MB < MA
Plot A ⇒ Ar
Plot B ⇒ He
Average
Speed of A
Average
Speed of B

46. Answer key – Ch. 5 𝑢𝑎𝑣𝑒= 8𝑅𝑇 𝜋𝑀 8𝑅𝑇𝐴𝑟 𝜋𝑀𝐴𝑟 = 8𝑅𝑇𝑁𝑒 𝜋𝑀𝑁𝑒
14. Calculate the temperature at which the average velocity of Ar (g) equals the average velocity of Ne (g) at 25°C.
a. 317°C
b. 151°C
c °C
d. 25°C
e. none of these
𝑢𝑎𝑣𝑒= 8𝑅𝑇 𝜋𝑀
uave Ar = uave Ne
8𝑅𝑇𝐴𝑟 𝜋𝑀𝐴𝑟 = 8𝑅𝑇𝑁𝑒 𝜋𝑀𝑁𝑒
8,R, and π constant
TAr MAr = TNe MNe ⇒ TAr = MArTNe MNe
TAr = (298K)(39.95 g/mol) (20.18 g/mol)
T = 590 K or 317°C

47. timeCO2 = (12 s) 44.01g/mol 2.016g/mol
Answer key – Ch. 5
15. It takes 12 seconds for a given volume of hydrogen gas to effuse through a porous barrier. How long will it take for the same volume of carbon dioxide?
timeCO2 timeH2 = MCO2 MH2
timeCO2 = timeH2 MCO2 MH2
timeCO2 = (12 s) g/mol 2.016g/mol
timeCO2 = 56 s

48. Molar mass can be used to identify
Answer key – Ch. 5
16. The diffusion rate of H2 gas is 6.45 times as great as that of a certain noble gas (both gases are at the same temperature). What is the noble gas?
a. Ne
b. He
c. Ar
d. Kr
e. Xe
Molar mass can be used to identify
rateH2 rateunk = Munk MH2
Munk =MH2 rateH2 rateunk
Munk = g/mol
Munk = g/mol
Unknown gas is Kr

49. Answer key – Ch. 5
17. Consider two 5 L flasks filled with different gases. Flask A has carbon monoxide at 250 torr and 0 °C while flask B has nitrogen at 500 torr and 0 °C.
a. Which flask has the molecules with the greatest average kinetic energy? According to KEavg = RT ⇒ we see the relationship ⇒ as T ↑ KEavg↑ ⇒ since both flasks are at the same T they will have the same KEavg
b. Which flask has the greatest collisions per second? According to
𝑍= 4𝑁 𝑑 2 𝑉 𝜋𝑅𝑇 𝑀
we see three relationships ⇒ as T↑ Z↑ or as molar mass↑ Z↓ or as N/V (or P)↑ Z↑ ⇒ so since both flasks have the same T and molar mass but the PB > PA ⇒ ZB > ZA

50. Answer key – Ch. 5
18. Under what conditions will a real gas behave like an ideal gas?
An “ideal” gas is one that in reality adheres to the ideal gas law ⇒ meaning experimental values agree with calculated values using PV = nRT
Gases are more likely to behave “idealy” when the pressure is low and/or the temperature is high
Deviations from the ideal gas law is due to the attractive forces between the gas particles and the volume of the gas particles relative to the volume of the container

51. Answer key – Ch. 5
19. Which of the following gases will have the lowest molar volume at STP?
a. He
b. CH2Cl2
c. CO2
The molar volume of an “ideal” gas is 22.4 L/mol
as the attractive forces of the gas particles ↑
the molar volume ↓ – later (in ch 16)
we will learn the specifics of attractive forces
however for now we can use the relationship that as molar
mass ↑ attractive forces ↑ (an exception is water – although water
is rather on the light side it has quite strong attractive
forces called H-Bonds which we’ll see further in ch 16)
Therefore since CH2Cl2 has the highest molar mass it has the
strongest attractive forces and the lowest molar volume