# Gases Chapter 5 E-mail: benzene4president@gmail.com Web-site: http://clas.sa.ucsb.edu/staff/terri/

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

Gases – 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 b c d e 3.25-L 8.64 atm O2 2.48-L 5.40 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 P1V1 n1T1 = P2V2 n2T2

Gases – Ch. 5 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.

Gases – 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 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 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 atm b atm c atm d atm e atm

Gases – Ch. 5 7. 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. 5 8. 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

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

Gases – Ch. 5 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 atm. What volume of carbon dioxide gas will be collected, assuming 100% yield? a L b L c L d L e L

Gases – Ch. 5 10. 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 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 12. The density of an unknown gas at STP is 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 = DRT P or M = mRT PV

Gases – Ch. 5 13. 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

Gases – Ch. 5 14. 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 𝑈𝑎𝑣𝑒= 8𝑅𝑇 𝜋𝑀

Gases – Ch. 5 15. 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 16. 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

Gases – Ch. 5 17. 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 18. 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?

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

Gases – Ch. 5 19. 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

Gases – Ch. 5 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? b. Which flask has the greatest collisions per second?

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

Gases – Ch. 5 21. Under what conditions will a gas behave the most like an ideal gas?

Gases – Ch. 5

Gases – Ch. 5 22. 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 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

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

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. 5 a. b. 1. 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.

Answer key – Ch. 5 2. 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

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 b c d e Each gas is affected by the valve opening P1V1 n1T1 = P2V2 n2T2 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

Answer key – Ch. 5 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 P1V1 n1T1 = P2V2 n2T2 solving for V2 ⇒ V2 = P1V1T2 P2T1 ⇒ V2 = (597 torr)(300 m3)(273 K) (760 torr)( K) 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 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 D2 D1 = (m2/V2) (m1/V1) where mass is constant D2 D1 = V1 V2 Using P1V1 n1T1 = P2V2 n2T where P and n are constant Solving for V1 V2 ⇒ V1 V2 = T1 T2 So D2 D1 = T1 T2 = (20+273K) ( K) ⇒ D2 D1 = or 93.6% ⇒ the density decreased by 6.4 %

Answer key – Ch. 5 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 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

Answer key – Ch. 5 7. 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

Answer key – Ch. 5 8. 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

Answer key – Ch. 5 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 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

Answer key – Ch. 5 10. 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

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

Answer key – Ch. 5 11. …continued 2 C (s) O2 (g)  2 CO (g) 3 mol
-3 -3( ) +3( 2 2 ) 1 mol Since there is 4 mol of gas in the container Ptotal = (4 𝑚𝑜𝑙)( 𝑎𝑡𝑚𝐿 𝑚𝑜𝑙𝐾 )(23+273𝐾) 3.5 𝐿 Ptotal = 27.8 atm

Answer key – Ch. 5 12. 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

Answer key – Ch. 5 13. 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

Answer key – Ch. 5 14. 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 15. 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

Answer key – Ch. 5 𝑢𝑎𝑣𝑒= 8𝑅𝑇 𝜋𝑀 8𝑅𝑇𝐴𝑟 𝜋𝑀𝐴𝑟 = 8𝑅𝑇𝑁𝑒 𝜋𝑀𝑁𝑒
16. 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

Answer key – Ch. 5 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 ( g/mol) Therefore the relative rates of effusion are Cl2 < NO2 < F2 < NO < CH4

timeCO2 = (12 s) 44.01g/mol 2.016g/mol
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 = MCO2 MH2 timeCO2 = timeH2 MCO2 MH2 timeCO2 = (12 s) g/mol 2.016g/mol timeCO2 = 56 s

Molar mass can be used to identify
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. 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

Answer key – Ch. 5 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 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

Answer key – Ch. 5 21. 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 22. 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