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Gas Laws BOYLE CHARLES AVOGADRO GAY-LUSSAC
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What happens to the volume of a gas when you increase the pressure? (e.g. Press a syringe that is stoppered) Consider This Push!
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What happens to the volume of a gas when you increase the pressure? (e.g. Press a syringe that is stoppered) Consider This Push!
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What happens to the Volume of a Gas When you Increase the Pressure? (e.g. Press a syringe that is stoppered) Consider This Why? There is lots of space between gas particles. Therefore, gases are compressible!
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Let’s investigate the relationship between pressure and volume if the quantity of gas and temperature are held constant. 100 kPa
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Volume = 50 L 10050
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200 kPa Volume = 25 L 10050 20025
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400 kPa Volume = 12.5 L 10050 20025 40012.5
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800 kPa Volume = 6.25 L 10050 20025 40012.5 8006.25 What is the mathematical relationship between P and V? P x V = constant 5000
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Boyle’s Law In the 17th Century Robert Boyle described this property as, “the spring of air”. Boyle showed that when temperature and amount of gas were constant then: P 1/V OR: PV = k
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Boyle’s Law For a fixed quantity of gas at a constant temperature, the volume and pressure are inversely proportional. Boyle showed that when temperature and amount of gas were constant then: P 1/V OR: PV = k
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Who Cares? Scuba Divers! At sea level air pressure = 100 kPa At 10 m deep in water pressure = 200 kPa At 20 m deep = 300 kPa At 30 m deep = 400 kPa SCUBA provides air at the same pressure
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What will the volume of air in the diver’s lungs be at the surface (100 kPa)? What will happen to the diver? A Scuba Diver goes to a depth of 90 m and takes a breath of 3 L volume from her tank. Suddenly! A Dolphin lunges at the diver and takes the SCUBA! The Diver holds her breath and quickly returns to the surface.
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A Scuba Diver goes to a depth of 90 m and takes a breath of 3 L volume from her tank. Assuming that T is constant we can use Boyle’s Law: PV = k At 90 m: k = P 1 V 1 At surface: k = P 2 V 2 Therefore! P 1 V 1 = P 2 V 2 (1000 kPa)(3 L) = 100 kPa(V 2 ) V 2 = 300 L Yikes Exploding lungs
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Consider This! Two balloons are filled with equal volumes of air. What happens to the Volume of each if one is heated and the other is frozen?
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HEATED BALLOON FROZEN BALLOON 50 o C, V=1.18 L
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FROZEN BALLOON HEATED BALLOON 60 o C, V=1.22 L40 o C, V=1.14 L
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FROZEN BALLOON HEATED BALLOON 70 o C, V=1.25 L30 o C, V=1.11 L
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FROZEN BALLOON HEATED BALLOON 80 o C, V=1.29 L20 o C, V=1.07 L
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FROZEN BALLOON HEATED BALLOON 90 o C, V=1.32 L10 o C, V=1.04 L
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FROZEN BALLOON HEATED BALLOON 100 o C, V=1.36 L0 o C, V=1.00 L
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To study this relationship let’s look at this data in a table. Graph this data using temperature as the independent variable
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If this line is extended backwards the volume of 0 L of gas is found to be -273 o C
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-273.15 o C is known as absolute zero. When using gas laws, temperature must be expressed using a temperature scale where 0 is -273.15 o C. This is called the Kelvin scale of absolute temperature. 0 K = -273.15 o C When changing o C to K simply add 273.15 What is 12.3 o C in K? 12.3 + 273.15 = 285.45 = 285.5 K
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Now let’s look at this table with temperatures in Kelvin (K). Can you spot a mathematical relationship between T and V.
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V/T in Kelvin is a constant. 0.0037
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Charles’ Law In the 17th Century Jacques Charles examined the relationship between Temperature and Volume Charles showed that when Pressure and amount of gas were constant then: V T OR: V/T = k
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Charles’ Law For a fixed quantity of gas at a constant pressure, absolute temperature and volume are directly proportional. Charles showed that when Pressure and amount of gas were constant then: V T OR: V/T = k
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Combined Gas Law If V1V1 T1T1 = V2V2 T2T2 V1P1V1P1 V2P2V2P2 and V1P1V1P1 T1T1 V2P2V2P2 T2T2 = = then or V 1 P 1 T 2 =V 2 P 2 T 1 What does P 2 = ? V2T1V2T1 V2T1V2T1
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RECAP Charles’ Law At Constant P and n: V/T = k Boyle’s Law At Constant T and n: VP = k Combined Gas Law For a fixed quantity of gas: VP/T = k
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If 12.5 L of a gas at a pressure of 125 kPa is placed in an elastic container at 15 o C what volume would it occupy if the pressure is increased to 145 kPa? Given: V 1 = 12.5 L P 1 = 125 kPa T 1 = 15 o C V 2 = ? P 2 = 145 kPa T 2 = 15 o C V1P1V1P1 T1T1 = V2P2V2P2 T2T2 Since T 1 = T 2 cancel them to get V1P1V1P1 V2P2V2P2 = (12.5 L)(125 kPa) = V 2 (145 kPa) V 2 = (12.5 L)(125 kPa)/145 kPa V 2 = 10.8 L
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Does this answer make sense? 12.5 L @ 125 kPa 125 kPa
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Does this answer make sense? 165 kPa
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Does this answer make sense? 165 kPa
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Does this answer make sense? 165 kPa V reduced to 10.8 L Yes, as the pressure increases at constant temperature the volume decreases.
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If 15.6 L of a gas at a pressure of 165 kPa is placed in an elastic container at 15 o C what volume would it occupy if the temperature is increased to 98 o C? Given: V 1 = 15.6 L P 1 = 165 kPa T 1 = 15 o C V 2 = ? T 2 = 98 o C P 2 = 165 kPa V1P1V1P1 T1T1 = V2P2V2P2 T2T2 Since P 1 = P 2 cancel them to get V 1 /T 1 V 2 / T 2 = (15.6 L)/(288 K) = V 2 /(371 K) V 2 = (15.6 L)(371 K) / 288K V 2 = 20.1 L 288 K 371 K
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If 5.3 L of a gas at a pressure of 75 kPa is placed in an elastic container at 24 o C what volume would it occupy if the temperature is increased to 62 o C and pressure to 155 kPa? Given: V 1 = 5.3 L P 1 = 75 kPa T 1 = 24 o C V 2 = ? T 2 = 62 o C P 2 = 155 kPa V1P1V1P1 T1T1 = V2P2V2P2 T2T2 (5.3 L)(75 kPa)(335 K) 297 K 335 K OR V 1 P 1 T 2 =V 2 P 2 T 1 (155 kPa)(297K) V 2 = V 2 = 2.9 L
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Ideal Gas Law An Ideal Gas is a hypothetical gas that obeys all the gas laws perfectly under all conditions. PV = nRT Where n is the number of moles of gas, P is pressure in kPa, R = 8.313 kPaL/molK and T is temperature in K. Find the mass of helium gas which would be introduced into a 0.95 L container to produce a pressure of 125 kPa at 25 o C.
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Find the volume occupied by 25 g of chlorine gas at SATP. 4.2 g of propane gas is introduced into a 325 mL container at 45 o C. What is the pressure of the container. Propane is C 3 H 8. What is the density of NH 3 gas at STP if 1.0 mol of this gas occupies 22.4 L. At what temperature does methane gas have a density of 1.2 mg/L if its pressure is 65 kPa. Methane is CH 4.
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This gas exerts a pressure of 100 kPa inside this container. If another gas is injected into the same container and it exerts a pressure of 70 kPa what is the total pressure in the container? 170 kPa
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This gas exerts a pressure of 100 kPa inside this container. If another gas is injected into the same container and it exerts a pressure of 70 kPa what is the total pressure in the container? 170 kPa
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This gas exerts a pressure of 100 kPa inside this container. If another gas is injected into the same container and it exerts a pressure of 70 kPa what is the total pressure in the container? 170 kPa
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The total pressure of a gas mixture is the sum of the partial pressures of each of the gases in the mixture. Example If a container of air has a pressure of 100 kPa and the % of N 2 in the container is 78%, % of O 2 is 21%, what are the partial pressures of each of these gases inside this container. P N 2 = 78 kPa, P O 2 = 21 kPa
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5.0 L 300 K 125 kPa 6.0 L, 400 K, 155 kPa 1 2 What would the total pressure be if the gas in container 1 was injected into container 2.
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5.0 L 300 K 125 kPa 6.0 L, 400 K, 155 kPa 1 2 The total pressure is the sum of the pressures of gas 1 and gas 2.Since gas 1 changed volume and temperature its pressure changed.
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5.0 L 300 K 125 kPa 6.0 L, 400 K, 155 kPa 1 2 V 1 =5.0 L, V 2 =6.0 L, T 1 =300 K, T 2 = 400 K P 1 =125 kPa, P 2 =? V 1 P 1 T 2 =V 2 P 2 T 1 P 2 = 5.0 L x 125 kPa x 400 K 6.0 L x 300 K = 139 kPa
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5.0 L 300 K 125 kPa 6.0 L, 400 K, 155 kPa 1 2 Total pressure is 155 kPa + 139 kPa = 294 kPa = 2.9 x 10 2 kPa 139 kPa
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5.0 L 300 K 125 kPa 6.0 L, 400 K, 155 kPa 1 2 Find the total pressure in container 1 if the gas in container 2 is injected into container 1.
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5.0 L 300 K 125 kPa 6.0 L, 400 K, 155 kPa 1 2 V 1 =6.0 L, V 2 =5.0 L, T 1 =400 K, T 2 = 300 K P 1 =155 kPa, P 2 =? V 1 P 1 T 2 =V 2 P 2 T 1 P 2 = 6.0 L x 155 kPa x 300 K 5.0 L x 400 K = 139.5 kPa
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5.0 L 300 K 125 kPa 6.0 L, 400 K, 155 kPa 1 2 Total pressure is 125 kPa + 139.5 kPa = 264.5 kPa = 2.6 x 10 2 kPa 139.5 kPa
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