- 241 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about 'Gases' - karsen

Download Now**An Image/Link below is provided (as is) to download presentation**

Download Now

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

Gases – Ch. 5

1. Draw the following:

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

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

Gases – Ch. 5

2. Which of the following relationships is not true?

a. PV = constant when temperature and moles of gas are held constant.

b. V/T = constant when pressure and moles of gas are held constant.

c. nT = constant when pressure and volume are held constant.

d. P/n = constant when volume and temperature are held constant.

e. All of these are true.

n1T1 n2T2

Gases – Ch. 5Combined Gas Law

If one variable (P, V, n or T) of a gas changes

at least one other variable must change

P1V1 P2V2

Gases – Ch. 5

3. In an experiment 300 m3 of methane is collected at 785 torr and 165 °C. What is the volume (in m3) at STP?

Gases – Ch. 5

4. Body temperature is about 308 K. On a cold day, what volume of air at 273 K must a person with a lung capacity of 2.00 L breathe in to fill the lungs?

a. 2.26 L

b. 1.77 L

c. 1.13 L

d. 3.54 L

e. none of these

Gases – Ch. 5

5. The volume of a helium balloon is 1.85 L at 24.0°C and 1.00 atm at sea level. The balloon is released and floats upward. At a certain altitude, the balloon has a volume of 2.14 L and the temperature is 15.2°C. What is the atmospheric pressure at this altitude?

a. 0.538 atm

b. 0.839 atm

c. 0.891 atm

d. 1.36 atm

e. none of these

Gases – Ch. 5

6. 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. 5

7. A gaseous mixture containing 1.5 mol Ar and 3.5 mol CO2 has a total pressure of 7.0 atm. What is the partial pressure of CO2?

a. 1.8 atm

b. 2.1 atm

c. 3.5 atm

d. 4.9 atm

e. 2.4 atm

Gases – Ch. 5

8. The valve between a 3.25-L tank containing O2 (g) at 8.64 atm and a 2.48-L tank containing Ne (g) at 5.40 atm is opened. Calculate the ratio of partial pressures (O2:Ne) in the container.

a. 1.31

b. 1.60

c. 2.10

d. 0.477

e. 0.615

Gases – Ch. 5

9. 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

Gases – Ch. 5

Ideal gas law

Considering one set of variables

for a gas under “ideal” conditions

PV = nRT

R ⇒ Universal gas constant

R = 0.08206 atmL/molK

R = 62.37 torrL/molK

R = 8.314 KPaL/molK or J/molK

Gases – Ch. 5

10. Potassium chlorate decomposes upon heating as follows:

2 KClO3 (s) → 2 KCl (s) + 3 O2 (g)

A 2.36-g sample of KClO3 decomposes, and the oxygen at 23.8°C and 0.930 atm is collected. What volume of oxygen gas will be collected, assuming 100% yield?

a. 0.504 L

b. 1.93 L

c. 0.756 L

d. 0.0607 L

e. 0.0404 L

Gases – Ch. 5

11. 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. 5

12. 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. 5

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

a. NH3

b. Ne

c. CH4

d. O2

Gases – Ch. 5

14. Air is 79% N2 and 21% O2 by volume. Calculate the density of air at 1.0 atm, 25°C.

a. 0.590 g/L

b. 1.18 g/L

c. 2.46 g/L

d. 14.1 g/L

e. none of these

Gases – Ch. 5

15. 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.

Gases – Ch. 5

Average Speed

Gases – Ch. 5

16. 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. 5

17. 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. 49.5°C

d. 25°C

e. none of these

Gases – Ch. 5

18. 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. 5

19. 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?

Gases – Ch. 5

20. 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

Gases – Ch. 5

21. 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?

c. Which flask has the greatest density?

Gases – Ch. 5

22. Under what conditions will a real gas behave like an ideal gas?

Gases – Ch. 5

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

a. He

b. CH2Cl2

c. CO2

Gases – Ch. 5

Van der Waal’s Equation

a ⇒ compensates for the attractive forces between gas particles

b ⇒ compensates for the volume of the gas particles

Answer key – Ch. 5

1. Draw the following:

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

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

a.

b.

Answer key – Ch. 5

2. Which of the following relationships is not true?

a. PV = constant when temperature and moles of gas are held constant.

b. V/T = constant when pressure and moles of gas are held constant.

c. nT = constant when pressure and volume are held constant.

d. P/n = constant when volume and temperature are held constant.

e. All of these are true.

Answer key – Ch. 5

3. In an experiment 300 m3 of methane is collected at 785 torr and 165 °C. What is the volume (in m3) at STP?

At STP the temperature and pressure are 273K and 760 torr respectively.

P1V1/n1T1 = P2V2/n2T2

n is constant

solving for V2⇒ V2 = P1V1T2/P2T1

V2 = (785 torr)(300 m3)(273 K)/(760 torr)(165 + 273K)

V2 = 193 m3

Answer key – Ch. 5

4. Body temperature is about 308 K. On a cold day, what volume of air at 273 K must a person with a lung capacity of 2.00 L breathe in to fill the lungs?

a. 2.26 L

b. 1.77 L

c. 1.13 L

d. 3.54 L

e. none of these

P1V1/n1T1 = P2V2/n2T2

n and P is constant

solving for V2⇒ V2 = V1T2/T1

V2 = (2 L)(273 K)/(308 K)

V2 = 1.77 L

Answer key – Ch. 5

5. The volume of a helium balloon is 1.85 L at 24.0°C and 1.00 atm at sea level. The balloon is released and floats upward. At a certain altitude, the balloon has a volume of 2.14 L and the temperature is 15.2°C. What is the atmospheric pressure at this altitude?

a. 0.538 atm

b. 0.839 atm

c. 0.891 atm

d. 1.36 atm

e. none of these

P1V1/n1T1 = P2V2/n2T2

n is constant

solving for P2⇒ P2 = P1V1T2/T1V2

P2 = (1 atm)(1.85 L)(15.2+273K)/(24+273K)(2.14 L)

P2 = 0.839 atm

Answer key – Ch. 5

6. 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.

Density = mass/volume

the mass is staying constant so we

need to determine what % the volume

is changing

P1V1/n1T1 = P2V2/n2T2

P and n is constant

Solving for V2/V1⇒ V2/V1 = T2/T1

V2/V1 = (40+273K)/(20 + 273K) ⇒ V2/V1 = 1.068 ⇒

since the volume went up by 6.8% ⇒ the density decreased by 6.8 %

Answer key – Ch. 5

7. A gaseous mixture containing 1.5 mol Ar and 3.5 mol CO2 has a total pressure of 7.0 atm. What is the partial pressure of CO2?

a. 1.8 atm

b. 2.1 atm

c. 3.5 atm

d. 4.9 atm

e. 2.4 atm

P1V1/n1T1 = P2V2/n2T2

V and T is constant

The combined gas law is not limited to just one

gas – in this case we can make the 1st set of

variables for CO2 and the 2nd set of variables for

the gas mixture

Solving for P1⇒ P1 = P2n1/n2

P1 = (7.0 atm)(3.5 mol)/(1.5 mol+3.5 mol)

P1 = 4.9 atm

Answer key – Ch. 5

8. The valve between a 3.25-L tank containing O2 (g) at 8.64 atm and a 2.48-L tank containing Ne (g) at 5.40 atm is opened. Calculate the ratio of partial pressures (O2:Ne) in the container.

a. 1.31

b. 1.60

c. 2.10

d. 0.477

e. 0.615

Each gas is affected by the valve opening

P1V1/n1T1 = P2V2/n2T2

n and T is 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)(2.48L)/(3.25L+2.48L) = 2.3 atm

The ratio of partial pressures = 4.9 atm/2.3 atm = 2.1

Answer key – Ch. 5

9. 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

# of molecules can be derived from moles

PV = nRT

n = PV/RT

n=(800torr)(4.5L)/(62.37torrL/molK)(300K)

n = 0.192 mol

0.192 mol x 6.022x1023 molecules/mol =

1.16x1023 molecules

Answer key – Ch. 5

10. Potassium chlorate decomposes upon heating as follows:

2 KClO3 (s) → 2 KCl (s) + 3 O2 (g)

A 2.36-g sample of KClO3 decomposes, and the oxygen at 23.8°C and 0.930 atm is collected. What volume of oxygen gas will be collected, assuming 100% yield?

a. 0.504 L

b. 1.93 L

c. 0.756 L

d. 0.0607 L

e. 0.0404 L

PV = nRT ⇒ V = nRT/P

2.36 g KClO3 x 1 mol KClO3 x 3mol O2 = 0.029 mol

2 mol KClO3

O2

122.55 g KClO3

V = (0.29mol)(0.08206atmL/molK)(296.8K)/(0.93atm)

V = 0.756 L

Answer key – Ch. 5

11. 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

…continue to next slide

Answer key – Ch. 5

11. …continued

3.54 g Pb(NO3)2 x 1mol Pb(NO3)2 x (4 mol NO2+1mol O2) = 0.0267 mol

331 g Pb(NO3)2

2 mol Pb(NO3)2

Ptotal = (0.0267 mol)(0.08206atmL/molK)(300K)/(1.6L)

Ptotal = 0.41 atm

Answer key – Ch. 5

12. 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 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

Answer key – Ch. 5

12. …continued

CO ⇒ 3 mol C x 2 mol CO = 3 mol CO are formed

O2⇒ 3 mol C x 1 mol O2 = 1.5 mol of O2 are consumed

2 mol C

2 mol C

2.5 mol – 1.5 mol = 1 mol O2 remains

Ptotal = (3mol + 1mol)(0.08206atmL/molK)(296 K)/(3.5 L)

Ptotal = 27.8 atm

Answer key – Ch. 5

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

a. NH3

b. Ne

c. CH4

d. O2

Molar mass can be useful to identify a substance

M = DRT/P

M = (0.715 g/L)(0.08206atmL/molK)(273K)/(1atm)

M = 16 g/mol

Unknown gas is CH4

Answer key – Ch. 5

14. Air is 79% N2 and 21% O2 by volume. Calculate the density of air at 1.0 atm, 25°C.

a. 0.590 g/L

b. 1.18 g/L

c. 2.46 g/L

d. 14.1 g/L

e. none of these

Density is in the equation ⇒

M = DRT/P

D = M P/RT

Since we have 2 gases ⇒

D = ((MN2 PN2 )+(MO2 PO2))/RT

D=((28g/mol)(0.79atm)+(32g/mol)(0.21 atm))/(0.08206 atmL/molK)(298K)

D = 1.18 g/L

Answer key – Ch. 5

15. 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

Answer key – Ch. 5

16. 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 ⇒ uavg = (8RT/π M)1/2

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

Answer key – Ch. 5

17. 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. 49.5°C

d. 25°C

e. none of these

uavg = (8RT/π M)1/2

uavg Ar = uavg Ne

(8RTAr/π MAr)1/2 =(8RTNe/π MNe)1/2

8,R, πand ½ are constant

TAr/MAr=TNe/MNe

TAr = TNeMAr/MNe

TAr = (298K)(39.95 g/mol)/(20.18 g/mol)

T = 590 K or 317°C

Answer key – Ch. 5

18. 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

19. 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)1/2

timeCO2 = (MCO2/ MH2)1/2timeH2

timeCO2 = (44.01g/mol/2.016g/mol)1/2(12 s)

timeCO2 = 56 s

Answer key – Ch. 5

20. 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)1/2

Munk = (rateH2/rateunk)2MH2

Munk = (6.45/1)(2.016 g/mol)

Munk = 83.87 g/mol

Unknown gas is Kr

Answer key – Ch. 5

21. 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 = 3/2RT ⇒ 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

…continue to next slide

Answer key – Ch. 5

21. …continued

c. Which flask has the greatest density?

According toM = DRT/Por D = MP/RT

we can see the following relationships

as M↑ D↑or as P↑ D↑or as T↑ D↓

since both flasks have the same molar mass and T but

PB > PA⇒ DB > DA

Answer key – Ch. 5

22. 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

Answer key – Ch. 5

23. 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 relationshipthat 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

Download Presentation

Connecting to Server..