Honors ChemistryName _______________________________ 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 ││ 8 x 104 mm) (12 L) (760 mm) = 600 L or 6. what is the partial pressure of CO? (vapor pressure of water at 28 oC is 28.0821 L ▪ atm) (298 K) = 0.0 cm3 when immersed in an ice-water bath at 0.0 cm3) = 238 K = – 35 oC 11.658 moles) (0. (Charles’ Law) T2 = T1V2 = (273 K) (87. A mixture of Ar and CO gases is collected over water at 28 oC and an atmospheric pressure of 1.0 L) (273 K) (301 K) (760.2 cm3) V1 (100.3 mmHg (Dalton’s Law of Partial Pressure) PT = PAr + PCO + PWater PCO = PT – (PAr + PWater) = 1.00 atm) = 14. what will the volume of that gas be? (Boyle’s Law) V2 = P1V1 P2 = (3.0 g HCl │ 1 mole HCl │ 22. Ammonia (NH3) is placed in 1.05 atm – (0. what is the density? (Ideal Gas Law and Density) n = PV = (0. torr) = 3. oC.22 atm 13.0372 atm) = 0.0 L at 28 oC and 900. torr.46 g HCl │ 1 mole HCl ││ or V = nRT = (0.0 x 102 L 9. the shock wave can be so strong that 12 liters of gas will reach a pressure of 3.05 atm. (Gay-Lussac’s Law) T2 = T1P2 = (373 K) (2 atm) P1 (1 atm) = 746 K = 700 K (Significantly) .939 g NH3 = 0. torr) (3. What would be the new volume for the gas if placed under STP? (Combined Gas Law) V2 = P1V1T2 T1P2 = (900.2 L 8.0 g of HCl at STP.7 L HCl │ 36.50 L flask at 25 oC. if the sample was placed under 2 atm of pressure.939 g NH3 │ 1 mole NH3 ││ 0. When immersed in boiling liquid chlorine.2 cm3. At 1 atm of pressure water boils at 100. Find the temperature of the boiling point of chlorine in °C.0821 L ▪ atm) (273 K) P (mole ▪ K) (1. Calculate the volume of 24.7 L HCl 10.0551 moles NH3 │ 17.50 L 12.04 g NH3 ││ 0.899 atm) (1.7. °C.8 x 104 mm Hg.50 L) (mole ▪ K) RT (0. 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. When the shock wave passes and the gas returns to a pressure of 760 mm Hg. If the partial pressure of Ar is 600.789 atm + 0. what would be the temperature? (This would be like a pressure cooker). While using explosives to knock down a building. A gas is placed in a balloon with a volume of 3. the volume of the hydrogen at the same pressure is 87. torr.4 L HCl ││ 14. If the pressure of the gas is 0. A hydrogen gas volume thermometer has a volume of 100.899 atm.626 g/L 1. (Ideal Gas Law) 24. 0100 = 0.0821 L ▪ atm) (345 K) = 10.8 kPa O2 98. On hot days. mL) (333 K) = 285 mL T1 (292 K) 15.0 kPa and -108 °C.0821 L ▪ atm) (233 K) 1500 moles He │ 4. Find the pressure of this gas in kilopascals. Some students believe that teachers are full of hot air.855 L) = 2680 kPa 19.000. What was the original pressure of the helium? (Boyle’s Law) P1 = P2V2 = (30.0 mL to a volume of 0.00 g He │ 1 kg ││ 6.20 L) (311 K) = 2. A sample of air collected at Bismarck at -22 °C and 98.1 kPa N2 98.0 kg He │ 1 mole He │ 1000 g ││ . If I have a 250.989 kPa Ar 20. At that altitude the atmospheric pressure is 210 mm Hg and the temperature is -40.00 atm? (Ideal Gas Law) v = nRT = (0.90 kPa x 0.780 = 77.35 L T1 (291 K) 18.0 L. 12. A lighter-than-air balloon is designed to rise to a height of 6 miles at which point it will be fully inflated.240 mL where its pressure was found to be 30.8 L of a certain gas are prepared at 100.0 % O2.0 % N2.8 L) (295 K) (165 K) (0.0 kPa) (12.14.750 mol of gas at 72 oC and 2.750 moles) (0.240 mL) = 0. The gas is then forced into an 855 mL cylinder in which it warms to room temperature. oxygen.28 atm) (100. What is the volume of 0. how many kilograms of helium will be needed to inflate the balloon? (Ideal Gas Law) n = PV = (0.00 % Ar.6 L P (mole ▪ K) (2. If I inhale 2. (Combined Gas Law) P2 = P1V1T2 T1V2 = (100. you may have noticed that potato chip bags seem to “inflate”. what is the new volume of the gas? (Charles Law) V2 = V1T2 = (2. what will the new volume of the bag be? (Charles’ Law) V2 = V1T2 = (250. °C. If the full volume of the balloon is 100. (Dalton’s Law of Partial Pressure) 98.90 kPa x 0. and 1.0 mL) 17.0 L) (mole ▪ K) = 1464 moles = 1500 moles (significantly) RT (0. Find the partial pressures of each of these gases.000. and I leave it in my car which has a temperature of 60. even though they have not been opened.0 mm Hg.0 mm) (0. and argon.90 kPa x 0. 22 °C.20 liters of gas at a temperature of 18 oC and it heats to a temperature of 38 oC in my lungs.90 kPa had 78. mL bag at a temperature of 19 oC. In a certain experiment a sample of helium in a vacuum system was compressed at 25 °C from a volume of 200. oC. Air from the prairies of North Dakota in winter contains essentially only nitrogen. 21.00 atm) 16.0360 mm V1 (200.210 = 20. 20 mole N2 = 0.3 kPa? (Dalton’s Law of Partial Pressure) 0. through heating.85 L cylinder exert a pressure of 3.30 moles 22. If the actual volume is 92.0821 L atm) 27.0 mm Hg and 15 °C is 6.2 g of a given gas occupies a volume of 93. What is the temperature in the cylinder (in °C)? (Ideal Gas Law) T = PV = (3.2 g) (10.97 kPa? (Ideal Gas Law) n = PV = (0.0 °C if its volume at 720.20 mole N2 and 0. °C in which there is 0.04 g NH3 ││ 12. mL) = 0. A quantity of potassium chlorate is selected to yield.0 mL) = 301 K = 28 oC .2 L at a particular temperature and pressure.84 L) (313 K) (288 K) (800. What volume of Ne at one atm and 25 °C would have to be added to a sign having a volume of 250.84 L.279 moles of O2 in a 1.0821 L ▪ atm) (304 K) │ 1 mole NH3 ││ 26.732 mole NH3 │ 17. what is the resulting temperature of oxygen in degrees Celsius? (Combined Gas Law) T2 = P2V2 T1 = (0.10 mole CO2 at a total pressure of 101.3 kPa = 68 kPa 0.9671 atm) (18.3 kPa = 33 kPa 0.4 L under the same conditions? (Avogadro’s Law) n2 = n1V2 = (23.59 g V1 (93. If 23.5 mL and the actual pressure is 0.0 mL of O2 when measured at STP.68 atm.9 L) (mole ▪ K) = 0. what mass of the gas occupies a volume of 10.894 atm.279 moles) (0.330 mL P1 (1.33 x 101. mL to create a pressure of one mm Hg at that temperature? (Boyle’s Law) V1 = P2V2 = (0.0 mm) = 6.68 atm) (1.10 mole CO2 = 0. What is the mass of 18.21.67 x 101.00 atm) 24. 75.69 L 25. (Combined Gas Law) V2 = P1V1T2 T1P2 = (720.5 mL) (273 K) P1V1 (1.894 atm) (92.5 g NH3 RT (0.0 mm) (6.0 mm Hg and 40.00 atm) (75. What would be the partial pressure of N2 in a container at 50.30 moles 0.4 L) = 2. 0.00132 atm) (250.85 L) (mole K) = 297 K = 24 oC nR (0. Find the volume of a gas at 800.9 L of NH3 at 31 °C and 97.2 L) 23. 275 atm) 30.00 atm of pressure.25 mole) 33. atm pushing on it. four moles of ethane. L at a particular temperature and pressure. A sample of gas is placed in a container at 25 oC and 2. A mixture of hydrocarbons contains three moles of methane.15 mole) = 435 L n1 (3.417 x 237 kPa = 98.0 moles) (0.00 x 106 atm) (1. atm) 32. If 3. then release the pressure until it is equal to 0.28. If the temperature is raised to 50. An experimental research submarine with a volume of 15.00 x 10-5 liter sample of a gas at that pressure.52 moles RT (0. How many moles of gas would occupy a volume of 14 L at a pressure of 700.0 kPa ethane 12 5 = 0.0821 L ▪ atm) (295 K) = 2.000 L) = 72 L P2 (250. in kPa. Find the partial pressures of the three gases. what volume does 14.275 atm.2 atm) (15.00 atm) (323 K) = 2.000 liters has an internal pressure of 1. what is the new pressure? (Gay-Lussac’s Law) P2 = P1T2 = (2.25 mol of argon gas occupies a volume of 100. The highest pressure ever produced in a laboratory setting was about 2.17 atm T1 (298 K) . and five moles of propane.3 kPa methane 12 4 = 0.921 atm) (14 L) (mole ▪ K) = 0.2 atm.0821 L ▪ atm) (303 K) 31.8 kPa propane 12 29. If the pressure of the ocean breaks the submarine forming a bubble with a pressure of 250.34 atm = 237 kPa V (mole ▪ K) (124 L) 3 = 0. Submarines need to be extremely strong to withstand the extremely high pressure of water pushing down on them.00 x 10-5 L) = 72. The container has a volume of 124 liters and the temperature is 22 °C. oC? (Ideal Gas Law) n = PV = (0.15 mol of argon occupy under the same conditions? (Avogadro’s Law) V2 = V1n2 = (100. how big will that bubble be? (Boyle’s Law) V2 = P1V1 = (1.333 x 237 kPa = 79. what would the new volume of that gas be? (Boyle’s Law) V2 = P1V1 = (2. L) (14. oC.250 x 237 kPa = 59. If we have a 1. (Ideal Gas Law and Dalton’s Law of Partial Pressure) P = nRT = (12. torr and a temperature of 30.00 x 106 atm.7 L P2 (0. T at constant P. P vs.0121 mole XeF4 │ 207.straight lines c. (Ideal Gas Law) 2. a – hyperbola b. T at constant V.00 liter container at 80.00 L) 35.50 grams of XeF4 is introduced into an evacuated 3. Find the pressure in atmospheres in the container. V vs.34.117 atm V (3. . For a mole of ideal gas. V at constant T.50 g XeF4 │ 1 mole XeF4 ││ 0. P vs.29 g XeF4 ││ P = nRT = (0. °C.0121 mole) (0. b&c .0821 L atm) (353 K) = 0. sketch graphs of a. 2. . at constant volume? It would increase the temperature.4 L for one mole of the gas. How are volume and temperature related (directly or inversely)? How would reducing the Kelvin temperature by one half effect the volume. How are volume and pressure related (directly or inversely)? How would tripling the pressure effect the volume. It can not be stated that the temperature would be double. at constant temperature? It would reduce the volume to one third of the original amount. Since it was not specified what unit of temperature. 37. How are temperature and pressure related (directly or inversely)? How would doubling the pressure effect the temperature. must be subtracted from the total pressure to determine the pressure of the gas. 38. If a gas is collected over water. at the specific temperature.36. 40. 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. the volume would reduce to one half of the original amount. what must be done to determine the pressure of the gas? The water vapor pressure. at constant pressure? Since the temperature is in Kelvin. 39. 2 L 8.Gas Law Worksheet Answer Key 1. 0. 6. 24. 6. 59.7 L HCl 10. 2.1 kPa N2. 0.17 atm 34. 0.59 g 23.8 kPa propane 29. 238 K = – 35 oC 11. 435 L 33. 1. 297 K = 24 oC 27. 12.5 g NH3 26. 3.8 kPa O2. 0. 72.0 kPa ethane.89 atm = 3720 torr 2. 20. 285 mL 15.69 L 25. 33 kPa CO2 22. 0.626 g/L 12. 0.52 moles 31. 301 K = 28 oC 28. 0.0 L F2 6.0360 mm 17.117 atm .0 kg He 21. 14. 2.7 L 30. 746 K = 700 K (Significantly) 14.3 kPa methane.429 atm NH3.0 x 102 L 9.18 atm 7. 6. 10. 77. 4. 98. 2680 kPa 19.33 moles 3.330 mL 24.214 atm F2 4. 79. 0. 0.857 atm Ne.35 L 18. 294 K 5.22 atm 13. 72 L 32. 0. 2. 68 kPa N2. 2.6 L 16.989 kPa Ar 20.