The Gas Laws

The Gas Laws Boyle’s Law Amonton’s Law Charles’s Law Combined Gas Law Gay-Lussac’s Law Avogadro’s Law Dalton’s Law

Boyle’s Law Pressure Volume At a constant temperature, pressure is inversely proportional to volume.

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

Boyle’s Law 1 P µ V PV = k P1V1 = P2V2 At a constant temperature, pressure is inversely proportional to volume. P µ 1 V PV = k P1V1 = P2V2 Year: 1662

Amonton’s Law P µ T P1 T1 = P2 T2 “Air thermometer”, 1695. This is not Gay-Lussac’s law. Diagram from http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch4/gaslaws3.html

Amonton’s Law P µ T P1 T1 = P2 T2 This is why you measure your tire pressure when the tire is cold. Tire pressures vary with temperature. P µ T P1 T1 = P2 T2

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

Finding Absolute Zero -273 C Pressure Temperature (C) Extrapolate to the x-axis -300 -150 0 100 200 300 Temperature (C)

Charles’s Law Volume Temperature At constant pressure, volume is directly proportional to temp.

Charles’s Law V µ T V T = k V1 T1 = V2 T2 At constant pressure, volume is directly proportional to temperature. V µ T V T = k V1 T1 = V2 T2

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

Charles’s Law Hydrogen balloon assent, 3000 m, 1783.

Combined Gas Law P µ 1 V T gives P µ T V

Combined Gas Law T P µ V kT P = V Next we convert to an equation by adding a constant. P = kT V

Combined Gas Law Rearranging the equation gives: = k PV T P = kT V

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

Combined Gas Law P2V2 P1V1 k k = = T2 T1 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.

Combined Gas Law P1V1 T1 = k P2V2 T2 = k P1V1 T1 = P2V2 T2

Combined Gas Law P1V1 T1 = P2V2 T2 Use the combined gas law whenever you are asked to find a new P, V or T after changes to a confined gas. P1V1 T1 = P2V2 T2

Sample Combined Gas Law Problem Consider a confined gas in a cylinder with a movable piston. The pressure is 0.950 atm. Find the new pressure when the volume is reduced from 100.0 mL to 65.0 mL, while the temperature remains constant? 100 mL 65 mL

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

Sample Combined Gas Law Problem P1V1 T1 = P2V2 T2 100 mL Since the temperature is constant, we can cancel out T1 and T2. 65 mL

Sample Combined Gas Law Problem P1V1 = P2V2 This becomes Boyle’s Law 100 mL P1V1 = V2 P2 Next, solve for P2. 65 mL

Sample Combined Gas Law Problem P1V1 = V2 P2 100 mL (0.950 atm)(100.0 mL) = 65.0 mL P2 = P2 1.46 atm 65 mL

Combined Gas Law Problems 1. A sample of neon gas has a volume of 2.00 L at 20.0 C and 0.900 atm. What is the new pressure when the volume is reduced to 0.750 L and the temperature increases to 24.0 C? The answer is 2.43 atm

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

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 0.988 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

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

2:1 Gay-Lussac’s Law 2 H2(g) + O2(g)  2 H2O(g) Gay-Lussac found that the volumes of gases in a reaction were in ratios of small, whole numbers. 2 H2(g) + O2(g)  2 H2O(g) 200 mL of hydrogen reacts with 100 mL of oxygen. 2:1

Gay-Lussac’s Law 2 H2(g) + O2(g)  2 H2O(g) The ratio of volumes of gases come from the ratios of the coefficients in the balanced equation.

Avogadro’s Law V µ n V n = k V1 n1 = V2 n2 Avogadro’s law followed Dalton’s atomic theory and Gay-Lussac’s law. Year: 1811. V µ n V n = k Equal volumes of gases at the same temperature and pressure have equal numbers of molecules. V1 n1 = V2 n2

Dalton’s Law Ptotal = P1 + P2 + P3 … Dalton’s law of partial pressures deals with mixtures of gases. The total pressure is the sum of the partial pressures: Ptotal = P1 + P2 + P3 … Use when dealing with the pressure of H2O(g) when collecting a gas over water.

Dalton’s Law “Collecting a gas over water” Inverted gas collecting bottle Rubber tubing carrying H2 gas Flask with metal and HC l Water in trough

Dalton’s Law “Collecting a gas over water” Inverted gas collecting bottle Rubber tubing carrying H2 gas Flask with metal and HC l Water in trough

Dalton’s Law Water is displaced through the mouth of the bottle as H2 gas bubbles in. H2 gas Flask with metal and HC l Water in trough

Dalton’s Law A mixture of hydrogen gas and water vapor is in the collection bottle. H2 gas Flask with metal and HC l Water in trough

Dalton’s Law The pressure of the mixture is the sum of the pressures of H2O gas and H2 gas. H2 gas Flask with metal and HC l Water in trough

Ptotal = PH2O + PH2 Dalton’s Law Flask with metal and HC l H2 gas Flask with metal and HC l Water in trough

Sample Problem The total pressure in the bottle is 712.7 torr. The temperature is 19 C. What is the H2 pressure? H2 gas Flask with metal and HC l Water in trough

Sample Problem At 19C, the vapor pressure of water is 16.5 torr. The H2 pressure is … H2 gas Flask with metal and HC l Water in trough

Sample Problem PH2 = Ptotal - PH2O PH2 = 712.7 torr – 16.5 torr = 696.2 torr H2 gas Flask with metal and HC l Water in trough

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.

Ptotal = P1 + P2 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. Ptotal = P1 + P2

Ptotal = Pair+ PHg 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. Ptotal = Pair+ PHg

PHg = Ptotal - Pair 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. PHg = Ptotal - Pair

PHg = 693 torr – 684 torr 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. PHg = 693 torr – 684 torr

PHg = 9 torr 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. PHg = 9 torr

More on the gas laws: http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch4/gaslaws3.html