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The Gas Laws

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Boyles Law Boyles Law Amontons Law Amontons Law Charless Law Charless Law Combined Gas Law Combined Gas Law Gay-Lussacs Law Avogadros Law Daltons Law

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Boyles Law At a constant temperature, pressure is inversely proportional to volume. Pressure Volume

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Boyles Law At a constant temperature, pressure is inversely proportional to volume. 1/Pressure Volume

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Boyles Law At a constant temperature, pressure is inversely proportional to volume. P 1 V 1 = P 2 V 2 PV = k P1V Year: 1662

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Amontons Law Air thermometer, This is not Gay-Lussacs law. PT P1P1P1P1 T1T1T1T1 = P2P2P2P2 T2T2T2T2 Diagram from

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Amontons Law PT P1P1P1P1 T1T1T1T1 = P2P2P2P2 T2T2T2T2 This is why you measure your tire pressure when the tire is cold. Tire pressures vary with temperature.

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Measure the pressures of a gas at various temperatures at a constant volume. Amontons air thermometer was used to find the value of absolute zero. Amontons Law

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Finding Absolute Zero Pressure Temperature (C) C Extrapolate to the x-axis

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Charless Law At constant pressure, volume is directly proportional to temp. Volume Temperature

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Charless Law At constant pressure, volume is directly proportional to temperature. VT VT = k V1V1V1V1 T1T1T1T1 = V2V2V2V2 T2T2T2T2

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Charless Law Studied gases during 1780s.Studied gases during 1780s. Hydrogen balloon assents, 3000 m, in 1783.Hydrogen balloon assents, 3000 m, in Collaborated with the Montgolfier brothers on hot air balloons, 1783.Collaborated with the Montgolfier brothers on hot air balloons, Charless gas studies published by Gay-Lussac in1802.Charless gas studies published by Gay-Lussac in1802.

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Charless Law Hydrogen balloon assent, 3000 m, 1783.

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Combined Gas Law gives PTV P1V PT

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PTV Next we convert equation by adding a constant. to an to an PkTV

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Combined Gas Law Rearranging the equation gives: = k PVT PkTV

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Combined Gas Law Combining the laws of Boyle, Amonton and Charles produces the combined gas law. = k PVT

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Combined Gas Law P1V1P1V1P1V1P1V1 T1T1T1T1 = k P2V2P2V2P2V2P2V2 T2T2T2T2 = k Consider a confined gas at two sets of conditions. Since the number of molecules is the same and the values of k are the same, then we can combine the two equations.

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Combined Gas Law P1V1P1V1P1V1P1V1 T1T1T1T1 = P2V2P2V2P2V2P2V2 T2T2T2T2 P1V1P1V1P1V1P1V1 T1T1T1T1 = k P2V2P2V2P2V2P2V2 T2T2T2T2 = k

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P1V1P1V1P1V1P1V1 T1T1T1T1 = P2V2P2V2P2V2P2V2 T2T2T2T2 Use the combined gas law whenever you are asked to find a new P, V or T after changes to a confined gas.

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Sample Combined Gas Law Problem Consider a confined gas in a cylinder with a movable piston. The pressure is atm. Find the new pressure when the volume is reduced from mL to 65.0 mL, while the temperature remains constant? 100 mL 65 mL

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Sample Combined Gas Law Problem Start with the equation for the combined gas law. 100 mL 65 mL

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Sample Combined Gas Law Problem 100 mL 65 mL P1V1P1V1P1V1P1V1 T1T1T1T1 = P2V2P2V2P2V2P2V2 T2T2T2T2 Since the temperature is constant, we can cancel out T 1 and T 2.

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Sample Combined Gas Law Problem 100 mL 65 mL P1V1P1V1P1V1P1V1= P2V2P2V2P2V2P2V2 Next, solve for P 2. This becomes Boyles Law P1V1P1V1P1V1P1V1= V2V2V2V2 P2P2P2P2

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Sample Combined Gas Law Problem 100 mL 65 mL P1V1P1V1P1V1P1V1= V2V2V2V2 P2P2P2P2 (0.950 atm)(100.0 mL) = 65.0 mL P2P2P2P2 = P2P2P2P atm

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Combined Gas Law Problems 1. A sample of neon gas has a volume of 2.00 L at 20.0 C and atm. What is the new pressure when the volume is reduced to L and the temperature increases to 24.0 C? The answer is 2.43 atm

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Combined Gas Law Problems 2. Some left over propane gas in a rigid steel cylinder has a pressure of 24.6 atm at a temperature of 20.C. When thrown into a campfire the temperature in the cylinder rises to 313C. What will be the pressure of the propane? The answer is 49.2 atm

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Combined Gas Law Problems 3. Consider the fuel mixture in the cylinder of a diesel engine. At its maximum, the volume is 816 cc. The mixture comes in at atm and 31 C. What will be the temperature (in C) when the gas is compressed to 132 cc and 42.4 atm? The answer is 1837 C

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Gay-Lussacs Law Year: 1802 At a given temperature and pressure, the volumes of reacting gases are in a ratio of small, whole numbers. Also known as the law of combining volumes.

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Gay-Lussacs Law Gay-Lussac found that the volumes of gases in a reaction were in ratios of small, whole numbers. 2 H 2 (g) + O 2 (g) 2 H 2 O(g) 200 mL of hydrogen reacts with 100 mL of oxygen. 2:1

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Gay-Lussacs Law 2 H 2 (g) + O 2 (g) 2 H 2 O(g) The ratio of volumes of gases come from the ratios of the coefficients in the balanced equation.

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Avogadros Law Equal volumes of gases at the same temperature and pressure have equal numbers of molecules. Vn = k Avogadros law followed Daltons atomic theory and Gay-Lussacs law. Year: Vn V1V1V1V1 n1n1n1n1 = V2V2V2V2 n2n2n2n2

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Daltons Law Daltons law of partial pressures deals with mixtures of gases. P total = P 1 + P 2 + P 3 … Use when dealing with the pressure of H 2 O(g) when collecting a gas over water. The total pressure is the sum of the partial pressures:

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Daltons Law Inverted gas collecting bottle Flask with metal and HC l Rubber tubing carrying H 2 gas Water in trough Collecting a gas over water

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Daltons Law Inverted gas collecting bottle Flask with metal and HC l Rubber tubing carrying H 2 gas Water in trough Collecting a gas over water

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Daltons Law Flask with metal and HC l Water in trough Water is displaced through the mouth of the bottle as H 2 gas bubbles in. H 2 gas

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Daltons Law Flask with metal and HC l Water in trough A mixture of hydrogen gas and water vapor is in the collection bottle. H 2 gas

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Daltons Law Flask with metal and HC l Water in trough The pressure of the mixture is the sum of the pressures of H 2 O gas and H 2 gas. H 2 gas

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Daltons Law Flask with metal and HC l Water in trough P total = P H 2 O + P H 2 H 2 gas

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Sample Problem Flask with metal and HC l Water in trough The total pressure in the bottle is torr. The temperature is 19 C. What is the H 2 pressure? H 2 gas

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Flask with metal and HC l Water in trough At 19C, the vapor pressure of water is 16.5 torr. The H 2 pressure is … H 2 gas Sample Problem

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Flask with metal and HC l Water in trough P H 2 = P total - P H 2 O H 2 gas P H 2 = torr – 16.5 torr = torr Sample Problem

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Another Problem A pressure sensor is attached to a sealed flask containing 80.0 mL of mercury at 50.0 C. The pressure sensor indicates 693 torr. The pressure of the air inside the empty flask at 50.0 C is 684 torr. Determine the vapor pressure of mercury at 50.0 C.

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Another Problem A pressure sensor is attached to a sealed flask containing 80.0 mL of mercury at 50.0 C. The pressure sensor indicates 693 torr. The pressure of the air inside the empty flask at 50.0 C is 684 torr. Determine the vapor pressure of mercury at 50.0 C. P total = P 1 + P 2

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Another Problem A pressure sensor is attached to a sealed flask containing 80.0 mL of mercury at 50.0 C. The pressure sensor indicates 693 torr. The pressure of the air inside the empty flask at 50.0 C is 684 torr. Determine the vapor pressure of mercury at 50.0 C. P total = P air + P Hg

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Another Problem A pressure sensor is attached to a sealed flask containing 80.0 mL of mercury at 50.0 C. The pressure sensor indicates 693 torr. The pressure of the air inside the empty flask at 50.0 C is 684 torr. Determine the vapor pressure of mercury at 50.0 C. P Hg = P total - P air

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Another Problem A pressure sensor is attached to a sealed flask containing 80.0 mL of mercury at 50.0 C. The pressure sensor indicates 693 torr. The pressure of the air inside the empty flask at 50.0 C is 684 torr. Determine the vapor pressure of mercury at 50.0 C. P Hg = 693 torr – 684 torr

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Another Problem A pressure sensor is attached to a sealed flask containing 80.0 mL of mercury at 50.0 C. The pressure sensor indicates 693 torr. The pressure of the air inside the empty flask at 50.0 C is 684 torr. Determine the vapor pressure of mercury at 50.0 C. P Hg = 9 torr

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More on the gas laws:

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