Extra Practice Mixed Gas Law Problems · PDF fileMixed Extra Gas Law Practice Problems (Ideal...
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Mixed Extra Gas Law Practice Problems (Ideal Gas, Dalton’s Law of Partial Pressures, Graham’s 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 35.1 o C. What is the pressure in the chamber after all of the dry ice has sublimed? = 1.28 ! 1 1 ! 44.009 ! = 0.29849 ! 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 = 35.1 o C = 308.1 K (4 significant figures) = = 0.290849 0.0821 − − 308.1 5.00 = 1.47 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 1 1 ! 22.4 ! = 0.011 ! 0.25 1 1 ! 22.4 ! 70.906 ! 1 ! = 0.79 ! OR – 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)
Transcript of Extra Practice Mixed Gas Law Problems · PDF fileMixed Extra Gas Law Practice Problems (Ideal...
Mixed Extra Gas Law Practice Problems (Ideal Gas, Dalton’s Law of Partial Pressures, Graham’s 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?
𝑃𝑉 = 𝑛𝑅𝑇
1.28 𝑔 𝐶𝑂!1
𝑥 1 𝑚𝑜𝑙𝑒 𝐶𝑂!44.009 𝑔 𝐶𝑂!
= 0.29𝟎849 𝑚𝑜𝑙𝑒𝑠 𝐶𝑂!
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)
𝑃 = 𝑛𝑅𝑇𝑉
= 0.290849 𝑚𝑜𝑙𝑒𝑠 0.0821 𝐿 − 𝑎𝑡𝑚𝑚𝑜𝑙𝑒 − 𝐾 308.1 𝐾
5.00 𝐿= 1.47 𝑎𝑡𝑚
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 𝐿1
𝑥 1 𝑚𝑜𝑙𝑒 𝐶𝑙!22.4 𝐿 𝐶𝑙!
= 0.011 𝑚𝑜𝑙𝑒𝑠 𝐶𝑙!
0.25 𝐿1
𝑥 1 𝑚𝑜𝑙𝑒 𝐶𝑙!22.4 𝐿 𝐶𝑙!
𝑥 70.906 𝑔 𝐶𝑙!1 𝑚𝑜𝑙𝑒 𝐶𝑙!
= 0.79 𝑔 𝐶𝑙!
OR – 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)
𝑛 = 𝑃𝑉𝑅𝑇
= 1 𝑎𝑡𝑚 (0.25 𝐿)
0.0821 𝐿 − 𝑎𝑡𝑚𝑚𝑜𝑙𝑒 − 𝐾 (273 𝐾)= 0.011 𝑚𝑜𝑙𝑒𝑠 𝐶𝑙!
0.011 𝑚𝑜𝑙𝑒 𝐶𝑙!1
𝑥 70.906 𝑔 𝐶𝑙!1 𝑚𝑜𝑙𝑒 𝐶𝑙!
= 0.78 𝑔 𝐶𝑙!
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 – need to calculate
92.4 𝑔𝑟𝑎𝑚𝑠 𝐶𝑙!1
𝑥 1 𝑚𝑜𝑙𝑒 𝐶𝑙!70.906 𝑔 𝐶𝑙!
= 1.3𝟎3133 𝑔𝑟𝑎𝑚𝑠 𝐶𝑙! − 𝑏𝑜𝑙𝑑𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 𝑖𝑠 𝑙𝑎𝑠𝑡 𝑠𝑖𝑔𝑛𝑖𝑓𝑖𝑐𝑎𝑛𝑡 𝑓𝑖𝑔𝑢𝑟𝑒
𝑉 = 𝑛𝑅𝑇𝑃
= 1.303133 𝑚𝑜𝑙𝑒𝑠 62.4 𝐿 −𝑚𝑚𝐻𝑔𝑚𝑜𝑙𝑒 − 𝐾 (319 𝐾)
694 𝑚𝑚𝐻𝑔= 37.4 𝐿
4. What is the molar mass of a gas that effuses 3.7 times faster than Krypton?
𝑅𝑎𝑡𝑒!𝑅𝑎𝑡𝑒!
= 𝑀!
𝑀!
A is the unknown gas (it effuses faster)
B is Krypton (it effuses slower)
𝑅𝑎𝑡𝑒! = 3.7 𝑅𝑎𝑡𝑒!" 𝑠𝑜 𝑅𝑎𝑡𝑒!𝑅𝑎𝑡𝑒!"
= 3.7
𝑅𝑎𝑡𝑒!𝑅𝑎𝑡𝑒!"
= 𝑀!"
𝑀!= 3.7
Square both sides
13.69 = 𝑀!"
𝑀!
Rearrange and solve for mass of the unknown
𝑀! = 𝑀!"
13.69= 84.80 𝑔
𝑚𝑜𝑙𝑒13.69
= 6.19𝑔
𝑚𝑜𝑙𝑒
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?
𝑃!"#$% = 1.3 𝑎𝑡𝑚
𝑃!"#$% !"#$% = 2.20 𝑘𝑃𝑎 = 2.20 𝑘𝑃𝑎
1 𝑥
1 𝑎𝑡𝑚101.3 𝑘𝑃𝑎
= 0.021𝟕1767 𝑎𝑡𝑚
𝑃!"#$%& !"#$"!% = 𝑃!"!#$ − 𝑃!"#$% !"#$% = 1.3 𝑎𝑡𝑚 − 0.02171767 𝑎𝑡𝑚 = 1.3 𝑎𝑡𝑚
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
𝑅𝑎𝑡𝑒!𝑅𝑎𝑡𝑒!
= 𝑀!
𝑀!
A – butane
B – iodine
𝑅𝑎𝑡𝑒!𝑅𝑎𝑡𝑒!
= 253.808 𝑔
𝑚𝑜𝑙𝑒58.123 𝑔
𝑚𝑜𝑙𝑒= 2.09
7. What is the molar mass of a gas that effuses 4.2 times faster than bromine?
𝑅𝑎𝑡𝑒!𝑅𝑎𝑡𝑒!
= 𝑀!
𝑀!
A is the unknown gas (it effuses faster)
B is Bromine (it effuses slower)
𝑅𝑎𝑡𝑒! = 4.2 𝑅𝑎𝑡𝑒!"! 𝑠𝑜 𝑅𝑎𝑡𝑒!𝑅𝑎𝑡𝑒!"!
= 4.2
𝑅𝑎𝑡𝑒!𝑅𝑎𝑡𝑒!"!
= 𝑀!"
𝑀!= 4.2
Square both sides
17.64 = 𝑀!"!𝑀!
Rearrange and solve for mass of the unknown
𝑀! = 𝑀!"!17.64
= 159.808 𝑔
𝑚𝑜𝑙𝑒17.64
= 9.06𝑔
𝑚𝑜𝑙𝑒
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?
𝑃!! = 3.7 𝑎𝑡𝑚
𝑃!!! = 382 𝑚𝑚 𝐻𝑔 = 382 𝑚𝑚 𝐻𝑔
1 𝑥
1 𝑎𝑡𝑚760 𝑚𝑚 𝐻𝑔
= 0.50𝟐63 𝑎𝑡𝑚
𝑃!"! = 234 𝑘𝑃𝑎 = 234 𝑘𝑃𝑎
1 𝑥
1 𝑎𝑡𝑚101.3 𝑘𝑃𝑎
= 2.3𝟎997 𝑎𝑡𝑚
𝑃!" = 143 𝑚𝑚 𝐻𝑔 = 143 𝑚𝑚 𝐻𝑔
1 𝑥
1 𝑎𝑡𝑚760 𝑚𝑚 𝐻𝑔
= 0.18𝟖157 𝑎𝑡𝑚
𝑃!"!#$ = 𝑃!! + 𝑃!!! + 𝑃!"! + 𝑃!" = 3.7 𝑎𝑡𝑚 + 0.50𝟐63 𝑎𝑡𝑚 + 2.3𝟎997 𝑎𝑟𝑚 + 0.18𝟖157 𝑎𝑡𝑚 = 6.7 𝑎𝑡𝑚
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 – need to calculate
83.28 𝑔𝑟𝑎𝑚𝑠 𝐶𝑂!1
𝑥 1 𝑚𝑜𝑙𝑒 𝐶𝑂!44.009 𝑔 𝐶𝑂!
= 1.89𝟐340 𝑔𝑟𝑎𝑚𝑠 𝐶𝑂! − 𝑏𝑜𝑙𝑑𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 𝑖𝑠 𝑙𝑎𝑠𝑡 𝑠𝑖𝑔𝑛𝑖𝑓𝑖𝑐𝑎𝑛𝑡 𝑓𝑖𝑔𝑢𝑟𝑒
𝑇 = 𝑃𝑉𝑛𝑅
= 1830 𝑡𝑜𝑟𝑟 (3.82 𝐿)
(1.892340 𝑚𝑜𝑙𝑒𝑠) 62.4 𝐿 − 𝑡𝑜𝑟𝑟𝑚𝑜𝑙𝑒 − 𝐾= 59.2 𝐿
