CHAPTER 5- Gases
Albert Kweyete
1. Perform the following conversions:
a) 0.357 atm to torr
b) 6.60x10-2 torr to atm c) 745 torr to pa d) 802 mmHg to atm
2. A barometer reads the atmospheric pressure 764.7 torr. A sample of gas is placed in
vessel attached to an open-end mercury monometer. The level of mercury in the open-end
arm of the monometer has a height of 136.4 mm, and that in the arm that is in contact
with the gas has a height 103.8 mm. What is the pressure of gas in atmosphere?
3. Write an equation or proportionality expression that expresses each of the following
statements:
a) Law of combining volumes explained by Avogadro,s hypothesis.
b) For a given quantity of gas, the product of pressure and volume is proportional to
absolute temperature.
c) For a given quantity of gas at constant temperature, the product of pressure and volume
is constant.
4. Calculate the following quantities for an ideal gas:
a) the pressure , in atms, if 8.25x10-2 mol occupies 174 ml at -15 0C;
b) the number of moles in 2.50 L at 37 0C and 725 torr.
c) the volume of a gas occupied by 6.72x10-3 mol at 145 0C and a pressure of 59.0 torr;
d) the absolute temperature of the gas at which 9.87x10-2 mol occupies 164 ml at 682
torr;
5. Consider 5.62 L chlorine gas, Cl2 ,at 740 torr and 33 0C.
a) What volume will the gas occupy at 107 0C and 680 torr?
b) What volume will the gas occupy at STP?
c) At what pressure will the volume be 5.00 L if the temperature is 67 0C?
d) At what pressure will the volume be twice as much as original, if temperature
remains constant?
e) How many moles Cl2 gas are in original condition?
6. Perform the following calculations;
a) the density of SO3(g) at 0.96 atm and 35 0C.
b) the density of SO3(g) at STP condition.
c) the molar mass of a gas if its density is 3.67 g/L at 15 0C and 825 torr.
d) the molar mass of gas if 4.40 g occupies 3.50 L at 560 torr and 41 0C.
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e) the molar mass of a gas if its density is 2.86 g/L at STP and identify the molecular formula
of the gas.
7. Consider the following reaction of ammonia gas and oxygen gas; at 850 0C and 5.00
atm.
4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(l)
a) How many liters NH3(g) at 850 0C and 5.00 atm are required to react with 1.00
mol of O2(g) in this reaction?
b) How many moles NO(g) can be prepared using 450 L of NH3(g) at a pressure of
5.00 atm and a temperature of 295 K?
c) How many grams of NH3(g) are needed to generate 10.0 L of NO(g) if pressure of
NO gas is 740 torr at 23 0C?
8. Perform the following calculations;
a) the partial pressure and total pressure of each of the following gases in the
mixture of 4.00 g of each gas CH4(g), C2H4(g) and C4H10(g), in 1.50 L flask at 0 0C.
b) the partial pressure of each component in a mixture of gases contains 0.55 mol
N2, 0.20 mol O2, and 0.10 mol CO2. The total pressure of the mixture is 1.32 atm.
c) the partial pressure of each component in a mixture of gases contains 3.50 g N2,
2.15 g H2, and 5.27 g of NH3. If the total pressure of the mixture is 2.50 atm.
9. Consider the following gases at 25 0C; CO2, He, Cl2, CO, CH4, N2
a) place them in order of increasing average molecular speed.
b) calculate and compare the rms speeds of CO(g) and CO2(g).
c) calculate the rate of diffusion of gases CH4(g) and He.
d) which gas effuses 1.41 times faster than O2(g)?
e) which two gases have the same rate of diffusion?
10. If the atmospheric pressure is 0.975 atm, what is the pressure of the enclosed
gas in each of the cases below?
h
(I) h = 45 mm
(closed)
h
h
(II) h = 78 mm
(open-end)
2
(III) h = 67 mm
(open-end)
CHAPTER 5- Gases (Answers)
1. a) (0.3570(760 torr/1 atm) = 271 torr
b) (6.60x10-2 torr )(1 atm /760 torr) = 8.68 x10-5 atm
c) (745 torr)( 1 atm /760 torr)(1.013x105 pa/1 atm) = 9.93x104 pa
d) ( 802 mmHg)(1 atm /760 mmHg) = 1.06 atm
2. 136.4mm – 103.8mm = 32.6mm  Pgas = 764.7 + 32.6 = 797.3 mmHg
3. a.) V = kn
4. a)
c.) PV = k
P = nRT = (8.25 x 10-2)(0.0821)(258K) = 10 atm
V
0.174L
b.) n = PV
RT
c.) V =
d.) T =
5. a)
b.) PV = kT
=
725 torr x 2.5 L
760 torr x 0.0821 x 310K
= 0.094 mole
nRT = 6.72 x 10-3mol x 0.0821 x 418K x 760 torr = 2.97L
P
59 torr
PV =
682 torr x 0.164 L
= 16.2 K = -257 0C
nR
760 torr x 9.87 x10-2 x 0.0821
P1V1 = P2V2  5.62 L x740 torr = V2 x 680  V2 = 7.57 L
T1
T2
306 K
380 K
b)
P1V1 = P2V2 = 5.62 L x740 torr = V2 x 760 ; V2 = 4.88 L
T1
T2
306 K
273 K
c)
P1V1 = P2V2 = 5.62 L x740 torr = 5.0 L x P2 ; P2 = 924 torr
T1
T2
306 K
273 K
n = PV =
740 torr x 5.62 L
= 0.22 mole
RT
760 torr x 0.0821 x 306K
e)
f)
P1V1 = P2V2  P1V1 = P2V2  P1V1 = P2 (2V1)  p2 = p1 /2  p2 = 740/2 = 370 torr
T1
T2
6. a.) Molecular wt. SO3 is 80g  D =
b)
D=
PM = 0.96 atm x 80g = 3.04 g/L
RT
0.0821 x 308 K
PM = 1.00 atm x 80g = 3.57 g/L
RT
0.0821 x 273 K
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c) M = DRT = (3.67)(0.0821)(288)(760) = 80 g/mol
p
825
d) M = DRT = (4.40)(0.0821)(314)(760) = 44 g/mol
p
(560(3.50)
e) at STP D = M  M = D(22.4) = 2.86 g/L)(22.4L/mol) = 64 g/mol  SO2 gas
22.4
7. a) Stoichiometricaly, every 1.00 mole O2 reacted  0.800 mole NH3 formed.
V=
nRT =
P
0.800 mol x 0.0821 x 1123 K = 15 L
5.00 atm
b) Stoichiometricaly, every 1.00 mole NH3 reacted  1.00 mole NO formed (1:1 ratio).
Moles NH3 =
PV =
RT
5.00 atm x 450 L = 93 moles NH3
0.0821 x 295 K
c) Stoichiometricaly, every 1.00 mole NH3 reacted  1.00 mole NO formed (1:1 ratio).
Moles NO = PV = 5.00 atm x 450 L = 0.40 moles NO 0.4 moles NH3
RT 0.0821x 295 K
0.4 moles NH3 x 17g/mol = 6.8g NH3
8. a) 4.00g of each gas divided by molar mass ofeach gas gives :
0.25mole CH4, 0.143mole C2H4, 0.069mole C4H10.
Total mole(ntotal )= 0.25 + 0.143 + 0.069 = 0.462 and Total volume (Vtotal) =1.5 L
Mole fraction (X) = mole of gas divided by total mole  Xgas = ngas / ntotal
Partial pressure, p = (Xgas )(Ptotal)
Total pressure (Ptotal) = nRT
0.462 x 0.0821 x 273 K = 6.9atm
V
1.5 L
Partial pressure of C2H4 ( PC2H4) = 0.143 x 6.9 atm = 2.14 atm
0.462
Partial pressure of CH4 ( PCH4) = 0.25 x 6.9 atm = 3.73 atm
0.462
Partial pressure of C4H10 ( PC4H10) = 0.069 x 6.9 atm = 1.03 atm
0.462
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b) Total moles = 0.55 + 0.20 + 0.10 = 0.85 mole, Ptotal = 1.32 atm
Partial pressure of N2 ( PN2) =
0.55 x 1.32atm = 0.854 atm
0.85
Partial pressure of O2 ( PO2) = 0.20 x 1.32atm
= 0.31atm
0.85
Partial pressure of CO2 ( PCO2) = 0.10 x 1.32 atm = 0.155atm
0.85
c) Total moles = (3.50/28) +(2.15/2) + (5.27/17)
= 0.125 + 1.08 + 0.310 = 1.52 mole, Ptotal = 2.50 atm
Partial pressure of N2 ( PN2) =
Partial pressure of O2 ( PO2) =
0.125 x 2.50atm = 0.210 atm
1.52
1.08 x 2.50atm
1.52
= 1.78 atm
Partial pressure of CO2 ( PCO2) = 0.310 x 2.50 atm = 0.510 atm
1.52
9. The larger the mass, the slower the gas.
a ) Cl2 < CO2 < CO , N2 < CH4 < He
b ) µ = √(3RT)/M ; M is in kilograms, R is 8.314 J/mol-K , √ means square root
For CO µ = √(3 x 8.314 x 298 K)/ 0.028 kg = 515.2 m/s
c) For CO2 µ = √(3 x 8.314 x 298 K)/ 0.044 kg = 411 m/s
d) r1/r2 = √M2/M1  rx/roxygen = 1.41 = √32g/Mx
(1.41)2
=
32g = 1.98  Mx = 16g/mole (CH4 gas)
Mx
e) CO & N2; Both have molar mass of 28g/mole.
10. 0.975 atm = 741 mmHg
a) Pgas = 45mmHg
b) Pgas = 741 + 78 = 819 mmHg
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c) Pgas = 741 – 67 = 674 mmHg