ADVANCED HONORS CHEMISTRY - CHAPTER 14 NAME: THE BEHAVIOR OF GASES DATE: GAS LAW WORKSHEET - ANSWERS TO 1 - 4 - V8 PAGE: 1. The gas in an aerosol can is at a pressure of 3.000 atm at 25.00 oC. Directions on the can warn the user not to keep the can in a place where the temperature exceeds 52.00 oC. What would the gas pressure, in atm, be in the can at 52.00 oC? DATA TABLE P1 = 3.000 atm P2 = ? atm V1 = - V2 = - * T1 = 25.00 oC = 298.0 K ** T2 = 52.00 oC = 325.0 K * K = oC + 273 ** K = oC + 273 K = 25.00 oC + 273 K = 52.00 oC + 273 K = 298.0 K K = 325.0 K This problem involves a change in temperature and pressure with volume staying constant. ∴ An Anmontons’ law problem that can be solved with the combined gas law: ( P )( V ) = ( P )( V ) (T ) (T ) 1 1 1 ( P )( V ) = ( P )( V ) (T ) (T ) 1 1 2 2 1 2 P1 P = 2 T1 T2 P2 = ( P )( T ) (T ) 1 2 1 P2 = (3.000 atm )(325.0 K ) ( 298.0 K ) P2 = 3.272 atm 1 2 2 2 2. A container of nitrogen monoxide has a volume of 242 cm3. The pressure of the gas in the container is measured and determined to be 87.6 kPa. What is the volume of the gas, in mL, at standard atmospheric pressure? DATA TABLE P1 = 87.6 kPa P2 = 101.3 kPa V1 = 242 cm3 = 242 mL V2 = ? mL T1 = - T2 = - This problem involves a change in volume and pressure with temperature staying constant. ∴ A Boyle's law problem that can be solved with the combined gas law: ( P1 )( V1 ) = ( P2 )( V2 ) ( T1 ) ( T2 ) ( P )( V ) = ( P )(V ) (T ) (T ) 1 1 2 2 1 2 ( P )( V ) = ( P )(V ) 1 1 V2 = V2 = 2 2 ( P )( V ) 1 1 P2 (87.6 kPa )( 242 mL ) 101.3 kPa V2 = 209 mL 2 - AHC - Chapter 14 - Gas Law Worksheet - Answers to Problems 1 - 4 - V8 3. You buy your honey a 3.0 L helium filled balloon. The Weather Channel states that the atmospheric pressure that day is 30.5 inches. When you buy the balloon, the temperature of the store is 22 oC. Your honey then puts the balloon inside a car on a sunny day where the temperature reaches 50. oC. Since the balloon can expand and contract, we will assume that the pressure of the balloon does not change. What is the volume of the balloon after it is stored in the car? DATA TABLE P1 = 30.5 inches P2 = 30.5 inches V1 = 3.0 L V2 = ? L * T1 = 22 oC = 295 K ** T2 = 50. oC = 323 K * K = oC + 273 ** K = oC + 273 K = 22 oC + 273 K = 50. oC + 273 K = 295 K K = 323 K This problem involves a change in volume and temperature with pressure staying constant. ∴ A Charles's law problem that can be solved with the combined gas law: ( P )( V ) = ( P )( V ) (T ) (T ) 1 1 1 ( P )( V ) = ( P )(V ) (T ) (T ) 1 1 2 2 1 2 V1 V2 = T1 T2 V2 = V2 = ( V )( T ) 1 2 T1 (3.0 L )(323 K ) 295 K V2 = 3.3 L 3 - AHC - Chapter 14 - Gas Law Worksheet - Answers to Problems 1 - 4 - V8 2 2 2 4. Before you started your vacation, early in the morning on the first day of February, you checked the tires on your car, which was parked in your driveway in Volga, South Dakota. The temperature in Volga that day was - 15 oF and the absolute pressure in your tires was 32 psi. You approximated the volume of your tires at 20. L. For your vacation you drove to Phoenix, Arizona. The temperature of your tires when you arrived in Phoenix was 160 oF and the volume of your tires had increased to 22. L. What was the pressure, in psi, in your tires? DATA TABLE P1 = 32 psi P2 = ? psi V1 = 20. L V2 = 22. L * T1 = - 15 oF = - 26 oC = 247 K ** T2 = 160 oF = 71 oC = 344 K o * oC = o C= oC o F - 32 1.8 ** oC = - 15 o F - 32 1.8 o = - 26 o C= oC F - 32 1.8 160 o F - 32 1.8 = 71 o K = oC + 273 K = oC + 273 K = - 26 oC + 273 K = 71 oC + 273 K = 247 K K = 344 K This problem involves a change in pressure, volume, and temperature. ∴ A combined gas law problem: ( P )( V ) = ( P )( V ) (T ) (T ) 1 1 1 P2 = 2 2 ( P )( V )( T ) ( T )( V ) 1 1 1 P2 = 2 2 2 (32 psi)( 20. L )(344 K ) ( 247 K )( 22 L ) P2 = 41 psi 4 - AHC - Chapter 14 - Gas Law Worksheet - Answers to Problems 1 - 4 - V8
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