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Gases - 4 min

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At a constant temperature, the pressure exerted by a gas varies inversely with its volume. At a constant temperature, the pressure exerted by a gas varies inversely with its volume.

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Used when only pressure changes. Whatever pressure does, volume does the opposite. Used when only pressure changes. Whatever pressure does, volume does the opposite.

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V 2 = V 1 P 1 P2P2 P2P2

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Practice Problem Practice Problem

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A gas occupies 15.5 dm 3 at a pressure of 30 mm Hg. What is its volume when the pressure is increased to 50 mm Hg. Assume temp does not change.

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V 2 = V 1 P 1 P2P2 P2P2 Write the proper equation.

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V 2 = 15.5 dm 3 30 mm Hg 50 mm Hg Substitute the given numbers.

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V 2 = 15.5 dm 3 30 mm Hg 50 mm Hg Since pressure increases, what will happen to volume?

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V 2 = 15.5 dm 3 30 mm Hg 50 mm Hg DECREASE V 2 will be less than V 1.

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V 2 = 15.5 dm 3 30 mm Hg 50 mm Hg Do the math. V 2 = 9.3 dm 3

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V 2 = V 1 P 1 P2P2 All Boyle's Law problems should look like this: = 15.5 dm 3 30 mm Hg 50 mm Hg = 9.3 dm 3 stop

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At a constant pressure, the volume of a gas varies directly with the Kelvin temperature. At a constant pressure, the volume of a gas varies directly with the Kelvin temperature.

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Used only when temperature changes. Whatever temperature does, volume does the same. Used only when temperature changes. Whatever temperature does, volume does the same.

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K e l v i n t e m p e r a t u r e K = o C + 2 7 3 K e l v i n t e m p e r a t u r e K = o C + 2 7 3 V 2 = V 1 T 2 T1T1 T1T1

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Practice Problem Practice Problem

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500 ml of air at 20 o C is heated to 60 o C with no pressure change. What is the new gas volume?

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V 2 = V 1 T 2 T1T1 T1T1 Write the proper equation.

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V 2 = 500 ml 60 o C 20 o C Substitute the given numbers.

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V 2 = 500 ml 333 K 293 K Temps must be in Kelvins. Add 273 to Celsius temps.

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V 2 = 500 ml 333 K 293 K Since temperature increases, what will happen to volume?

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V 2 = 500 ml 333 K 293 K INCREASE V 2 will be greater than V 1.

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V 2 = 500 ml 333 K 293 K Do the math. V 2 = 600 ml

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V 2 = V 1 T 2 T1T1 All Charles' Law problems should look like this: = 500 ml 333 K 293 K = 600 ml stop

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In the real world, pressure and temperature BOTH will change. In the real world, pressure and temperature BOTH will change.

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What do we do then?? What do we do then??

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Used when both temperature and pressure change Used when both temperature and pressure change

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V 2 = V 1 P 1 T 2 P 2 T 1

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Practice Problem Practice Problem

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A gas has a volume of 810 ml at 44 o C and 325 mm Hg. What would be the gas volume at 227 o C and 650 mm Hg?

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V 2 = V 1 P 1 T 2 P 2 T 1 Write the proper equation.

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V 2 = 810 ml 325 mm Hg 500 K 650 mm Hg 317 K Substitute the given numbers.

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V 2 = 810 ml 325 mm Hg 500 K 650 mm Hg 317 K Do the math. = 640 ml

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Calculate the volume of a gas at STP if 5.05 dm 3 of the gas are collected at 27.5 o C and 95.0 kPa.

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V 2 = V 1 P 1 T 2 P 2 T 1 5.05 dm 3 95.0 kPa 273 K 101 kPa 300.5 K = = 4.32 dm 3

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The total pressure in a container is the sum of the partial pressures of all the gases in the container The total pressure in a container is the sum of the partial pressures of all the gases in the container

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This law is used most often when doing calculations with gases collected over water. This law is used most often when doing calculations with gases collected over water.

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Practice Problem #1 Practice Problem #1

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A container holds three gases: O 2, CO 2, and N 2. The partial pressures are 2 atm, 3 atm, and 4 atm respectively. What is the total pressure inside the container? A container holds three gases: O 2, CO 2, and N 2. The partial pressures are 2 atm, 3 atm, and 4 atm respectively. What is the total pressure inside the container?

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O 2 - 2 atm CO 2 - 3 atm N 2 - 4 atm O 2 - 2 atm CO 2 - 3 atm N 2 - 4 atm Total Pressure = 9 atm

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Collecting a Gas Over Water

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As gas bubbles through the water... As gas bubbles through the water...

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water vapor mixes with the gas being collected. water vapor mixes with the gas being collected.

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Gas pressure in container is equal to air pressure... Gas pressure in container is equal to air pressure...

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when water levels are the SAME. when water levels are the SAME.

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Practice Problem #2 Practice Problem #2

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If 60 liters of nitrogen gas is collected over water when the temperature is 40 o C and atmospheric pressure is 760 mm Hg, what is the partial pressure of the nitrogen? If 60 liters of nitrogen gas is collected over water when the temperature is 40 o C and atmospheric pressure is 760 mm Hg, what is the partial pressure of the nitrogen?

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From research, the vapor pressure of H 2 O at 40 o C is 7.4 kPa. From research, the vapor pressure of H 2 O at 40 o C is 7.4 kPa. 101.0 kPa Total - 7.4 kPa H 2 O 93.6 kPa N 2

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At equal temperature and pressure, equal volumes of gases contain the same number of molecules. At equal temperature and pressure, equal volumes of gases contain the same number of molecules.

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At STP, 22.4 dm 3 of any gas contains one mole of molecules, 6.02 X 10 23 At STP, 22.4 dm 3 of any gas contains one mole of molecules, 6.02 X 10 23

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Theoretical gas molecules that have mass, but no volume Theoretical gas molecules that have mass, but no volume

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P V = n R T

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P - standard pressure P V = n R T P - standard pressure

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P V = n R T V - molar volume P V = n R T V - molar volume

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P V = n R T n - number of moles P V = n R T n - number of moles

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P V = n R T R - 8.31 dm 3. kPa P V = n R T R - 8.31 dm 3. kPa mole. K

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P V = n R T T - standard temp (K) P V = n R T T - standard temp (K)

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How many moles of gas are found in a 500 dm 3 container if the conditions inside the container are 25 o C and 200 kPa?

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PV = nRT n = PV RT 200 kPa 500 dm 3 mole K 298 K 8.31 dm 3 kPa = 40.4 moles

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Substituting for n: moles = mass (m) Substituting for n: moles = mass (m) molecular mass (M)

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P V = m R T M

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The relative rates at which two gases diffuse under identical conditions vary inversely as the square roots of their molecular masses. The relative rates at which two gases diffuse under identical conditions vary inversely as the square roots of their molecular masses.

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Fuel in a Butane Lighter

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