Gas Laws REVIEW GAME. Question 1 A 4.3 liter tank of hydrogen is at a pressure of 6.2 atmospheres....

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Gas Laws Gas Laws REVIEW GAME REVIEW GAME

Transcript of Gas Laws REVIEW GAME. Question 1 A 4.3 liter tank of hydrogen is at a pressure of 6.2 atmospheres....

Page 1: Gas Laws REVIEW GAME. Question 1 A 4.3 liter tank of hydrogen is at a pressure of 6.2 atmospheres. What volume of hydrogen will be available if the hydrogen.

Gas LawsGas LawsREVIEW GAMEREVIEW GAME

Page 2: Gas Laws REVIEW GAME. Question 1 A 4.3 liter tank of hydrogen is at a pressure of 6.2 atmospheres. What volume of hydrogen will be available if the hydrogen.

Question 1Question 1

A 4.3 liter tank of hydrogen is at a A 4.3 liter tank of hydrogen is at a pressure of 6.2 atmospheres. pressure of 6.2 atmospheres. What volume of hydrogen will be What volume of hydrogen will be available if the hydrogen is used available if the hydrogen is used at pressure of 0.48 atmospheres?at pressure of 0.48 atmospheres? Boyle’s LawBoyle’s Law (4.3)(6.2) = 0.48V(4.3)(6.2) = 0.48V22

VV2 2 = 56 L = 56 L

Page 3: Gas Laws REVIEW GAME. Question 1 A 4.3 liter tank of hydrogen is at a pressure of 6.2 atmospheres. What volume of hydrogen will be available if the hydrogen.

Question 2Question 2

What mass of oxygen gas must What mass of oxygen gas must be placed in a container with a be placed in a container with a volume of 123.0L to produce a volume of 123.0L to produce a pressure of 1.87 atmospheres at pressure of 1.87 atmospheres at 28.0°C?28.0°C? Ideal Gas LawIdeal Gas Law 1.87(123) = n(0.0821)(301)1.87(123) = n(0.0821)(301) n = 9.3 mol x 32 = 298 g n = 9.3 mol x 32 = 298 g

Page 4: Gas Laws REVIEW GAME. Question 1 A 4.3 liter tank of hydrogen is at a pressure of 6.2 atmospheres. What volume of hydrogen will be available if the hydrogen.

Question 3Question 3

What is the pressure of 20.9 What is the pressure of 20.9 grams of neon gas at -19°C in a grams of neon gas at -19°C in a rigid container whose volume is rigid container whose volume is 19.0 L?19.0 L? Ideal Gas LawIdeal Gas Law 20.9 g Ne / 20 = 1.045 mol20.9 g Ne / 20 = 1.045 mol P(19) = 1.045(0.0821)(254)P(19) = 1.045(0.0821)(254) P = 1.2 atmP = 1.2 atm

Page 5: Gas Laws REVIEW GAME. Question 1 A 4.3 liter tank of hydrogen is at a pressure of 6.2 atmospheres. What volume of hydrogen will be available if the hydrogen.

Question 4Question 4

A 4.45 L balloon at 14°C contains A 4.45 L balloon at 14°C contains carbon dioxide gas. If the balloon carbon dioxide gas. If the balloon is taken outside where the is taken outside where the temperature is -22°C, what temperature is -22°C, what volume will the balloon occupy?volume will the balloon occupy? Charles LawCharles Law 4.45 / 287 = V4.45 / 287 = V22 / 251 / 251

VV2 2 = 3.9 L= 3.9 L

Page 6: Gas Laws REVIEW GAME. Question 1 A 4.3 liter tank of hydrogen is at a pressure of 6.2 atmospheres. What volume of hydrogen will be available if the hydrogen.

Question 5Question 5

A gas occupies a volume of 0.65 L A gas occupies a volume of 0.65 L at 118.3 kPa and a temperature of at 118.3 kPa and a temperature of 0.00°C. What volume will the gas 0.00°C. What volume will the gas occupy at 2.50 atm and 50.0°C?occupy at 2.50 atm and 50.0°C? Combined Gas LawCombined Gas Law 118.3 kPa x (1 atm/101.325 kPa) = 1.17 atm118.3 kPa x (1 atm/101.325 kPa) = 1.17 atm 1.17(0.65)/273 = 2.5V1.17(0.65)/273 = 2.5V22/323/323

VV22 = 0.36 L = 0.36 L

Page 7: Gas Laws REVIEW GAME. Question 1 A 4.3 liter tank of hydrogen is at a pressure of 6.2 atmospheres. What volume of hydrogen will be available if the hydrogen.

Question 6Question 6

A sample of gas occupies A sample of gas occupies 25 mL at -142°C. What 25 mL at -142°C. What volume does the sample volume does the sample occupy at 65°C?occupy at 65°C? CharlesCharles 25/131 = V25/131 = V22/338/338 VV22 = 65 mL = 65 mL

Page 8: Gas Laws REVIEW GAME. Question 1 A 4.3 liter tank of hydrogen is at a pressure of 6.2 atmospheres. What volume of hydrogen will be available if the hydrogen.

Question 7Question 7

When a container is filled with 6.000 When a container is filled with 6.000 grams of Hgrams of H22, 64.00 grams of O, 64.00 grams of O22, and , and 28.00 grams of N28.00 grams of N22, the pressure in the , the pressure in the container is 5412 kPa. What is the container is 5412 kPa. What is the partial pressure of Opartial pressure of O22?? Dalton’s LawDalton’s Law 6 g H6 g H22/2 = 3mol 64g O/2 = 3mol 64g O22/32 = 2mol 28 g N/32 = 2mol 28 g N22/28 = 1mol/28 = 1mol 3+2+1 = 6 total moles3+2+1 = 6 total moles PPO2O2 = 2/6(5412) = 1804 kPa = 2/6(5412) = 1804 kPa

Page 9: Gas Laws REVIEW GAME. Question 1 A 4.3 liter tank of hydrogen is at a pressure of 6.2 atmospheres. What volume of hydrogen will be available if the hydrogen.

Question 7.5Question 7.5

Magnesium reacts with Magnesium reacts with hydrochloric acidhydrochloric acid Mg + 2HCl Mg + 2HCl MgCl MgCl22 + H + H22

How many grams of magnesium How many grams of magnesium would I have to start with to make would I have to start with to make 3.5L of hydrogen gas at 130 kPa 3.5L of hydrogen gas at 130 kPa and 50. degrees celsius?and 50. degrees celsius? 130(3.5) = n (8.314)(323)130(3.5) = n (8.314)(323) n = 0.17 mol Hn = 0.17 mol H22 x 1/1 x 24.3 = 4.1 g x 1/1 x 24.3 = 4.1 g

Page 10: Gas Laws REVIEW GAME. Question 1 A 4.3 liter tank of hydrogen is at a pressure of 6.2 atmospheres. What volume of hydrogen will be available if the hydrogen.

Question 8Question 8

What is the volume occupied What is the volume occupied by 71.0 grams of chlorine gas at by 71.0 grams of chlorine gas at STP?STP? Ideal Gas LawIdeal Gas Law 71 g Cl71 g Cl22 / 71 = 1 mol / 71 = 1 mol 1V = 1(0.0821)(273)1V = 1(0.0821)(273) V = 22.4 LV = 22.4 L

Page 11: Gas Laws REVIEW GAME. Question 1 A 4.3 liter tank of hydrogen is at a pressure of 6.2 atmospheres. What volume of hydrogen will be available if the hydrogen.

Question 9Question 9

A rigid container of OA rigid container of O22 has a has a

pressure of 185 kPa at a pressure of 185 kPa at a temperature of 458 K. What is temperature of 458 K. What is the pressure at 283 K?the pressure at 283 K? Gay Lussac’s LawGay Lussac’s Law 185/458 = P185/458 = P22 / 283 / 283

PP22 = 114 kPa = 114 kPa

Page 12: Gas Laws REVIEW GAME. Question 1 A 4.3 liter tank of hydrogen is at a pressure of 6.2 atmospheres. What volume of hydrogen will be available if the hydrogen.

Question 10Question 10

An unknown gas moves three An unknown gas moves three times as fast as sulfur dioxide times as fast as sulfur dioxide gas. What is the mass of the gas. What is the mass of the unknown gas?unknown gas? SSuu = 3 m = 3 muu = ? = ? SSso2so2 = 1 m = 1 mso2so2 = 64 = 64 1/3 = Sqrt(m/64)1/3 = Sqrt(m/64) m = 7.1 g/molm = 7.1 g/mol

Page 13: Gas Laws REVIEW GAME. Question 1 A 4.3 liter tank of hydrogen is at a pressure of 6.2 atmospheres. What volume of hydrogen will be available if the hydrogen.

Question 11Question 11

How many moles of NHow many moles of N22 are in a are in a flask with a volume of 125 mL flask with a volume of 125 mL at a pressure of 2550 mm Hg at a pressure of 2550 mm Hg and a temperature of 300.0 K?and a temperature of 300.0 K? Ideal Gas LawIdeal Gas Law 2550 mm Hg ( 1 atm / 760 mm Hg) = 3.36atm2550 mm Hg ( 1 atm / 760 mm Hg) = 3.36atm 3.36 (0.125) = n (0.0821) ( 300)3.36 (0.125) = n (0.0821) ( 300) 0.0171 mol N0.0171 mol N22

Page 14: Gas Laws REVIEW GAME. Question 1 A 4.3 liter tank of hydrogen is at a pressure of 6.2 atmospheres. What volume of hydrogen will be available if the hydrogen.

Question 12Question 12

The volume of a gas is 125 mL at The volume of a gas is 125 mL at 254 kPa pressure. What will the 254 kPa pressure. What will the volume be when the pressure is volume be when the pressure is reduced to 1.78 psi, assuming the reduced to 1.78 psi, assuming the temperature remains constant?temperature remains constant? Boyle’s LawBoyle’s Law 254 kPa ( 14.7psi / 101.325 kPa) = 36.85 psi254 kPa ( 14.7psi / 101.325 kPa) = 36.85 psi 36.85(125) = 1.78V36.85(125) = 1.78V V = 2590 mLV = 2590 mL

Page 15: Gas Laws REVIEW GAME. Question 1 A 4.3 liter tank of hydrogen is at a pressure of 6.2 atmospheres. What volume of hydrogen will be available if the hydrogen.

Question 13Question 13

A mixture of gases at a total pressure A mixture of gases at a total pressure of 95 kPa contains Nof 95 kPa contains N22, CO, CO22, and O, and O22. . The partial pressure of COThe partial pressure of CO22 is 24 kPa is 24 kPa and the partial pressure of the Nand the partial pressure of the N22 is is 48 kPa. What is the partial pressure 48 kPa. What is the partial pressure of the Oof the O22?? Dalton’s LawDalton’s Law 95 = 24 + 48 + P95 = 24 + 48 + Po2o2

PPo2o2 = 23 kPa = 23 kPa

Page 16: Gas Laws REVIEW GAME. Question 1 A 4.3 liter tank of hydrogen is at a pressure of 6.2 atmospheres. What volume of hydrogen will be available if the hydrogen.

Question 14Question 14

A gas, collected over water, has a A gas, collected over water, has a measured pressure of 1.23 atm and has measured pressure of 1.23 atm and has a volume of 590 mL at a temperature of a volume of 590 mL at a temperature of 47°C. What volume will the dry gas 47°C. What volume will the dry gas occupy at 25°C and standard pressure?occupy at 25°C and standard pressure? Dalton and CombinedDalton and Combined 1.23 atm (101.325 kPa / 1atm ) = 124.6 kPa1.23 atm (101.325 kPa / 1atm ) = 124.6 kPa 124.6 kPa = P124.6 kPa = Pgasgas + 10.62 + 10.62 PPgasgas = 113.98 kPa = 113.98 kPa 113.98(590) / 320 = 101.325(V113.98(590) / 320 = 101.325(V22) / 298) / 298 620 mL620 mL

Page 17: Gas Laws REVIEW GAME. Question 1 A 4.3 liter tank of hydrogen is at a pressure of 6.2 atmospheres. What volume of hydrogen will be available if the hydrogen.

Question 15Question 15

Nitrogen gas diffuses into an Nitrogen gas diffuses into an empty container at a rate of empty container at a rate of 254 m/s. At what velocity will 254 m/s. At what velocity will carbon monoxide gas move at carbon monoxide gas move at the same temperature?the same temperature? GrahamGraham VVCOCO / 254 = Sqrt (28/28) / 254 = Sqrt (28/28) VVCOCO = 254 m/s = 254 m/s

Page 18: Gas Laws REVIEW GAME. Question 1 A 4.3 liter tank of hydrogen is at a pressure of 6.2 atmospheres. What volume of hydrogen will be available if the hydrogen.

Question 16Question 16

A gas initially is present, in a balloon, at A gas initially is present, in a balloon, at 25°C under 1 atmosphere of pressure. 25°C under 1 atmosphere of pressure. The volume of the gas under these The volume of the gas under these conditions is 2.5L. What will be the new conditions is 2.5L. What will be the new volume of the gas if it is taken outside on volume of the gas if it is taken outside on a hot day, where the temperature is a hot day, where the temperature is 85°C and the pressure is 1.08 85°C and the pressure is 1.08 atmospheres?atmospheres? CombinedCombined 1(2.5)/298 = 1.08V1(2.5)/298 = 1.08V22/358/358 VV22 = 2.8 L = 2.8 L

Page 19: Gas Laws REVIEW GAME. Question 1 A 4.3 liter tank of hydrogen is at a pressure of 6.2 atmospheres. What volume of hydrogen will be available if the hydrogen.

Question 17Question 17

93.0 mL of O93.0 mL of O22 gas is collected over gas is collected over

water at 0.930 atm and 10.0°C. What water at 0.930 atm and 10.0°C. What would be the volume of this dry gas at would be the volume of this dry gas at standard conditions?standard conditions? Dalton and CombinedDalton and Combined 0.930 atm x (101.325 kPa / 1atm) = 94.23 kPa0.930 atm x (101.325 kPa / 1atm) = 94.23 kPa 94.23 kPa = P94.23 kPa = Pgasgas + 1.2281 + 1.2281

PPgasgas = 93 kPa = 93 kPa

93(93)/283 = 101.325V93(93)/283 = 101.325V22/273/273

VV22 = 82.3 mL = 82.3 mL

Page 20: Gas Laws REVIEW GAME. Question 1 A 4.3 liter tank of hydrogen is at a pressure of 6.2 atmospheres. What volume of hydrogen will be available if the hydrogen.

Question 18Question 18

If 9.0 moles of nitrogen gas will If 9.0 moles of nitrogen gas will fill a balloon that is 2.0 L in fill a balloon that is 2.0 L in volume at 293 K, what volume volume at 293 K, what volume will 28 moles of nitrogen gas fill will 28 moles of nitrogen gas fill at the same temperature?at the same temperature? Avogadro’s LawAvogadro’s Law 2/9 = V2/9 = V22/ 28/ 28 6.2 L6.2 L

Page 21: Gas Laws REVIEW GAME. Question 1 A 4.3 liter tank of hydrogen is at a pressure of 6.2 atmospheres. What volume of hydrogen will be available if the hydrogen.

Question 19Question 19

A large cylinder of He gas, such as A large cylinder of He gas, such as that used to inflate balloons, has a that used to inflate balloons, has a volume of 25.0 L at 22°C and 5.6 volume of 25.0 L at 22°C and 5.6 atm. How many grams of He are atm. How many grams of He are in such a cylinder?in such a cylinder? IdealIdeal 5.6(25) = n(0.0821)(295)5.6(25) = n(0.0821)(295) 5.78 mol x 4 = 23 g5.78 mol x 4 = 23 g

Page 22: Gas Laws REVIEW GAME. Question 1 A 4.3 liter tank of hydrogen is at a pressure of 6.2 atmospheres. What volume of hydrogen will be available if the hydrogen.

Question 20Question 20

I begin with 460 mL of oxygen I begin with 460 mL of oxygen gas at a pressure of 740 mm gas at a pressure of 740 mm Hg. How many liters of gas are Hg. How many liters of gas are present at standard pressure?present at standard pressure? BoyleBoyle 740(460) = 760(V740(460) = 760(V22))

VV22 = 450 mL = 450 mL

Page 23: Gas Laws REVIEW GAME. Question 1 A 4.3 liter tank of hydrogen is at a pressure of 6.2 atmospheres. What volume of hydrogen will be available if the hydrogen.

Question 21Question 21

Calculate the pressure of a gas Calculate the pressure of a gas whose temperature is increased whose temperature is increased from 15°C to 25°C and whose from 15°C to 25°C and whose original pressure is 0.75 atm.original pressure is 0.75 atm. Gay LussacGay Lussac .75/288 = P.75/288 = P22/298/298

PP22 = 0.78 atm = 0.78 atm

Page 24: Gas Laws REVIEW GAME. Question 1 A 4.3 liter tank of hydrogen is at a pressure of 6.2 atmospheres. What volume of hydrogen will be available if the hydrogen.

Question 22Question 22

A sample of gas at standard A sample of gas at standard temperature and pressure is put into temperature and pressure is put into an expandable container. The an expandable container. The original volume of the gas is 250 mL. original volume of the gas is 250 mL. What is the new volume if the gas is What is the new volume if the gas is cooled to -15°C and the pressure is cooled to -15°C and the pressure is increased to 780 mm Hg?increased to 780 mm Hg? CombinedCombined (760)(250)/273 = 780V(760)(250)/273 = 780V22/258/258 VV22 = 230 mL = 230 mL

Page 25: Gas Laws REVIEW GAME. Question 1 A 4.3 liter tank of hydrogen is at a pressure of 6.2 atmospheres. What volume of hydrogen will be available if the hydrogen.

Question 23Question 23

A balloon is filled with 150. mL of A balloon is filled with 150. mL of carbon dioxide gas at 900. mm Hg carbon dioxide gas at 900. mm Hg and 25.0°C. If the pressure is held and 25.0°C. If the pressure is held constant, what will the new constant, what will the new volume be if the temperature is volume be if the temperature is raised to 75.0°C?raised to 75.0°C? CharlesCharles 150/298 = V150/298 = V22/348/348 VV22 = 175 mL = 175 mL

Page 26: Gas Laws REVIEW GAME. Question 1 A 4.3 liter tank of hydrogen is at a pressure of 6.2 atmospheres. What volume of hydrogen will be available if the hydrogen.

Question 24Question 24

Sulfur dioxide gas can move Sulfur dioxide gas can move with a velocity of 150 m/s. How with a velocity of 150 m/s. How fast will carbon dioxide gas fast will carbon dioxide gas move at the same temperature?move at the same temperature? GrahamGraham VV11/150 = Sqrt(64/44)/150 = Sqrt(64/44) 180 m/s180 m/s

Page 27: Gas Laws REVIEW GAME. Question 1 A 4.3 liter tank of hydrogen is at a pressure of 6.2 atmospheres. What volume of hydrogen will be available if the hydrogen.

Question 25Question 25

6 moles of nitrogen gas is 6 moles of nitrogen gas is placed in a 3 L container. placed in a 3 L container. What volume will 18 moles of What volume will 18 moles of gas occupy under the same gas occupy under the same conditionsconditions AvogadroAvogadro 3/6 = V3/6 = V22/18/18 9 L9 L

Page 28: Gas Laws REVIEW GAME. Question 1 A 4.3 liter tank of hydrogen is at a pressure of 6.2 atmospheres. What volume of hydrogen will be available if the hydrogen.

Question 26Question 26

40.0g of neon gas is put into a 40.0g of neon gas is put into a 500. mL container at 15.0°C. 500. mL container at 15.0°C. What is the pressure within the What is the pressure within the container in kPa?container in kPa? IdealIdeal 40 g Ne / 20 = 2 mol40 g Ne / 20 = 2 mol P(.5) = 2(8.314)(288)P(.5) = 2(8.314)(288) 9580 kPa9580 kPa

Page 29: Gas Laws REVIEW GAME. Question 1 A 4.3 liter tank of hydrogen is at a pressure of 6.2 atmospheres. What volume of hydrogen will be available if the hydrogen.

Question 27Question 27

If a gas begins at a temperature If a gas begins at a temperature of 27°C and a pressure of 150 of 27°C and a pressure of 150 kPa, what will the new pressure kPa, what will the new pressure be when the temperature is be when the temperature is raised to 50.°C?raised to 50.°C? Gay LussacGay Lussac 150/300 = P150/300 = P22/323/323 160 kPa160 kPa

Page 30: Gas Laws REVIEW GAME. Question 1 A 4.3 liter tank of hydrogen is at a pressure of 6.2 atmospheres. What volume of hydrogen will be available if the hydrogen.

Question 28Question 28

Hydrogen gas moves with a Hydrogen gas moves with a velocity of 400. m/s at 25°C. velocity of 400. m/s at 25°C. What is the mass of a gas that What is the mass of a gas that moves at 250. m/s at the same moves at 250. m/s at the same temperature?temperature? GrahamGraham 400/250 = Sqrt(m400/250 = Sqrt(m22/2)/2) 5.12 g/mol5.12 g/mol

Page 31: Gas Laws REVIEW GAME. Question 1 A 4.3 liter tank of hydrogen is at a pressure of 6.2 atmospheres. What volume of hydrogen will be available if the hydrogen.

Question 29Question 29

Oxygen gas is collected over water Oxygen gas is collected over water at 18°C at a pressure of 800. mm at 18°C at a pressure of 800. mm Hg and a volume of 100. mL. Hg and a volume of 100. mL. What is the volume of the dry gas What is the volume of the dry gas at STP?at STP? Dalton and CombinedDalton and Combined 800 mmHg x (101.325 kPa/760 mmHg) = 106.7 kPa800 mmHg x (101.325 kPa/760 mmHg) = 106.7 kPa 106.7 = P106.7 = Pgasgas + 2.0644 + 2.0644 104.6 kPa104.6 kPa 104.6(100)/291 = 101.325V104.6(100)/291 = 101.325V22/273/273 VV22 = 97 mL = 97 mL

Page 32: Gas Laws REVIEW GAME. Question 1 A 4.3 liter tank of hydrogen is at a pressure of 6.2 atmospheres. What volume of hydrogen will be available if the hydrogen.

Question 29Question 29

6Li + N6Li + N22 2Li 2Li33NN I have 40.0 grams of lithium metal. How many I have 40.0 grams of lithium metal. How many

liters of nitrogen gas will react with it at 230. kPa liters of nitrogen gas will react with it at 230. kPa and 17.0 degrees celsius?and 17.0 degrees celsius?

40g/7 x 1/6 = 0.95 mol40g/7 x 1/6 = 0.95 mol 230(v) = 0.95(8.314)(290)230(v) = 0.95(8.314)(290) V = 9.95 LV = 9.95 L

I have 25.0 L of nitrogen gas at 1.34 atm and 80.0 I have 25.0 L of nitrogen gas at 1.34 atm and 80.0 degrees celsius. How many grams of Lidegrees celsius. How many grams of Li33N can I N can I make?make?

1.34(25) = n (0.0821)(353)1.34(25) = n (0.0821)(353) n = 1.15 mol Nn = 1.15 mol N22 x 2/1 x 35 = 80.5 g x 2/1 x 35 = 80.5 g

Page 33: Gas Laws REVIEW GAME. Question 1 A 4.3 liter tank of hydrogen is at a pressure of 6.2 atmospheres. What volume of hydrogen will be available if the hydrogen.

Question 30Question 30

I have 80. grams of carbon monoxide gas, 55 I have 80. grams of carbon monoxide gas, 55 grams of dinitrogen monoxide gas, and 100. grams of dinitrogen monoxide gas, and 100. grams of diphosphorus pentoxide gas. The total grams of diphosphorus pentoxide gas. The total pressure of the mixture of gases is 1.67 atm. pressure of the mixture of gases is 1.67 atm. What is the partial pressure of each gas?What is the partial pressure of each gas? Dalton’s LawDalton’s Law 80 g CO / 28 = 2.86 mol80 g CO / 28 = 2.86 mol 55 g N55 g N22O / 44 = 1.25 molO / 44 = 1.25 mol 100 g P100 g P22OO55 / 142 = 0.70 mol / 142 = 0.70 mol TOTAL MOLES = 4.81 molesTOTAL MOLES = 4.81 moles PPCOCO = 2.86/4.81*(1.67) = 0.99 atm = 2.86/4.81*(1.67) = 0.99 atm PPN2ON2O = 1.25/4.81*(1.67) = 0.43 atm = 1.25/4.81*(1.67) = 0.43 atm PPP2O5P2O5 = 0.70/4.81*(1.67) = 0.24 atm = 0.70/4.81*(1.67) = 0.24 atm