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GasesChapter 5

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Gases – Ch. 51. Draw the following:

a. A closed monometer attached to a flask filled with CO at 250 torr

b. An open monometer at sea level attached to a flask filled with N2O at 600 torr

Gases – Ch. 5

Gases – Ch. 52. Determine if the following are directly or inversely proportional –

assume all other variables are constanta. Pressure and volumeb. Pressure and temperature

3. 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. 1.31b. 1.60c. 2.10d. 0.477e. 0.615

Gases – Ch. 5

3.25-L8.64 atm

O2

2.48-L5.40 atm

Ne

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

=

Gases – Ch. 54. 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.

Gases – Ch. 55. Consider a sample of neon gas in a container fitted with a movable

piston (assume the piston is mass-less and frictionless). The temperature of the gas is increased from 20.0°C to 40.0°C. The density of neon

a. increases less than 10%.b. decreases less than 10%.c. increases more than 10%.d. decreases more than 10%.e. does not change.

Gases – Ch. 56. 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. 1.4 atmb. 2.2 atmc. 3.8 atmd. 4.6 atme. 2.7 atm

Gases – Ch. 57. 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?

Gases – Ch. 58. A sample of oxygen gas has a volume of 4.50 L at 27°C and 800.0

torr. How many oxygen molecules are in the sample?

a. 1.16 × 1023

b. 5.8 × 1022

c. 2.32 × 1024

d. 1.16 × 1022

e. none of these

Gases – Ch. 5

Ideal gas lawConsidering one set of variables

for a gas under “ideal” conditionsPV = nRT

R ⇒ Universal gas constantR = 0.08206 atmL/molKR = 62.37 torrL/molK

R = 8.314 KPaL/molK or J/molK

Gases – Ch. 59. 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 0.930 atm. What volume of carbon dioxide gas will be collected, assuming 100% yield?

a. 0.504 Lb. 1.93 Lc. 2.23 Ld. 0.607 Le. 4.04 L

Gases – Ch. 510. 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.

Gases – Ch. 511. 2.5 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?

Gases – Ch. 512. The density of an unknown gas at STP is 0.715 g/L. Identify the

gas.

a. NO

b. Ne

c. CH4

d. O2

Gases – Ch. 5

Molar mass (M) can be used to

identify an unknown substance

M =

or

M =

Gases – Ch. 513. Air is 79% N2 and 21% O2 by volume. Calculate the density of air

at 1.0 atm, 25°C.

a. 0.590 g/Lb. 1.18 g/Lc. 2.46 g/Ld. 14.1 g/Le. none of these

Gases – Ch. 514. 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.

Average Speed

Gases – Ch. 5

𝑈 𝑎𝑣𝑒=√ 8𝑅𝑇𝜋 𝑀

Gases – Ch. 515. 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.

Gases – Ch. 516. Calculate the temperature at which the average velocity of Ar (g)

equals the average velocity of Ne (g) at 25°C.

a. 317°Cb. 151°Cc. 49.5°Cd. 25°Ce. none of these

Gases – Ch. 517. Order the following according to increasing rate of effusion if all

gases are at the same T and P.

F2, Cl2, NO, NO2, CH4

Gases – Ch. 518. 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?

Gases – Ch. 5

Graham’s Law

If two or more gases are

effusing under the same conditions

=

Gases – Ch. 519. 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. Neb. Hec. Ard. Kre. Xe

Gases – Ch. 520. 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?

Gases – Ch. 5

Useful equations

KEavg = 3/2RT

KE = 1/2mu2

𝐶𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦⇒ 𝑍=4𝑁 𝑑2

𝑉 √ 𝜋 𝑅𝑇𝑀

Gases – Ch. 521. Under what conditions will a gas behave the most like an ideal gas?

Gases – Ch. 5

Gases – Ch. 522. Which of the following gases will have the lowest molar volume at

STP?

a. He

b. CH2Cl2

c. CO2

Gases – Ch. 5

The molar volume can be derived from the ideal gas law:

= At STP the molar volume of an ideal gas is 22.41 L/mol

As a gas deviates from ideal behavior the molar volume decreases

+ a2](V – nb) = nRT

a ⇒ compensates for the attractive forces between gas particlesb ⇒ compensates for the volume of the gas particles

Gases – Ch. 5

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

Gases – Ch. 5

You have completed ch. 5

Answer key – Ch. 51. Draw the following:

a. A closed monometer attached to a flask filled with CO at 250 torr

b. An open monometer at sea level attached to a flask filled with N2O at 600 torr

a. b.

2. Determine if the following are directly or inversely proportional – assume all other variables are constanta. Pressure and volume inverselyb. Pressure and temperature directly

Answer key – Ch. 5

3. 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. 1.31b. 1.60c. 2.10d. 0.477e. 0.615

Answer key – Ch. 5

Each gas is affected by the valve opening

= where n and T are constant

solving for P2 ⇒ P2 =

for O2 ⇒ P2 =

P2 = 4.9 atm

for Ne ⇒ P2 = = 2.3 atmThe ratio of partial pressures ⇒ = = 2.1

4. 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

=

solving for V2 ⇒ V2 = ⇒ V2 =

V2 = 190 m3

Answer key – Ch. 5

5. Consider a sample of neon gas in a container fitted with a movable piston (assume the piston is mass-less and frictionless). The temperature of the gas is increased from 20.0°C to 40.0°C. The density of neona. increases less than 10%.b. decreases less than 10%.c. increases more than 10%.d. decreases more than 10%.e. does not change.

Answer key – Ch. 5

Density = = where mass is constant

= Using = where P and n are constant

Solving for ⇒ =

So = = ⇒ = 0.936 or 93.6% ⇒ the density decreased by 6.4 %

6. 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. 0.95 atmb. 2.2 atmc. 3.8 atmd. 2.7 atme. 4.8 atm

Answer key – Ch. 5

PNe = XNePtotal

XNe =

XNe = = 0.318

PNe = (0.318)(7.0 atm) = 2.2 atm

Answer key – Ch. 57. 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

= 2.88 mol O2

= 1.92 mol He

PO2 = XO2

Ptotal = 745 torr = 447 torr4.80 total moles

8. A sample of oxygen gas has a volume of 4.50 L at 27°C and 800.0 torr. How many oxygen molecules does it contain?

a. 1.16 × 1023

b. 5.8 × 1022

c. 2.32 × 1024

d. 1.16 × 1022

e. none of these

Answer key – Ch. 5

# of molecules can be derived from molesPV = nRT

n = n =

n = 0.192 mol0.192 mol x 6.022x1023 molecules/mol =

1.16x1023 molecules

9. 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 0.930 atm. What volume of carbon dioxide gas will be collected, assuming 100% yield?a. 0.504 Lb. 1.93 Lc. 2.23 Ld. 0.607 Le. 4.04 L

Answer key – Ch. 5

Need to determine the limiting reagentnC6H14

= = 0.0177 mol vs.

nO2 = = 0.135 mol

0.135/19 < 0.0177/2 so O2 is the LR0.135 mol O2 = 0.0853 mol

V = = = 2.23 L

Answer key – Ch. 510. 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 =

= 0.0276 mol gas

Ptotal =

Answer key – Ch. 511. 2.5 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 LRSince 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

Answer key – Ch. 5

2 C (s) O2 (g) 2 CO (g)

3 mol 2.5 mol 0 mol

-3 -3() +3()

0 1 mol 3 mol

11. …continued

Since there is 4 mol of gas in the container

Ptotal = Ptotal = 27.8 atm

12. The density of an unknown gas at STP is 0.715 g/L. Identify the gas.

a. NO

b. Ne

c. CH4

d. O2

Answer key – Ch. 5

Molar mass can be useful to identify a substanceM = M =

M = 16 g/molUnknown gas is CH4

13. Air is 79% N2 and 21% O2 by volume. Calculate the density of air at 1.0 atm, 25°C.a. 0.590 g/Lb. 1.18 g/Lc. 2.46 g/Ld. 14.1 g/Le. none of these

Answer key – Ch. 5

Density is in the equation ⇒ M = D =

Since we have 2 gases ⇒D =

D=

D = 1.18 g/L

Answer key – Ch. 514. 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 speedequation ⇒ 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 ⇒ 300KPlot B ⇒ 1000KAverage

Speed of AAverage

Speed of B

Answer key – Ch. 515. 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.

AverageSpeed of A

AverageSpeed of B

According to the average speedequation ⇒

we can see the relationship between average speed and molar mass

as molar mass ↑ uavg ↓since uavg B > uavg A ⇒ MB < MA

Plot A ⇒ ArPlot B ⇒ He

Answer key – Ch. 516. Calculate the temperature at which the average velocity of Ar (g)

equals the average velocity of Ne (g) at 25°C.a. 317°Cb. 151°Cc. 49.5°Cd. 25°Ce. none of these

uave Ar = uave Ne

=

8,R, and π constant = ⇒ TAr =

TAr = T = 590 K or 317°C

17. Order the following according to increasing rate of effusion:

F2, Cl2, NO, NO2, CH4

As molar mass ↑ average speed ↓ rate of effusion ↓

Since the relative molar masses are

Cl2 (70.9 g/mol) > NO2 (46.01 g/mol) > F2 (38 g/mol) >

NO (30.01 g/mol) > CH4 (16.042 g/mol)

Therefore the relative rates of effusion are

Cl2 < NO2 < F2 < NO < CH4

Answer key – Ch. 5

18. 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

timeCO2 = (12 s)

timeCO2 = 56 s

Answer key – Ch. 5

19. 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. Neb. Hec. Ard. Kre. Xe

Answer key – Ch. 5

Molar mass can be used to identify =

Munk =MH2

Munk = 2.016 g/mol Munk = 83.87 g/molUnknown gas is Kr

20. 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

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

Answer key – Ch. 5

Answer key – Ch. 521. 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

22. Which of the following gases will have the lowest molar volume at STP?

a. He

b. CH2Cl2

c. CO2

Answer key – Ch. 5

The molar volume of an “ideal” gas is 22.4 L/molas 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