Gas Laws Worksheet III Answer Key 11-12

Honors Chemistry

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Chapter 11: Gas Law Worksheet Answer Key

Date _____/_____/_____

Period _____

Complete the following calculation by list the given information, rewriting the formula to solve for the unknown, and plugging in the information (number and units), and writing the answer significantly. 1. A sample of 0.500 moles of gas is placed in a container of volume of 2.50 L. What is the pressure of the gas in torr if the gas is at 25 oC? (Ideal Gas Law) P = nRT V

=

(0.500 moles) (0.0821 L ▪ atm) (298 K) (mole ▪ K) (2.50 L)

=

P

=

4.89 atm

=

3720 torr

2. If 2.00 mol of gas occupies 4.50 L at STP. How much of the same gas will occupy 3.00 L at STP? (Avogadro’s Law) n2

=

V2n1 V1

=

(3.00 L) (2.00 moles) (4.50 L)

=

1.33 moles

3. Determine the partial pressures of each of the gases in the following mixture: 17.04 g NH3, 40.36 g Ne and 19.00 g F2. The gases are at 1.50 atm of pressure. (Dalton’s Law of Partial Pressure & Mole Ratios) 17.04 g NH3 │ 1 mole NH3 ││ 1.000 mole NH3 │ 17.04 g NH3 ││

1.000 mole NH3 = 0.2857 x 1.50 atm = 0.429 atm NH3 3.500 moles

40.36 g Ne │ 1 mole Ne ││ 2.000 moles Ne │ 20.18 g Ne ││

2.000 moles Ne = 0.5714 x 1.50 atm = 0.857 atm Ne 3.500 moles

19.00 g F2 │ 1 mole F2 ││ 0.5000 moles F2 │ 38.00 g F2 ││

0.5000 moles F2 = 0.1429 x 1.50 atm = 0.214 atm F2 3.500 moles

1.000 moles + 2.000 moles + 0.5000 moles = 3.500 moles

4. If at 1.00 atm of pressure water boils at 100. oC, at what temperature would water boil if the pressure is 600. torr? (This shows why food doesn't cook well at higher elevations) (Gay-Lussac’s Law) T2 = T1P2 P1

=

(373 K) (0.789 atm) (1.00 atm)

=

294 K

5. Calculate the volume of 40.6 g of F2 at STP. (Ideal Gas Law) 40.6 g F2 │ 1 mole F2 ││ 22.4 L F2 ││ 23.9 L F2 │ 38.00 g F2 ││ 1 mole F2 ││ or 40.6 g F2 │ 1 mole F2 ││ 1.07 mole F2 │ 38.00 g F2 ││ V = nRT P

=

(1.07 moles) (0.0821 L ▪ atm) (273 K) (mole ▪ K) (1.00 atm))

=

24.0 L F2

6. If the winter pressure that most car tire tires should be at to wear evenly is 32.0 psi, what is the pressure in atm? (Pressure Conversion) 32.0 psi │ 1.00 atm ││ 2.18 atm │ 14.7 psi ││

7. A gas is placed in a balloon with a volume of 3.0 L at 28 oC and 900. torr. What would be the new volume for the gas if placed under STP? (Combined Gas Law) V2 = P1V1T2 T1P2

=

(900. torr) (3.0 L) (273 K) (301 K) (760. torr)

=

3.2 L

8. Part of the reason that conventional explosives cause so much damage is that their detonation produces a strong shock wave that can knock things down. While using explosives to knock down a building, the shock wave can be so strong that 12 liters of gas will reach a pressure of 3.8 x 104 mm Hg. When the shock wave passes and the gas returns to a pressure of 760 mm Hg, what will the volume of that gas be? (Boyle’s Law) V2 = P1V1 P2

=

(3.8 x 104 mm) (12 L) (760 mm)

=

600 L or 6.0 x 102 L

9. Calculate the volume of 24.0 g of HCl at STP. (Ideal Gas Law) 24.0 g HCl │ 1 mole HCl │ 22.4 L HCl ││ 14.7 L HCl │ 36.46 g HCl │ 1 mole HCl ││ or V = nRT = (0.658 moles) (0.0821 L ▪ atm) (273 K) P (mole ▪ K) (1.00 atm)

=

14.7 L HCl

10. A hydrogen gas volume thermometer has a volume of 100.0 cm3 when immersed in an ice-water bath at 0. °C. When immersed in boiling liquid chlorine, the volume of the hydrogen at the same pressure is 87.2 cm3. Find the temperature of the boiling point of chlorine in °C. (Charles’ Law) T2 = T1V2 = (273 K) (87.2 cm3) V1 (100.0 cm3)

=

238 K = – 35 oC

11. Ammonia (NH3) is placed in 1.50 L flask at 25 oC. If the pressure of the gas is 0.899 atm, what is the density? (Ideal Gas Law and Density) n = PV = (0.899 atm) (1.50 L) (mole ▪ K) RT (0.0821 L ▪ atm) (298 K)

=

0.0551 moles NH3 │ 17.04 g NH3 ││ 0.939 g NH3 │ 1 mole NH3 ││

0.939 g NH3 = 0.626 g/L 1.50 L 12. A mixture of Ar and CO gases is collected over water at 28 oC and an atmospheric pressure of 1.05 atm. If the partial pressure of Ar is 600. torr, what is the partial pressure of CO? (vapor pressure of water at 28 oC is 28.3 mmHg (Dalton’s Law of Partial Pressure) PT = PAr + PCO + PWater PCO = PT – (PAr + PWater)

=

1.05 atm – (0.789 atm + 0.0372 atm) = 0.22 atm

13. At 1 atm of pressure water boils at 100. oC, if the sample was placed under 2 atm of pressure, what would be the temperature? (This would be like a pressure cooker). (Gay-Lussac’s Law) T2 = T1P2 = (373 K) (2 atm) P1 (1 atm)

=

746 K = 700 K (Significantly)

14. On hot days, you may have noticed that potato chip bags seem to “inflate”, even though they have not been opened. If I have a 250. mL bag at a temperature of 19 oC, and I leave it in my car which has a temperature of 60. oC, what will the new volume of the bag be? (Charles’ Law) V2 = V1T2 = (250. mL) (333 K) = 285 mL T1 (292 K) 15. What is the volume of 0.750 mol of gas at 72 oC and 2.00 atm? (Ideal Gas Law) v = nRT = (0.750 moles) (0.0821 L ▪ atm) (345 K) = 10.6 L P (mole ▪ K) (2.00 atm) 16. In a certain experiment a sample of helium in a vacuum system was compressed at 25 °C from a volume of 200.0 mL to a volume of 0.240 mL where its pressure was found to be 30.0 mm Hg. What was the original pressure of the helium? (Boyle’s Law) P1 = P2V2 = (30.0 mm) (0.240 mL) = 0.0360 mm V1 (200.0 mL) 17. Some students believe that teachers are full of hot air. If I inhale 2.20 liters of gas at a temperature of 18 oC and it heats to a temperature of 38 oC in my lungs, what is the new volume of the gas? (Charles Law) V2 = V1T2 = (2.20 L) (311 K) = 2.35 L T1 (291 K) 18. 12.8 L of a certain gas are prepared at 100.0 kPa and -108 °C. The gas is then forced into an 855 mL cylinder in which it warms to room temperature, 22 °C. Find the pressure of this gas in kilopascals. (Combined Gas Law) P2 = P1V1T2 T1V2

=

(100.0 kPa) (12.8 L) (295 K) (165 K) (0.855 L)

=

2680 kPa

19. Air from the prairies of North Dakota in winter contains essentially only nitrogen, oxygen, and argon. A sample of air collected at Bismarck at -22 °C and 98.90 kPa had 78.0 % N2, 21.0 % O2, and 1.00 % Ar. Find the partial pressures of each of these gases. (Dalton’s Law of Partial Pressure) 98.90 kPa x 0.780 = 77.1 kPa N2 98.90 kPa x 0.210 = 20.8 kPa O2 98.90 kPa x 0.0100 = 0.989 kPa Ar 20. A lighter-than-air balloon is designed to rise to a height of 6 miles at which point it will be fully inflated. At that altitude the atmospheric pressure is 210 mm Hg and the temperature is -40. °C. If the full volume of the balloon is 100,000.0 L, how many kilograms of helium will be needed to inflate the balloon? (Ideal Gas Law) n = PV = (0.28 atm) (100,000.0 L) (mole ▪ K) = 1464 moles = 1500 moles (significantly) RT (0.0821 L ▪ atm) (233 K) 1500 moles He │ 4.00 g He │ 1 kg ││ 6.0 kg He │ 1 mole He │ 1000 g ││

21. What would be the partial pressure of N2 in a container at 50. °C in which there is 0.20 mole N2 and 0.10 mole CO2 at a total pressure of 101.3 kPa? (Dalton’s Law of Partial Pressure) 0.20 mole N2 = 0.67 x 101.3 kPa = 68 kPa 0.30 moles 0.10 mole CO2 = 0.33 x 101.3 kPa = 33 kPa 0.30 moles 22. If 23.2 g of a given gas occupies a volume of 93.2 L at a particular temperature and pressure, what mass of the gas occupies a volume of 10.4 L under the same conditions? (Avogadro’s Law) n2 = n1V2 = (23.2 g) (10.4 L) = 2.59 g V1 (93.2 L) 23. What volume of Ne at one atm and 25 °C would have to be added to a sign having a volume of 250. mL to create a pressure of one mm Hg at that temperature? (Boyle’s Law) V1 = P2V2 = (0.00132 atm) (250. mL) = 0.330 mL P1 (1.00 atm) 24. Find the volume of a gas at 800.0 mm Hg and 40.0 °C if its volume at 720.0 mm Hg and 15 °C is 6.84 L. (Combined Gas Law) V2 = P1V1T2 T1P2

=

(720.0 mm) (6.84 L) (313 K) (288 K) (800.0 mm)

=

6.69 L

25. What is the mass of 18.9 L of NH3 at 31 °C and 97.97 kPa? (Ideal Gas Law) n = PV = (0.9671 atm) (18.9 L) (mole ▪ K) = 0.732 mole NH3 │ 17.04 g NH3 ││ 12.5 g NH3 RT (0.0821 L ▪ atm) (304 K) │ 1 mole NH3 ││ 26. 0.279 moles of O2 in a 1.85 L cylinder exert a pressure of 3.68 atm. What is the temperature in the cylinder (in °C)? (Ideal Gas Law) T = PV = (3.68 atm) (1.85 L) (mole K) = 297 K = 24 oC nR (0.279 moles) (0.0821 L atm) 27. A quantity of potassium chlorate is selected to yield, through heating, 75.0 mL of O2 when measured at STP. If the actual volume is 92.5 mL and the actual pressure is 0.894 atm, what is the resulting temperature of oxygen in degrees Celsius? (Combined Gas Law) T2 = P2V2 T1 = (0.894 atm) (92.5 mL) (273 K) P1V1 (1.00 atm) (75.0 mL)

=

301 K = 28 oC

28. A mixture of hydrocarbons contains three moles of methane, four moles of ethane, and five moles of propane. The container has a volume of 124 liters and the temperature is 22 °C. Find the partial pressures of the three gases, in kPa. (Ideal Gas Law and Dalton’s Law of Partial Pressure) P = nRT = (12.0 moles) (0.0821 L ▪ atm) (295 K) = 2.34 atm = 237 kPa V (mole ▪ K) (124 L) 3 = 0.250 x 237 kPa = 59.3 kPa methane 12 4 = 0.333 x 237 kPa = 79.0 kPa ethane 12 5 = 0.417 x 237 kPa = 98.8 kPa propane 12 29. The highest pressure ever produced in a laboratory setting was about 2.00 x 106 atm. If we have a 1.00 x 10-5 liter sample of a gas at that pressure, then release the pressure until it is equal to 0.275 atm, what would the new volume of that gas be? (Boyle’s Law) V2 = P1V1 = (2.00 x 106 atm) (1.00 x 10-5 L) = 72.7 L P2 (0.275 atm) 30. How many moles of gas would occupy a volume of 14 L at a pressure of 700. torr and a temperature of 30. oC? (Ideal Gas Law) n = PV = (0.921 atm) (14 L) (mole ▪ K) = 0.52 moles RT (0.0821 L ▪ atm) (303 K) 31. Submarines need to be extremely strong to withstand the extremely high pressure of water pushing down on them. An experimental research submarine with a volume of 15,000 liters has an internal pressure of 1.2 atm. If the pressure of the ocean breaks the submarine forming a bubble with a pressure of 250. atm pushing on it, how big will that bubble be? (Boyle’s Law) V2 = P1V1 = (1.2 atm) (15,000 L) = 72 L P2 (250. atm) 32. If 3.25 mol of argon gas occupies a volume of 100. L at a particular temperature and pressure, what volume does 14.15 mol of argon occupy under the same conditions? (Avogadro’s Law) V2 = V1n2 = (100. L) (14.15 mole) = 435 L n1 (3.25 mole) 33. A sample of gas is placed in a container at 25 oC and 2.00 atm of pressure. If the temperature is raised to 50. oC, what is the new pressure? (Gay-Lussac’s Law) P2 = P1T2 = (2.00 atm) (323 K) = 2.17 atm T1 (298 K)

34. 2.50 grams of XeF4 is introduced into an evacuated 3.00 liter container at 80. °C. Find the pressure in atmospheres in the container. (Ideal Gas Law) 2.50 g XeF4 │ 1 mole XeF4 ││ 0.0121 mole XeF4 │ 207.29 g XeF4 ││ P = nRT = (0.0121 mole) (0.0821 L atm) (353 K) = 0.117 atm V (3.00 L) 35. For a mole of ideal gas, sketch graphs of a. P vs. V at constant T. a – hyperbola

b. P vs. T at constant V. b&c - straight lines

c. V vs. T at constant P.

36. How are temperature and pressure related (directly or inversely)? How would doubling the pressure effect the temperature, at constant volume? It would increase the temperature. Since it was not specified what unit of temperature, It can not be stated that the temperature would be double. 37. How are volume and temperature related (directly or inversely)? How would reducing the Kelvin temperature by one half effect the volume, at constant pressure? Since the temperature is in Kelvin, the volume would reduce to one half of the original amount. 38. How are volume and pressure related (directly or inversely)? How would tripling the pressure effect the volume, at constant temperature? It would reduce the volume to one third of the original amount. 39. How are volume and moles related (directly or inversely)? What is the volume of any gas at STP? The volume of any gas at STP is equal to 22.4 L for one mole of the gas. 40. If a gas is collected over water, what must be done to determine the pressure of the gas? The water vapor pressure, at the specific temperature, must be subtracted from the total pressure to determine the pressure of the gas.

Gas Law Worksheet Answer Key 1. 4.89 atm = 3720 torr 2. 1.33 moles 3. 0.429 atm NH3, 0.857 atm Ne, 0.214 atm F2 4. 294 K 5. 24.0 L F2 6. 2.18 atm 7. 3.2 L 8. 6.0 x 102 L 9. 14.7 L HCl 10. 238 K = – 35 oC 11. 0.626 g/L 12. 0.22 atm 13. 746 K = 700 K (Significantly) 14. 285 mL 15. 10.6 L 16. 0.0360 mm 17. 2.35 L 18. 2680 kPa 19. 77.1 kPa N2, 20.8 kPa O2, 0.989 kPa Ar 20. 6.0 kg He 21. 68 kPa N2, 33 kPa CO2 22. 2.59 g 23. 0.330 mL 24. 6.69 L 25. 12.5 g NH3 26. 297 K = 24 oC 27. 301 K = 28 oC 28. 59.3 kPa methane, 79.0 kPa ethane, 98.8 kPa propane 29. 72.7 L 30. 0.52 moles 31. 72 L 32. 435 L 33. 2.17 atm 34. 0.117 atm