Mixed Extra Gas Law Practice Problems (Ideal Gas, Daltonβs Law of Partial Pressures, Grahamβs Law)
1. Dry ice is carbon dioxide in the solid state. 1.28 grams of dry ice is placed in a 5.00 L chamber that is maintained at 35.1oC. What is the pressure in the chamber after all of the dry ice has sublimed?
ππ = ππ π
1.28 π πΆπ!1
π₯ 1 ππππ πΆπ!44.009 π πΆπ!
= 0.29π849 πππππ πΆπ!
P = ?
V = 5.00 L (3 significant figures)
n = 0.290849 moles (3 significant figures)
R = 0.0821 L*atm/mole*K (infinite significant figures)
T = 35.1oC = 308.1 K (4 significant figures)
π = ππ ππ
= 0.290849 πππππ 0.0821 πΏ β ππ‘πππππ β πΎ 308.1 πΎ
5.00 πΏ= 1.47 ππ‘π
If you used a different R, then the answers are:
1120 torr
1120 mm Hg
149 kPa
2. A sample of chlorine gas is loaded into a 0.25 L bottle at standard temperature of pressure. How many moles of bromine gas are in the container? How many grams?
At STP 1 mole = 22.4 L
Molar Mass of Chlorine (remember, it is a diatomic) = 70.906 g/mole
Use factor label
0.25 πΏ1
π₯ 1 ππππ πΆπ!22.4 πΏ πΆπ!
= 0.011 πππππ πΆπ!
0.25 πΏ1
π₯ 1 ππππ πΆπ!22.4 πΏ πΆπ!
π₯ 70.906 π πΆπ!1 ππππ πΆπ!
= 0.79 π πΆπ!
OR β use the ideal gas law, PV = nRT
P = 1 atm (infinite significant figures)
V = 0.25 L (2 significant figures)
R = 0.0821 L*atm/mole*K (infinite significant figures)
T = 273 K (infinite significant figures)
π = πππ π
= 1 ππ‘π (0.25 πΏ)
0.0821 πΏ β ππ‘πππππ β πΎ (273 πΎ)= 0.011 πππππ πΆπ!
0.011 ππππ πΆπ!1
π₯ 70.906 π πΆπ!1 ππππ πΆπ!
= 0.78 π πΆπ!
3. What is the volume of 92.4 grams of chlorine gas that is at a temperature of 46oC and a pressure of 694 mmHg?
P = 694 mmHg (3 significant figures)
V = ?
R = 62.4 L*mm Hg/mole*K (infinite significant figures)
T = 46oC = 319 K (3 significant figures)
n β need to calculate
92.4 πππππ πΆπ!1
π₯ 1 ππππ πΆπ!70.906 π πΆπ!
= 1.3π3133 πππππ πΆπ! β ππππππ ππ’ππππ ππ πππ π‘ π ππππππππππ‘ ππππ’ππ
π = ππ ππ
= 1.303133 πππππ 62.4 πΏ βπππ»πππππ β πΎ (319 πΎ)
694 πππ»π= 37.4 πΏ
4. What is the molar mass of a gas that effuses 3.7 times faster than Krypton?
π ππ‘π!π ππ‘π!
= π!
π!
A is the unknown gas (it effuses faster)
B is Krypton (it effuses slower)
π ππ‘π! = 3.7 π ππ‘π!" π π π ππ‘π!π ππ‘π!"
= 3.7
π ππ‘π!π ππ‘π!"
= π!"
π!= 3.7
Square both sides
13.69 = π!"
π!
Rearrange and solve for mass of the unknown
π! = π!"
13.69= 84.80 π
ππππ13.69
= 6.19π
ππππ
5. Carbon dioxide is collected over water at a temperature of 18oC. The pressure of water vapor at 18oC is 2.20 kPa. If the pressure of the gas collected is 1.3 atm, what is the pressure of the dry gas alone?
π!"#$% = 1.3 ππ‘π
π!"#$% !"#$% = 2.20 πππ = 2.20 πππ
1 π₯
1 ππ‘π101.3 πππ
= 0.021π1767 ππ‘π
π!"#$%& !"#$"!% = π!"!#$ β π!"#$% !"#$% = 1.3 ππ‘π β 0.02171767 ππ‘π = 1.3 ππ‘π
6. What diffuses more quickly: tetracarbon decahydride or iodine? By how much?
C4H10 versus I2
Molar mass of C4H10 = 58.123 g/mole
Molar mass of I2 = 253.808 g/mole
Butane will effuse more quickly because it has a smaller molar mass
π ππ‘π!π ππ‘π!
= π!
π!
A β butane
B β iodine
π ππ‘π!π ππ‘π!
= 253.808 π
ππππ58.123 π
ππππ= 2.09
7. What is the molar mass of a gas that effuses 4.2 times faster than bromine?
π ππ‘π!π ππ‘π!
= π!
π!
A is the unknown gas (it effuses faster)
B is Bromine (it effuses slower)
π ππ‘π! = 4.2 π ππ‘π!"! π π π ππ‘π!π ππ‘π!"!
= 4.2
π ππ‘π!π ππ‘π!"!
= π!"
π!= 4.2
Square both sides
17.64 = π!"!π!
Rearrange and solve for mass of the unknown
π! = π!"!17.64
= 159.808 π
ππππ17.64
= 9.06π
ππππ
8. A mixture of gases contains oxygen, water vapor, carbon dioxide, and argon. If the partial pressure of oxygen is 3.7 atm, the partial pressure of the water vapor is 382 mmHg, the partial pressure of the carbon dioxide is 234 kPa, and the partial pressure of the argon is 143 mmHg, what is the total pressure?
π!! = 3.7 ππ‘π
π!!! = 382 ππ π»π = 382 ππ π»π
1 π₯
1 ππ‘π760 ππ π»π
= 0.50π63 ππ‘π
π!"! = 234 πππ = 234 πππ
1 π₯
1 ππ‘π101.3 πππ
= 2.3π997 ππ‘π
π!" = 143 ππ π»π = 143 ππ π»π
1 π₯
1 ππ‘π760 ππ π»π
= 0.18π157 ππ‘π
π!"!#$ = π!! + π!!! + π!"! + π!" = 3.7 ππ‘π + 0.50π63 ππ‘π + 2.3π997 πππ + 0.18π157 ππ‘π = 6.7 ππ‘π
9. What is the temperature of 83.28 grams of carbon dioxide at a pressure of 1830 torr contained in a 3.82 L container?
P = 1830 torr (3 significant figures)
V = 3.82 L (3 significant figures)
R = 62.4 L*torr/mole*K (infinite significant figures)
T = ?
n β need to calculate
83.28 πππππ πΆπ!1
π₯ 1 ππππ πΆπ!44.009 π πΆπ!
= 1.89π340 πππππ πΆπ! β ππππππ ππ’ππππ ππ πππ π‘ π ππππππππππ‘ ππππ’ππ
π = ππππ
= 1830 π‘πππ (3.82 πΏ)
(1.892340 πππππ ) 62.4 πΏ β π‘πππππππ β πΎ= 59.2 πΏ
